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Algebra 1, Spanish Homework Practice Workbook
Description: TheSpanish Homework Practice Workbookcontains two Spanish worksheets for every lesson in the Student Edition. This workbook helps students: Practice the skills of the lesson, Use their skills to solve word problems.
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TheSpanish Homework Practice Workbookcontains two Spanish worksheets for every lesson in the Student Edition. This workbook helps students: Practice the skills of the lesson, Use their skills to solve word problems |
State curriculum standards are mandating more coverage of geometry, as are the curricula for pre-service mathematics education and in-service teaching. Yet many secondary teachers know just enough geometry to stay one chapter ahead of their students! What's more, most college-level geometry texts don't address their specific needs.Advanced Euclidean Geometry fills this void by providing a thorough review of the essentials of the high school geometry course and then expanding those concepts to advanced Euclidean geometry, to give teachers more confidence in guiding student explorations and questions. The text contains hundreds of illustrations created in The Geometer's Sketchpad Dynamic Geometry? software, and it is packaged with a CD-ROM (for Windows?/Macintosh? formats) containing over 100 interactive sketches using Sketchpad(TM) (assumes that the user has access to the program). |
This is a thesis templet for Beijing Normal University, coded in LaTeX. It is designed for bachelor's degree, master's degree as well as doctor's degree. As planned, it is adapted to various platformsEin Programm zur schrittweisen Bestimmung der Lage von Punkten, Geraden und Ebenen. A german program for school mathematics: It helps you to calculate common points of planes and lines. It works step-by-step so you can find your mistakes easily. |
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The Math Foundation Advantage While traditional math programs rely on static numbers on a page, Math Foundation's online course ware uses unique animated lessons which allow you to visualize and grasp math concepts more effectively. This innovative approach makes math come alive and can generate those "lightbulb moments" that elude many people who are trying to learn math by conventional methods.
The graphics and animations also reduce the anxiety that many people experience when learning math, and help to maintain interest. Math Foundation's course ware is self paced and allows you to learn math online at your own rate and comfort level. The lessons are easy to follow with step-by-step examples. Inside Math Foundation you'll find clearly written, narrated tutorials that cover Basic Math, Pre-Algebra, Basic Algebra and Geometry including: |
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UL offers algebra-free math class
Yipee! No more algebra for non-science majors.
UL Lafayette has added a new math course, Math 102, for freshmen majoring in non-science areas. Beginning this fall, students can take the course, which emphasizes quantitative reasoning, as an alternative to college algebra.
Three faculty members created the course: Dr. Kathleen Lopez, an associate professor of mathematics; Melissa Myers, a master instructor; and Christy Sue Langley, a mathematics instructor.
In a press release, Myers says universities across the country are moving away from formula-based teaching to concept-oriented, practical applications of mathematics. Myers, who is also director of freshman mathematics at the university, says this course will likely appeal to the majority of students enrolled in the College of the Arts and the College of Liberal Arts.
In the past, college algebra was the first math course taken by all undergraduates.
There are two paths students follow upon completion of their initial math course. One path is taken primarily by business and science-oriented majors, who are required to take advanced mathematics courses. The other path is for non-science majors, with the courses they take emphasizing applied mathematics.
"Creating a college algebra alternative designed specifically for non-science majors will enable us to better serve these two different populations," notes Myers.
Course topics in Math 102 include traditional concepts, such as linear and exponential functions, as well as topics designed to increase students' ability to reason quantitatively. The course emphasizes critical thinking.
"Our primary goal is to make students better educated," Myers continues. "As consumers, for instance, we're constantly exposed to advertisements. Upon completion of this course, our students will be better educated consumers who are able to recognize misleading advertisements."
Some of the skills students will learn include:
mentally estimating the sale price of an item that has been discounted;
determining which of two possible financial situations is most advantageous;
applying deductive and inductive reasoning, such as understanding the error of statements such as "All services not available in all areas."
reading and interpreting graphs, particularly recognizing graphs which have been designed to intentionally mislead;
computing the consequences of not paying off a "no-money down, two-years interest free" purchase during the interest-free period;
comparing a flat-rate subscription movie site to a service which charges an initial fee plus per-movie charge;
using proportional reasoning to determine which size of product is the best deal; and
recognizing when it is appropriate to use the phrase "growing exponentially."
"Students will be able to use the reasoning and mathematical skills they learn in this course to enhance their decision-making skills, both personally and professionally," Myers |
This first volume, a three-part introduction to the subject, is intended for students with a beginning knowledge of mathematical analysis who are motivated to discover the ideas that shape Fourier analysis. It begins with the simple conviction that Fourier arrived at in the early nineteenth century when studying problems in the physical sciences--that an arbitrary function can be written as an infinite sum of the most basic trigonometric functions.
The first part implements this idea in terms of notions of convergence and summability of Fourier series, while highlighting applications such as the isoperimetric inequality and equidistribution. The second part deals with the Fourier transform and its applications to classical partial differential equations and the Radon transform; a clear introduction to the subject serves to avoid technical difficulties. The book closes with Fourier theory for finite abelian groups, which is applied to prime numbers in arithmetic progression.
In organizing their exposition, the authors have carefully balanced an emphasis on key conceptual insights against the need to provide the technical underpinnings of rigorous analysis. Students of mathematics, physics, engineering and other sciences will find the theory and applications covered in this volume to be of real interest.
The Princeton Lectures in Analysis represents a sustained effort to introduce the core areas of mathematical analysis while also illustrating the organic unity between them. Numerous examples and applications throughout its four planned volumes, of which Fourier Analysis is the first, highlight the far-reaching consequences of certain ideas in analysis to other fields of mathematics and a variety of sciences. Stein and Shakarchi move from an introduction addressing Fourier series and integrals to in-depth considerations of complex analysis; measure and integration theory, and Hilbert spaces; and, finally, further topics such as functional analysis, distributions and elements of probability theory |
High School Algebra
Description: ThisMore...
This Logarithms, Inequalities, Probabilities and Determinants. Each chapter is follows a simple structure. A definition of terms and detailed explanation of concepts; followed by derivation of useful formulas through first principles. A few problems are solved in order to give a flavor of the problem solving process. This book is intended to help the student to understand ways to build an airtight reasoning on how the problem solving process moves towards the final answer. This serves as a demonstration of our understanding of the subject - basics, formulas and methods of manipulation. Students interested in solving additional problems may refer to the books "Problems in High School Algebra" and "Challenges in High School Algebra |
Suitable for the scientific, engineering, medical and teaching professions, this free program offers financial and statistical functions that go above and beyond other programs like it. It contains a comprehensive set of conversions, constants and physical property data, a built-in periodic table of the elements, number base conversions, vectors, matrices and complex numbers. This could very possibly be the only calculator you need! |
2. Number properties positive and megative numbers adding substracting properties of odd and even numbers rules for odd and even numbers multiplying and dividing positive and negative numbers prime numbers pemdas
12. Triangles sides and angles of a triangle perimeter of a triangle area of a triangle isosceles triangle equilateral triangles right triangles pythagorean theorem pythagorean triplets special right triangles isosceles right triangles isosceles right triangles - side ratio 30-60-90 right triangles 30-60-90 right triangles - side ratio
13. Quadliterals what is a quadliteral parallelograms rectangles squares and other four-sided objects perimeter perimeter of a square area of a rectangle area of a square area of a parallelogram rectangular solids - volume and surface area cubes
14. Circles components of a circle circumference circumference example arc length area of a circle area of a sector cylinders
Good compilation. One thing though, does the OG's math review section cover all these topics? If the OG covers these topics I can breathe a sigh of relief, else I'll have to go looking for material on something. _________________
Good compilation. One thing though, does the OG's math review section cover all these topics? If the OG covers these topics I can breathe a sigh of relief, else I'll have to go looking for material on something.
dear Sidhu
Mind it that relying purely on OG quant for scoring in high 40s is a risky proposition. The level of questions that are asked in real Gmat are far ahead as compared to that provided in OG. Only few of them (around last 50) might be of the level. Best would be to go through quant practice with forums such as Gmatclub and constantly review the shortcuts mentioned therein. Also try to prepare the error log of questions that you get wrong or take time to solve. It might help you a lot to return back to these solutions at regular intervals untill you become completely familiar to them |
Math and Soccer Video - Sharp Tack Productions
A video revealing the role played by numbers, graphs, geometry, algebra, angles, area, surveys, and chance data in the game of soccer; for example, how a goalie relies on the properties of angles to know where to stand when defending the goals, and on
...more>>
MathApplications - Cathleen V. Sanders
An interactive mathematics course offered via the Internet to students throughout the Hawaiian islands. Cathi Sanders, a teacher at Punahou School in Honolulu, teaches this course through ESchool, a pilot project of the Hawaii Department of Education
...more>>
Math Courses Online - Nancy Parham
Online courses offered by Cal State Bakersfield, Fresno, Los Angeles, San Bernardino, San Marcos, and Cal Poly San Luis Obispo: designed for students preparing to take the math exams ELM, GRE, CBEST, or SAT, or adults who are reentering college after
...more>>
The Math Dude - Mike DeGraba
Mike DeGraba is the Math Dude, bringing engaging explanations to Algebra I students in this series of videos. The 5- to 7-minute episodes are available via Flash on the web, podcast, RSS feed, and if you live in Montgomery County, MD, cable TV. Episodes
...more>>
Mathematical Gems
Brief articles include Irrationality of the Square Root of 2; The Best Card Trick; Boltzmann in Berkeley; Four of a kind and two jokers; Mathematical Paint (on Gabriel's Horn); How Many Squares, Mr. Franklin? Goldbach's Proof of the Infinitude of Primes;
...more>>
Mathematics Illuminated - Annenberg Media
A video course for adult learners, high school and college teachers; 13 half-hour video programs, online text, course guide, and Web site; graduate credit available. "Rather than a series of problems to be solved, mathematics is presented as play we engage
...more>>
Math for Morons Like Us - ThinkQuest 1998
Students talk to students about math: a site designed to help you understand math concepts better. Tutorials, sample problems, and quizzes for Pre-Algebra, Algebra, Geometry, Algebra II, and Pre-Calc/Calculus, designed assuming you know some of the basic
...more>>
Math Fundamentals Problem of the Week - Math Forum
Math problems for students working with concepts of number, operation, and measurement, as well as introductory geometry, data, and probability. The goal is to challenge students with non-routine problems and encourage them to put their solutions into
...more>>
Mathie x Pensive - Gregory Taylor
Musings on mathematics, teaching, and more by an Ottawa (Canada) high school math teacher who majored in computer science and graduated from the mathematics program at the University of Waterloo before receiving his B. Ed from Queen's University. Posts,
...more>>
Math in Daily Life - Annenberg Media
Students often question how they will use basic mathematical concepts, algebra, and geometry throughout their lives - but the average person uses math at least three times a day. Read how math affects daily decision-making in this series of short articles
...more>>
mathletics - Wayne Winston
From the author of the book by the same title, which reveals "how professional baseball, football and basketball teams use math to improve their performance" à la Moneyball, and also discusses line-up evaluations, the effectiveness of running and
...more>>
MathMagic on the Web - Alan A. Hodson; The Math Forum
MathMagic was a K-12 telecommunications project developed in El Paso, Texas. The intent of the project was to provide motivation for students to use computer technology while increasing problem-solving strategies and communications skills. MathMagic posted
...more>> |
Solving Mathematical Problems: A Personal Perspective
Book Description: Authored by a leading name in mathematics, this engaging and clearly presented text leads the reader through the various tactics involved in solving mathematical problems at the Mathematical Olympiad level. Covering number theory, algebra, analysis, Euclidean geometry, and analytic geometry, Solving Mathematical Problems includes numerous exercises and model solutions throughout. Assuming only a basic level of mathematics, the text is ideal for students of 14 years and above in pure mathematics |
This book is great. With reasonable knowledge of linear algebra and vector analysis I was able to quickly grasp the concepts of some of the mathematics behind Special and General Relativity. The author seems to do a near perfect job at explaining the theory in a way that seems to flow beautifully from one concept to the other without leaving you with too many unanswered questions. The mathematics behind Einstein's theories can be very subtle at times. I am studying the mathematics of Special and General relativity for a hobby (it's like doing a big puzzle. A bit of a crude analogy, but alright) , and am by no means a physicist as such. If you have the right background (Mine being Computer Science, Computer Graphics), and you are willing to spend a lot of thought on the matter, you will be able to follow this text, and be a considerably more knowledgeable person at the end.
This book approaches the idea of curvature in a gentle way. The introductory chapters are quite accessible to a student with limited mathematical backround. I found the combination of this book, Schutz's 'Geometrical Methods of Mathematical Physics' and Foster and Nightingale's 'First Course in GR' as suitable preparation for more weighty tomes such as Misner Thorne and Wheeler's 'Gravitation'.
4.0 out of 5 starsThe greatest approach of General Relativity for dummies, 1 Mar 2001
By A Customer
This review is from: A First Course in General Relativity (Paperback)
I think this is an obligated reference for undergraduate beginners in General Relativity. Almost all the mathematical requirements are included, but it would be the best book if it also treated the Hilbert action and the variational deduction of Einstein's field equation. I suggest buying the other Schut's title "Geometrical Methods for Mathematical Physics" which supplies more mathematical background. |
AGS Consumer Mathematics - Revised
AGS Consumer Mathematics - Revised
Teach your students to become well-informed consumers
Consumer Mathematics presents basic math skills used in everyday situations—paying taxes, buying food, banking and investing, and managing a household. The full-color text helps students and young adults become wiser, more informed consumers. |
PBIS 188: Modern Mathematics and Its Applications
This course is designed to bring the excitement of contemporary mathematical ideas to the non-specialist and to help develop the capacity to engage in logical thinking and the ability to read mathematical information critically. We will investigate a broad range of mathematical topics that may be encountered in your day-to-day life, including voting systems, fair division, networks, growth and symmetry, and statistics. Many of the mathematical topics that we will consider may be completely new to you. In fact, much of it has only been discovered within the last 20 years! Nevertheless, the course material will be accessible to anyone with an active curiosity, a willingness to work hard, and a decent background in basic algebra. Interesting and deep mathematics often occurs in places where you might least expect it! |
Fundamentals of Math Online Only
Fundamentals of Math for Distance Learning
Fundamentals of Math (2nd edition) focuses on problem solving and real-life uses of math with special features in each chapter while reinforcing computational skills and building a solid math foundation. Dominion through Math problems regularly illustrates how mathematics can be used to manage God's creation to His glory.
This course includes an abridged electronic version of the teacher's edition and student textbook that can be viewed while logged on to bjupressonline.com.
>>Click the Resources tab to view technical requirements for Distance Learning Online and information about the course's instructor.
Register now! You will receive an email from BJU Press Distance Learning Online regarding login information for the online courses which will be available June 1, 2014. You will have through December 2015 to complete your online learning program.
About the Instructor
Mr. Bill Harmon, BS
Bill Harmon has loved science for as long as he can remember. After completing his B.S. in Chemistry, he returned to Florida where he gained experience teaching a variety of subjects: science, math, Latin, and computer courses. Now he works as a chemist in the Safety Services Office at BJU, teaches Distance Learning Physics and Algebra, and teaches Chemistry at Bob Jones Academy. He is currently pursuing an M.Ed. in Secondary Education. He and his wife Mary Ann have two children, Brian and Janette. His favorite Bible verse is II Timothy 3:14. |
Pembroke Pines GeometryAdvanced functions such as Ln and Exponential functions are also explained in the subject. The focus on differences become crucial when dealing with advanced mathematics. Calculus branches into two sections, differential and integral calculusStudents will learn how to solve linear equations, including multistep equations, equations with multiple variables and equations involving decimals, as well as write a linear equation based on the graph of a line. Algebra 1 course also gives the students a thorough introduction of functions and... |
A discussion of the many ways to classify curves, how they are named, a curve family tree, and interconnection between curves, with related Web sites about fractals and curves. Hosted by the Math Forum.
An online course: learning units presented in worksheet format review the most important results, techniques and formulas in college and pre-college calculus. Logarithms and Exponential; Sequences; Series; Techniques of...
An online course: learning units presented in worksheet format review the most important results, techniques and formulas in matrix algebra. Introduction to Matrices; Systems of Linear Equations; Determinant;...
A short article designed to provide an introduction to geometry, including classical Euclidean geometry and synthetic (non-Euclidean) geometries; analytic geometry; incidence geometries (including projective planes);...
A short article designed to provide an introduction to sequences and series, the most common examples of limiting processes; convergence criteria and rates of convergence are as important as finding "the answer."... |
Reviews
"The book is a first, very valuable contribution to the field, and should be recommended to everyone interested, and willing to go deeper into the subject (...)". Alberto Parmeggiani, Mathematical reviews, 2004. |
To me, math is less important than the principles of our country and our Constitution.--that's what needs to be taught to American students. After all we have trillions of calculators all around us but what do we have to remind us of how precious our Constitution is, including the Second Amendment that the Liberals would love to abolish! Maybe electronic gadjets should automatically include a reprint of the U.S. Constitution! That's the last thing we want to subtract!
While I completely agree that there needs to be more emphasis placed on the principles of the Founding Fathers and the Constitution, you are seriously off base on math. Beyond the simplest forms of math, a calculator is useless if the person using it doesn't understand advanced math. There's a good reason math is called the language of science and engineering. A failure to teach advanced math will translate into an inability to teach science and engineering.
Just like a knowledge of history is necessary to understand the present, an understanding of math is necessary to create the future.
There was some talk on national TV about a year and a half ago about TX reforming the textbooks. Actually what happened was that the libs wanted to replace all info on the Constitution and on Jefferson with info on the man who invented the fireman's hat and other "important" people in US history. When the majority on the state education committee refused, the media claimed that we were "reforming" the textbooks ... because we didn't use the ones the libs wants used |
Modeling, Functions, and Graphs: Algebra for College Students (with iLrn(TM) Printed Access Card)
Book Description: The Fourth Edition of Yoshiwara and Yoshiwara's MODELING, FUNCTIONS, AND GRAPHS: ALGEBRA FOR COLLEGE STUDENTS includes content found in a typical algebra course, along with introductions to curve-fitting and display of data. Yoshiwara and Yoshiwara focus on three core themes throughout their textbook: Modeling, Functions, and Graphs. In their work of modeling and functions, the authors utilize the Rule of Four, which is that all problems should be considered using algebraic, numerical, graphical, and verbal methods. The authors motivate students to acquire the skills and techniques of algebra by placing them in the context of simple applications that use real-life data |
You can see where we have exercises and where we have gaps. We're working on filling in all the places we have gaps, which should be just a few months away. Until then, you should be able to use Khan Academy and supplement it if there's anything you're looking for that we don't have. |
·Review
·Review key math vocabulary, basic functions, and equations
·Use proven techniques to solve complicated questions
The 2007-2008 edition of Cracking the SAT Math 1 and 2 Subject Tests is revised and updated to include the most current information |
What You'll Learn
In the 17th century, Isaac Newton and Gottfried Liebnitz — two giants of the mathematic world — both developed the first real theory of the infinite. They fought over who was first for the rest of their lives, but it created what we call calculus.
Calculus lies at the heart of that idea, and learning how it can be applied to everyday situations will change your world.
What will I study as a math major?
At JCSU, we won't just teach you theories and equations. Our program emphasizes problem-based learning.
So what does that mean for you?
It means you'll learn how to apply what you learn in calculus and analytic geometry to situations you encounter every day. Math can be found in everything we do. Art, music, sports, even fashion all rely on the use of math.
If you have a head for numbers, or just a passion for math, our program will help you strengthen your ability to use logic and creativity to answer some pretty interesting questions.
Popular classes in the math program
Course
Description
MTH 362 Applications of Mathematics
Explore the various ways math is used in finance, basic logic, art and music, spreadsheets, games and puzzles, number systems and culture.
MTH 432 History and Foundations of Mathematics
Examine the work of pioneers in math and its history through creative problem solving.
MTH 434 Differential Equations I
Learn how to solve ordinary differential equations, like the predator-prey model, and how these techniques can be used in science.
MTH 333 Probability and Statistics I
Study data analysis techniques, elements of probability theory and how to deal with uncertainty.
Where will I study math?
Math is best learned when you can see how to apply it. At JCSU, you'll be taught in our smart classrooms, which feature:
Wireless internet access
The latest audio-visual equipment
Plasma TVs
As a math major, you'll also have access to the latest software on supercomputers in our new MACMAS lab.
What will I learn outside the math program?
JCSU is a private liberal arts university. What this means to you is that your education will broadly cover all areas of study. The diversity of your classes will help you look at "the big picture" in any career you choose.
Classes in computer science and communicationarts will give you a definite advantage in the job market and workplace.
Taking several English and foreign language classes will help you interact with a global community that is getting smaller each day.
By taking history and science classes, you'll have a better understanding of the world around you.
Health and physical education courses give you the knowledge to cope with everyday stress and lead a better, longer life.
Having a college education will give you the tools to succeed in life. Having a liberal arts education will give you the opportunity to take your success further. |
Finite Mathematics for Business, Economics, Life Sciences and Social Sciences - 11th edition
Summary: This book covers mathematics of finance, linear algebra, linear programming, probability, and descriptive statistics, with an emphasis on cross-discipline principles and practices. Designed to be reader-friendly and accessible, it develops a thorough, functional understanding of mathematical concepts in preparation for their application in other areas. Each chapter concentrates on developing concepts and ideas followed immediately by developing computational skills a...show morend problem solving. Two-part coverage presents a library of elementary functions and finite mathematics. For individuals looking for a view of mathematical ideas and processes, and an illustration of the relevance of mathematics to the real world. Illustrates relevance of mathematics to the real world2007 Hardcover This book is in good condition and clean copy. Same day shipping. Thank you.
$14.99 +$3.99 s/h
Good
Weirdbooks Napa, CA
Good, some shipping damage to cover- ding to right cover edges and bump to corner2007-02-03 |
This web page provides a preliminary look at the pedagogy behind a vision to improve the teaching of mathematics and to provide math relevant to students studying emerging technologies. Recommendations stress critical...
Are the technical math courses needed for your technology courses teaching the right math requirements? Please see this web site for ways to improve and enhance your math offerings. AMATYC has funding for the NSF CCLIThis section of Crossroads in Mathematics, a guideline for improving introductory postsecondary mathematics courses, outlines the standards for the content of mathematics courses. While mathematical content has...
This learning community, created by Juan Morata and Miguel Montanez, integrates biology and algebra through joint group projects, joint case studies, and class examples and exercises. Through the integrated approach,... |
Beginning and Intermediate Algebra - 4th edition
Summary: Get Better Results with high quality content, exercise sets, and step-by-step pedagogy! The Miller/O'Neill/Hyde author team continues to offer an enlightened approach grounded in the fundamentals of classroom experience in Beginning and Intermediate Algebra 4e. The text reflects the compassion and insight of its experienced author team with features developed to address the specific needs of developmental level students. Throughout the text, the authors communicate to students the ve...show morery points their instructors are likely to make during lecture, and this helps to reinforce the concepts and provide instruction that leads students to mastery and success. Also included are Problem Recognition Exercises, designed to help students recognize which solution strategies are most appropriate for a given exercise. These types of exercises, along with the number of practice problems and group activities available, permit instructors to choose from a wealth of problems, allowing ample opportunity for students to practice what they learn in lecture to hone their skills. In this way, the book perfectly complements any learning platform, whether traditional lecture or distance-learning; its instruction is so reflective of what comes from lecture, that students will feel as comfortable outside of class as they do inside class with their instructor |
Related Directories
Learn or re-learn the algebra you need to know to solve math problems found on your certification exam or during your daily work. This 8 lecture course will show you how to solve problems and give you plenty of guided practice to master the techniques |
MATH 220 - LINEAR ALGEBRA
This section of Math 220 teaches elementary linear algebra with
MAPLE. That is, it is a "computer oriented" linear algebra section. You
don't need to be a computer expert to register in this section. You will
learn all what you need for the course during the semester. However you
must be aware that this means that there are going to be computer LABS
concerning the topics we discuss in class. The final grade will be based
on the results from 4 exams PLUS the labs.
SYLLABUS
This course will cover the basics of matrices and their applications:
solving systems of equations, finding inverse matrices, determinants, applications
to vector spaces, eigenvalues and more. We will solve problems by hand
and deal with more complicated systems by using maple. We will get familiar
with maple while solving problems.
There will be four exams. I will count the best three towards your final
grade. There will also be assignments to be handed in, at least once a
week. Your grade for the course will be the sum of all these exams plus
the grade from your assignments. There will be no cumulative final exam.
Attendance is optional, but it is your responsibility to be aware of
the dates of the exams and the content and due date of assignments. If
you miss a class, it is your responsibility to be aware of the topics discussed
during that class, the assigned homework and the possibly given assignment.
There is no reason to miss an exam other than getting sick (bring note
from doctor), being on a team that has a game at the same time an exam
is given (bring a note from your coach), or a death or serious illness
in your family. In the event you can not attend an exam, you need to notify
me IN ADVANCE. You can call me (2-4641), e-mail me (reinhold@math.albany.edu),
or leave a message under my door or with the secretaries.
EXAMS' SCHEDULE
The schedule for the exams is subject to change. Please, keep abreast
of any possible changes. The current plan is as follows: |
This course focuses on linear ordinary differential equations (or ODEs) and will introduce several other subclasses and their...
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This course focuses on linear ordinary differential equations (or ODEs) and will introduce several other subclasses and their respective properties. Despite centuries of study, numerical approximation is the only practical approach that has emerged to the solution of complicated ODEs ; this course will introduce you to the fundamentals behind numerical solutions. This free course may be completed online at any time. See course site for detailed overview and learning outcomes. (Mathematics 221; Mechanical Engineering 003; Computer Science)
Partial differential equations (PDEs) describe the relationships among the derivatives of an unknown function with respect to...
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Partial differential equations (PDEs) describe the relationships among the derivatives of an unknown function with respect to different independent variables, such as time and position. A very large fraction of solvable PDEs are either linear first- or second-order PDEs, or are related to such PDEs by transformation or perturbation theory. Fortunately, these PDEs also make up the language for much of the mathematical description of nature. Most of this class will concentrate on those equations that have proven to be of great importance to real-world applications. This free course may be completed online at any time. See course site for detailed overview and learning outcomes. (Mathematics 222)
Undergraduate computer lab designed to teach quantitative thinking in the context of neurobiology. Ideally, this lab would...
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Undergraduate computer lab designed to teach quantitative thinking in the context of neurobiology. Ideally, this lab would be taught as a supplement to a concurrent lecture course. Students are assumed to have completed one year of undergraduate calculus.Topics include Nernst equation, GHK equation, passive electrical spread, voltage clamp, action potentials, synaptic currents, reciprocal inhibition, lateral inhibition, and the swimming circuit of Tritonia. Math skills used include graphing, logarithms, differential equations, and discrete-time models.The modules are designed to be self-contained lab exercises. They are Mathcad documents that the students complete for credit. Thus, students must have access to Mathcad (version 13 or higher). PDF versions of the modules are also provided for demonstration purposes.
Differential Equations are the language in which the laws of nature are expressed. Understanding properties of solutions of...
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Differential Equations are the language in which the laws of nature are expressed. Understanding properties of solutions of differential equations is fundamental to much of contemporary science and engineering. Ordinary differential equations (ODE's) deal with functions of one variable, which can often be thought of as time. |
... read more
Introductory Statistical Mechanics for Physicists by D. K. C. MacDonald This concise introduction is geared toward those concerned with solid state or low temperature physics. It presents the principles with simplicity and clarity, reviewing issues of critical interest. 1963 edition.
Counterexamples in Topology by Lynn Arthur Steen, J. Arthur Seebach, Jr. Over 140 examples, preceded by a succinct exposition of general topology and basic terminology. Each example treated as a whole. Numerous problems and exercises correlated with examples. 1978 edition. Bibliography.
Differential Topology: First Steps by Andrew H. Wallace Keeping mathematical prerequisites to a minimum, this undergraduate-level text stimulates students' intuitive understanding of topology while avoiding the more difficult subtleties and technicalities. 1968 edition.
Elementary Concepts of Topology by Paul Alexandroff Concise work presents topological concepts in clear, elementary fashion, from basics of set-theoretic topology, through topological theorems and questions based on concept of the algebraic complex, to the concept of Betti groups. Includes 25 figures.
General Topology by Stephen Willard Among the best available reference introductions to general topology, this volume is appropriate for advanced undergraduate and beginning graduate students. Includes historical notes and over 340 detailed exercises. 1970 edition. Includes 27 figuresReal Variables with Basic Metric Space Topology by Robert B. Ash Designed for a first course in real variables, this text encourages intuitive thinking and features detailed solutions to problems. Topics include complex variables, measure theory, differential equations, functional analysis, probability. 1993 edition.
Topology for Analysis by Albert Wilansky Three levels of examples and problems make this volume appropriate for students and professionals. Abundant exercises, ordered and numbered by degree of difficulty, illustrate important topological concepts. 1970 editionA Modern View of Geometry by Leonard M. Blumenthal Elegant exposition of the postulation geometry of planes, including coordination of affine and projective planes. Historical background, set theory, propositional calculus, affine planes with Desargues and Pappus properties, much more. Includes 56 figuresProduct Description:
applied mathematics. It assumes no detailed background in topology or geometry, and it emphasizes physical motivations, enabling students to apply the techniques to their physics formulas and research. "Thoroughly recommended" by The Physics Bulletin, this volume's physics applications range from condensed matter physics and statistical mechanics to elementary particle theory. Its main mathematical topics include differential forms, homotopy, homology, cohomology, fiber bundles, connection and covariant derivatives, and Morse theory |
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Featured Research
from universities, journals, and other organizations
Math goes viral in the classroom
Date:
December 14, 2009
Source:
University of Alberta
Summary:
At least a dozen Alberta high-school calculus classrooms were exposed to the West Nile virus recently. Luckily, it wasn't literally the illness. Educators used the virus as a theoretical tool when they designed materials for use in an advanced high-school math course.
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At least a dozen Alberta high-school calculus classrooms were exposed to the West Nile virus recently.
Luckily, however, it wasn't literally the illness. University of Alberta education professor Stephen Norris and mathematics professor Gerda de Vries used the virus as a theoretical tool when they designed materials for use in an advanced high-school math course. The materials allow students to use mathematical concepts learned in their curriculum to determine the disease's reproductive number, which determines the likelihood of a disease spreading.
The approach is a marriage of science and math, subjects the researchers say seem to exist in separate worlds at a secondary-school level, but that when brought together can effectively bring real-world scenarios into the classroom to enhance learning and understanding.
Not to mention answering that ages old high-school student question: "why do I need to know this?"
"This piece was designed to satisfy an optional unit in Math 31 (Calculus), for which there are no materials, so we said, 'let's fill the gap,'" said Norris. "These materials show a real application of mathematics in the biology curriculum for high-school students."
Norris and de Vries chose a published academic math paper on the transmission of the West Nile virus and modified it -keeping the science intact, but making it readable and practical for high-school calculus students.
The information and equations in the original paper dealing with disease transmission were then used as the basis for calculus math problems to be solved by the students. Students were presented with a variety of materials that covered topics and concepts such as rate of change, exponential growth-decay models, and models for the carriers of the virus, including mosquitoes and infectious and susceptible birds. The students' mathematical skills were then put to use in determining the spread of the disease using various parameters, which included variables such as biting rate and the probability of infection.
Norris underlines that the project challenged the students to see and understand science in a different fashion from what they learn inside the science curricula. He points out that high-school classroom scientific experiments are "proven" science and have been around for at least 300 years, in many cases. For the students to discover that real scientists often work with some assumptions that they know to be false in order to reach their conclusions was certainly an eye-opening realization for them, he says.
"There's no way out of the fact that the knowledge you gain from science is imperfect; it's tentative and subject to change," said Norris. "I think that's what struck the students between the eyes."
Both researchers agree that this form of collaborative, interdisciplinary learning can take place across all subject areas. De Vries and Norris are currently working on another project that focuses on population genetics that will fit into Grade 12 biology and math courses.
"It's mathematics in the real world. Kids are always asking, 'why am I learning this,'" she said. "All of a sudden the mathematics that kids have learned comes together in a project like 3, 2014 — Single-sex education does not educate girls and boys any better than coed schools, according to research analyzing 184 studies of more than 1.6 million students from around the ... full story
Oct. 11, 2013 — Writing instruction in US classrooms is "abysmal" and the Common Core State Standards don't go far enough to address glaring gaps for students and teachers, an education scholar |
Beginning Algebra - With CD - 5th edition
Summary: KEY MESSAGE:Elayn Martin-Gay'sdevelopmental math textbooks and video resources are motivated by her firm belief that every student can succeed. Martin-Gay's focus on the student shapes her clear, accessible writing, inspires her constant pedagogical innovations, and contributes greatly to the popularity and effectiveness of her video resources. This revision of Martin-Gay's algebra series continues this focus on students and what they need to be successful. Martin-Gay also strives t...show moreo provide the highest level of instructor and adjunct support. Review of Real Numbers; Equations, Inequalities, and Problem Solving; Graphing; Solving Systems of Linear Equations and Inequalities; Exponents and Polynomials; Factoring Polynomials; Rational Expressions; Roots and Radicals; Quadratic Equations For all readers interested in algebra, and for all readers interested in learning or revisiting essential skills in beginning algebra through the use of lively and up-to-date applications. ...show less
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of The Theory of Interest is significantly revised and expanded from previous editions. The text covers the basic mathematical theory of interest as traditionally developed. The book is a thorough treatment of the mathematical theory and practical applications of compound interest, or mathematics of finance. The pedagogical approach of the second edition has been retained in the third edition. The textbook narrative emphasizes both the importance of conceptual understanding and the ability to apply the techniques to practical problems. The third edition has considerable updates that make this book relevant to students in this course area. |
New Almaden Calculus...Many times the instructor has assumptions and pieces of knowledge that he/she is unaware of that enable him/her to understand the concept he/she is explaining; bad explanation is the result of the teacher assuming that the student already has this knowledge because it seems so obvious to the teac... |
Vector and Geometric Calculus
This textbook for the undergraduate vector calculus course presents a unified treatment of vector and geometric calculus. It is a sequel to the text Linear and Geometric Algebra by the same author. That text is a prerequisite for this one.
Linear algebra and vector calculus have provided the basic vocabulary of mathematics in dimensions greater than one for the past one hundred years. Just as geometric algebra generalizes linear algebra in powerful ways, geometric calculus generalizes vector calculus in powerful ways.
Traditional vector calculus topics are covered, as they must be, since readers will encounter them in other texts and out in the world.
Differential geometry is used today in many disciplines. A final chapter is devoted to it.
Visit the book's web site: to download the table of contents, preface, and index.
From a review of Linear and Geometric Algebra:
Alan Macdonald's text is an excellent resource if you are just beginning the study of geometric algebra and would like to learn or review traditional linear algebra in the process. The clarity and evenness of the writing, as well as the originality of presentation that is evident throughout this text, suggest that the author has been successful as a mathematics teacher in the undergraduate classroom. This carefully crafted text is ideal for anyone learning geometric algebra in relative isolation, which I suspect will be the case for many readers. |
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Synopses & Reviews
Publisher Comments:
Would you order a multi-course gourmet buffet and just eat salad? If you have a TI-83 Plus graphing calculator, you have a veritable feast of features and functions at your fingertips, but chances are you don't take full advantage of them. This friendly guide will help you explore your TI-83 Plus Graphing Calculator and use it for all it's worth, and that's a lot. With easy-to-follow, step-by-step instructions plus screen shots, TI-83 Plus Graphing Calculator For Dummies shows you how to:
Perform basic arithmetic operations
Use Zoom and panning to get the best screen display
Use all the functions in the Math menu, including the four submenus: MATH, NUM, CPS, and PRB
Use the fantastic Finance application to decide whether to lease or get a loan and buy, calculate the best interest, and more
Graph and analyze functions by tracing the graph or by creating a table of functional values, including graphing piecewise-defined and trigonometric functions
Explore and evaluate functions, including how to find the value, the zeros, the point of intersection of two functions, and more
Draw on a graph, including line segments, circles, and functions, write text on a graph, and do freehand drawing
Work with sequences, parametric equations, and polar equations
Use the Math Probability menu to evaluate permutations and combinations
Enter statistical data and graph it as a scatter plot, histogram, or box plot, calculate the median and quartiles, and more
Deal with matrices, including finding the inverse, transpose, and determinant and using matrices to solve a system of linear equations
Once you discover all you can do with your TI-83 Plus Graphing Calculator, you'll find out how to make it do more! This guide shows you how to download and install the free TI Connect software to connect your calculator to your computer, and how to link it to other calculators and transfer files. It shows you how to help yourself to more than 40 applications you can download from the TI Web site, and most of them are free. You can choose from Advanced Finance, CellSheet, that turns your calculator into a spread sheet, NoteFolio that turns it into a word processor, Organizer that lets you schedule events, create to-do lists, save phone numbers and e-mail addresses, and more.
Get this book and discover how your TI-83 Plus Graphing Calculator can solve all kinds of problems for you.
Synopsis:
Use this cool tool to perform dozens of tasks!About the Author
C.C. Edwards is an instructor at Coastal Carolina University. Previously, she was the editor of Eightysomething, Texas Instruments' newsletter that explored the features of graphing calculators for parents and educators. |
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Having trouble understanding algebra? Do algebraic concepts, equations, and logic just make your head spin? We have great news: Head First Algebra is designed for you. Full of engaging stories and practical, real-world explanations, this book will help you learn everything from natural numbers and exponents to solving systems of equations and graphing polynomials.
Along the way, you'll go beyond solving hundreds of repetitive problems, and actually use what you learn to make real-life decisions. Does it make sense to buy two years of insurance on a car that depreciates as soon as you drive it off the lot? Can you really afford an XBox 360 and a new iPhone? Learn how to put algebra to work for you, and nail your class exams along the way.
Your time is way too valuable to waste struggling with new concepts. Using the latest research in cognitive science and learning theory to craft a multi-sensory learning experience, Head First Algebra uses a visually rich format specifically designed to take advantage of the way your brain really works.
Have a killer app idea for iPhone and iPad? Head First iPhone and iPad Development will help you get your first application up and running in no time. You'll not only learn how to design for Apple's devices, you'll also master the iPhone SDK tools -- including Interface Builder, Xcode, and Objective-C programming principles -- to create eye-catching, top-selling apps |
Hialeah Gardens, FL ACTI am looking forward to working with you soon.Factoring is the main emphasis after Prealgebra concerns. Prerequisites include solving 1 and 2 step linear equations; adding, subtracting, multiplying, and dividing integers; combining like terms; the Order of Operations. Prerequisites include fact |
Free Algebra Lessons from Purplmath
Lessons: "How do you really do this stuff?" — Purplemath's algebra lessons are written with the student in mind. These lessons emphasize the practicalities rather than the technicalities, demonstrating dependable techniques, warning of likely "trick" questions, and pointing out common mistakes. The lessons are cross-referenced to help you find related material, and a "search" box is on every page to help you find what you're looking for.
You can access their preliminary through advanced Algebra lessons here |
Euclidean And Transformational Geometry: A Deductive Inquiry
Book Description: Ideal for mathematics majors and prospective secondary school teachers, Euclidean and Transformational Geometry provides a complete and solid presentation of Euclidean geometry with an emphasis on solving challenging problems. The author examines various strategies and heuristics for approaching proofs and discusses the process students should follow to determine how to proceed from one step to the next through numerous problem solving techniques. A large collection of problems, varying in level of difficulty, are integrated throughout the text and suggested hints for the more challenging problems appear in the instructor's solutions manual and can be used at the instructor's discretion |
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Starting at $73 Lial Series has helped thousands of students succeed in developmental mathematics by providing the best learning and teaching support to students and instructors.
Author Biography
Marge Lial became interested in math at an early age—it was her favorite subject in the first grade! Marge's intense desire to educate both her students and herself inspired the writing of numerous best-selling textbooks. Marge, who received Bachelor's and Master's degrees from California State University at Sacramento, was affiliated with American River College. An avid reader and traveler, her travel experiences often found their way into her books as applications, exercise sets, and feature sets. She was particularly interested in archeology; trips to various digs and ruin sites produced some fascinating problems for her textbooks involving such topics as the building of Mayan pyramids and the acoustics of ancient ball courts in the Yucatan. We dedicate the new editions of the paperback developmental math series to Marge in honor of her contributions to the field in which she helped thousands of students succeed.
When John Hornsby enrolled as an undergraduate at Louisiana State University, he was uncertain whether he wanted to study mathematics education or journalism. His ultimate decision was to become a teacher, but after twenty-five years of teaching at the high school and university levels and fifteen years of writing mathematics textbooks, both of his goals have been realized. His love for both teaching and for mathematics is evident in his passion for working with students and fellow teachers as well. His specific professional interests are recreational mathematics, mathematics history, and incorporating graphing calculators into the curriculum. John's personal life is busy as he devotes time to his family (wife Gwen, and sons Chris, Jack, and Josh). He has been a rabid baseball fan all of his life. John's other hobbies include numismatics (the study of coins) and record collecting. He loves the music of the 1960s and has an extensive collection of the recorded works of Frankie Valli and the Four Seasons.
A native Midwesterner, Terry McGinnis received her Bachelor's of Science in Elementary Education with a concentration in Mathematics from Iowa State University. She has taught elementary and middle school mathematics, and developed and implemented the curriculum used with her students. Terry has been involved in college mathematics publishing for over 20 years, working with a variety of authors on textbooks in developmental mathematics and precalculus. After working behind the scenes on many of the Lial/Hornsby textbooks and supplements for over 10 years, Terry joined Margaret Lial and John Hornsby in 2002 as coauthor of their developmental mathematics series that includes Introductory Algebra, Intermediate Algebra, and Introductory and Intermediate Algebra. When not working, Terry enjoys spinning at a local health club, walking, and reading fiction. She is the devoted mother of two sons, Andrew and Tyler.
Diana Hestwood lives in Minnesota and has taught at Metropolitan Community College in Minneapolis for two decades. She has done research on the student brain and is an expert on study skills. She is the author of Lial/Hestwood's Prealgebra and coauthor of Lial/Salzman/Hestwood's Basic Math and Lial/Hestwood/Hornsby/McGinnis's Prealgebra and Introductory Algebra.
Stan Salzman is a long time resident of Sacramento, California. Stan has taught at American River College for many years, where he was a member of the business department. He is the author of Business Math and Essential Math, published by Pearson Education, Inc., and is coauthor of Basic Math.
Table of Contents
1. Whole Numbers
1.1 Reading and Writing Whole Numbers
1.2 Adding Whole Numbers
1.3 Subtracting Whole Numbers
1.4 Multiplying Whole Numbers
1.5 Dividing Whole Numbers
1.6 Long Division
1.7 Rounding Whole Numbers
1.8 Exponents, Roots, and Order of Operations
1.9 Reading Pictographs, Bar Graphs, and Line Graphs
1.10 Solving Application Problems
2. Multiplying and Dividing Fractions
2.1 Basics of Fractions
2.2 Mixed Numbers
2.3 Factors
2.4 Writing a Fraction in Lowest Terms
2.5 Multiplying Fractions
2.5 Applications of Multiplication
2.7 Dividing Fractions
2.8 Multiplying and Dividing Mixed Numbers
3. Adding and Subtracting Fractions
3.1 Adding and Subtracting Like Fractions
3.2 Least Common Multiples
3.3 Adding and Subtracting Unlike Fractions
3.4 Adding and Subtracting Mixed Numbers
3.5 Order Relations and the Order of Operations
4. Decimals
4.1 Reading and Writing Decimal Numbers
4.2 Rounding Decimal Numbers
4.3 Adding and Subtracting Decimal Numbers
4.4 Multiplying Decimal Numbers
4.5 Dividing Decimal Numbers
4.6 Fractions and Decimals
5. Ratio and Proportion
5.1 Ratios
5.2 Rates
5.3 Proportions
5.4 Solving Proportions
5.5 Solving Application Problems with Proportions
6. Percent
6.1 Basics of Percent
6.2 Percents and Fractions
6.3 Using the Percent Proportion and Identifying the Components in a Percent Problem
6.4 Using Proportions to Solve Percent Problems
6.5 Using the Percent Equation
6.6 Solving Application Problems with Percent
6.7 Simple Interest
6.8 Compound Interest
7. Geometry
7.1 Lines and Angles
7.2 Rectangles and Squares
7.3 Parallelograms and Trapezoids
7.4 Triangles
7.5 Circles
7.6 Volume and Surface Area
7.7 Pythagorean Theorem
7.7 Congruent and Similar Triangles
8. Statistics
8.1 Circle Graphs
8.2 Bar Graphs and Line Graphs
8.3 Frequency Distributions and Histograms
8.4 Mean, Median, and Mode
9. The Real Number System
9.1 Exponents, Order of Operations, and Inequality
9.2 Variables, Expressions, and Equations
9.3 Real Numbers and the Number Line
9.4 Adding Real Numbers
9.5 Subtracting Real Numbers
9.6 Multiplying and Dividing Real Numbers
9.7 Properties of Real Numbers
9.8 Simplifying Expressions
10. Equations, Inequalities, and Applications
10.1 The Addition Property of Equality
10.2 The Multiplication Property of Equality
10.3 More on Solving Linear Equations
10.4 An Introduction to Applications of Linear Equations
10.5 Formulas and Additional Applications from Geometry
10.6 Solving Linear Inequalities
11. Graphs of Linear Equations and Inequalities in Two Variables
11.1 Linear Equations in Two Variables; The Rectangular Coordinate System |
This
chapter presents the answers to the four research questions proposed for this
study in chapter one, based on the analyses carried out in chapter four.Conclusions, limitations of the study and
recommendations follow in order to complete this chapter.
1.How did the
students in the two groups, experimental and control, compare in their prior
achievement and attitudes, and their experiences with technology?
2.What
relationships appear to exist between attitudes and achievement in the learning
of linear functions activities?
3.At which level
and in what ways, can the use of multiple representations be supported by
spreadsheets learning activities promote to better promote understanding of
linear functions in students at college level algebra?
4.How well does the
medium of a powerful spreadsheet like Excel, lend itself to promoting
instruction through multiple representations?
The following sections will answer each one of
these questions separately.
Answering the First Research Question
1.How did the
students in the two groups, experimental and control, compare in the prior
achievement and attitudes, and their experiences with technology?
Prior achievement on mathematics based on
grades reported by students from both groups was examined.The students' achievement in mathematics is
represented by grades referred to a general knowledge in this field, not
necessarily in the topic of linear functions.It is important to point out here that students in both groups had
completely different prior experiences, as well as, performances in
mathematics.This was attributed to the
fact of receiving the course from different schools programs (public and
private).The research question was
formulated, as the first one, since the prior achievement in mathematics was
the only variable that students exhibited before treatment for the study
started.
The results dealing with this variable,
reported on chapter four revealed no significant difference (p > .05)
between the control and experimental groups on prior achievement in
mathematics.Based on these findings,
prior achievement in mathematics based on grades seemed not to be determinant
factor in helping student attain a broad understanding about linear
functions.These results show that a
good prior achievement in mathematics did not necessarily imply a better
achievement in linear functions.Different mathematics topics were taught in a college level algebra
course.In this study, linear functions
were strongly emphasized.According to
the results, a good performance in prior mathematics classes did not guarantee
a good performance in linear functions.
In
terms of technology use in prior mathematics courses, the students' profiles
revealed that the use of computers, calculators, spreadsheets, Internet, among
other tools, were almost nonexistent.Further, the profiles indicated that for the majority of the students in
the study, it was their first experience using technology, particularly,
spreadsheets, in their mathematics courses. The findings indicate that in the
experimental group, previous mathematics achievement , particularly those
supported by technology were not a decisive factor in promoting better
understanding in linear functions using spreadsheets and multiple
representations.
Answering the Second Research Question
2.What
relationships appear to exist between attitudes and achievement in the learning
of linear functions activities?
Attitudes
toward mathematics were explored in this study by collecting data through a
scale that was administered in the beginning and at the end of the treatment,
for control and experimental groups.The results on this variable, summarized on the previous chapter,
indicate that no significant (p > .05) was found on the comparisons
made.
The items of the attitudes scale toward
mathematics were organized in two clusters: attitudes toward technology and its
uses, and opinion or feelings toward mathematics as a subject.In terms of technology uses in mathematics
and by inspecting students' responses on items 1 and 2 of the attitudes scale,
the experimental group had not significant (p > .05) more positive
experiences than the control group with the use of calculators to perform
routine calculations in their mathematics courses prior the study.In terms of the attitudes toward technology,
the control group exhibited a very slight (1 point of difference) positive
attitude at the beginning of the study, according to the inspection on item 3
of the scale.Once the treatment
concluded, the experimental group seemed to have a gain in positive attitudes
toward technology.These results agreed
with the treatments that each group received, as described on chapter three.
The treatment received by the experimental
group as part of the study was based on intensive use of spreadsheets while
emphasizing multiple representations of linear functions.The improvement in attitudes toward mathematics
from this group is noticeable.It could
be inferred that the traditional approach, based strictly in the use of
textbook and not technology allowed to enhance learning, did not promote an
improvement in the attitudes of the control group at the end of the study.
These
results on attitudes toward mathematics agreed with the findings on achievement
discussed in a previous section of this chapter.Based on this information, it would be sensible to conclude in
this study, that attitudes toward mathematics seemed to have limited effects on
mathematical achievement.
Answering the Third Research Question
3.At which level
and in what ways, can the use of multiple representations be supported by the
spreadsheets learning activities to better promote understanding of linear
functions in students at college level algebra?
Since the capabilities of spreadsheets to
illustrate the multiple representations (symbolic, graphical, tabular, and
verbal) of a linear function, this technology was intensively used in this
study, particularly with the experimental group, to teach the mathematics
concepts in a college level algebra course.The use of spreadsheets allowed students to work with multiple
representations of linear functions.As
Kaput (1992) affirms, linking one representation with the others together in
the same computer screen was essential to understanding the advantages of
technology in learning.This capability
also provided student with the understanding of how all the representations
referred to the same concept (Keller and Hirsch, 1998). Through the use of spreadsheets on the
mathematics lessons, students interacted with all representations and they
examined all the effects caused on representations when the x variable
from a linear functions assumed different values.
Mathematics
achievement, specifically on linear functions, was one of the variables
explored in this study.This variable
was measured through an achievement test, emphasizing multiple representations
administered at the beginning and at the end of the treatment.The results, reported on chapter four,
indicate that at the beginning of the study, the control group exhibited a
significant (p < .05) higher achievement than the experimental
group.It is assumed that the
experimental group's intensive use of spreadsheets and multiple representations
in the mathematical lessons through the treatment reflected a significant (p
< .05) improvement in achievement at the end of the study.
Moreover, in order to explore students'
performance on achievement in specific content topics related to linear
functions, the research instrument on achievement was divided in a cluster
dealing with these topics.A
significant (p < .05) higher achievement was observed in favor of the
control group in all of these areas at the beginning of the study.Once the treatment concluded, no significant
(p > .05) was found between groups. It was observed in the
experimental group a significant gain (p < .05) in achievement in
mathematics in the areas of graphs and slope.Both groups exhibited significant (p < .05) improvements in
the content of Cartesian coordinates once the teaching experiment
concluded.Interestingly, the two-way
ANOVA reported significant (p < .05) interactions between effects
(pre-post administrations and groups) in the topic of slope.
In terms of the representations of the linear
function, the results revealed that in three of these representations
(symbolic, tabular, and verbal) significant differences(p < .05) were found at the end of
the study in the experimental group.That is, students in the experimental group performed higher at the end
of this study in achievement on linear functions through symbolic, tabular, and
verbal representations.In terms of the
graphical representations, both groups exhibited a significant improvement (p
< .05) on achievement at the post administration of the test.The analysis of variance (ANOVA) carried out
between effects (groups and administrations of the test) reported significant
interactions (p < .05) in graphical and verbal representations.
The
results of this study on achievement in mathematics, through the emphasis
placed on multiple representations supported by the use of spreadsheets,
suggest that the approach presented served to promote a better understanding of
linear functions on students in a college algebra course.It was observed that students, who used
multiple representations and spreadsheets as part of their mathematics course,
performed higher in achievement in mathematics (linear functions) at the end of
the study that did the students in the control group.The data from this study suggest that the multiple
representations approach supported by technology was more successful in promote
achievement gain than the traditional approach.The findings of this research are supported by previous research
in the field of representations done by Porzio (1994).His studies revealed that the emphasis
placed in multiple representations and technology was more adequate to promote
understanding and connections between representations.
Answering the Fourth Research Question
4.How well does the
medium of a powerful spreadsheets like Excel, lend itself to promoting
instruction through multiple representations?
Spreadsheets
were selected as the technology medium to be used in this research project
because their capabilities to promote and show the multiple representations of
the linear functions.The software
provided an engaging environment in where students explored all the
representations separately first, and then, all together in the same
worksheet.The linking process between
representations strongly recommended by Kaput (1992), Keller and Hirsch (1998),
and Dufour-Janvier, et al. (1987) was particularly shown on the spreadsheets.
Mathematical lessons based on spreadsheets
developed in this study, allowed students in the experimental group to explore
the effects of different values on the representations. This technology
fulfilled the objectives set for this study and supported the instructional
activities.Not only multiple
representations were supported by the use of spreadsheets,the mathematical concepts (Cartesian
coordinates, graphs, and slope) taught during the course, were introduced
through this technology. Recognized
scholars such as Fey (1989), Goldenberg (1987), Kaput (1992), and Porzio (1994)
have noticed that technology has contributed to increase the access of multiple
representations ofmathematical
concepts.
In
terms of promoting instruction, the results on achievement for this study
showed that the spreadsheets approach using multiple representations was more
adequate than the traditional approach.Furthermore, this approach based on spreadsheets seemed to serve to
enhance higher attitudes toward mathematics and technology.
Comparing the Two Approaches Used in this
Study: The Multiple Representations and the Traditional
The
purpose of this section is to present how the same mathematical concept was
taught using two different approaches.The multiple representations, supported with spreadsheets approach used
with the experimental group and the traditional used with the control
group.The topic discussed in this
class was Cartesian coordinates.
Multiple Representation Approach
In
the class dealing with Cartesian coordinates, students in the experimental
group used a worksheet.In the first
part of this activity, six different coordinates were given and students had to
fill in the corresponding blanks the location of each one of these points.In the second part of the activity, students
offered their own coordinates, different from the presented in part one, and
satisfying certain given locations. Then, using spreadsheets they were asked to
locate the coordinates and explore the effects of points locations when the
values of x and y in (x, y) changed.To end the activity, students were
encouraged to get printouts in order to show their work.Finally, theyfind an equation related to these coordinates and told a story
about the application of Cartesian coordinates in their fields of study.Figures 19, 20, and 21 show the first and
the last parts of this activity done by Student 22.The answers to the questions on the worksheet appear in italic
font and printouts from spreadsheets appear in the following set of figures.
Give an example about an equation that can
relate one of the coordinates stated above.
The equation y = is satisfied by the point (2,4). Because
if you substitute the values on the equation: 4 = you get a true
statement: 4 = 4.
Tell a story about the applications that
Cartesian coordinates may have in daily situations or another fields:
I think that in medicine they serve to
imagine two cut points in a surgery.Also, to make a map of the head where it is divided in sections.Finally, coordinates may be used to
measure distances between different bones of the body.
Traditional Approach
This
approach was based on instructor lectures.The only resource used in this lesson was the course textbook.Figures 22 and 23 show samples of a
student's notes from the control group.All notes are in Spanish.
Mathematics
Reasoning (MRSG 1010) is a college mathematics core course, intended primarily
for freshmen students, where technology has been slightly used throughout the semesters
that it has been taught.Although this
course has been reviewed in multiple times in terms of its objectives and how
they have been fulfilled or not, the use of more technology has been limited
and unfortunately, out of the realm of discussion. Nevertheless, regular
instructors have supported mainly the use of calculators as a supplement to
instruction.The researcher was
convinced that another kind of technology (beyond calculators) could be used in
this course to enhance achievement in mathematics.Therefore, the main challenge for the investigator of this
research was to incorporate the intensive use of computers, particularly the
use of spreadsheets to the instructional activities dealing with linear functions
of this course.
The incorporation of this technology did not
constitute an easy activity in this project.In the beginning of the study, the students exhibited surprise and
sometimes lack of belief about how through the use of spreadsheets they could
learn the same mathematical content knowledge as did the traditional
students.Furthermore, the researcher
had to deal with the lack of expertise on students in the use of
spreadsheets.As stated earlier, this
experience using technology represented to the majority of the students their
first time using computers in a mathematics course.In order to correct this deficiency, the instructor of this study
spent some class periods teaching the basic features of spreadsheets and how
they can be used in mathematics.
In this research project, the investigator
dealt mainly with two important aspects: (a) the mathematics topics included in
the course syllabus taught during the length of the study, and (b) students
attitudes dealing particularly with the concern if spreadsheets will work or
not in order to learn linear functions and related themes.The first aspect was fulfilled when the
instructional topics were taught parallel with the traditional group and in
accordance with the syllabus.With the
purpose to fulfill the second aspect, the investigator had to motivate,
encourage and over all, show and introduce each content topic, in each class
session using spreadsheets.In this
way, students realized, once the treatment concluded, that this technology
constituted a really invaluable tool to learn mathematics.
Some Comments from Students
This
section presents some sample of comments from students in the experimental
group about their experiences using spreadsheets to learn linear functions in
mathematics.These data was collected
through a weekly electronic journal that students sent to the instructor of the
course. The electronic journal contained two questions: a) What is the big idea
that you learned in the class? and b) What topic was difficult or unclear for
you?.
During
the first week of instruction.Here
are some comments from students.
Student
27 said:
The most important thing was how do the graphs
in Excel.
Student 25 said:
The most important idea in this class was how
work with the computer and learn something new with the computer.
Student 22 said:
In my opinion, the most important thing in our
class discussion was the explanation of the spreadsheets as tool for the
course.
During
the second week of instruction. Here are some additional comments from students
in the experimental group.
Student 22 said:
The most important idea for me was the
different representations of the linear equation and how works with them based
in paper and pencil and using spreadsheets.
Student
13 said:
The Cartesian plane has been explained
perfectly.Although I realize that I
found it a little difficult to understand.It is not that it was not well explained, but I am not skillful with
computers.
Student 26 said:
The class with the computer becomes more easy
for me that the class just explaining on the board.
Other
colleagues who were teaching the course during the semester that the study took
place, or who had taught the course before, expressed interest in knowing how
students were performing in the lessons activities based on spreadsheets.For the majority of the regular instructors
of MRSG 1010, the use of spreadsheets and multiple representations to teach
linear functions constituted something new and innovative.
As
a result of the intensive use of technology in a college level mathematics
course, the researcher's approach to teaching changed rather dramatically and
was reinforced in terms of promoting technology use to enhance achievement in
mathematics.As the report Shaping
the Future from the National Science Foundation (George, et al., 1996)
affirms, technology should be available to all students, and they need the
opportunity to work with it, and get expertise using it as a tool of their
learning.It is reasonable to interpret
here that technology in mathematics is intended to first; to provide an
environment to promote understanding and achievement, to get help, offer
alternatives, and sometimes solutions to certain problems.
This
report discusses some barriers found in educational settings.The ineffective use of instructional
technology is one of them, pertinent to this study.This problem consists in "a specific lack of knowledge about the
hardware and technology that has been spreading into increasing use, and to
which many students are already attracted" (p. 44).Discussing on this issue, the report cited the Director of the
Science and Technology Resource Center at the Prince Georges County Community
College.She said:
I see the following as serious problems… the
challenge of teaching faculty and students how to access, utilize, and
incorporate the vast amounts of information available in print and
electronically, and learning how to utilize technology in making education more
attractive to students who might otherwise lack motivation or interest in
science, mathematics, engineering, and technology courses. (p. 44)
These problems identified in the report were
also found in this research experience.The first situation was discussed at the beginning of this section.
Technology has been erroneously used if it
promotes misconceptions, confusion, and unclear solutions.It is important to point out here, that
certain students who participated in this study developed an excessive
dependency to the calculators or computers use to perform simple mathematics
computations and they were unable to perform them without these technological
equipments.This represents an example
about the inappropriate use and promotion of technology, where it is believed
that computers and calculators are magic boxes to solve all the problems in
mathematics.As cited in the NSF
report, Noam (1995), said: "Technology would augment, not substitute" (p. 32).
This
study through the design of mathematics lessons on linear functions, examined first,
three major variables: prior achievement in mathematics (based in reported
grades), achievement in mathematics, and attitudes toward mathematics.Second, it was compared two teaching
approaches: the multiple representation supported by technology (spreadsheets)
and the traditional.
Based
on the analyses of the data provided by the various instruments, this
researcher draws the following conclusions regarding the variables of concern.
With
respect to the prior achievement in mathematics based on reported grades, it is
concluded that this variable did not play a significant role in determining how
much mathematics was learned in the two groups, experimental and control.
Regarding
students attitudes toward mathematics, it is concluded that if positive and
higher attitudes are observed, they are a possible factor to enhance
achievement in mathematics.Students in
the experimental group had somewhat positive more attitudes toward mathematics
and performed higher in achievement.
In
regard to the achievement in mathematics, this researcher concludes that
mathematics lessons emphasizing multiple representations supported by the use
of spreadsheets constitute an appropriate teaching approach to enhance a broad
achievement on linear functions.Students taught with the multiple representations approach achieved
higher on linear functions than students taught with the traditional
approach.
Finally,
regarding the comparison of the two teaching approaches, it was found in this
study that the approach based on multiple representations and spreadsheets is
more effective than the traditional, in promoting and enhancing achievement in
linear functions as well as more positive attitudes toward mathematics.
The
following section presents the limitations of this study.The short period of time (just four weeks of
instruction) devoted to the teaching experiment, instead the whole academic
semester, was the first limitation of this research project.Other findings could be expected if more time
was added to the treatment.
The
numbers of topics taught during this study was other limitation of this
project.Content topics strongly
related with linear functions were only emphasized throughout the study.Future studies could explore the use of
representations with other content topics discussed in a college level algebra
course.
The
researcher was the instructor of the two groups of this study: control and
experimental.Additional research in
this field can determine the effects, and other results of using more than one
teacher in similar conditions.
The
"Hawthorne effect" constituted the another limitation of the study.The changes observed at the end of the
experiment on attitudes and achievement in linear functions between control and
experimental groups could be as a result of other factors involved during the
length of the treatment, rather than the emphasis placed on multiple
representations of linear functions and the intensive use of spreadsheets.Examples of these factors could be: teacher
style, class environment, and students' interests.
3.To provide an
attractive curriculum in undergraduate mathematics that students can feel that mathematics
is useful in their fields of specialization.
4.To encourage
faculty to use in the teaching of this course other technological tools, such
as, spreadsheets.
5.To provide and
promote curricular innovations in the teaching of this course.
6.To encourage
faculty members to do research in all mathematics courses and the publication
of their findings.
To the mathematics faculty:
7.To believe that
all students can learn mathematics in different ways and to create and
harmonious and attractive environment that can engage students in their
learning.
8.To suggest a
curricular review, specifying the mathematical topics that should be taught in
MRSG 1010.To recommend what content
should receive more emphasis and what should receive less.
9.To believe that
technology has changed the education in mathematics and the use of it should
not be omitted.
10.To promote and
encourage the use of technology (such as spreadsheets) in all mathematics
courses, not only used as supplement of instruction.
11.To explore the
use of other teaching approaches, such as multiple representations.Using this approach, each representation of
the same concept is taught and emphasized letting students effectively manage
these representations. |
Intermediate Algebra with Student CD-Rom Windows mandatory package
Book Description: Intermediate Algebra is designed to provide your students with the algebra background needed for further college-level mathematics courses. The unifying theme of this text is the development of the skills necessary for solving equations and inequalities, followed by the application of those skills to solving applied problems. The primary goal in writing the third edition of Intermediate Algebra has been to retain the features that made the second edition so successful, while incorporating the comments and suggestions of second-edition users. Many new features have been provided that will help instructors reach the goals that they have set for their students. As always, the author endeavors to write texts that students can read, understand, and enjoy, while gaining confidence in their ability to use mathematics |
What is Numeracy/QL/QR?
Some call it Numeracy, an expression first used in the UK's 1959 "Crowther Report" to include secondary school students' ability to reason and solve sophisticated quantitative problems, their basic understanding of the scientific method, and their ability to communicate at a substantial level about quantitative issues in everyday life. Others call it Quantitative Literacy (QL), and describe this comfort, competency, and "habit of mind" in working with numerical data as being as important in today's highly quantitative society as reading and writing were in previous generations. Still others refer to it as Quantitative Reasoning (QR), emphasizing the higher-order reasoning and critical thinking skills needed to understand and to create sophisticated arguments supported by quantitative data.
Seminal Books and Articles
In this brief volume (115 pages) Lynn Steen offers a synopsis of major issues raised at the National Forum on QL that was held at the National Academy of the Sciences in Washington, DC in December 2001. Steen describes the place for QL in higher education -- preparing students for civic and economic life in an age dominated by computers and quantitative data. Order from MAA Online Bookstore
This volume, edited by Bernie Madison and Lynn Steen, presents the proceedings of the National Forum on QL held at the National Academy of Sciences in Washington DC in December 2001. The forum, hosted by the NCED, in cooperation with the National Research Council and the Mathematics Association of America, offered background papers and forum papers. The background papers are divided into three types: need of QL for work and learning, curriculum issues, and policy challenges. The forum papers are of four types: need of QL for work and learning, policy perspectives, international perspectives, and reflections and observations. On-line version
Mathematics and Democracy: The Case for Quantitative Literacy, 2001
This publication, produced by the National Council on Education and the Disciplines (NCED) and edited by Lynn Steen, begins with the now oft quoted line "The world of the twenty-first century is a world awash in numbers." (This phrase inspired Beth Fratesi's "Wave of Numbers" used in the NNN's Web site banner.) Mathematics and Democracy makes a strong case for the universality of QL, which is essentially defined as "the use of mathematical and logical tools to solve common problems" in real world contexts. The case statement, a collaborative work by the sixteen-person design team, details the importance of quantitative skills in a wide array of academic fields, in various professions, and in everyday life (particularly in citizenship, personal finance, and personal health). This first chapter also describes the various elements and skills of QL, painting a rich picture of QL. Twelve chapters follow the case statement, each a response by invited authors. On-line version
Why Numbers Count: Quantitative Literacy for Tomorrow's America, 1997
This collection of essays written by diverse consumers of quantitative information represents one of the early attempts to distinguish between mathematics (which was well-defined) and quantitative literacy (which clearly meant different things to different people). As Robert Orill describes in the foreword of this volume, the authors' observations about QL "challenge traditional approaches to orienting the mathematics curriculum" in the US. Order from The College Board |
Tables, Charts, and Graphs
Description: Integrate a real-world, problem-solving focus into math classes! Covers key middle school and high school topics in context of everyday life scenarios Teachs students to create, read, and interpret a variety of visual presentations
Integrate a real-world, problem-solving focus into math classes! Covers key middle school and high school topics in context of everyday life scenarios Teachs students to create, read, and interpret a variety of visual presentations |
This site covers everything from elementary math through calculus, with
detailed, image-rich explanations that go beyond static content with check-able quizzes, a Question and Answer forum and interaction with tutors.
This web site provides web pages that
describe some properties and physical applications of vectors.
Each section builds on the previous ones to make a logical sequence
and there are hot links within sections so that it is easy to
refer back if you want to. This particular link will connect you
to a page that has information on basic trig.
The QuickMath site provides a means
for getting help with common math problems over the Internet.
Think of it as an online calculator that solves equations and
does all sorts of algebra and calculus problems. It contains sections
on algebra, equations, inequalities, calculus, and matrices.
The Math League site offers a 'Help
Facility' that covers decimals, fractions, geometry, introductory
algebra, and much more. While the site targets students in grades
4-8, the information is useful to those who need a refresher on
basic math. This particular link will bring you to a page on ratio
and proportions."
AlgebraHelp.com uses some of the latest
technology to help users learn and understand algebra. The site
features lessons, calculators that show how to solve problems
step-by-step, and interactive worksheets to test skills.
S.O.S. MATHematics is a free resource
for math review material on subjects such as algebra, trigonometry,
calculus, and differential equations. A good site for high school,
college students and adult learners.
This web site provides math reference
tables for General Math, Algebra, Geometry, Trig., Stat., Calc.,
and more. It also contains a math message board that allows users
to have web-based discussion about math. |
More About
This Textbook
Overview
CALCULUS: APPLICATIONS AND TECHNOLOGY is a modern text that is guided by four basic principles: The Rule of Four, technology, the Way of Archimedes, and an exploratory teaching method. Where appropriate, each topic is presented graphically, numerically, algebraically, and verbally, helping students gain a richer, deeper understanding of the material. A pronounced emphasis in the text on technology, whether graphing calculators or computers, permits instructors to spend more time teaching concepts. Additionally, applications play a central role in the text and are woven into the development of the material. More than 500 referenced exercises and hundreds of data sets contained in the text make this text useful and practical for students. Most importantly, this text lets students investigate and explore calculus on their own, and discover concepts |
Book Description: This book can be used individually or as a set with Chenier's Practical Math Dictionary. This book is designed to parallel and enhance any practical math class from general education through college level programs. Many of these math concepts are left out of traditional math books and are relevant to many different trades, occupations, do-it-yourselfers, home owners, home schools, etc. This book includes testing material, economical hands-on projects that simulate industry (use with sticks of wood, chalk lines, flip chart paper, etc.), the answers, and many different unique modules for projects, classroom situations, self-study, industry, etc. All have been proven in the classroom and on-the-job. It's size is 8 1/2" x 10 1/4", has perforated pages and 3 hole drilled |
courses in Algebra-based Physics. Walker's suite of pedagogical tools fosters understanding of physical principles and active involvement in the learning process. Because one-size-fits-all examples do not sufficiently address the needs of students, Walker employs a variety of pedagogical elements, each used where it can contribute most to developing conceptual insight and problem-solving skills. |
Vector Algebra Worksheet
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THEA Math
All about THEA Math
Every student who lives in the state of Texas and is about to finish school, has to give the THEA test, which stands for Texas Higher Education Test. This standardized test was introduced by the Texas Higher Education Coordinating Board in the 1980s. The main idea behind developing this set of tests was to facilitate the admission process for both the applicants as well as for the various colleges and universities across Texas who offer diploma, certificate and graduation programs. The test is conducted in two main formats: Quick test, which is a paper-based format and the Internet-based test, which is in the online format; however, the structure of the test is similar for both the formats. There are three sections in the test: Reading, Writing and Math. This particular article discusses the THEA Math section in detail along with some important points that you must know about this section.
Structure of the THEA Math Section
The THEA Math section, in both the Quick test format and the Internet-based test format, is composed of 50 questions. Each of these questions is objective and multiple-choice based. This means that you need to select one of the given four answer options for each question. There are four main areas that are tested under this particular section:
Problem Solving: These are simple word-problems in which you need to use your knowledge in the basic areas of mathematics.
Geometry: The questions from Geometry can be based on solid geometry (e.g. area, volume, surface area etc) or coordinate geometry (e.g. line equations, symmetry etc).
Algebra: Most of such questions will be about different equations - binomial theorems, quadratic equations, roots of equation etc.
Important Points Regarding the THEA Math
Here a few important points regarding the THEA Math section that you cannot afford to miss:
In the Internet-based format of the test, you will be given an online calculator to help you with lengthy calculations.
In the THEA Math section in the Quick-test format, you can bring your own simple 4-function calculator that is non-programmable. This will help you in solving huge calculations in a short period of time. However, to be on the safe side, you must remember to bring extra batteries for the calculator to the exam.
You will also be given a reference sheet. This sheet will contain a brief list of all important formulae that you need to know for solving the THEA Math section.
Basically, the THEA Math section is a part of this test because it can evaluate your ability to solve various mathematical operations and apply reasoning and problem-solving skills. All these concepts are tested across various freshman-level courses.
Preparation Guidelines for THEA Math Section
Here are some useful tips that you can follow to prepare for the THEA Math section:
The first tip that you can use in your preparation would be to clarify the basic fundamentals of all the concepts tested in Mathematics. For this, you may consult your school books and practice the questions again.
In case you are weak in a particular concept, you can talk to a teacher and consult him or her regarding that particular topic.
You can also make use of many THEA Math self-study guides that are available for online purchase such as Most of these study guides contain a lot of practice questions and also contain a few practice tests.
Hence, it can be asserted at the end that the THEA Math section is an important part of your THEA test. You need to put in a lot of hard work to pass this section. Also, you need to practice a lot for it, so that one can develop a good pace for solving mathematical problems. All the preparation that you put in will be subsequently useful in the courses that you take up in your college courses. So, start preparing for this right away |
Roussas's Introduction to Probability features exceptionally clear explanations of the mathematics of probability theory and explores its diverse applications through numerous interesting and motivational examples. It provides a thorough introduction to the subject for professionals and advanced students taking their first course in probability. The content is based on the introductory chapters of Roussas's book, An Intoduction to Probability and Statistical Inference, with additional chapters and revisions.
• Written by a well-respected author known for great exposition and readability • Boasts many real world examples • Pedagogy includes chapter summaries, tables of distributions and formulas, and answers to even-numbered exercises |
This first volume, a three-part introduction to the subject, is intended for students with a beginning knowledge of mathematical analysis who are motivated to discover the ideas that shape Fourier analysis. It begins with the simple conviction that Fourier arrived at in the early nineteenth century when studying problems in the physical sciences--that an arbitrary function can be written as an infinite sum of the most basic trigonometric functions.
The first part implements this idea in terms of notions of convergence and summability of Fourier series, while highlighting applications such as the isoperimetric inequality and equidistribution. The second part deals with the Fourier transform and its applications to classical partial differential equations and the Radon transform; a clear introduction to the subject serves to avoid technical difficulties. The book closes with Fourier theory for finite abelian groups, which is applied to prime numbers in arithmetic progression.
In organizing their exposition, the authors have carefully balanced an emphasis on key conceptual insights against the need to provide the technical underpinnings of rigorous analysis. Students of mathematics, physics, engineering and other sciences will find the theory and applications covered in this volume to be of real interest.
The Princeton Lectures in Analysis represents a sustained effort to introduce the core areas of mathematical analysis while also illustrating the organic unity between them. Numerous examples and applications throughout its four planned volumes, of which Fourier Analysis is the firstAll the exercises plus their solutions for Serge Lang's fourth edition of "Complex Analysis," ISBN 0-387-98592-1. The problems in the first 8 chapters are suitable for an introductory course at undergraduate level and cover power series, Cauchy's theorem, Laurent series, singularities and meromorphic functions, the calculus of residues, conformal mappings, and harmonic functions. The material in the remaining 8 chapters is more advanced, with problems on Schwartz reflection, analytic continuation, Jensen's formula, the Phragmen-Lindeloef theorem, entire functions, Weierstrass products and meromorphic functions, the Gamma function and Zeta function. Also beneficial for anyone interested in learning complex analysis. [via]
Real Analysis is the third volume in the Princeton Lectures in Analysis, a series of four textbooks that aim to present, in an integrated manner, the core areas of analysis. Here the focus is on the development of measure and integration theory, differentiation and integration, Hilbert spaces, and Hausdorff measure and fractals. This book reflects the objective of the series as a whole: to make plain the organic unity that exists between the various parts of the subject, and to illustrate the wide applicability of ideas of analysis to other fields of mathematics and science.
After setting forth the basic facts of measure theory, Lebesgue integration, and differentiation on Euclidian spaces, the authors move to the elements of Hilbert space, via the L2 theory. They next present basic illustrations of these concepts from Fourier analysis, partial differential equations, and complex analysis. The final part of the book introduces the reader to the fascinating subject of fractional-dimensional sets, including Hausdorff measure, self-replicating sets, space-filling curves, and Besicovitch sets. Each chapter has a series of exercises, from the relatively easy to the more complex, that are tied directly to the text. A substantial number of hints encourage the reader to take on even the more challenging exercises.
As with the other volumes in the series, Real Analysis is accessible to students interested in such diverse disciplines as mathematics, physics, engineering, and finance, at both the undergraduate and graduate levels.
Also available, the first two volumes in the Princeton Lectures in Analysis: |
two-volume text in harmonic analysis introduces a wealth of analytical results and techniques. It is largely self-contained and will be useful to graduate students and researchers in both pure and applied analysis. Numerous exercises and problems make the text suitable for self-study and the classroom alike. This first volume starts with classical one-dimensional topics: Fourier series; harmonic functions; Hilbert transform. Then the higher-dimensional Calderon-Zygmund and Littlewood-Paley theories are developed. Probabilistic methods and their applications are discussed, as are applications of harmonic analysis to partial differential equations. The volume concludes with an introduction to the Weyl calculus. The second volume goes beyond the classical to the highly contemporary and focuses on multilinear aspects of harmonic analysis: the bilinear Hilbert transform; Coifman-Meyer theory; Carleson's resolution of the Lusin conjecture; Calderon's commutators and the Cauchy integral on Lipschitz curves. The material in this volume has not previously appeared together in book form. |
This eBook comprehensively introduces the reader to our Key Stage 2 (KS2) maths eBook publications. It encompasses a curriculum overview, a list of publications and a set of overviews for each modular maths eBook, each principle publication eBook as well as individual Key Stage 2 (KS2) eBooks.
The 1st Century Traveller's Guide to Palestine is an ebook by Professor Hawkes that gives the reader an impression of what it was like in New Testament times. Set in the year AD 60 it covers topics such as travelling there, what to see, what people eat, cost of living, what accommodation is like etc. The book is well researched with a comprehensive bibliography angles, bearings and scale drawings. To angles as it relates to angular turn about a point, angles in polygons, angle facts and inter-relational aspects with parallel as well as crossing lines, to bearings as they relate to navigation and scale drawings as an aspect of technical drawing.
This eBook introduces the subject of transformation as it relates to translations, reflections, rotations and enlargements either as individual operations or composite operations. In this eBook we illustrate each of these translations using right-angled triangles, but the principles developed extend to all 2D shapes as well as to 3D shapes using extensions.
This eBook introduces the related subjects of Pythagoras' theorem, trigonometry and similarity, as Pythagoras' theorem relates to all right-angles triangles, trigonometry as it relates to angles and ratios of sine, cosine and tangent in right-angled triangles, angles of elevation and depression as well as similarity and congruence.
This eBook introduces the subject of measures and measurement, and looks at both metric and imperial units of measurement, the process and accuracy of reading scales, limits on the accuracy of measurements and compound measurements.
Fractions, percentages and ratio introduces the student to fractions, their representation and their arithmetic, percentages, their representation and their conversion to or from fractions or decimals, as well as ratio, its representation and arithmetic manipulation using ratio.
Decimals considers the significance of the position of the decimal point, compares & converts 'fractions-decimals-percentages', considers 'multiplying & dividing' decimals by 10 and 100, rounds decimals to the nearest whole number, tenth, or hundredth, considers 'less than', 'greater than' and '=' signs in arithmetic and walks the student through the addition and subtraction methods with decimals
This eBook introduces the significant scientific notation of the very large, the intermediate and the very small in terms of numbers and algebra through an exploration of indices, the rules of indices and standard index form.
This eBook introduces the subject of algebra to the student encompassing, inverse operators, equations, the order of precedence, algebraic conventions, BODMAS, expressions, formulae, factorising, rearranging and solving linear, quadratic and simultaneous equations as well as inequalities.
This eBook introduces the student to number patterns and sequences, including odd and even numbers, square numbers, square roots, cube numbers, cube roots, factors, prime numbers, multiples, linear sequences, square number and cube number sequences, Fibonacci number sequences, triangular number sequences and sequences of the powers of 2, 3, 4, 5 and 10Many people believe that the time of physical content such as DVD and Bluray is coming to an end.
With the ability to purchase a movie or TV show online, and watch it in a matter of seconds, I can see the reasoning behind these thoughts, but nothing is further than the truth.
Interesting and easy to read guide to managing your time in the collegiate environment. Also contains easy study methods to maximize your study time. At the very least, you can print the cover and have an excellent coaster for your beer. |
Do the Math: Secrets, Lies, and Algebra
In the eighth grade, 1 math whiz < 1 popular boy, according to Tess's calculations. That is, until she has to factor in a few more variables, like:
1 stolen test (x),
3 cheaters (y),
and 2 best friends (z) who can't keep a secret.
Oh, and she can't forget the winter dance (d)!
Then there's the suspicious guy Tess's parents know, but that's a whole different problem—
Sydney (Fair Oaks Ranch, TX)
Do the Math: Secrets, Lies, and Algebra by Wendy Lichtman was awesome! I really enjoyed the various chacters and the fast paced plot! I might have learned some math along the way too! :) I rate this book an 8/10!
—
Allie (Forest Hill, MD)
This was a very interesting book. It had a new way of looking at life: through math. As the main character discovers, math is so logical that it can often help to solve problems in real life--and she has some big ones. Any math lover would instantly love this book, and anyone else would love it also for its unique perspective on life. I would highly recommend it to anyone, even those who think math is useless (maybe this will change their minds).
—
Molly (Agua Dulce, CA)
This wonderful, witty book puts things in a refreshingly new perspective, relating everyday things to math in a way that will have you thinking. This book evokes an interest in math without being a textbook and also allows us to enter the world of a typical teenage girl. This book combines typical teenage life and math in a way that will make you excited for math class. |
Description:Looking for a tutorial on How To Solve For Covariance? This practical instructional video explains accurately how it's done, and will help you get good at math. Enjoy this advice video from the world's most comprehensive library of free factual video content online.(7:36) |
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MPJ's Ultimate Math Lessons is a resource for teachers of Algebra, Pre-Algebra, and Geometry in grades 6-12. It contains 80 innovative lessons and 27 thought-provoking articles taken directly from the Math Projects Journal, a periodical that, for more than six years, has helped teachers around the world improve student performance in mathematics. |
Mathematics 2, Essential
When we paint a room, put up a fence, buy a rug, or wrap a present, we are using shapes. Essential Math 2 deals with the nature and property of shapes such as circles, triangles and squares. In doing so, this course provides an introduction to geometry and algebra. Essential Math 2 also acquaints students with the metric system of measurement |
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BEGINNING ALGEBRA
An introduction to algebra with a review of basic arithmetic. Includes decimals, fraction, percentage, ratio, proportion, signed numbers, algebraic expressions, factoring, exponents and radicals, linear equations, and graphs. MATH 098 is offered through Extended Studies and a fee is assessed. Credit does not count toward graduation. Graded Satisfactory/Unsatisfactory only. |
More About
This Textbook
Overview
A Friendly Introduction to Number Theory, Fourth Edition is designed to introduce readers to the overall themes and methodology of mathematics through the detailed study of one particular facet—number theory. Starting with nothing more than basic high school algebra, readers are gradually led to the point of actively performing mathematical research while getting a glimpse of current mathematical frontiers. The writing is appropriate for the undergraduate audience and includes many numerical examples, which are analyzed for patterns and used to make conjectures. Emphasis is on the methods used for proving theorems rather than on specific results.
Editorial Reviews
Booknews
Silverman (Brown U.) originally wrote the book as a text for a course designed to attract non-science majors with little interest in pursuing the standard calculus sequence, and convince them to study some college mathematics. He expects readers to have some facility with high school algebra and access to a calculator, though he points out that those who know how to program a computer have great fun generating reams of data and implementing assorted algorithms. He mentions concepts from calculus now and then, but does not lay them down as barriers to cross. The first edition appeared in 1997. Annotation c. Book News, Inc., Portland, OR (booknews.com)
Product Details
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Meet the Author
Joseph H. Silverman is a Professor of Mathematics at Brown University. He received his Sc.B. at Brown and his Ph.D. at Harvard, after which he held positions at MIT and Boston University before joining the Brown faculty in 1988. He has published more than 100 peer-reviewed research articles and seven books in the fields of number theory, elliptic curves, arithmetic geometry, arithmetic dynamical systems, and cryptography. He is a highly regarded teacher, having won teaching awards from Brown University and the Mathematical Association of America, as well as a Steele Prize for Mathematical Exposition from the American Mathematical Society. He has supervised the theses of more than 25 Ph.D. students, is a co-founder of NTRU Cryptosystems, Inc., and has served as an elected member of the American Mathematical Society Council and Executive Committee |
Complex Analysis
9780387950693
ISBN:
0387950699
Pub Date: 2001 Publisher: Springer Verlag
Summary: The book provides an introduction to complex analysis for students with some familiarity with complex numbers from high school. The first part comprises the basic core of a course in complex analysis for junior and senior undergraduates. The second part includes various more specialized topics as the argument principle the Poisson integral, and the Riemann mapping theorem. The third part consists of a selection of to...pics designed to complete the coverage of all background necessary for passing Ph.D. qualifying exams in complex analysis.
Gamelin, Theodore W. is the author of Complex Analysis, published 2001 under ISBN 9780387950693 and 0387950699. Six hundred thirty Complex Analysis textbooks are available for sale on ValoreBooks.com, one hundred fifty nine used from the cheapest price of $36.68, or buy new starting at $51.35 |
Learn GCSE Maths on your Smartphone
A Unit Study to use while Reading the novel.
This unit study DOES NOT include the novel.
This unit study offers many wonderful activities to use while having students read the book. There are between 6 and 10 lessonsDo you need some yawn-proof and sleep-proof writing exercises to get your students to actively participate? Then this book is for you, as it is very complete and is designed to be taught over 16 weeks. It can, of course, be taught in less, by combining some of the lessons into one week. So it would work with an 8-week or 12-week course schedule eBook reviews simultaneous equations and inequalities. We introduce simultaneous equations as systems of equations, and consider some relatively simple pairs of simultaneous equations, one pair involving a pair of linear equations, and another pair involving one linear equation and one quadratic equation. We go on to introduce the two methods of solving simultaneous equations, elimim. ...A Unit Study to use while Reading the novel.
This unit study DOES NOT include the novel.
This unit study offers many wonderful activities to use while having students read the book. There are between 6 and 10 lessons ... title is part of our ongoing series of academic research manuscripts. This study was conducted in 2006 by St. Edward's Researcher, Dr. Michael J. Hollis. It examines the communications approaches that universities have used to reduce the rates of sexual violence on campuses and the effectiveness of each approach. |
How is algebra used as a biological scientist?
how is algebra used in biological science?
Asked By: Hello - 10/21/2010
Best Answer - Chosen by Asker
Mathematics is used a lot in modern biology in subjects like biometrics, studying the spread and distribution of parasites and other diseases, in population growth and decline, and the usefulness of various methods of biological control. Mathematical models are used to predict animal and plant behaviour and assess the...
More
Answered By: jonal - 10/22/2010
Additional Answers (1)
algebra uses a problem solving area of the brain that normal problems dont, so applying this techniqe with biology could yeild some really usefull and insightfull knoweledge and findings..
Where are the jobs? Is productivity and globalization creating a permanent "recession" of jobs?
My main issue is I am doing a paper for school and have no idea where to begin. I was hoping suggestions... |
Intermediate Algebra Remedial math course designed to prepare students for Math Problem Solving or College Algebra. Mathematical thought and reasoning developed through the study of polynomials, factoring, rational expressions, exponents, roots and radicals, quadratic equations, functions and graphing.
MATH 110 4.00 cr.
Math Problem Solving A liberal arts mathematical course designed specifically to focus on the improvement of problem solving skills and mathematical reasoning in many different areas. Topics discussed will include mathematical modeling, probability, statistics, logic, exponential growth, matrices, and chaos. Student needs to be proficient in Intermediate Algebra.
MATH 111 4.00 cr.
College Algebra A study of functions, starting with the definition and focusing on the use of functions in all forms to model the real world. Includes comparing linear and nonlinear functions, transforming functions, looking at polynomial and rational functions globally and locally, models of growth and decline and systems of equations. Student needs to be proficient in Intermediate Algebra.
Calculus I A study of limits and continuity of functions, derivatives, rules and applications of differentiation, hyperbolic and inverse trigonometric functions, rates of change, single-variable optimization, Newton's method, and indefinite integrals. A wide variety of applications from the physcial, natural, and social sciences is explored. Prerequisite: MATH112 or equivalent.
Foundations of Abstract Mathematics This course is an introduction to the theory and methods of mathematical proof, including the methods of contradiction and contraposition. The primary objectives are for students to be able to read and write mathematical proofs. Subject material covered may include set theory, logic and number theory. Prerequisite: MATH142.
Numerical Analysis This course introduces students to the design, analysis, and implementation of numerical algorithms designed to solve mathematical problems that arise in the real-world modeling of physical processes. Topics will include several categories of numerical algorithms such as solving systems of linear equations, root-finding, approximation, interpolation, numerical solutions to differential equations, numerical integration, and matrix methods. Prerequisites: MATH243 and MATH351.
History of Mathematics An introduction to the historical development of fundamental mathematical concepts. Emphasis is placed on the development of numeration systems, geometry and formal axiomatic systems, solutions of polynomial equations, the development of calculus, and the impact of global events on the development and proliferation of mathematical ideas. Prerequisite: MATH142.
MATH 440 3.00 cr.
Real Analysis An extension of MATH341, the primary topics of this course include rigorous developments of mulivariate differentiation, Riemann and Riemann-Stieltjes integration, sequences, series, continuous functions, and the topology of Euclidean space. Prerequisite: MATH341.
MATH 450 4.00 cr.
Abstract Algebra The three primary topics of this course are groups, rings, and fields. Groups will be studied, including homomorphisms, normal subgroups, and the symmetric and alternating groups. The theorems of Lagrange, Cauchy, and Sylow will be developed and proven. Rings, including subrings, ideals, quotient rings, homomorphisms, and integral domains will be covered. Lastly, finite and infinite fields will be discussed. Prerequisite: MATH295.
MATH 460 4.00 cr.
Partial Differential Equations The primary topics of this course include Fourier series, Sturm-Liouville and boundary value problems, Cauchy problems and the method of characteristics, separation of variables and Laplace transform methods. Numerical methods and selected topics are also included. Prerequisites: MATH243 and MATH260.
Topics in Mathematics A course designed to include topics outside the scope of our other course offerings. Topics may include, but are not limited to, mathematical biology, point-set and algebraic topology, graph theory, combinatorics, differential geometry, set theory, number theory, advanced linear algebra, advanced abstract algebra, and Galois theory. Prerequisite: Consent of instructor.
MATH 491 1.00 cr.
Mathematics Colloquium A one-credit capstone course intended to introduce students to topics in mathematics that are not covered in other courses. This is done through faculty and visiting professor presentations as well as student presentations of selected topics or research. Prerequisite: MATH295 or consent of instructor.
MATH 495 2.00 cr.
Senior Thesis Satisfies the mathematics major capstone requirement and is composed of a written report based on student research. Each student will be expected to present their thesis to the Bethany community through a presentation in Mathematics Colloquium. Prerequisite: Consent of instructor (senior status normally required).
MATH 499 3.00 cr.
Mathematics Internship A mathematics-related field experience with an approved agency fulfilling an individual learning contract negotiated between student, department, Internship Coordinator and worksite. Each student will be expected to give a presentation of their internship to the Bethany community in Mathematics Colloquium. Prerequisite: Consent of mathematics internship coordinator. |
2003 undergraduate introduction to analytic number theory develops analytic skills in the course of studying ancient questions on polygonal numbers, perfect numbers and amicable pairs. The question of how the primes are distributed amongst all the integers is central in analytic number theory. This distribution is determined by the Riemann zeta function, and Riemann's work shows how it is connected to the zeroes of his function, and the significance of the Riemann Hypothesis. Starting from a traditional calculus course and assuming no complex analysis, the author develops the basic ideas of elementary number theory. The text is supplemented by series of exercises to further develop the concepts, and includes brief sketches of more advanced ideas, to present contemporary research problems at a level suitable for undergraduates. In addition to proofs, both rigorous and heuristic, the book includes extensive graphics and tables to make analytic concepts as concrete as possible. less |
IAI GECC Course Description: M1901
M1 901: Quantitative Literacy (3-4 Semester Credits)
Develops conceptual understanding, problem-solving, decision-making and analytic skills dealing with quantities and their magnitudes and interrelationships, using calculators and personal computers as tools. Includes: representing and analyzing data through such statistical measures as central tendency, dispersion, normal and chi-square distributions, and correlation and regression to test hypotheses (maximum of one-third of course); using logical statements and arguments in a real-world context; estimating, approximating and judging the reasonableness of answers; graphing and using polynomial functions and systems of equations and inequalities in the interpretation and solutions of problems; and selecting and using appropriate approaches and tools in formulating and solving real-world problems. Prerequisite: C or better in intermediate algebra and geometry. |
MA 125 Intermediate Algebra Klagge, Neil ONone listed in catalog, however a good grasp of fractions, decimals, negative numbers, order of operations and many "PreAlgebra" concepts is necessary in order to succeed in this course.
Credit Hours
3
Textbook: INTERMEDIATE ALGEBRA, by Lial, Hornsby and Mcginnis, 10th ed. Pearson/Addison-Wesley, Publishers. ISBN: 0-321-50721-5. A scientific calculator, such as a TI-30, (not a graphing calculator) is recommended for this class. The calculator should support the entry of fractions (the fraction key looks like "A b/c" resembling a mixed number like "2 ¾" )
Educational Philosophy: I believe every student has the right to learn and to hear what is being said. Part of your grade includes your level of participation. "Participation" refers to attending and being attentive during class, as well as not distracting others. The participation grade is about 25% of the course grade and is affected by attendance if the absence is not made up as described elsewhere in this document. Also, chronic or unnecessary tardiness may affect your participation grade as well.
Class Assessment:
Three graded examinations will be given. The exams will cover assigned reading, homework and quizzes. Reviews will be conducted to ensure students fully grasp the concepts that will be tested. The exams will cover 75% of the grade while classroom participation will cover the other 25%. Also, there may be "pop" quizes from time to time which would increase the total points above 400.
Grading:
Each of the 3 exams will count 100 points of the grade with class participation being the remaining 100 points. Participation points may be deducted for behavior that disrupts the flow of learning. Absences will affect the participation grade if not made up as outlined below.
Grading scale: A=90-100%, B=80-89, C=70-79, D=60-69, F= BELOW 59.50%
Late Submission of Course Materials: All absences must be made up by HANDING IN the completed assignment as soon as possible after the absence. Work must be shown that indicates how each answer was obtained wherever possible. Assignments can be completed in one of two ways: (1) by doing the ODD problems in the "CHAPTER REVIEW EXERCISES" for each section you missed, or (2) by doing the "NOW TRY...." exercises listed at the end of each example in each secton. If extenuating circumstances arise, please make them known to the instructor so that he can work with you in your situation.
Classroom Rules of Conduct: Distractions that disrupt the flow of learning will affect your "participation" grade. If you need to have a conversation with someone in the class or via cell phone, please take him or her out of the classroom and finish your conversation. All conversation in the classroom should relate directly to the topic being discussed. Your cooperation in this will be appreciated by all. The classroom will be free of sexual harassment. Unacceptable behaviors include unwelcome sexual advances, or sexually related or racially related jokes or comments
Course Topic/Dates/Assignments:
Classes will begin promptly. If a class is cancelled (i.e., military exercises, inclement weather) a make up class may occur. The following is an approximate schedule. Watch for updates and changes on the class web site If you are absent, or arrive late, or have to leave class early, you must hand in an assignment for EACH SECTION you missed. Assignments are discussed above in the section entitled "LATE SUBMISSION OF COURSE MATERIAL". Each section that you fail to make up will reduce your total grade by about ½%, and each chapter by about 2.5%.
PLEASE REFER TO THE FOLLOWING SCHEDULE FOR ANY MAKE UP WORK.
CALENDARFALL 2MA125N. KLAGGE
WK
Monday
Wednesday
Thursday
1
Oct 18
1.1-1.2
20
1.3-1.4
21
2.1-2.2
2
25
2.3-2.4
27
2.5-2.6
28
Review for test 1
3
Nov. 1 Test over ch. 1,2
3
3.1-3.2
4
3.3-3.4
4
8
3.5, Review ch.3
10
4.1 & 4.3 only. Take home PT
11Veterans Day No Class
5
15 Test over ch. 3,4
17
5.1-5.2
18
5.3-5.4, (+5.1.1)
6
22
6.1-6.2
24
6.3-6.4
25Thanksgiving Day No Class
7
29
6.5, review of ch. 5
Dec. 1
9.1-9.2
2
9.3, review of ch. 6
8
6
9.4, Take home PT 3
8
review of chs 5, 6, 9.
9 Test over ch. 5, 6, 9 Failure to communicate about your absence will lead to an UNEXCUSED ABSENCE! This will affect your participation points. Keep me informed about absences!!
Copyright:
This material is copyright and can not be reused without author permission. |
Students will be required to use MATLAB occasionally and should know how to set up vectors, perform mathematical operations on vectors, write simple programmes and plot functions. Demos will be given in examples classes throughout the term and examples given on handouts. Useful MATLAB resources and tutorials can be found on the web, including, HERE. An extensive range of MATLAB manuals are also available at the library.
Online Lecture Notes
Students are required to take their own notes in the lectures. Additional online lecture notes (to be read between lectures) are available below. Note that these are supplementary to lectures. I will usually give out paper copies at the end of lectures.
Week 0: Classical PDEs (This gives an overview of some of the PDEs you will meet in the course)
The material on solving differential equations via finite differences gives a first taste of numerical analysis. This is a branch of applied mathematics with many important applications in the real world. For more details, and a list of other courses on numerical analysis, see the Numerical Analysis undergraduate student pathway.
Example Sheets
On average, there is one example sheet per week. HOWEVER - questions are grouped according to topics. Some sheets have more questions than you will be able to do in one week but can be used for revision later. Example classes start in week 2, so you should aim to have done most of sheet 1 for week 2. You will get the most out of the example classes if you try the questions beforehand. You can then ask questions about the problems you are unable to do. You should attend ONE example class per week (not both).
MATLAB codes
For certain lectures (e.g. the ones on finite difference methods in weeks 9 and 10) students will need the following MATLAB codes. Download the files and save to your P-drive. Open them in the MATLAB editor and read the instructions.
fourierN_demo.m (illustrates convergence of a Fourier Series - example from Fourier Series notes)
Exam resources
Past papers are avaliable from the main School of Mathematics website . Solutions to these exam papers are not provided. Solutions to examples sheets and the sample exam will help you revise for the exam. |
Secondary Curricula
Personalized Math Instruction for Grades 6-8.
The Carnegie Learning Math Series is a Common Core math program for grades 6-8. The Carnegie Learning Math Series is available in any of our various solution offerings, including blended curricula, software, and textbook solutions.
I'm very happy that Carnegie Learning also provides 24/7 Math Help for students where in the middle of the night or wee hours of the morning can get online and get their questions answered so they're not stuck and frustrated, and can move on in their learning.
– Suzanne Etheridge, Mathematics Instructor Pellissippi State Community College |
Description:
Mathlets are a collection of over 40 Java applets written by Dr. Tom Leathrum, a professor of mathematics at Jacksonville State University. These handy utilities perform basic calculator and graphing functions, as well as demonstrating several concepts from precalculus and calculus. The three-dimensional graphing applets are especially useful; by simply clicking on the graph and moving the mouse, the entire plot rotates, allowing the user to see it from any angle. This can greatly ease the visualization process, which is often very difficult for students. For Java programmers interested in how Dr. Leathrum created these applets, many examples of the source code are given on the Web site. |
basic ideas of matrix and linear algebra in such a way that users from diverse backgrounds (who have had some exposure to calculus) will understand, by utilizing both algebraic and geometric reasoning. A spiral approach gradually introduces the abstract foundations of the topics involved—linear combination, closure, subspaces, linear independence/dependence, and bases. Opportunities for a variety of applications, and the optional use of MATLAB, provide hands-on explorations of computations and concepts. Chapter topics include matrices, linear systems and their solutions, Eigen information, vector spaces, inner product spaces, and linear transformations. For individuals who want to learn abstract concepts and deal with a wide variety of applications that can be drawn from fields such as physics, chemistry, biology, geology, economics, engineering, computer science, psychology, and sociology. |
Katzensprung 1
Textbook and Student CD-Rom
Katzensprung 1 by Pauline Rogan
Book Description
The title of this book is Katzensprung 1 and is written by author Pauline Rogan. The book Katzensprung 1 is published by Pearson Education Limited. The ISBN of this book is 9781740850322 and the format is Paperback. The publisher has not provided a book description for Katzensprung 1 by Pauline Rogan.
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Suitable for sixth graders, this book helps them learn about connecting ratio and rate to whole number multiplication and division and using concepts of ratio and rate to solve problems; and an understanding of division of fractions and extending the notion of number to the system of rational numbers, which includes negative numbers.
Includes modules that deepen and extend students' understanding of linear and exponential relationships by contrasting them with each other and by applying linear models to data that exhibit a linear trend. This book engages them in methods for analyzing, solving, and using quadratic functions.
Suitable for seventh graders, this book helps them learn about developing an understanding of and applying proportional relationships; and developing an understanding of operations with rational numbers and working with expressions and linear equations |
This book was written by an experienced maths tutor to help parents, and carers, to be able to tutor their child in general maths. This book contains 15 lesson plans for hourly tuition sessions (which would cost £20-£25 if you paid a tutor)on maths units ranging from 2D shapes to Pythagoras theory. This ebook is also suitable for adults taking basic and functional skills maths courses |
presents the state-of-the-art in tackling differential equations using advanced methods and software tools of symbolic computation. It focuses on the symbolic-computational aspects of three kinds of fundamental problems in differential equations: transforming the equations, solving the equations, and studying the structure and properties of their solutions. The 20 chapters are written by leading experts and are structured into three parts. The book is worth reading for researchers and students working on this interdisciplinary subject but may also serve as a valuable reference for everyone interested in differential equations, symbolic computation, and their interaction. |
Mathematics
Topics in Geometry
Class Level: Junior
Credits: 3
Department: Mathematics and Computing
Term:
Description: An introduction to modern geometric concepts, including foundations of geometry and non-Euclidean geometry. Incorporates use of computer software to illustrate and explore geometric concepts. Prerequisite: MAT 142, MAT 182, and CMP 150 or consent of instr |
Quadratic equations with real coefficients; Relations between roots and coefficients; Nature of roots; Formation of a quadratic equation, sign and magnitude of the quadratic expression ax2+bx+c (a,b,c are rational numbers and a
0).
Permutation and combination :
Permutation of n different things taken r at a time (r
n). Permutation
of n things not all different. Permutation with repetitions (circular permutation excluded).
Combinations of n different things taken r at a time (r
n). Combination of n things not all different. Basic properties.
Problems involving both permutations and combinations.
Principle of Mathematical Induction :
Statement of the principle. Proof by induction for the sum of squares, sum of cubes of first n natural numbers, divisibility properties like 2 2n – 1 is divisible by 3 (n ≥ 1), 7 divides 32n+1+2 n+2(n ≥ 1).
Concepts of m.Sets, Relations and Mappings :
Idea of sets, subsets, power set, complement, union, intersection and difference of sets, Venn diagram, De Morgan's Laws, Inclusion / Exclusion formula for two or three finite sets, Cartesian product of sets.
Straight line : Slope of a line. Equation of lines in different forms, angle between two lines. Condition of perpendicularity and parallelism of two lines. Distance of a point from a line. Distance between two parallel lines. Lines through the point of intersection of two lines.
Circle : Equation of a circle with a given center and radius. Condition that a general equation of second degree in x, y may represent a circle. Equation of a circle in terms of endpoints of a diameter . Parametric equation of a circle. Intersection of a line with a circle. Equation of common chord of two intersecting circles.
Parabola : Standard equation. Reduction of the form x = ay²+by+c or y = ax²+bx+c to the standard form y² = 4ax or x² = 4ay respectively. Elementary properties and parametric equation of a parabola.
Ellipse and Hyperbola : Reduction to standard form of general equation of second degree when xy term is absent. Conjugate hyperbola. Simple properties. Parametric equations. Location of a point with respect to a conic.
Differential calculus: Functions, composition of two functions and inverse of a function, limit, continuity, derivative, chain rule, derivatives of implicit functions and of functions defined parametrically. |
In advanced math students are asked to open their mind to new and challenging mathematics. Students who are in advanced math are typically one full grade level above their peers in their math skills. We will be covering all of the 7th grade math standards and approximately half of the 8th grade standards in just one year. Due to this, the workload is more than a typical math class. Students are expected to keep their grade at a B- or above or the student will be placed on probation for three weeks. If at that time the grade is not improved, the student will be placed in a general math classroom.
Use the links on the right to find information about the curriculum, as well as specific help on problems. For every lesson there are extra examples, a personal tutor, and a self check quiz. There is also a sample chapter test, and the Hotmath site which will guide you through any odd numbered problem in the entire book. An online version of the book is also available. If you need any help navigating through the site, please do not hesitate to contact me. The code to access the 7th grade book online is B6A0E2E11D and the code for the 8th grade book is C67BB97154 |
Algebra Word Problem: Problems Involving Geometry Movie Description
This program teaches students how to solve word problems that involve geometry. Students are taught how to read the problem and what keywords to look for to gain clues on how to proceed and solve the problem. Students are then taught, for each proble how to set up the appropriate algebraic equation and solve for the unknown. All problem are fully worked with each step shown so that students learn how to proceed from the problem statement to the solution and every step in between.
Movie Details
EAN:
8901736032629
Studio / Distributor:
Excel Home Entertainment
Format:
DVD
Language(s):
English
Release Year:
2009 Algebra Word Problem: Problems Involving Geometry |
More About
This Textbook
Overview
Here. This book explains everything you need to know to begin using MATLAB. Intermediate and advanced users will find useful information here, especially if they are making the switch to MATLAB 6 from an earlier version.
Editorial Reviews
From the Publisher
Review of previous edition: 'Major highlights of the book are completely transparent examples of classical yet always intriguing mathematical, statistical, engineering, economics, and physics problems. In addition, the book explains a seamless use with Microsoft Word for integrating MATLAB outputs with documents, reports, presentations, or other on-line processes. Advanced topics with examples include: Monte Carlo simulation, population dynamics, and linear programming. Overall, it is an outstanding textbook, and, likewise, should be an integral part of the technical reference shelf for most IT professionals. It is a great resource for wherever MATLAB is available!' ACM Ubiquity
Review of previous edition: 'This is a short, focused introduction to MATLAB, a comprehensive software system for mathematical and technical computing. For the beginner it explains everything needed to start using MATLAB, while experienced users ... will find much useful information here.' L'enseignement mathematique |
Elementary Real Analysis - 01 edition
Summary: For undergraduate courses in Advanced Calculus and Real Analysis, as well as for beginning graduate students in mathematics. For a one or two semester sequence.
Elementary Real Analysis is written in a reader friendly style with motivational and historical material that emphasizes the ''big picture'' and makes proofs seem natural rather than mysterious. It introduces key concepts such as point set theory, uniform continuity of functions and uniform converg...show moreence of sequences of functions. It is designed to prepare students for graduate work in mathematical analysis. ...show less
University mathematics departments have for many years offered courses with titles such as Advanced Calculus or Introductory Real Analysis. These courses are taken by a variety of students, serve a number of purposes, and are written at various levels of sophistication. The students range from ones who have just completed a course in elementary calculus to beginning graduate students in mathematics. The purposes are multifold:
To present familiar concepts from calculus at a more rigorous level.
To introduce concepts that are not studied in elementary calculus but that are needed in more advanced undergraduate courses. This would include such topics as point set theory, uniform continuity of functions, and uniform convergence of sequences of functions.
To provide students with a level of mathematical sophistication that will prepare them for graduate work in mathematical analysis, or for graduate work in several applied fields such as engineering or economics.To develop many of the topics that the authors feel all students of mathematics should know.
There are now many texts that address some or all of these objectives. These books range from ones that do little more than address objective (1) to ones that try to address all four objectives. The books of the first extreme are generally aimed at one-term courses for students with minimal background. Books at the other extreme often contain substantially more material than can be covered in a one-year course.
The level of rigor varies considerably from one book to another, as does the style of presentation. Some books endeavor to give a very efficient streamlined development; others try to be more user friendly. We have opted for the user-friendly approach. We feel this approach makes the concepts more meaningful to the student.
Our experience with students at various levels has shown that most students have difficulties when topics that are entirely new to them first appear. For some students that might occur almost immediately when rigorous proofs are required. For others, the difficulties begin with elementary point set theory, compactness arguments, and the like.
To help students with the transition from elementary calculus to a more rigorous course, we have included motivation for concepts most students have not seen before and provided more details in proofs when we introduce new methods. In addition, we have tried to give students ample opportunity to see the new tools in action.
For example, students often feel uneasy when they first encounter the various compactness arguments (Heine-Borel theorem, Bolzano-Weierstrass theorem, Cousin's lemma, introduced in Section 4.5). To help the student see why such theorems are useful, we pose the problem of determining circumstances under which local boundedness of a function f on a set E implies global boundedness of f on E. We show by example that some conditions on E are needed, namely that E be closed and bounded, and then show how each of several theorems could be used to show that closed and boundedness of the set E suffices. Thus we introduce students to the theorems by showing how the theorems can be used in natural ways to solve a problem.
We have also included some optional material, marked as ''Advanced'' or ''Enrichment'' and flagged with a scissors symbol.
Enrichment
We have indicated as ''Enrichment''' some relatively elementary material that could be added to a longer course to provide enrichment and additional examples. For example, in Chapter 3 we have added to the study of series a section on infinite products. While such a topic plays an important role in the representation of analytic functions, it is presented here to allow the instructor to explore ideas that are closely related to the study of series and that help illustrate and review many of the fundamental ideas that have played a role in the study of series.
Advanced
We have indicated as ''Advanced'' material of a more mathematically sophisticated nature that can be omitted without loss of continuity. These topics might be needed in more advanced courses in real analysis or in certain of the marked sections or exercises that appear later in this book. For example, in Chapter 2 we have added to the study of sequence limits a section on lim sups and lim infs. For an elementary first course this can be considered somewhat advanced and skipped. Later problems and text material that require these concepts are carefully indicated. Thus, even though the text carries on to relatively advanced undergraduate analysis, a first course can be presented by avoiding these advanced sections.
We apply these markings to some entire chapters as well as to some sections within chapters and even to certain exercises. We do not view these markings as absolute. They can simply be interpreted in the following ways. Any unmarked material will not depend, in any substantial way, on earlier marked sections. In addition, if a section has been flagged and will be used in a much later section of this book; we indicate where it will be required.
The material marked ''Advanced'' is in line with goals (2) and (3). We resist the temptation to address objective (4). There are simply too many additional topics that one might feel every student should know (e.g., functions of bounded variation, Riemann-Stieltjes and Lebesgue integrals). To cover these topics in the manner we cover other material would render the book more like a reference book than a text that could reasonably be covered in a year. Students who have completed this book will be in a good position to study such topics at rigorous levels.
We include, however, a chapter on metric spaces. We do this for two reasons: to offer a more general framework for viewing concepts treated in earlier chapters, and to illustrate how the abstract viewpoint can be applied to solving concrete problems. The metric space presentation in Chapter 13 can be considered more advanced as the reader would require a reasonable level of preparation. Even so, it is more readable and accessible than many other presentations of metric space theory, as we have prepared it with the assumption that the student has just the minimal background. For example, it is easier than the corresponding chapter in our graduate level text (Real Analysis, Prentice Hall, 1997) in which the student is expected to have studied the Lebesgue integral and to be at an appropriately sophisticated level.
The Exercises
The exercises form an integral part of the book. Many of these exercises are routine in nature. Others are more demanding. A few provide examples that are not usually presented in books of this type but that students have found challenging, interesting, and instructive.
Some exercises have been flagged with the scissors symbol to indicate that they require material from a flagged section. For example, a first course is likely to skip over the section on lim sups and lim infs of sequences. Exercises that require those concepts are flagged so that the instructor can decide whether they can be used or not. Generally, that symbol on an exercise warns that it might not be suitable for routine assignments.
The exercises at the end of some of the chapters can be considered more challenging. They include some Putnam problems and some problems from the journal American Mathematical Monthly. They do not require more knowledge than is in the text material but often need a bit more persistence and some clever ideas. Students should be made aware that solutions to Putnam problems can be found on various Web sites and that solutions to Monthly problems are published; even so, the fun in such problems is in the attempt rather than in seeing someone else's solution.
Designing a Course
We have attempted to write this book in a manner sufficiently flexible to make it possible to use the book for courses of various lengths and a variety of levels of mathematical sophistication.
Much of the material in the book involves rigorous development of topics of a relatively elementary nature, topics that most students have studied at a nonrigorous level in a calculus course. A short course of moderate mathematical sophistication intended for students of minimal background can be based entirely on this material. Such a course might meet objective (1).
We have written this book in a leisurely style. This allows us to provide motivational discussions and historical perspective in a number of places. Even though the book is relatively large (in terms of number of pages), we can comfortably cover most of the main sections in a full-year course, including many of the interesting exercises.
Instructors teaching a short course have several options. They can base a course entirely on the unmarked material of Chapters 1, 2, 4, 5, and 7. As time permits, they can add the early parts of Chapters 3 and 8 or parts of Chapters 11 and 12 and some of the enrichment material.
Background
We should make one more point about this book. We do assume that students are familiar with nonrigorous calculus. In particular, we assume familiarity with the elementary functions and their elementary properties. We also assume some familiarity with computing derivatives and integrals. This allows us to illustrate various concepts using examples familiar to the students. For example, we begin Chapter 2, on sequences, with a discussion of approximating Ö2 using Newton's method. This is merely a motivational discussion, so we are not bothered by the fact that that we don't treat the derivative formally until Chapter 7 and haven't yet proved that d/dx(x2 - 2) = 2x. For students with minimal background we provide an appendix that informally covers such topics as notation, elementary set theory, functions, and proofs.
Acknowledgments
A number of friends, colleagues, and students have made helpful comments and suggestions while the text was being written. We are grateful to the reviewers of the text: Professors Eugene Allgower (Colorado State University), Stephen Breen (California State University, Northridge), Robert E. Fennell (Clemson University), Jan E. Kucera (Washington State University), and Robert F. Lax (Louisiana State University). The authors are particularly grateful to Professors Steve Agronsky (California Polytechnic State University), Peter Borwein (Simon Fraser University), Paul Humke (St. Olaf College), T. H. Steele (Weber State University), and Clifford Weil (Michigan State University) for using preliminary versions of the book in their classes.
A.M.B. J.B.B. B.S.T.
View Author Bio
Thomson, Brian S. : Simon Frasier University
Bruckner, Andrew M. : University of California-Santa Barbara
View Table of Contents
1. Properties of the Real Numbers
Introduction The Real Number System Algebraic Structure Order Structure Bounds Sups and Infs The Archimedean Property Inductive Property of IN The Rational Numbers Are Dense The Metric Structure of R Challenging Problems for Chapter 1
Introduction Cauchy's First Method Properties of the Integral Cauchy's Second Method Cauchy's Second Method (Continued) The Riemann Integral Properties of the Riemann Integral The Improper Riemann Integral More on the Fundamental Theorem of Calculus Challenging Problems for Chapter 8
Introduction Power Series: Convergence Uniform Covergence Functions Represented by Power Series The Taylor Series Products of Power Series Composition of Power Series Trigonometric Series
11. The Euclidean Spaces Rn
The Algebraic Structure of Rn The Metric Structure of Rn Elementary Topology of Rn Sequences in Rn Functions and Mappings Limits of Functions from Rn to Rm Continuity of Functions from Rn to Rm Compact Sets in Rn Continuous Functions on Compact Sets Additional Remarks |
Resources
Below are some of the resources that students have at their disposal.
The navigation panel on the left also lists more specific resources that students can take advantage of.
Department Office: The Mathematics Department Office, wonderfully staffed by Ms. Melanie Chamberlin, is located in SCI 361. Her extension is x3148. The office contains information about summer work and research opportunities, actuarial opportunities, and information about graduate studies.
Common room (SCI 362): The department common room is located in SCI 362. Whenever it is not being used for a department event, the common room is available to students for study or discussion. If you need to consult someone on a mathematical idea, the common room is one place to try. This room is also used in the evenings as the Math Help Room.
The Mathematics Computer Laboratory (SCI 257): Outfitted with Macintosh computers, the computer lab is open all day and evening most days, and the computers are available for students to use whenever the lab is not being used for a class. Each of the computers is equipped with Mathematica and Joy of Mathematica, as well as a variety of other software. Both Mathematica and Joy of Mathematica are also available at various sites around campus, including the Science Center Minifocus.
Science Library: The Science Library is a wonderful resource. Explore it for suggested alternative texts, for popular mathematics books, or biographies of famous mathematicians. Check out on their webpage of interesting mathematics sources |
Graphing
Author:
Unknown
ISBN-13:
9780768202335
ISBN:
0768202337
Publisher: Schaffer Publications, Frank
Summary: Help students succeed in math! Math Minders provide students with the self-confidence they need to succeed in math. Students learn one step at a time, reviewing skills learned in earlier grades, then moving to skills appropriate for their grade level. They progress gradually, giving them the constant feeling of success! Vocabulary is kept at a level appropriate for each grade level to help ensure success. Fun and sim...ple formats help maintain a high level of student interest. Perfect for home or school, or to reinforce any existing math program |
and Trigonometry
This best selling author team explains concepts simply and clearly, without glossing over difficult points. Problem solving and mathematical modeling ...Show synopsisThis best selling author team explains concepts simply and clearly, without glossing over difficult points. Problem solving and mathematical modeling are introduced early and reinforced throughout, providing students with a solid foundation in the principles of mathematical thinking. Comprehensive and evenly paced, this book provides complete coverage of the function concept, and integrates a significant amount of graphing calculator material to help students develop insight into mathematical ideas. The authors' attention to detail and clarity, the same as found in James Stewart's market-leading "Calculus text," is what makes this text the market leader.Hide synopsis
Description:Good. Hardcover. May include moderately worn cover, writing,...Good. Hardcover. May include moderately worn cover, writing, markings or slight discoloration. SKU: 9780840068132-4-0-3 Orders ship the same or next business day. Expedited shipping within U.S. will arrive in 3-5 days. Hassle free 14 day return policy. Contact Customer Service for questions. ISBN: 9780840068132READ CAREFULLY BEFORE ORDERING. This is the instructor edition...READ CAREFULLY BEFORE ORDERING. This is the instructor edition-content is identical to student edition. Has tape and stickers on front and back cover. Ships same or next working day with free signature confirmation |
Southwestern College Spring 2014 Schedule of Classes
Math
To contact instructors, click on their names to get the email addresses of full-time professors or the appropriate school contact information for adjunct faculty. 104TRIGONOMETRY3 Units[Prerequisite: MATH 70 or equivalent skill level as determined by the Southwestern College Mathematics Assessment or equivalent. Recommended Preparation: RDG 158 or the equivalent skill level as determined by the Southwestern College Reading Assessment or equivalent.] Emphasizes graphic and numerical applications of trigonometry, circular and inverse trigonometric functions, proving and applying identities, solutions and practical applications of right and oblique triangles, and applications of DeMoivre's Theorem. Requires graphing calculator. A student can earn a maximum of six units by successfully completing MATH 244 or both 101 and 104. [D; CSU]
EXTENSION SITE - DAY
Begins: 1/21/2014 Ends: 4/30/2014This section is for the "Southwest Regional Apprenticeship Program."*Open seats as of Wednesday, March 12, 2014 8:07 AM
MATH 112CHILDREN'S MATHEMATICAL THINKING2 UnitsPass/No Pass only. [Prerequisite: MATH 60 or the equivalent skill level as determined by the Southwestern College Mathematics Assessment or equivalent. Recommended Preparation: RDG 158 or the equivalent skill level as determined by the Southwestern College Reading Assessment or equivalent. Recommended Concurrent Enrollment: MATH 110 (may be taken previously).] Assists students in undertaking an in-depth analysis of children's understanding of operations, place values and fractions. Helps students understand how children approach mathematics and how children best learn mathematics. Designed for elementary education majors. [D; CSU] [EFFECTIVE FALL 2014: REMOVE MATH 65 FROM PREREQ.]
MATH 241MMATH SOFTWARE WORKSHOP USING MATLAB1 Unit[Prerequisite: MATH 121 or 250 or equivalent.] Focuses on an introduction to mathematical software using MATLAB with a particular emphasis on problems from engineering and the sciences. Serves as a companion course for the linear algebra class or the differential equations class. [D; CSU] |
The primary purposes of College Algebra are to develop problem-solving capabilities that follow logical patterns and to provide the essential algebraic background for work in other fields or courses. The main mathematical topics in this course are functions and graphs, polynomial and rational functions, exponential and logarithmic functions, conic sections, sequences, and series. The historical development of these topics, as well as applications to life and culture, will receive emphasis where appropriate.
College Algebra is taught in a lecture setting. However, there is much interaction between students and the teacher through examples and problems, worked and presented in class. The teacher presents situations to the students that require reasoning intended to produce better problem-solving skills. Problem sets in the textbook constitute the main source of assignments to be completed outside of class, but the students may be asked to complete reading assignments from sources other than the textbook, write on topics of a mathematical nature related to the history of the solution of a particular problem, or use computer based programs to develop solutions to problems.
Student Expectations
Students are expected to provide and use a graphing calculator (similar to the TI83), to participate in class discussions, and to work problems both in and out of class. Normally at least 2 hours of work is needed to complete each class assignment. Performance on scheduled tests constitutes the major part of the course grade.
Prerequisites
Credit for MA101/102 and a passing score on the Intermediate Algebra Assessment, MA 095 with a grade of 'C' or higher, or ACT Math subscore of 18-20 with MA 095 placement score of 14 or higher, or ACT Math subscore of 21 or higher. |
Deals with pricing and hedging financial derivatives.… Computational methods are introduced and the text contains the Excel VBA routines corresponding to the formulas and procedures described in the book. This is valuable since computer simulation can help readers understand the theory….The book…succeeds in presenting intuitively advanced derivative modelling… it provides a useful bridge between introductory books and the more advanced literature." --MATHEMATICAL REVIEWS |
Geometry Demystified
1 rating:
1.0
A book by Stan Gibilisco
Test provides an effective, and totally painless way to learn the fundamentals and general concepts of geometry. Self-teaching guide offers multiple-choice questions at the end of each chapter, and a 100-question self-test. Softcover.
Solid for review or refreshment, weak choice for self-study
The talented and extremely dedicated person can use almost any reference for successful self-study; the differences between references are in the degree of difficulty. This book is touted as "A Self-teaching Guide", but it is higher on the scale of difficulty of use. Some time ago, I tutored a high school student in geometry and her text was a large hardback over 600 pages in length. The reasons for the size of that textbook were clear, it takes that much surface area to completely describe the subject at the level of the beginner. Therefore, this book with half that many pages would have to be either efficiently compact or inadequate in explanation. While there are hints of the first, the second is more dominant, it would be very difficult for the beginner to learn geometry using this book. The fundamentals are quickly reviewed, but there is the assumption that the reader has had algebra at the level where they know how to rearrange expressions. Furthermore, it is assumed that the reader understands functions and shapes such as parabolas. There are eleven chapters with a short quiz at the end of each one and a comprehensive final at the end. All questions are multiple choice and all solutions are included. This would make the book an excellent candidate for a review of geometry, but the swift pace and brevity would make it difficult to use as a reference in a self-study environment. |
Doral, FL Calculuscalculus, a combination of algebra and trigonometry, is an introduction to Analysis, that explores topics that will be applied in calculus. If you understand that a^2 + b^2 = c^2, where c is the hypotenuse of a right-triangle, and 'a', 'b' are the legs, then you are well on your way to under |
Quantitative Aptitude eBook for CAT from Quantitative Reasoning (Math) 3 Refresher books covering all relevant topics - starting from Numbersystems, Percentages to Permutation & Combination to Geometry toFunctions. • A total of 27 topics from Arithmetic, Algebra and Geometry. • The refresher books contain introduction and explanation on concepts in each topic followed by adequate number of solved examples which cover a wide range of questions that appear from these chapters in CAT. • Solved examples in the refresher books include questions that are replicas of questions that appeared in previous CATs. Such questions are separately identified for your convenience. • Solved examples are followed by several exercise problems. These problems are provided with answers and detailed explanatory solutions. • Shortcuts or alternate methods to solve quant questions are provided alongside the solved examples wherever possible. A book of Chapter wise Tests in Math. • Each test comprises 30 to 60 oft-repeated questions. • The Speed Tests are to be taken after you complete each chapter. • The tests have been designed to help you consolidate what you have learnt in the respective topic. • Speed Tests help you identify ways of solving a question in the quickest possible time, when multiple choices are provided. • Explanatory answers along with correct answers to each question is provided Quant Proficiency Tests in select Math topics • These tests test your proficiency in 1 to 3 chapters in mathematics. • These tests are to be taken as part of your final revision, about two months before CAT. • These tests are designed to acquaint you with a variety of CAT - like questions and to help you master concepts in Math and skillfully employ smart techniques in answering these questions. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Course MaterialsCAT Quantitative Ability (Quant / Math)Syllabus, Topics tested in IIMs Common Admissions TestThe Quant (Math) section in CAT usually accounts for a third of the questions in CAT.For instance, in CAT 2006 the quant section had 100 marks worth questions out ofthe total of 300 marks worth questions. More often than not students who take CATfind the quant section as the toughest one. Albeit, CAT 2006 was an exceptionNumber Questions & AnswersAscents CAT Math Refresher Books cover the following topicsBroadly categorized as Arithmetic, Algebra and Geometry CAT typically tests astudents quantitative ability from over 25 topics. These topics that appear in CATare of high school level. Click on the links that follow each topic for details of what iscovered in Ascents Quant Refresher books on these topics and for accesing anarchive of sample questions from these topics. I. ArithmeticNumber Theory Question bank - CAT 2007 Sample QuestionsAn oft repeated topic in CAT since CAT 2000. Questions include simple wordproblems testing ones understanding of applied class="text"cation of LCM, HCF,Factors, Divisibility class="text"ty to questions that would require knowledge ofremainders, remainder theorem, factorials, different bases to which numbers can beexpressed.Number Theory: Remainders, Finding Divisors World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Choice - (4). Correct Answer is 37Explanatory AnswerLet the original number be aLet the divisor be dLetCorrect Choice - (4). Correct Answer is 37Explanatory AnswerLet the original number be aLet the divisor be'd' World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Number Theory: Counting Methods : CombinatoricsNumber of two-digit, three-digit positive integersQuestionHow many keystrokes are needed to type numbers from 1 to 1000?(1) 3001(2) 2893(3) 2704(4) 2890Correct Choice is (2) and Correct Answer is 2893Explanatory Answer 1. While typing numbers from 1 to 1000, you have 9 single digit numbers from 1 to 9. Each of them require one keystroke. That is 9 key strokes. There are 90 two-digit numbers, from 10 to 99. Each of these numbers require 2 keystrokes. Therefore, one requires 180 keystrokes to type the 2 digit numbers. There are 900 three-digit numbers, from 100 to 999. Each of these numbers require 3 keystrokes. Therefore, one requires 2700 keystrokes to type these 3 digit numbers. Then 1000 is a four-digit number which requires 4 keystrokes. Totally, therefore, one requires 9 + 180 + 2700 + 4 = 2893 keystrokes. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Theory: Remainders, DivisorsRemainders of division of two different numbers and their sum by the samedivisorQuestionWhen 242 is divided by a certain divisor the remainder obtained is 8. When 698 isdivided by the same divisor the remainder obtained is 9. However, when the sum ofthe two numbers 242 and 698 is divided by the divisor, the remainder obtained is 4.What is the value of the divisor?(1) 11(2) 17(3) 13(4) 23Correct Choice is (3) and Correct Answer is 13Explanatory AnswerLet the divisor be d.When 242 is divided by the divisor, let the quotient be x and we know that theremainder is 8.Therefore, 242 = xd + 8Similarly, let y be the quotient when 698 is divided by d.Then, 698 = yd + 9.242 + 698 = 940 = xd + yd + 8 + 9940 = xd + yd + 17As xd and yd are divisible by d, the remainder when 940 is divided by d should havebeen 17.However, as the question states that the remainder is 4, it would be possible onlywhen leaves a remainder of 4.If the remainder obtained is 4 when 17 is divided by d, then d has to be 1Number Theory : Division of PolynomialRemainders of division of a polynomialQuestion World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from number should be subtracted from x3 + 4x2 - 7x + 12 if it is to be perfectlydivisible by x + 3?(1) 42(2) 39(3) 13(4) None of theseCorrect Choice is (1) and Correct Answer is 42Explanatory AnswerAccording to remainder theorem when f(x) / x+a, then the remainder is f(-a).In this case, as x + 3 divides x3 + 4x2 - 7x + 12 - k perfectly (k being the number tobe subtracted), the remainder is 0 when the value of x is substituted by -3.i.e., (-3)3 + 4(-3)2 - 7(-3) + 12 - k = 0or -27 + 36 + 21 + 12 = kor k = 42Number Theory : HCF, GCD, Factors, DivisorsWord problem in number theory, using the concept of HCF / GCDQuestionWhat is the minimum number of square marbles required to tile a floor of length 5metres 78 cm and width 3 metres 74 cm?(1) 176(2) 187(3) 54043(4) 748Correct Choice is (2) and correct answer is 187Explanatory Answer 2. The marbles used to tile the floor are square marbles. Therefore, the length of the marble = width of the marble. As we have to use whole number of marbles, the side of the square should a factor of both 5 m 78 cm and 3m 74. And it should be the highest factor of 5 m 78 cm and 3m 74. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from 5 m 78 cm = 578 cm and 3 m 74 cm = 374 cm. The HCF of 578 and 374 = 34. Hence, the side of the square is 34. The number of such square marbles required = 578*374 / 34*34 = 17*11= 187 marbles.Number Theory : Division of factorials, remaindersThe highest power of 10 that can divide a factorial. Number of trailingzeroes.QuestionA person starts multiplying consecutive positive integers from 20. How manynumbers should he multiply before the will have result that will end with 3 zeroes?(1) 11(2) 10(3) 6(4) 5Correct Choice is (3) and correct answer is 6Explanatory Answer 3. A number will end in 3 zeroes when it is multiplied by 3 10s. To get a 10, one needs a 5 and a 2. Therefore, this person should multiply till he encounters three 5s and three 2s. 20 has one 5 (5 * 4) and 25 has two 5s (5 * 5). 20 has two 2s (5 * 2 * 2) and 22 has one 2 (11 * 2). Therefore, he has to multiply till 25 to get three 5s and three 2s, that will make three 10s. So, he has to multiply from 20 to 25 i.e. 6 numbers.Number Theory : Remainders of divisionFinding remainders when the same power of two numbers leave the sameremainder when divided by a common integer. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from what value of n will the remainder of 351^n and 352^n be the same whendivided by 7?(1) 2(2) 3(3) 6(4) 4Correct Choice is (2) and the Correct Answer is 3Explanatory AnswerWhen 351 is divided by 7, the remainder is 1.When 352 is divided by 7, the remainder is 2.Let us look at answer choice (1), n = 2When 3512 is divided by 7, the remainder will be 12 = 1.When 3522 is divided by 7, the remainder will be 22 = 4.So when n = 2, the remainders are different.When n = 3,When 3513 is divided by 7, the remainder will be 13 = 1.When 3523 is divided by 7, the remainder will be 23 = 8.As 8 is greater than 7, divide 8 again by 7, the new remainder is 1.So when n = 3, both 351n and 352n will have the same remainder when divided by 7.Number Theory : Remainders of division by 6Finding remainders when sum of powers of 9 are divided by 6QuestionWhat is the remainder when 9^1 + 9^2 + 9^3 + .... + 9^8 is divided by 6?(1) 3(2) 2(3) 0 World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from 5Correct Choice is (3) and Correct Answer is 0Explanatory Answer6 is an even multiple of 3. When any even multiple of 3 is divided by 6, it will leave aremainder of 0. Or in other words it is perfectly divisible by 6.On the contrary, when any odd multiple of 3 is divided by 6, it will leave a remainderof 3. For e.g when 9 an odd multiple of 3 is divided by 6, you will get a remainder of3.9 is an odd multiple of 3. And all powers of 9 are odd multiples of 3.Therefore, when each of the 8 powers of 9 listed above are divided by 6, each ofthem will leave a remainder of 3.The total value of the remainder = 3 + 3 + .... + 3 (8 remainders) = 24.24 is divisible by 6. Hence, it will leave no remainder.Hence, the final remainder when the expression 9^1 + 9^2 + 9^3 + ..... + 9^8 isdivided by 6 will be equal to 0. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from .Percent Questions & AnswersPercentages, Ratio Sample Question - CAT 2007Question 4 the day: March 20, 2006The question for the day is a sample practice problem in percentages, an ArithmeticTopic.Question: 1If the price of petrol increases by 25% and Raj intends to spend only an additional15% on petrol, by how much % will he reduce the quantity of petrol purchased? 1. 10% 2. 12% 3. 8% 4. 6.67%Correct Answer - 8%. Choice (3)Explanatory AnswerLet the price of 1 litre of petrol be Rs.x and let Raj initially buy y litres of petrol.Therefore, he would have spent Rs. xy on petrol.When the price of petrol increases by 25%, the new price per litre of petrol is 1.25x.Raj intends to increase the amount he spends on petrol by 15%.i.e., he is willing to spend xy + 15% of xy = 1.15xyLet the new quantity of petrol that he can get be q.Then, 1.25x * q = 1.15xyOr q = (1.15xy/1.25x)=(1.15y)/1.25 = 0.92y.As the new quantity that he can buy is 0.92y, he gets 0.08y lesser than what heused to get earlier.Or a reduction of 8%.Percentages - Quant/Math - CAT 2007Question 4 the day: February 14, 2005The CAT Math sample question for the day is from the topic Percentages inArithmetic .Question: 2A shepherd has 1 million sheeps at the beginning of Year 2000. The numbers growby x% (x > 0) during the year. A famine hits his village in the next year and many of World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from sheeps die. The sheep population decreases by y% during 2001 and at thebeginning of 2002 the shepherd finds that he is left with 1 million sheeps. Which ofthe following is correct? 1. x>y 2. y>x 3. x=y 4. Cannot be determinedCorrect choice (1). Correct Answer - (x > y)Solution:Let us assume the value of x to be 10%.Therefore, the number of sheep in the herd at the beginning of year 2001 (end of2000) will be 1 million + 10% of 1 million = 1.1 millionIn 2001, the numbers decrease by y% and at the end of the year the number sheepin the herd = 1 million.i.e., 0.1 million sheep have died in 2001.In terms of the percentage of the number of sheep alive at the beginning of 2001, itwill be (0.1/1.1)*100 % = 9.09%.From the above illustration it is clear that x > y.Percentages - Quant/Math - CAT 2007Question 4 the day:April 15, 2004The question for the day is from the topic Percentages.Question: 3In an election contested by two parties, Party D secured 12% of the total votes morethan Party R. If party R got 132,000 votes, by how many votes did it lose theelection? (1) 300,000 (2) 168,000 (3) 36,000 (4) 24,000Correct Answer - (3)Solution:Let the percentage of the total votes secured by Party D be x%Then the percentage of total votes secured by Party R = (x – 12)% World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from there are only two parties contesting in the election, the sum total of the votessecured by the two parties should total up to 100%i.e., x + x – 12 = 1002x – 12 = 100or 2x = 112 or x = 56%.If Party D got 56% of the votes, then Party got (56 – 12) = 44% of the total votes.44% of the total votes = 132,000i.e., (44/100)*T = 132,000 T = (132,000*100)/44 = 300,000 votes.The margin by which Party R lost the election = 12% of the total votes= 12% of 300,000 = 36,000Percentages - Quant/Math - CAT 2007Question 4 the day:April 23, 2003The question for the day is from the topic Percentages.Question: 3A candidate who gets 20% marks fails by 10 marks but another candidate who gets42% marks gets 12% more than the passing marks. Find the maximum marks. (1) 50 (2) 100 (3) 150 (4) 200Correct Answer - (2)Solution:From the given statement pass percentage is 42% - 12% = 30%By hypothesis, 30% of x – 20% of x = 10 (marks)i.e., 10% of x = 10Therefore, x = 100 marks. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from 20, 2002The question for the day is from the topic Percentages.Question: 720%August 20, 2002The question for the day is from the topic Percentages.Question: 8 World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from June 13, 2002The question for the day is from the topic Percentages.Question: 9If the price of petrol increases by 25%, by how much must a user cut down hisconsumption so that his expenditure on petrol remains constant? (1) 25% (2) 16.67% (3) 20% (4) 33.33%Correct Answer - (3)Solution:Let the price of petrol be Rs.100 per litre. Let the user use 1 litre of petrol.Therefore, his expense on petrol = 100 * 1 = Rs.100Now, the price of petrol increases by 25%. Therefore, the new price of petrol =Rs.125.As he has to maintain his expenditure on petrol constant, he will be spending onlyRs.100 on petrol.Let 'x' be the number of litres of petrol he will use at the new price.Therefore, 125*x = 100 => x = 100/125=4/5 =0.8 litres.He has cut down his petrol consumption by 0.2 litres = (0.2/1)*100 = 20%reduction.There is a short cut for solving this problem.If the price of petrol has increased by 25%, it has gone up 1/4th of its earlier price.Therefore, the % of reduction in petrol that will maintain the amount of money spenton petrol constant = 1/(4+1) = 1/5 = 20%i.e. Express the percentage as a fraction. Then add the numerator of the fraction to World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from denominator to obtain a new fraction. Convert it to percentage - that is theanswer.Percentages - Quant/Math - CAT 2007Question 4 the day:May 30, 2002The question for the day is from the topic Percentages.Question: 10Peter got 30% of the maximum marks in an examination and failed by 10 marks.However, Paul who took the same examination got 40% of the total marks and got15 marks more than the passing marks. What was the passing marks in theexamination? (1) 35 (2) 250 (3) 75 (4) 85Correct Answer - (4)Solution:Let 'x' be the maximum marks in the examination.Therefore, Peter got 30% of x = (30/100)x = 0.3xAnd Paul got 40% of x = (40/100)*x = 0.4x.In terms of the maximum marks Paul got 0.4x - 0.3x = 0.1x more than Peter. ——(1)The problem however, states that Paul got 15 marks more than the passing markand Peter got 10 marks less than the passing mark.Therefore, Paul has got 15 + 10 = 25 marks more than Peter. —— (2)Equating (1) and (2), we get0.1x = 25 => x = 25/0.1= 250'x' is the maximum mark and is equal to 250 marks.We know that Peter got 30% of the maximum marks. Therefore, Peter got(30/100)*250 = 75 marks.We also know that Peter got 10 marks less than the passing mark. Therefore, thepassing mark will be 10 marks more than what Peter got = 75 + 10 = 85. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Profit Questions & AnswersProfit, Loss and Discounts - CAT 2007Question 4 the day : May 2, 2006The question for the day is a sample practice problem in profit, loss, discounts, anArithmetic Topic and the problem provides an understanding of the concept ofdiscount and mark up.QuestionIf a merchant offers a discount of 40% on the marked price of his goods and thusends up selling at cost price, what was the % mark up? 1. 28.57% 2. 40% 3. 66.66% 4. 58.33%Correct Answer - 66.66%. Choice (3)Explanatory AnswerIf the merchant offers a discount of 40% on the marked price, then the goods aresold at 60% of the marked price.The question further states that when the discount offered is 40%, the merchantsells at cost price.Therefore, selling @ 40% discount = 60% of marked price (M) = cost price (C)i.e., (60/100) M=Cor M =(60/100)C or M = 1.6666 Ci.e., a mark up 66.66%Profit, Discounts, List Price - CAT 2007 QuantQuestion 4 the day : April 3, 2006The question for the day is a sample practice problem in profit, loss, discounts, anArithmetic Topic and the problem provides an understanding of the different termssuch as Cost Price, List Price, Selling price and margins.Question:If a merchant offers a discount of 30% on the list price, then she makes a loss of World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from What % profit or % loss will she make if she sells at a discount of 10% of thelist price? 1. 6% loss 2. 0.8% profit 3. 6.25% loss 4. 8% profitCorrect Answer - 8% profit. Choice (4)Explanatory AnswerLet the cost price of the article be Rs.100.Let the List price of the article by "x".Then, when the merchant offers a discount of 30%, the merchant will sell the articleat x - 30% of x = 70% of x = 0.7x. (1)Note: Discount is measured as a percentage of list price.The loss made by the merchant when she offers a discount of 30% is 16%.Therefore, the merchant would have got 100 - 16% of 100 = Rs.84 when she offereda discount of 30%. (2)Note: Loss is always measured as a percentage of cost price.Therefore, equating equations (1) and (2), we get0.7x = 84or x = 120.If the list price is Rs.120 (our assumption of cost price is Rs.100), then when themerchant offers a discount of 10%, she will sell the article at120 - 10%o of 120 = Rs.108.As the cost price of the article was Rs.100 and the merchant gets Rs.108 whileoffering a discount of 10%, she will make a profit of 8%.Profit, Loss and Discounts - CAT 2007Question 4 the day : September 2, 2004The question for the day is a sample practice problem in profit, loss, discounts, anArithmetic Topic and the problem provides an understanding of the different termssuch as Cost Price, List Price, Selling price, Marked Price, Label price, profit marginsand loss made, if any.QuestionA merchant marks his goods up by 60% and then offers a discount on the markedprice. If the final selling price after the discount results in the merchant making noprofit or loss, what was the percentage discount offered by the merchant? 1. 60% 2. 40% 3. 37.5% World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from 4. Depends on the cost priceCorrect Answer - 37.5% discount. Choice (3)Explanatory AnswerAssume the cost price to be 100.Therefore, the merchants marked price will be 100 + 60% of 100 = 160Now, the merchant offers a discount on the marked price. The discount results in themerchant selling the article at no profit or loss or at the cost price.That is the merchant has sold the article at 100.Therefore, the discount offered = 60.Discount offered is usually measured as a percentage of the marked price.Hence, % discount =(60/160)*100= 37.5%Profit, Loss and Discounts - CAT 2007Question 4 the day : July 22, 2003The question for the day is from the topic of Profit & Loss. It provides anunderstanding of the different terms such as Cost Price, List Price, Selling price andmargins.QuestionA merchant marks his goods up by 75% above his cost price. What is the maximum% discount that he can offer so that he ends up selling at no profit or loss? 1. 75% 2. 46.67% 3. 300% 4. 42.85%Correct Answer - 42.85%. Choice (4)Explanatory AnswerLet us assume that the cost price of the article = Rs.100Therefore, the merchant would have marked it to Rs.100 + 75% of Rs.100 = 100 +75 = 175.Now, if he sells it at no profit or loss, he sells it at the cost price.i.e. he offers a discount of Rs.75 on his selling price of Rs.175Therefore, his % discount = (75/175)*100 = 42.85% World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Loss and Discounts - CAT 2007Question 4 the day : July 3, 2003The question for the day is a sample practice problem in profit, loss, discounts, anArithmetic Topic and the problem provides an understanding of the different termssuch as Cost Price, List Price, Selling price and margins.QuestionTwo merchants sell, each an article for Rs.1000. If Merchant A computes his profit oncost price, while Merchant B computes his profit on selling price, they end up makingprofits of 25% respectively. By how much is the profit made by Merchant B greaterthan that of Merchant A? 1. Rs.66.67 2. Rs.50 3. Rs.125 4. Rs.200Correct Answer - Rs.50. Choice (2)Explanatory AnswerMerchant B computes his profit as a percentage of selling price. He makes a profit of25% on selling price of Rs.1000. i.e. his profit = 25% of 1000 = Rs.250Merchant A computes his profit as a percentage of cost price.Therefore, when he makes a profit of 25% or 1/4th of his cost price, then his profitexpressed as a percentage of selling price = 1/(1+4) = 1/5th or 20% of selling price.So, Merchant A makes a profit of 20% of Rs.1000 = Rs.200.Merchant B makes a profit of Rs.250 and Merchant A makes a profit of Rs.200Hence, Merchant B makes Rs.50 more profit than Merchant A.P & L : CP, SP, Ratios - CAT 2007Question 4 the day : May 12, 2003The question for the day is a sample practice problem in profit, loss, discounts, anArithmetic Topic and the problem provides an understanding of the different termssuch as Cost Price, List Price, Selling price and margins.QuestionOne year payment to the servant is Rs. 200 plus one shirt. The servant leaves after9 months and recieves Rs. 120 and a shirt. Then find the price of the shirt. 1. Rs. 80 2. Rs. 100 3. Rs. 120 4. Cannot be determined World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Answer - Rs. 120. Choice (3)Explanatory AnswerThe servant worked for 9 months instead of 12 months, he should receive 9/12 of hisannual paymenti.e., ¾ (200 + 1S).However, the question states that the servant receive Rs. 120 + 1S where S is theprice of the shirt.By equating the two equations we get ¾ (200 + S) = 120 + S.Therefore Price of the shirt S = Rs. 120.Profit, Loss and Discounts - CAT 2007Question 4 the day : April 29, 2003The question for the day is a sample practice problem in profit, loss, discounts, anArithmetic Topic.QuestionIf apples are bought at the rate of 30 for a rupee. How many apples must be sold fora rupee so as to gain 20%? 1. 28 2. 25 3. 20 4. 22Correct Answer - 25 apples. Choice (2)Explanatory AnswerThe merchant makes a profit of 20%.This means that the merchant sells 30 apples for Rs.1.20Therefore, selling price of 1 apple = (1.20/30) = Rs.0.04 or 4 paiseThe number of apples that can be sold for Rs.1.00 = Rs.1.00/0.04 = 25 apples.Profit, Loss and Discounts - CAT 2007Question 4 the day : September 17, 2002The question for the day is a sample practice problem in profit, loss, discounts.QuestionA trader buys goods at a 19% discount on the label price. If he wants to make a World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from of 20% after allowing a discount of 10%, by what % should his marked pricebe greater than the original label price? 1. +8% 2. -3.8% 3. +33.33% 4. None of theseCorrect Answer is 8% profit. Correct Choice is (1)Explanatory AnswerLet the label price be = Rs.100. The trader buys at a discount of 19%.Hence, his cost = 100 - 19 = 81.He wants to make a profit of 20%. Hence his selling price = 1.2 (81) = 97.2However, he wants to get this Rs.97.2 after providing for a discount of 10%. i.e. hewill be selling at 90% of his marked price.Hence, his marked price M = 97.2/0.9= 108 which is 8% more than the original labelprice.Profit, Loss, Cost Price, Selling Price - CAT 2007Question 4 the day : August 26, 2002The question for the day is a sample practice problem in profit, loss, discounts, anArithmetic Topic and the problem provides an understanding of the different termssuch as Cost Price, List Price, Selling price and margins.QuestionRajiv sold an article for Rs.56 which cost him Rs.x. If he had gained x% on hisoutlay, what was his cost? 1. Rs.40 2. Rs.45 3. Rs.36 4. Rs.28Correct Answer - Rs.40. Choice (1)Explanatory Answerx is the cost price of the article and x% is the profit margin.Therefore, s.p = X* ( 1 + X / 100)= 56 => X*((100+X) / 100) = 56So, 100x + x2 = 5600.Solving for x , we get x = 40 or x = -140.As the price cannot be a -ve quantity, x = 40. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from cost price is 40 and the markup is 40.It is usually easier to solve such questions by going back from the answer choices asit saves a considerable amount of time.Profit, Loss, Margins - CAT 2007Question 4 the day : July 9 professes to sell his goods at a loss of 8% but weights 900 grams in place ofa kg weight. Find his real loss or gain per cent. 1. 2% loss 2. 2.22% gain 3. 2% gain 4. None of theseCorrect Answer is 2.22% gain. Correct Choice is (2)Explanatory AnswerThe trader professes to sell his goods at a loss of 8%.Therefore, Selling Price = (100 - 8)% of Cost Priceor SP = 0.92CPBut, when he uses weights that measure only 900 grams while he claims to measure1 kg.Hence, CP of 900gms = 0.90 * Original CPSo, he is selling goods worth 0.90CP at 0.92CPTherefore, he makes a profit of 0.02 CP on his cost of 0.9 CPProfit % = (S.P-C.P / C.P) * 100i.e., ( ( 0.92-0.90) / 0.90)*100=(0.02/0.90)*100=2 2/9% or 2.22%Profit, Loss, CP, SP, Margins - CAT 2007Question 4 the day : April 4, 2002The question for the day is a sample practice problem in profit, loss, discounts, anArithmetic Topic and the problem provides an understanding of the different terms World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from as Cost Price, List Price, Selling price and margins.QuestionA merchant buys two articles for Rs.600. He sells one of them at a profit of 22% andthe other at a loss of 8% and makes no profit or loss in the end. What is the sellingprice of the article that he sold at a loss? 1. Rs.404.80 2. Rs.440 3. Rs.536.80 4. Rs.160Correct Answer - Rs.404.80 . Choice (1)Explanatory AnswerLet C1 be the cost price of the first article and C2 be the cost price of the secondarticle.Let the first article be sold at a profit of 22%, while the second one be sold at a lossof 8%.We know, C1 + C2 = 600.The first article was sold at a profit of 22%. Therefore, the selling price of the firstarticle = C1 + (22/100)C1 = 1.22C1The second article was sold at a loss of 8%. Therefore, the selling price of the secondarticle = C2 - (8/100)C2 = 0.92C2.The total selling price of the first and second article = 1.22C1 + 0.92C2.As the merchant did not make any profit or loss in the entire transaction, hiscombined selling price of article 1 and 2 is the same as the cost price of article 1 and2.Therefore, 1.22C1 + 0.92C2 = C1+C2 = 600As C1 + C2 = 600, C2 = 600 - C1. Substituting this in 1.22C1 + 0.92C2 = 600, weget1.22C1 + 0.92(600 - C1) = 600or 1.22C1 - 0.92C1 = 600 - 0.92*600or 0.3C1 = 0.08*600 = 48or C1 = 48/(0.3) = 160.If C1 = 160, then C2 = 600 - 160 = 440.The item that is sold at loss is article 2. The selling price of article 2 = 0.92*C2 =0.92*440 = 404.80.Note: When you actually solve this problem in CAT, you should be using thefollowing steps only1.22C1 + 0.92C2 = C1+C2 = 6001.22C1 + 0.92(600 - C1) = 600C1 = 48/(0.3) = 160.C2 = 600 - 160 = 440. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from the final step of the answer which is 0.92*440 which you should not actuallycompute. As two of the answer choices (2) and (3) are either 440 or more, theycannot be the answers. The last one is way too low to be 92% of 440, therefore, theanswer should be choice (1)Profit, Loss, Discounts, Markups - CAT 2007Question 4 the day : April 1 makes a profit equal to the selling price of 75 articles when he sold 100 ofthe articles. What % profit did he make in the transaction? 1. 33.33% 2. 75% 3. 300% 4. 150%Correct Answer - 300% profit. Choice (3)Explanatory AnswerLet S be the selling price of 1 article.Therefore, the selling price of 100 articles = 100 S. --(1)The profit earned by selling these 100 articles = selling price of 75 articles = 75 S --(2)We know that Selling Price (S.P.) = Cost Price (C.P) + Profit -- (3)Selling price of 100 articles = 100 S and Profit = 75 S from (1) and (2). Substitutingthis in eqn (3), we get]100 S = C.P + 75 S. Hence, C.P = 100 S - 75 S = 25 S.Profit % = (Profit/Cost.Price )*100 =( 755/255)*100=300%Typically, you should take about 25 to 30 seconds to crack a problem of this kind. Inreality, you should not be writing down all the steps that I have used to explain thesolution. You should probably be framing equation (3) directly and compute the laststep. The rest of the steps should be done mentally as you read the question for thefirst time. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Questions & AnswersInterest - Quant/Math - CAT 2008Question 4 the day:April 30, 2002 A father left a will of Rs.35 lakhs between his two daughters aged 8.5 and 16 such that they may get equal amounts when each of them reach the age of 21 years. The original amount of Rs.35 lakhs has been instructed to be invested at 10% p.a. simple interest. How much did the elder daughter get at the time of the will? (1) Rs. 17.5 (2) Rs. 21 (3) Rs. 15 (4) Rs. 20 lakhs lakhs lakhs lakhsCorrect Answer - (2)Solution:Let Rs.x be the amount that the elder daughter got at the time of the will. Therefore,the younger daughter got (3,500,000 - x).The elder daughter's money earns interest for (21 - 16) = 5 years @ 10% p.a simpleinterestThe younger daughter's money earns interest for (21 - 8.5) = 12.5 years @ 10% p.asimple interest.As the sum of money that each of the daughters get when they are 21 is the same,x + (5 * 10 * x ) / 100= (3,500,000 - x) + (125*10*(3,500,000-x))/100=> x + (50*x) / 100= 3,500,000 - x + (125/100)*3,500,000- 125x/100=> 2x + = 3,500,000 (1 + 5/4)=> (200x+50x+125x) / 100 = (3,500,000)=> x = 2,100,000 = 21 lakhsInterests - Quant/Math - CAT 2008Question 4 the day: June 5, 2002The question for the day is from the topic of compound interest. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from What will Rs.1500 amount to in three years if it is invested in 20% p.a. compound interest, interest being compounded annually? (1) 2400 (2) 2592 (3) 2678 (4) 2540Correct Answer - (2)Solution:The usual way to find the compound interest is given by the formula A =.p(1+(r/100))^nIn this formula, A is the amount at the end of the period of investmentP is the principal that is investedr is the rate of interest in % p.aAnd n is the number of years for which the principal has been invested.In this case, it would turn out to be A = 1500(1+(20/100))^3So great. How do you find the value of the above term? It is time consuming.Let us look at another alternative.What happens in compound interest?Interest is paid on interest.In the first year, interest is paid only on the principal. That is very similar to simpleinterest.However, from the second year onwards things change. In the second year, you payinterest on the principal and also interest on interest.Therefore, the Amount at the end of 2nd year in compound interest can be computedas follows1 * Principal + 2* Simple interest on principal + 1 * interest on interest.Similarly, if you were to find the Amount at the end of 3 years in compound interestuse the following method1*Principal + 3 * Simple interest on principal + 3 * interest on interest + 1 *interest on interest on interestLet us see how it works in our example.The principal is Rs.1500. The rate of interest is 20%. Therefore, the simple intereston principal is 20% of 1500 = Rs.300The interest on interest = 20% interest on the interest of Rs.300 = 20% of Rs.300 =Rs.60.Interest on interest on interest = 20% of Rs.60 = Rs.12.Now add all theseAmount at the end of 3 years = 1*Principal + 3 * Simple interest on principal + 3 *interest on interest + 1 * interest on interest on interest= 1500 + 3*300 + 3*60 + 1*12 = 1500 +900 + 1800 +12 = 2592.You will get the same answer if you had used the formula. However, the calculationin this case was far easier than using the formula. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from out the same method for four and five years and remember the 1-2-1, 1-3-3-1,1-4-6-4-1 etc method which you can use comfortably in the exam.Interest - Quant/Math - CAT 2008Question 4 the day: July 30, 2002The question for the day is from the topic Interest. If a sum of money grows to 144/121 times when invested for two years in a scheme where interest is compounded annually, how long will the same sum of money take to treble if invested at the same rate of interest in a scheme where interest is computed using simple interest method? (1) 9 years (2) 22 years (3) 18 years (4) 33 yearsCorrect Answer - (2)Solution:The sum of money grows to times in 2 years.If P is the principal invested, then it has grown to P in two years when invested incompound interest.In compound interest, if a sum is invested for two years, the amount is found usingthe following formulaA=P(1+(r/100))^2 = P in this case.=>(1+ (r/100) )^2=144/121 => (1+ (r/100))^2=(12/11)^2 => 1+(r/100)=12/11 => r/100 = 1/11 => r=100/11If r = (100/11) %, then in simple interest the time it will take for a sum of money totreble is found out as follows:Let P be the principal invested. Therefore, if the principal trebles = 3P, the remaining2P has come on account of simple interest.Simple Interest = Pnr/100, where P is the simple interest, r is the rate of interestand 'n' is the number of years the principal was invested.Therefore, 2P = (Pn*100) / (11*100) => 2 = or n = 22 years.Interest - Quant/Math - CAT 2008Question 4 the day: August 08, 2002 World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from question for the day is from the topic of compound interest. The population of a town was 3600 three years back. It is 4800 right now. What will be the population three years down the line, if the rate of growth of population has been constant over the years and has been compounding annually? (1) 6000 (2) 6400 (3) 7200 (4) 9600Correct Answer - (2)Solution:The population grew from 3600 to 4800 in 3 years. That is a growth of 1200 on 3600during three year span.Therefore, the rate of growth for three years has beenThe rate of growth during the next three years will also be the same.Therefore, the population will grow from 4800 by 4800 * 1/3= 1600Hence, the population three years from now will be 4800 + 1600 = 6400.Interest - Quant/Math - CAT 2008Question 4 the day: August 22, 2002The question for the day is from the topic of compound interest. A man invests Rs.5000 for 3 years at 5% p.a. compound interest reckoned yearly. Income tax at the rate of 20% on the interest earned is deducted at the end of each year. Find the amount at the end of the third year. (1) 5624.32 (2) 5630.50 (3) 5788.125 (4) 5627.20Correct Answer - (1)Solution:5% is the rate of interest. 20% of the interest amount is paid as tax. That is 80% ofthe interest amount stays back. Therefore, if we compute the rate of interest as 80%of 5% = 4% p.a., we will get the same value.The interest accrued for 3 years in compound interest = 3*simple interest onprincipal + 3*interest on simple interest + 1*interest on interest on interest. =3*(200) + 3*(8) + 1*0.32 = 600 + 24 + 0.32 = 624.32The amount at the end of 3 years = 5000 + 624.32 = 5624.32 25, 2002The question for the day is from the topic of Interest. The difference between the compound interest and the simple interest on a certain sum at 12% p.a. for two years is Rs.90. What will be the value of the amount at the end of 3 years? (1) 9000 (2) 6250 (3) 8530.80 (4) 8780.80Correct Answer - (4)Solution:The difference in the simple interest and compound interest for two years is onaccount of the interest paid on the first years interest, when interest is reckonedusing compound interest, interest being compounded annually.Hence 12% of simple interest = 90 => simple interest = 90/0.12 =750.As the simple interest for a year = 750 @ 12% p.a., the principal = 750 / 0.12=Rs.6250.If the principal is 6250, then the amount outstanding at the end of 3 years = 6250 +3(simple interest on 6250) + 3 (interest on simple interest) + 1 (interest on intereston interest) = 6250 + 3(750) + 3(90) + 1(10.80) = 8780.80.Simple & Compound Interest - Quant - CAT 2008Question 4 the day: February 10, 2003The question for the day is from the topic of Ratio and Proportion. Vijay invested Rs.50,000 partly at 10% and partly at 15%. His total income after a year was Rs.7000. How much did he invest at the rate of 10%? (1) Rs.40,000 (2) Rs.40,000 (3) Rs.12,000 (4) Rs.20,000Correct Answer - (2)Solution:The best way to solve this problem is by using the concept in Mixtures and Alligation.Vijay earned a total income of Rs.7000 for his investment of 50,000. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from his average rate of return =By rule of alligation, if the value of one of the products is 10 (cheaper) and the otheris 15 (dearer) and if the two are mixed such that the average value of the mixture is14 (mean price), then the two products have been mixed in the following ratio.The ratio of Cheaper product : Dearer product (Dearer product price - mean price) : (Mean price - cheaper product price)In our example, the cheaper product is the investment at 10%, the dearer product isthe investment at 15% and the mean price is the average return of 14%.Therefore, the amount invested @10% interest = (1/5) * 50,000= 10,000.Simple Interest - Quant/Math - CAT 2008Question 4 the day: May 27, 2003The question for the day is from the topic of Simple Interest. A sum of money invested for a certain number of years at 8% p.a. simple interest grows to Rs.180. The same sum of money invested for the same number of years at 4% p.a. simple interest grows to Rs.120 only. For how many years was the sum invested? (1) 25 years (2) 40 years (3) 33 years and 4 months (4) Cannot be determinedCorrect Answer - (1)Solution:From the information provided we know that,Principal + 8% p.a. interest on principal for n years = 180 …….. (1)Principal + 4% p.a. interest on principal for n years = 120 ……… (2)Subtracting equation (2) from equation (1), we get4% p.a. interest on principal for n years = Rs.60.Now, we can substitute this value in equation (2),i.e Principal + 60 = 120= Principal = Rs.60.We know that SI = pnr / 100, where p is the principal, n the number of years and rthe rate percent of interest. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from equation (2), p = Rs.60, r = 4% p.a. and the simple interest = Rs.60.Therefore, 60 = (60*n*4) / 100=> n = 100/4 = 25 years.Interest - Quant/Math - CAT 2008Question 4 the day: July 9, 2003The question for the day is from the topic of Interest. How long will it take for a sum of money to grow from Rs.1250 to Rs.10,000, if it is invested at 12.5% p.a simple interest? (1) 8 years (2) 64 years (3) 72 years (4) 56 yearsCorrect Answer - (4)Solution:Simple interest is given by the formula SI = (pnr/100), where p is the principal, n isthe number of years for which it is invested, r is the rate of interest per annumIn this case, Rs. 1250 has become Rs.10,000.Therefore, the interest earned = 10,000 – 1250 = 8750.8750 = [(1250*n*12.5)/100]=> n = 700 / 12.5 = 56 years.Interest - Quant/Math - CAT 2008Question 4 the day:November 13, 2003The question for the day is from the topic of Interest. Rs. 5887 is divided between Shyam and Ram, such that Shyams share at the end of 9 years is equal to Rams share at the end of 11 years, compounded annually at the rate of 5%. Find the share of Shyam. (1) 2088 (2) 2000 (3) 3087 (4) None of these World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Answer - (3)Solution:Shyams share * (1+0.05)9 = Rams share * (1 + 0.05)11Shyams share / Rams share = (1 + 0.05)11 / (1+ 0.05)9 = (1+ 0.05)2 = 441/400Therefore Shyams share = (441/841) * 5887 = 3087.Quant/Math - CAT 2008Question 4 the day: March 29, 2004The question for the day is from the topic simple and compound interest. Shawninvested one half of his savings in a bond that paid simple interest for 2 years andreceived Rs.550 as interest. He invested the remaining in a bond that paid compoundinterest, interest being compounded annually, for the same 2 years at the same rateof interest and received Rs.605 as interest. What was the value of his total savingsbefore investing in these two bonds?(1) Rs.5500 (2) Rs.11000(3) Rs.22000 (4) Rs.2750Correct choice - (4)Solution:Explanatory AnswerShawn received an extra amount of (Rs.605 – Rs.550) Rs.55 on his compoundinterest paying bond as the interest that he received in the first year also earnedinterest in the second year.The extra interest earned on the compound interest bond = Rs.55The interest for the first year =550/2= Rs.275Therefore, the rate of interest =(55/275) * 100 = 20% p.a.20% interest means that Shawn received 20% of the amount he invested in thebonds as interest.If 20% of his investment in one of the bonds = Rs.275, then his total investment ineach of the bonds = (275/20)*100 = 1375.As he invested equal sums in both the bonds, his total savings before investing =2*1375 = Rs.2750. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Time & Distance Question AnswersSpeed, Time and Distance - Quant/Math - CAT 2008Question 4 the day: September 05, 2002The question for the day is from the topic of Speed, Time and Distance. A ship develops a leak 12 km from the shore. Despite the leak, the ship is able to move towards the shore at a speed of 8 km/hr. However, the ship can stay afloat only for 20 minutes. If a rescue vessel were to leave from the shore towards the ship, and it takes 4 minutes to evacuate the crew and passengers of the ship, what should be the minimum speed of the rescue vessel in order to be able to successfully rescue the people aboard the ship? (1) 53 km/hr (2) 37 km/hr (3) 28 km/hr (4) 44 km/hrCorrect Answer - (2)Solution:The distance between the rescue vessel and the ship, which is 12 km has to becovered in 16 minutes. (The ship can stay afloat only 20 minutes and it takes 4minutes to evacuate the people aboard the ship). Therefore, the two vessels shouldmove towards each other at a speed of km/hr = = 45 km/hr.The ship is moving at a speed of 8 km/hr. Therefore, the rescue vessel should moveat a speed of 45 - 8 = 37 km/hr.Speed, Time and Distance - Quant/Math - CAT 2008Question 4 the day: September 09, 2002The question for the day is from the topic of Speed, Time and Distance. A man driving his bike at 24 kmph reaches his office 5 minutes late. Had he driven 25% faster on an average he would have reached 4 minutes earlier than the scheduled time. How far is his office? (1) 24 km (2) 72 km (3) 18 km (4) Data InsufficientCorrect Answer - (3) World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from x km be the distance between his house and office.While traveling at 24kmph, he would take hours. While traveling at 30 kmph, hewould take hours. Therefore, (given in the problem. 5 min late + 4 min early = 9min)=> x = 18 kmSpeed, Time and Distance - Quant/Math - CAT 2008Question 4 the day: September 23, 2002The question for the day is from the topic of Speed, Time and Distance. When an object is dropped, the number of feet N that it falls is given by the formula N = ½gt2 where t is the time in seconds from the time it was dropped and g is 32.2. If it takes 5 seconds for the object to reach the ground, how many feet does it fall during the last 2 seconds? (1) 64.4 (2) 96.6 (3) 161.0 (4) 257.6Correct Answer - (4)Solution:In 5 seconds it travels½ * 32.2 * 52 = 16.1 * 25 = 402.5In first 3 seconds it travels½ * 32.2 * 32 = 16.1 * 9 = 144.9Hence in the last 2 seconds it traveled 402.5 - 144.9 = 257.6Speed, Time and Distance - Quant/Math - CAT 2008Question 4 the day: October 8, 2002The question for the day is from the topic of Speed, Time and Distance. Rajesh traveled from city A to city B covering as much distance in the second part as he did in the first part of this journey. His speed during the second part was twice as that of the speed during the first part of the journey. What is his average speed of journey during the entire travel? His average speed is the harmonic mean of the individual speeds for the (1) two parts. His average speed is the arithmetic mean of the individual speeds for the (2) two parts. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from His average speed is the geometric mean of the individual speeds for the (3) two parts. (4) Cannot be determined.Correct Answer - (2)Solution:The first part is 1/3rd of the total distance and the second part is 2/3rd of the totaldistance. He travels at s km/hr speed during the first half and 2s km/hr speed duringthe second half.If 3 km is the total distance, then 1 km was traveled at s km/hr and 2 kms wastraveled at 2s km/hr speed.Hence average speed = Total Distance / Total Time = [ 3 / (1/s + 2/s) ] = 3 / ( 4 /2s ) = 3s / 2. This, however, = s+2s / 2 = 3s / 2 which is the arithmetic mean ofthe speeds of the two parts.Speed, Time and Distance - Quant/Math - CAT 2008Question 4 the day: February 24, 2003The question for the day is from the topic of Speed, Time and Distance. Two boys begin together to write out a booklet containing 535 lines. The first boy starts with the first line, writing at the rate of 100 lines an hour; and the second starts with the last line then writes line 534 and so on, backward proceeding at the rate of 50 lines an hour. At what line will they meet? (1) 356 (2) 277 (3) 357 (4) 267Correct Answer - (3)Solution:Writing ratio = 100:50= 2:1Since equal quantities are taken,Therefore in a given time, first boy will be writing the line number World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from X 535 Or 356 2/3 Or 357 th LineHence, both of them shall meet on 357th lineTime and Distance - Quant/Math - CAT 2008Question 4 the day: April 08, 2003The question for the day is from the topic of Time and Distance. A man and a woman 81 miles apart from each other, start travelling towrds each other at the same time. If the man covers 5 miles per hour to the womens 4 miles per hour, how far will the woman have travelled when they meet? (1) 27 (2) 36 (3) 45 (4) None of these.Correct Answer - (2)Solution:Time taken to meet = Distance between them / Relative speed= 81/(4 + 5) = 9 hoursTherefore, woman travells = 9 x 4 = 36 miles.Speed, Time and Distance - Quant/Math - CAT 2008Question 4 the day: April 28, 2003The question for the day is from the topic of Speed, Time and Distance. Two friends A and B run around a circular track of length 510 metres, starting from the same point, simultaneously and in the same direction. A who runs faster laps B in the middle of the 5th round. If A and B were to run a 3 km race long race, how much start, in terms of distance, should A give B so that they finish the race in a dead heat? (1) 545.45 metres (2) 666.67 metres (3) 857.14 metres (4) Cannot be determinedCorrect Answer - (2)Solution: World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from and B run around a circular track. A laps B in the middle of the 5thlap. i.e. when Ahas run four and a half laps he has covered a distance which is 1 lap greater thanthat covered by Bs.Therefore, when A runs 9/2 laps, B runs 7/2 laps.Which is the same as saying when A runs 9 laps, B runs 7 laps.i.e in a race that is 9 laps long, A can give B a start of 2 laps.So, if the race is of 3000 metres long, then A can give B a start of 2/9 * 3000=666.67 metres.The information with regard to the length of the circular track is redundantinformation.Speed, Time and Distance - Quant/Math - CAT 2008Question 4 the day: September 30, 2002The question for the day is from the topic of Speed, Time and Distance. If the wheel of a bicycle makes 560 revolutions in travelling 1.1 km, what is its radius? (1) 31.25 cm (2) 37.75 cm (3) 35.15 cm (4) 11.25 cmCorrect Answer - (1)Solution:The distance covered by the wheel in 560 revolutions = 1100 m . Hence, thedistance covered per revolution = metres. The distance covered in one revolution =circumference of the wheel.Circumference = => r = 31.25 cm. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from and Cisterns - Quant/Math - CAT 2008Question 4 the day: April 8, 2002 A tank is fitted with 8 pipes, some of them that fill the tank and others that are waste pipe meant to empty the tank. Each of the pipes that fill the tank can fill it in 8 hours, while each of those that empty the tank can empty it in 6 hours. If all the pipes are kept open when the tank is full, it will take exactly 6 hours for the tank to empty. How many of these are fill pipes? (1) 2 (2) 4 (3) 6 (4) 5Correct Answer - (2)Solution:Let the number of fill pipes be 'n. Therefore, there will be 8-n, waste pipes.Each of the fill pipes can fill the tank in 8 hours. Therefore, each of the fillpipes will fill 1/8th of the tank in an hour.Hence, n fill pipes will fill n/8th of the tank in an hour.Similarly, each of the waste pipes will drain the full tank in 6 hours. That is,each of the waste pipes will drain 1/6th of the tank in an hour.Therefore, (8-n) waste pipes will drain ((8-n)/6)th of the tank in an hour.Between the fill pipes and the waste pipes, they drain the tank in 6 hours.That is, when all 8 of them are opened, 1/6th of the tank gets drained in anhour.(Amount of water filled by fill pipes in 1 hour - Amount of water drained bywaste pipes 1 hour) = 1/6th capacity of the tank drained in 1 hour.n/8 - 8-n / 6 = -1/6 => 6n-64+8n / 48 = -1/6 => 14n-64= -8 or 14n= 56 or n=4Note: In problems pertaining to Pipes and Cisterns, as a general rule findout the amount of the tank that gets filled or drained by each of the pipes inunit time (say in 1 minute or 1 hour).Work and Time - Quant/Math - CAT 2008Question 4 the day: April 15, 2002 If A and B work together, they will complete a job in 7.5 days. However, if A works alone and completes half the job and then B takes over and completes the remaining half alone, they will be able to complete the job in 20 days. How long will B alone take to do the job if A is more efficient than B? (1) 20 days (2) 40 days (3) 30 days (4) 24 daysCorrect Answer - (3) World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from a be the number of days in which A can do the job alone. Therefore,working alone, A will complete 1/a th of the job in a day.Similarly, let b the number of days in which B can do the job alone. Hence, Bwill complete 1/b th of the job in a day.Working together, A and B will complete (1/a + 1/b) th of the job in a day.The problem states that working together, A and B will complete the job in7.5 or 15/2 days. i.e they will complete 2/15th of the job in a day.Therefore,1/a + 1/b = 2/15 -(1)From statement 2 of the question, we know that if A completes half the jobworking alone and B takes over and completes the next half, they will take20 days.As A can complete the job working alone in 'a' days, he will complete halfthe job, working alone, in a/2 days.Similarly, B will complete the remaining half of the job in b/2 days.Therefore,a/2+b/2 = 20 => a+b = 40 or a = 40 - b - (2)From (1) and (2) we get, 1 / 40-b + 1/b = 2/15 => 600=2b(40-b)=> 600 = 80b - 2b2=> b2 - 40b + 300 = 0=> (b - 30)(b - 10) = 0=> b = 30 or b = 10.If b = 30, then a = 40 - 30 = 10 orIf b = 10, then a = 40 - 10 = 30.As A is more efficient then B, he will take lesser time to do the job alone.Hence A will take only 10 days and B will take 30 days.Note: Whenever you encounter work time problems, always find out howmuch of the work will be completed by A in unit time (an hour, a day, amonth etc). Find out how much of the work will be completed by B in unittime and add those to find the amount of work that will be completed in unittime.If 'A' takes 10 days to do a job, he will do 1/10th of the job in a day.Similarly, if 2/5ths of the job is done in a day, the entire job will be done in5/2 days.Work & Time - Quant/Math - CAT 2008Question 4 the day: April 25, 2002 1. Working together, A and B can do a job in 6 days. B and C can do the same job in 10 days, while C and A can do it in 7.5 days. How long will it take if all A, B and C work together to complete the job? World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from (1) 8 days (2) 5 days (3) 3 days (4) 7 days 2. 3. How long will it take for A alone to complete the job? (1) 8 days (2) 6 days (3) 10 days (4) 20 daysCorrect Answers1. (2) 2. (3)Solution:Even before you start working on the problem, check out if you can eliminate someanswer choices as impossible.In question (1), we know that if A and B alone work, they can complete the job in 6days. Therefore, if all three of them A, B and C work together the number of days itwill take to complete the job will surely be less than 6 days. Hence, we can eliminateanswer choices (1) and (4) right away.Similarly in question (2), we know that A and B together take 6 days to complete thejob. Therefore, A alone will take more than 6 days to complete the job. Therefore,we can eliminate answer choice (2).In any question, as a rule spend about 5 seconds to see if the answer choicesprovide any clue to solve the question or help in eliminating one or more obviouslyabsurd choices. This will help you (1) in reducing the time it will take to do theproblem and (2) in increasing your probability of success should you choose to takea guess without actually solving the problem.Question 1Let A be the number of days that A will take to complete the job alone, B days for Bto complete the job alone and C days for C to complete the job alone.A and B can do a job in 6 days. They complete 1/6 th of the job in a day. i.e.1/A +1/B = 1/6 -- (1)Similarly, B and C will complete 1/10th of the job in a day. i.e 1 / B + 1/C = 1/10-- (2) World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from C and A will complete 1/7.5 or 2/15th of the job in a day i.e 1/C + 1/ A =2/15-- (3).Adding (1), (2) and (3) we get 1/A + 1/B + 1/B + 1/C + 1/C + 1/A = 1/6 +1/10 + 2/15==>2/A + 2/B +2/C = 5+3+4 / 30 or 1/A + 1/B + 1/C = 6/30 = 1/5. i.eworking together, A, B and C complete 1/5th of the job in a day. Therefore, they willcomplete the job in 5 days.Question 2Subtracting eqn (2) from eqn (1) we get1/A - 1/C = 1/6 -1/10 = 1/15 ---(4)Adding eqn (4) and eqn (3) we get,1/A - 1/C + 1/A + 1/C = 2/A = 1/15 +2/15 = 1/5 or 1/A = 1/10 . i.e. A does 1/10 of the job in a day andtherefore, will take 10 days to complete the job working alone.Pipes and Cisterns - Quant/Math - CAT 2008Question 4 the day: May 23, 2002The question for the day is from the topic Pipes and Cisterns. The problems from thetopic Pipes and Cisterns and Work and Time are very similar in nature. So, if youunderstand the nature of one of these types, you will be able to attempt the otherquite comfortably. Pipe A fills a tank of 700 litres capacity at the rate of 40 litres a minute. Another pipe B fills the same tank at the rate of 30 litres a minute. A pipe at the bottom of the tank drains the tank at the rate of 20 litres a minute. If pipe A is kept open for a minute and then closed and pipe B is kept open for a minute and then closed and then pipe C is kept open for a minute and then closed and the cycle repeated, how long will it take for the empty tank to overflow? (1) 42 minutes (2) 14 minutes (3) 39 minutes (4) None of theseCorrect Answer - (4)SolutionPipe A fills the tank at the rate of 40 litres a minute. Pipe B at the rate of 30 litres aminute and Pipe C drains the tank at the rate of 20 litres a minute.If each of them is kept open for a minute in the order A-B-C, the tank will have 50litres of water at the end of 3 minutes.After 13 such cycles, the tank will have 13 * 50 = 650 litres of water.It will take 13 * 3 = 39 minutes for the 13 cycles to be over.At the end of the 39th minute, Pipe C will be closed and Pipe A will be opened. It willadd 40 litres to the tank. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from at the end of the 40th minute, the tank will have 650 + 40 = 690 litres ofwater.At the end of the 40th minute, Pipe A will be closed and Pipe B will be opened. It willadd 30 litres of water in a minute.Therefore, at the end of the 41st minute, the tank will have 690 + 30 = 720 litres ofwater.But then at 700 litres, the tank will overflow. Therefore, Pipe B need not be keptopen for a full minute at the end of 40 minutes.Pipe B needs to add 10 more litres of water at the end of 40 minutes. It will take1/3rd of a minute to fill 10 litres of water.Therefore, the total time taken for the tank to overflow = 40 minutes + 1/3 of aminuteor 40 minutes 20 seconds.Pipes and Cisterns - Quant/Math - CAT 2008Question 4 the day: May 27, 2002The question for the day is from the topic Pipes and Cisterns. The problems from thetopic Pipes and Cisterns and Work and Time are very similar in nature. So, if youunderstand the nature of one of these types, you will be able to attempt the otherquite comfortably. There are 12 pipes that are connected to a tank. Some of them are fill pipes and the others are drain pipes. Each of the fill pipes can fill the tank in 8 hours and each of the drain pipes can drain the tank completely in 6 hours. If all the fill pipes and drain pipes are kept open, an empty tank gets filled in 24 hours. How many of the 12 pipes are fill pipes? (1) 6 (2) 8 (3) 7 (4) 5Correct Answer - (3)SolutionLet there be 'n' fill pipes attached to the tank.Therefore, there will be 12 - n drain pipes attached to the tankEach fill pipe fills the tank in 8 hours. Therefore, each of the fill pipes will fill 1/8th ofthe tank in an hour.Hence, n fill pipes will fill n*1/8 = n/8 th of the tank in an hour.Each drain pipe will drain the tank in 6 hours. Therefore, each of the drain pipes willdrain 1/6 th of the tank in an hour.Hence, 12 - n drain pipes will drain(12-n) / 6 = 12-n / of the tank in an hour.When all these 12 pipes are kept open, it takes 24 hours for an empty tank tooverflow. Therefore, in an hour 1/24 thof the tank gets filled. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from - 12 - n/6 = 1/24.i.e.3n - 4(12-n) / 24 = 1/24 or 7n - 48 = 1 => 7n = 49 or n = 7.Work and Time - Quant/Math - CAT 2008 Question 4 the day: June 6, 2002The question for the day is from the topic work and time. Four men and three women can do a job in 6 days. When five men and six women work on the same job, the work gets completed in 4 days. How long will a woman take to do the job, if she works alone on it? (1) 18 days (2) 36 days (3) 54 days (4) None of theseCorrect Answer – (3)Solution:Let the amount of work done by a man in a day be 'm' and the amount of work doneby a woman in a day be 'w'.Therefore, 4 men and 3 women will do 4m + 3w amount of work in a day. If 4 menand 3 women complete the entire work in 6 days, they will complete 1/6th of thework in a day.Hence eqn (1) will be 4m + 3w = 1/6and from statement (2), eqn (2) will be 5m + 6w = ¼Solving eqn (1) and eqn (2), we get 3m =1/12 or m = 1/36. i.e. a man does 1/36thof the work in a day. Hence he will take 36 days to do the work.Substituting the value of m in eqn (1), we get 4 * 1/36 + 3w = 1/6=> 3w =1/6-1/9 = 3-2 / 18 = 1/18 or w = 1/54. i.e. a woman does 1/54th of thework in a day. Hence she will take 54 days to do the entire work.Pipes and Cisterns – Quant/Math – CAT 2008Question 4 the day: June 17, 2002The question for the day is from the topic – Pipes and Cisterns. A pump can be used either to fill or to empty a tank. The capacity of the tank is 3600 m3. The emptying capacity of the pump is 10 m3/min higher than its filling capacity. What is the emptying capacity of the pump if the pump needs 12 more minutes to fill the tank than to empty it? World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from (1) 50 m3 / min (2) 60 m3 / min (3) 45 m3 / min (4) 90 m3 /minCorrect Answer - (2)Solution:Let 'f' m3/min be the filling capacity of the pump. Therefore, the emptying capacity ofthe pump will be = (f + 10 ) m3 / min.The time taken to fill the tank will be = 3600 / f minutesAnd the time taken to empty the tank will be = 3600 / f+10.We know that it takes 12 more minutes to fill the tank than to empty iti.e 3600/f - 3600 / f+10 = 12 => 3600 f + 36000 - 3600 f = 12 (f2 + 10 f)=> 36000 = 12 (f2 + 10 f) => 3000 = f2 + 10 f => f2 + 10 f - 3000 = 0.Solving for positive value of 'f' we get, f = 50.Therefore, the emptying capacity of the pump = 50 + 10 = 60 m3 / minWork and Time - Quant/Math - CAT 2008 Question 4 the day: June 19, 2002The question for the day is from the topic - Work and Time. Shyam can do a job in 20 days, Ram in 30 days and Singhal in 60 days. If Shyam is helped by Ram and Singhal every 3rd day, how long will it take for them to complete the job? (1) 12 days (2) 16 days (3) 15 days (4) 10 daysCorrect Answer - (3)Solution:As Shyam is helped by Ram and Singhal every third day, Shyam works for 3 dayswhile Ram and Singhal work for 1 day in every 3 days.Therefore, the amount of work done in 3 days by Shyam, Ram and Singhal = 3/20+ 1/30 + 1/60 = 9+2+1/60 = 12/60 =1/5 th of the job. Hence, it will takethem 5 times the amount of time = 3*5 = 15 days.Pipes and Cisterns - Quant/Math - CAT 2008Question 4 the day: July 29, 2002The question for the day is a pipes and cisterns problem. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Pipe A usually fills a tank in 2 hours. On account of a leak at the bottom of the tank, it takes pipe A 30 more minutes to fill the tank. How long will the leak take to empty a full tank if pipe A is shut? (1) 2 hours 30 minutes (2) 5 hours (3) 4 hours (4) 10 hoursCorrect Answer - (4)Solution:Pipe A fills the tank normally in 2 hours. Therefore, it will fill ½ of the tank in anhour.Let the leak take x hours to empty a full tank when pipe A is shut. Therefore, theleak will empty 1/4 of the tank in an hour.The net amount of water that gets filled in the tank in an hour when pipe A is openand when there is a leak = 1/2 -1/x of the tank. — (1)When there is a leak, the problem states that Pipe A takes two and a half hours to fillthe tank. i.e. 5/2 hours. Therefore, in an hour, 5/2 th of the tank gets filled. – (2)Equating (1) and (2), we get 1/2 - 1/x =2/5=> 1/x = 1/2 - 2/5 = 1/10 => x= 10 hours.The problem can also be mentally done as follows.Pipe A takes 2 hours to fill the tank. Therefore, it fills half the tank in an hour or 50%of the tank in an hour.When there is a leak it takes 2 hours 30 minutes for the tank to fill. i.e 5/2 hours tofill the tank or 2/5 th or 40% of the tank gets filled.On account of the leak, (50 - 40)% = 10% of the water gets wasted every hour.Therefore, the leak will take 10 hours to drain a full tank.Pipes and Cisterns - Quant/Math - CAT 2008Question 4 the day: August 01, 2002The question for the day is from the topic Pipes and Cisterns. There are 12 pipes attached to a tank. Some of them are fill pipes and some are drain pipes. Each of the fill pipes can fill the tank in 12 hours, while each of the drain pipes will take 24 hours to drain a full tank completely. If all the pipes are kept open when the tank was empty, it takes 2 hours for the tank to overflow. How many of these pipes are drain pipes? (1) 6 (2) 11 (3) 4 (4) 7Correct Answer - (3) World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from are 12 pipes attached to the tank. Let 'n' of them be fill pipes. Therefore, therewill be 12-n drain pipes.Each fill pipe, fills the tank in 12 hours. Therefore, 1/12th of the tank gets filledevery hour by one fill pipe.'n' fill pipes will, therefore, fill n/12 th of the tank in an hour.Each drain pipe drains the tank in 24 hours. That is, 1/24 th of the tank gets drainedby one drain pipe every hour.12-n drain pipes, will therefore, drain 12-n / 24 th of the tank in an hour.When all the pipes are open when the tank is empty, it takes 2 hours for the tank tooverflow. i.e. ½ the tank gets filled every hour.Equating the information, we get n/12 - 12-n/24 = 1/2=> 2n+n-12 / 24 = 1/2 => 3n - 12 = 12 or 3n = 24 or n = 8.Therefore, there are 8 fill pipes and (12 - 8) = 4 drain pipes.Work and Time - Quant/Math - CAT 2008Question 4 the day: September 27, 2002The question for the day is from the topic of Work and Time. Two workers A and B manufactured a batch of identical parts. A worked for 2 hours and B worked for 5 hours and they did half the job. Then they worked together for another 3 hours and they had to do (1/20)th of the job. How much time does B take to complete the job, if he worked alone? (1) 24 hours (2) 12 hours (3) 15 hours (4) 30 hoursCorrect Answer - (3)Solution:Let a hours be the time that worker A will take to complete the job. Let b hours bethe time that worker B takes to complete the job. When A works for 2 hours and Bworks for 5 hours half the job is done. i.e. 2/a + 5/b + 1/2. --- (1)When they work together for the next three hours, 1/20th of the job is yet to becompleted. They have completed half the job earlier and 1/20th is still left. So byworking for 3 hours, they have completed 1 - 1/2- 1/20 = 9/20th of the job. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from 3/a + 3/b = 9/20--- (2).Solving for (1) and (2), we get b = 15 hours.Work and Time - Quant/Math - CAT 2008Question 4 the day: October 16, 2002The question for the day is from the topic of Work and Time. A and B working together can finish a job in T days. If A works alone and completes the job, he will take T + 5 days. If B works alone and completes the same job, he will take T + 45 days. What is T? (1) 25 (2) 60 (3) 15 (4) None of theseCorrect Answer - (3)Solution:The time it will take when A and B work together is given by the formula(5*45)^(1/2) = 225^(1/2)= 15 days. Where 5 and 45 are the extra time that Aand B take to complete the job if the work alone as against working together.Work, Time - Quant/Math - CAT 2008Question 4 the day: March 13, 2003The question for the day is from the topic Work and Time A man can do a piece of work in 60 hours. If he takes his son with him and both work together then the work is finished in 40 hours. How long will the son take to do the same job, if he worked alone on the job? (1) 20 hours (2) 120 hours (3) 24 hours (4) None of theseCorrect Answer - (2)Solution:If the man takes 60 hours to complete the work, then he will finish 1/60 th of thework in 1 hour.Let us assume that his son takes x hours to finish the same work.If they work together for 1 hour they will finish 1/60 + 1/x = 1/40 th of the work. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from 1/x = 1/120The son, working alone would take 120 hours to complete the work.Work and Time - Quant/Math - CAT 2008Question 4 the day: October 10, 2002The question for the day is from the topic of Work and Time. Pipe A can fill a tank in a hours. On account of a leak at the bottom of the tank it takes thrice as long to fill the tank. How long will the leak at the bottom of the tank take to empty a full tank, when pipe A is kept closed? (1) (3/2)a hours (2) (2/3)a (3) (4/3)a (4) (3/4)aCorrect Answer - (1)Solution:Pipe A fills the tank in a hours. Therefore, of the tank gets filled in an hour. Onaccount of the leak it takes 3a hours to fill the tank. Therefore, of the tank getsfilled in an hour. Let the leak at the bottom of the tank take x hours to empty thetank. Hence, 1/x of the tank gets emptied every hour.1/a - 1/x =1/3a => 1/x =1/a - 1/3a = 2/3aHence, x = 3a/2Work and Time - Quant/Math - CAT 2008Question 4 the day: April 15, 2003The question for the day is from the topic of Work and Time. A, B and C can do a work in 5 days, 10 days and 15 days respectively. They started together to do the work but after 2 days A and B left. C did the remaining work (in days) (1) 1 (2) 3 (3) 5 (4) 4Correct Answer - (4)Solution:If A, B and C work together for a day then they will finish (1/5 + 1/10 + 1/15)th work= 11/30th work. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from working together for two days they will finish 2 * 11/30 = 11/15th work.C alone does remain (11/15 – 1) 4/15th work.But C finishes 1/15th work in one day. Therefore C will finish 4/15th work in 4 days.Time and Work - Quant/Math - CAT 2008Question 4 the day: June 13, 2003The question for the day is from the topic of Time and Work. X alone can do a piece of work in 15 days and Y alone can do it in 10 days. X and Y undertook to do it for Rs. 720. With the help of Z they finished it in 5 days. How much is paid to Z? (1) Rs. 360 (2) Rs. 120 (3) Rs. 240 (4) Rs. 300Correct Answer - (2)Solution:In one day X can finish 1/15th of the work.In one day Y can finish 1/10th of the work.Let us say that in one day Z can finish 1/Zth of the work.When all the three work together in one day they can finish 1/15 + 1/10 + 1/Z =1/5th of the work.Therefore, 1/Z = 1/30.Ratio of their efficiencies = 1/15: 1/10: 1/30 = 2: 3: 1.Therefore Z receives 1/6th ofthe total money.According to their efficiencies money is divided as 240: 360: 120.Hence, the share of Z = Rs. 120.Work and Time - Quant/Math - CAT 2008Question 4 the day: June 27, 2003The question for the day is from the topic of Work and Time. Ram starts working on a job and works on it for 12 days and completes 40% of the work. To help him complete the work, he employs Ravi and together World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from they work for another 12 days and the work gets completed. How much more efficient is Ram than Ravi? (1) 50% (2) 200% (3) 60% (4) 100%Correct Answer - (4)Solution:Ram completes 40% of work in 12 days.i.e. another 60% of the work has to be completed by Ram and Ravi. They havetaken 12 days to complete 60% of the work.Therefore, Ram and Ravi, working together, would have completed the entire work in(12/60)*100 = 20 days.As Ram completes 40% of the work in 12 days, he will take (12/40)*100 = 30 daysto complete the entire workWorking alone, we know Ram takes 30 days to complete the entire work. Let usassume that Ravi takes x days to complete the entire work, if he works alone. Andtogether, they complete the entire work in 20 days.Therefore, (1/30) + (1/x) = (1/20) => (1/x) = (1/20) - (1/30) = (1/60)Therefore, Ravi will take 60 days to complete the work, if he works alone.Hence Ram is 100% more efficient than Ram.Work & Time - Quant/Math - CAT 2008Question 4 the day: August 08, 2003The question for the day is from the topic of Work & Time. A red light flashes 3 times per minute and a green light flashes 5 times in two minutes at regular intervals. If both lights start flashing at the same time, how many times do they flash together in each hour? (1) 30 (2) 24 (3) 20 (4) 60Correct Answer - (1)Solution:Red light flashes every 20 seconds World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from light flashes every 24 secondsTherefore, they will flash together every 120 secondsIn every hour they will flash 3600/120 = 30 timesRaces Questions & AnswersRaces - Quant/Math - CAT 2008Question 4 the day: June 18, 2002The question for the day is from the topic - Races. A takes 3 min 45 seconds to complete a kilometre. B takes 4 minutes to complete the same 1 km track. If A and B were to participate in a race of 2 kms, how much start can A give B in terms of distance? (1) 30 m (2) 62.5 m (3) 125 m (4) 250 mCorrect Answer - (3)Solution:A can give B a start of 15 seconds in a km race.B takes 4 minutes to run a km. i.e 1000/4= 250 m/min = 250/60 m/secTherefore, B will cover a distance of 250/60 * 15 = 62.5 meters in 15 seconds.The start that A can give B in a km race therefore, is 62.5 meters, the distance thatB run in 15 seconds. Hence in a 2 km race, A can give B a start of 62.5 * 2 = 125 mor 30 seconds.Races - Quant/Math - CAT 2008Question 4 the day: July 12, 2002The question for the day is the from the topic - Races. In a kilometre race, A can give B a start of 100 m or 15 seconds. How long does A take to complete the race? 150 165 135 66.67 (1) (2) (3) (4) seconds seconds seconds secondsCorrect Answer - (3)Solution: World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from a 1000 metre race A gives B a start of 100 m or 15 seconds.This essentially means that B takes 15 seconds to run 100 m.Therefore, B will take 150 seconds to run the stretch of 1000 metres. (1000 m = 10times 100 m and therefore the time taken will also be 10 times 15 seconds = 150seconds).As A takes 15 seconds less than B, he will take 135 seconds to run the 1000 m.Races - Quant/Math - CAT 2008Question 4 the day: August 07, 2002The question for the day is from the topic Races. A gives B a start of 10 metres in a 100 metre race and still beats him by 1.25 seconds. How long does B take to complete the 100 metre race if A runs at the rate of 10 m/sec? (1) 8 seconds (2) 10 seconds (3) 16.67 seconds (4) 12.5 secondsCorrect Answer - (4)Solution:A gives B a start of 10 metres in a 100 metre race. This means that when A runs 100metres, B runs only 90 metres.Despite that start, A beats B by 1.25 seconds.As A is running at the speed of 10 m/sec, he will take 10 seconds to complete the100 metre race. And B takes 10 + 1.25 = 11.25 seconds to cover 90 metres.Therefore, the speed at which B is running = 8 m/sec.Running at 8 m/sec, B will take 100/8= 12.5 seconds to complete the 100 metrerace.Hence the correct answer is (4).Ratio, Proportion - Quant/Math - CAT 2008Question 4 the day: August 28, 2002The question for the day is from the topic of Ratio, Proportion. A predator is chasing its prey. The predator takes 4 leaps for every 6 leaps of the prey and the predator covers as much distance in 2 leaps as 3 leaps of the prey. Will the predator succeed in getting its food? World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from (1) Yes (2) In the 6th leap (3) Never (4) Cannot determineCorrect Answer - (4)Solution:Distance covered in 2 leaps by predator = 3 leaps of the prey.Distance covered in 1 leap of predator = 3/2 leaps of prey. ----(1)4 leaps of predator : 6 leaps of prey ----(2)Using (1) and (2), we get4*3/2 leaps of predator : 6 leaps of prey.=> 1:1If the predator and prey start simultaneously at the same point, the predator willcatch the prey immediately. If not so, then the predator will never catch the prey asit was running at the same speed.As it was not mentioned in the question that they start simultaneously from thesame point or not, we cant determine the answer. Therefore, the answer choice is(4).Races - Quant/Math - CAT 2008Question 4 the day: February 19, 2003The question for the day is from the topic of Races. A skating champion moves along the circumference of a circle of radius 21 meters in 44 seconds. How many seconds will it take her to move along the perimeter of a hexagon of side 42 meters? (1) 56 (2) 84 (3) 64 (4) 48Correct Answer - (2)Solution:Circumference 2*22/7 r = 2 * 22/7 *21= 132 meters World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from of the skater = Distance covered / Time taken132/44=3 m/sPerimeter of hexagon = 6a = 6 X 42= 252mTime taken to cover the perimeter of the hexagon = Distance (perimeter) / Speed252/3=84SecondsRaces - Quant/Math - CAT 2008Question 4 the day:May 22, 2003The question for the day is from the topic of Races. A runs 13/5 times as fast as B. If A gives a start of 240m, how far must the post be so that A and B might reach at the same time. (1) 390 m (2) 330 m (3) 600 m (4) 720 mCorrect Answer - (1)Solution:A runs 13/5 times fast as B which means A runs 13 metres for every 5 meters of B.Therefore, A gains 8 metre in a 11m race or if A gives a start of 8m in a 13m racethen the race might end in a dead heat.Therefore, if A gives a start of 240m (8* 30) then the length of the race should beequal to 13*30 = 390mOr the length of the race after A gives a start of 240 start so that A and B reach atthe same time is given by (240*13) /8 = 390m.Races - Quant/Math - CAT 2008Question 4 the day:June 2, 2003The question for the day is from the topic of Races. P can give Q a start of 20 seconds in a kilometer race. P can give R a start of 200 meters in the same kilometer race. And Q can give R a start of 20 World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from seconds in the same kilometer race. How long does P take to run the kilometer? (1) 200 seconds (2) 240 seconds (3) 160 seconds (4) 140 secondsCorrect Answer - (3)Solution:P can give Q a start of 20 seconds in a kilometer race. So, if Q takes xseconds to run a kilometer, then P will take x – 20 seconds to run thekilometer.Q can give R a start of 20 seconds in a kilometer race. So, if R takes yseconds to run a kilometer, then Q will take y – 20 seconds to run thekilometer.We know Q takes x seconds to run a kilometerTherefore, x = y – 20Therefore, P will take x – 20 = y – 20 – 20 = y – 40 seconds to run akilometer.i.e. P can give R a start of 40 seconds in a kilometer race, as R takes yseconds to run a kilometer and P takes only y – 40 seconds to run thekilometer.We also know that P can give R a start 200 meters in a km race.This essentially means that R runs 200 meters in 40 seconds.Therefore, R will take 200 seconds to run a km.If R takes 200 seconds to run a km, then P will take 200 – 40 = 160 secondsto run a km.Races - Quant/Math - CAT 2008Question 4 the day:June 30, 2003The question for the day is from the topic of Races. A gives B a start of 30 seconds in a km race and still beats him by 20 m. However, when he gives B a start of 35 seconds, they finish the race in a dead heat. How long does A take to run the km? (1) 250 seconds (2) 285 seconds (3) 220 seconds (4) 215 secondsCorrect Answer - (4) World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from A gives B a start of only 30 seconds, he beats him by 20 m.But, when he gives him a start of 35 seconds, they finish the race in a dead heatEssentially, B is able to run 20 m in the extra 5 second start that he gets in thesecond instance.Hence, Bs speed = 20/5 = 4 m/sec.As B runs at 4 m/sec speed, he will take 1000 / 4 = 250 seconds to complete a km.A can give B a start of 35 seconds in a km race. Hence, A will take only 215 secondsto run the km.Races - Quant/Math - CAT 2008Question 4 the day: August 13, 2003The question for the day is from the topic of Races. Three runners A, B and C run a race, with runner A finishing 12 meters ahead of runner B and 18 meters ahead of runner C, while runner B finishes 8 meters ahead of runner C. Each runner travels the entire distance at a constant speed. What was the length of the race? (1) 36 meters (2) 48 meters (3) 60 meters (4) 72 metersCorrect Answer - (2)Solution:Go from answer choicesLet the race be of length 48 metersSo when A runs 48m B run 36m and C runs 30mRatio of distance covered by B : C = 6 : 5So when B covers 48m C would have covered 40m => B can give C a start of 8m 09, 2003The question for the day is from the topic of Races. A can give B a start of 50 metres or 10 seconds in a kilometer race. How long does A take to complete the race? (1) 200 seconds (2) 140 seconds (3) 220 seconds (4) 190 secondsCorrect Answer - (4)Solution:A can give B a start of 50 metres or 10 seconds in a 1000 m race.That is, B takes 10 seconds to run 50 metres.Therefore, B will take (10/50) * 1000 = 200 seconds to run 1000 metres.A who can give B a start of 10 seconds will take 10 seconds lesser to run the 1000m.Hence, the time taken by A = 190 seconds.Races - Quant/Math - CAT 2008 Question 4 the day: November 10, 2003The question for the day is from the topic of Races. A can give B 20 points, A can give C 32 points and B can give C 15 points. How many points make the game? (1) 150 (2) 200 (3) 100 (4) 170Correct Answer - (3)Solution: World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from x points make the game, according to the statement if A has x points. Then Bhas (x – 20) pointsB has x points, then C has (x – 15) points, if C has (x – 32), then A has x pointsHence by chain rule x * x (x – 32) = (x – 20) * (x – 15) * xx2 – 32x = x2 – 35x + 300 => x = 100. Hence 100 points make the game.Arithmetic Mean Question AnswersWeighted Average - CAT 2007 Math PreparationQuestion 4 the day : November 10, 2006The question for the day is a sample practice problem in Aritbmetic Mean, weightedAverage, an Arithmetic Topic and the problem provides an understanding of simpleand weighted average.QuestionThe average monthly salary of 12 workers and 3 managers in a factory was Rs. 600.When one of the manager whose salary was Rs. 720, was replaced with a newmanager, then the average salary of the team went down to 580. What is the salaryof the new manager? 1. 570 2. 420 3. 690 4. 640Correct Answer - 420. Choice (2)Explanatory AnswerThe total salary amount = 15 * 600 = 9000The salary of the exiting manager = 720.Therefore, the salary of 12 workers and the remaining 2 managers = 9000 - 720 =8280When a new manager joins, the new average salary drops to Rs.580 for the totalteam of 15 of them.The total salary for the 15 people i.e., 12 workers, 2 old managers and 1 newmanager = 580 *15 = 8700Therefore, the salary of the new manager is 9000 - 8700 = 300 less than that of theold manager who left the company, which is equal to 720 - 300 = 420. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from alternate method of doing the problem is as follows:The average salary dropped by Rs.20 for 15 of them. Therefore, the overall salaryhas dropped by 15*20 = 300.Therefore, the new managers salary should be Rs.300 less than that of the oldmanager = 720 - 300 = 420.CAT Sample Questions : Arithmetic MeanQuestion 4 the day : June 11, 2004The question for the day is from the topic average in arithmetic.QuestionThe average wages of a worker during a fortnight comprising 15 consecutive workingdays was Rs.90 per day. During the first 7 days, his average wages was Rs.87/dayand the average wages during the last 7 days was Rs.92 /day. What was his wage onthe 8th day? 1. 83 2. 92 3. 90 4. 97Correct Answer - 97. Choice (4) is correct.Explanatory AnswerThe total wages earned during the 15 days that the worker worked = 15 * 90 = Rs.1350.The total wages earned during the first 7 days = 7 * 87 = Rs. 609.The total wages earned during the last 7 days = 7 * 92 = Rs. 644.Total wages earned during the 15 days = wages during first 7 days + wage on 8thday + wages during the last 7 days. 1350 = 609 + wage on 8th day + 644 wage on 8th day = 1350 - 609 - 644 = Rs.97. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Sample Questions in Math - AveragesQuestion 4 the day : April 13, 2004The question for the day is from the topic averages. It is a question based on theconcept of simple averageQuestionThe average of 5 quantities is 6. The average of 3 of them is 8. What is the averageof the remaining two numbers? 1. 6.5 2. 4 3. 3 4. 3.5Correct choice is (3) and the Correct Answer is 3Explanatory AnswerThe average of 5 quantities is 6.Therefore, the sum of the 5 quantities is 5 * 6 = 30.The average of three of these 5 quantities is 8.Therefore, the sum of these three quantities = 3 * 8 = 24The sum of the remaining two quantities = 30 - 24 = 6.Average of these two quantities = = 3.Note:From the answer choices, you can eliminate choice (1) and choice (2) even withoutsolving the question.As the average of the 5 quantities is 6 and the average of three of these five is 8, theaverage of the remaining two should be a value that is less than 6. So choice (1)which is more than 6 can be eliminated.If there were a total of four quantities and the overall average was 6 and theaverage of 2 of the four were 8, then the average of the remaining two would havebeen 4. i.e., the simple average of 4 and 8 is 6. However, we have unequal numberof quantities. Hence, choice (2) can be eliminated.Averages, Mean - CAT 2007 PreparationQuestion 4 the day : November 12, 2003The question for the day is a question in Simple Average, Arithmetic Mean - an World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Topic and the problem provides an understanding of the differentconcepts related to Averages.QuestionThe average temperature on Wednesday, Thursday and Friday was 250. The averagetemperature on Thursday, Friday and Saturday was 240. If the temperature onSaturday was 270, what was the temperature on Wednesday? 1. 240 2. 210 3. 270 4. 300Correct Answer is 300. Correct Choice is (4)Explanatory AnswerTotal temperature on Wednesday, Thursday and Friday was 25 * 3 = 750Total temperature on Thursday, Friday and Saturday was 24 * 3 = 720Hence, difference between the temperature on Wednesday and Saturday = 30If Saturday temperature = 270, then Wednesdays temperature = 27 + 3 = 300Averages, Arithmetic Mean - CAT 2007 PreparationQuestion 4 the day : October 21, 2003The question for the day is a sample practice problem in Simple Average, ArithmeticMean on change in the average with change in number of elements - an ArithmeticTopic and the problem provides an understanding of the different concepts related toAverages.QuestionThe average age of a group of 12 students is 20 years. If 4 more students join thegroup, the average age increases by 1 year. The average age of the new students is 1. 24 2. 26 3. 26 4. 22Correct Answer is 24 years. Correct Choice is (1)Explanatory AnswerTotal age of 12 students = 12 * 20 = 240 and the total age of 16 students = 21*16= 336. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from the average age of 4 new students be x.Therefore total age of the new students = 4x.Hence the total age of 16 students = 240 + 4x = 336 => x = 24Arithmetic Mean Questions, Answers - CAT 2007 Maths PreparationQuestion 4 the day : August 19, 2003The question for the day is from the topic of Averages. It is a weighted averagequestion and helps understand the basic concepts in averages and weighted average.QuestionWhen a student weighing 45 kgs left a class, the average weight of the remaining 59students increased by 200g. What is the average weight of the remaining 59students? 1. 57 2. 56.8 3. 58.2 4. 52.2Correct Answer is 57 kgs. Choice (1) is right.Explanatory AnswerLet the average weight of the 59 students be A.Therefore, the total weight of the 59 of them will be 59A.The questions states that when the weight of this student who left is added, the totalweight of the class = 59A + 45When this student is also included, the average weight decreases by 0.2 kgs. 59A + 45 / 60 = A - 0.2=> 59A + 45 = 60A - 12=> 45 + 12 = 60A - 59A=> A = 57.Weighted Average Questions : CAT 2007 Online PreparationQuestion 4 the day : August 12, 2003The question for the day is from the topic Averages. It is a question on weightedaverages.QuestionThree math classes: X, Y, and Z, take an algebra test.The average score in class X is 83. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from average score in class Y is 76.The average score in class Z is 85.The average score of all students in classes X and Y together is 79.The average score of all students in classes Y and Z together is 81.What is the average for all the three classes? 1. 81 2. 81.5 3. 82 4. 84.5Correct Answer is 81.5. Choice (2) is right.Explanatory AnswerAverage score of class X is 83 and that of class Y is 76 and the combined average ofX and Y is 79.By rule of alligation ratio of students in X : Y is given by X : Y 79 / 83 76 3 : 4Similarly, average score of class Y is 76 and that of class Z is 85 and the combinedaverage is 81.By rule of alligation ratio of students in Y : Z is Y : Z 81 / 76 85 4 : 5 World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from average for X, Y and Z = ( 3*83+4*76+5*85) / (3+4+5)= (249 + 304 + 425) / 12 = 81.5Averages, Mean - CAT 2007Question 4 the day : July 16, 2003The question for the day is a sample practice problem in Simple Average - anArithmetic Topic and the problem provides an understanding of the differentconcepts related to Averages.QuestionThe average weight of a class of 24 students is 36 years. When the weight of theteacher is also included, the average weight increases by 1kg. What is the weight ofthe teacher? 1. 60 kgs 2. 61 kgs 3. 37 kgs 4. None of theseCorrect Answer - 61 kgs. Correct Choice is (2)Explanatory AnswerThe average weight of a class of 24 students = 36 kgs.Therefore, the total weight of the class = 24 * 36 = 864 kgsWhen the weight of the teacher is included, there are 25 individuals.The average weight increases by 1kg. That is the new average weight = 37 kgs.Therefore, the total weight of the 24 students plus the teacher = 25 * 37 = 925Weight of the teacher = Weight of 24 students + teacher - weight of 24 students= 925 - 864 = 61 kgs.Averages Practice Questions : CAT 2007 Quant PreparationQuestion 4 the day : June 26, 2003The question for the day is from the topic Averages. It is a question on simpleaverage. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from average of 5 quantities is 10 and the average of 3 of them is 9. What is theaverage of the remaining 2? 1. 11 2. 12 3. 11.5 4. 12.5Correct Answer is 11.5. Choice (3) is right.Explanatory AnswerThe average of 5 quantities is 10.Therefore, the sum of all 5 quantities is 50.The average of 3 of them is 9.Therefore, the sum of the 3 quantities is 27.Therefore, the sum of the remaining two quantities = 50 - 27 = 23.Hence, the average of the 2 quantities = 23/2 = 11.5.Averages, Mean - CAT 2007Question 4 the day : June 12, 2003The question for the day is a sample practice problem in Simple Average, ArithmeticMean - an Arithmetic Topic and the problem provides an understanding of thedifferent concepts related to Averages.QuestionThe average age of a family of 5 members is 20 years. If the age of the youngestmember be 10 years then what was the average age of the family at the time of thebirth of the youngest member? 1. 13.5 2. 14 3. 15 4. 12.5Correct Answer is 12.5. Correct Choice is (4)Explanatory AnswerAt present the total age of the family = 5 * 20 = 100The total age of the family at the time of the birth of the youngest member = [100-10-(10*4)] = 50 World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from average age of the family at the time of birth of the youngest member =50/4 = 12.5.Averages Questions Answers - CAT 2007 Online PreparationQuestion 4 the day : March 27, 2003The question for the day is from the topic Averages. The question is a CAT 2002question.QuestionA student finds the average of 10 positive integers. Each integer contains two digits.By mistake, the boy interchanges the digits of one number say ba for ab. Due tothis, the average becomes 1.8 less than the previous one. What was the differenceof the two digits a and b? 1. 8 2. 6 3. 2 4. 4Correct Answer - 2. Choice (3) is right.Explanatory AnswerLet the original number be ab i.e., (10a + b).After interchanging the digits, the new number becomes ba i.e., (10b + a).The question states that the average of 10 numbers has become 1.8 less than theoriginal average. Therefore, the sum of the original 10 numbers will be 10*1.8 morethan the sum of the 10 numbers with the digits interchanged.i.e., 10a + b = 10b + a + 18, 9a - 9b = 18, a - b = 2.Averages questions, answers: CAT 2007 Quant PreparationQuestion 4 the day: March 06, 2003The question for the day is from the topic Averages.QuestionAverage cost of 5 apples and 4 mangoes is Rs. 36. The average cost of 7 apples and8 mangoes is Rs. 48. Find the total cost of 24 apples and 24 mangoes. 1. 1044 2. 2088 3. 720 World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from and Proportions - Quant/Math - CAT 2008Question 4 the day: June 3, 2002The question for the day is from Ratio and Proportions. Rs.432 is divided amongst three workers A, B and C such that 8 times A's share is equal to 12 times B's share which is equal to 6 times C's share. How much did A get? (1) Rs.192 (2) Rs.133 (3) Rs.144 (4) Rs.128Correct Answer - (3)Solution:8 times A's share = 12 times B's share = 6 times C's share.Note that this is not the same as the ratio of their wages being 8 : 12 : 6In this case, find out the L.C.M of 8, 12 and 6 and divide the L.C.M by each of theabove numbers to get the ratio of their respective shares.The L.C.M of 8, 12 and 6 is 24.Therefore, the ratio A:B:C :: 24/8 : 24/12 : 24/6=> A : B : C :: 3 : 2 : 4The sum of the total wages = 3x + 2x + 4x = 432 => 9x = 432 or x = 48.Hence A gets 3 * 48 = Rs. 144.Ratio and Proportions - Quant/Math - CAT 2008Question 4 the day: June 7, 2002The question for the day is from the topi Ratio and Proportions. If 20 men or 24 women or 40 boys can do a job in 12 days working for 8 hours a day, how many men working with 6 women and 2 boys take to do a job four times as big working for 5 hours a day for 12 days? (1) 8 men (2) 12 men (3) 2 men (4) 24 menCorrect Answer - (3)Solution:Amount of work done by 20 men = 24 men = 40 boys or 1 man = 1.2 woman = 2boys.Let us therefore, find out the amount of men required, if only men were working onthe job, to complete the new job under the new conditions and then makeadjustments for the women and children working with the men.The man hours required to complete the new job = 4 times the man hours requiredto complete the old job. (As the new job is 4 times as big as the old job)Let 'n' be the number of men required. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from * 12 * 8 = n * 5 * 12 * 4. n = 8.8 men working will be able to complete the given job.However, the problem states that 6 women and 2 boys are working on the job.6 women = 6/12= 5 men and 2 boys = 1 man. The equivalent of 5 + 1 = 6 men arealready working.Therefore, 2 men a required to work with 6 women and 2 boys to complete the job.Ratio and Proportion - Quant/Math - CAT 2008Question 4 the day: July 11, 2002The question for the day is the from the topic - Ratio and Proportion. Two cogged wheels of which one has 32 cogs and other 54 cogs, work into each other. If the latter turns 80 times in three quarters of a minute, how often does the other turn in 8 seconds? (1) 48 (2) 135 (3) 24 (4) None of theseCorrect Answer - (3)Solution:Less Cogs => more turns and less time => less turns cogs time turnsA 54 45 80B 32 8 ?Number of turns required = 80 * 54/32 * 8/45= 24 timesRatio and Proportion - Quant/Math - CAT 2003Question 4 the day: February 4, 2003The question for the day is from the topic of Ratio and Proportion. The monthly incomes of A and B are in the ratio 4 : 5, their expenses are in the ratio 5 : 6. If A saves Rs.25 per month and B saves Rs.50 per month, what are their respective incomes? (1) Rs.400 and Rs.500 (2) Rs.240 and Rs.300 (3) Rs.320 and Rs.400 (4) Rs.440 and Rs.550 World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Answer - (1)Solution:SolutionLet As income be = 4xAs expenses, therefore = 4x - 25Let Bs income be = 5xBs expenses, therefore = 5x - 50We know that the ratio of their expenses = 5 : 6=> 24x - 150 = 25x - 250=> Therefore, x = 100.=> As income = 4x = 400 and Bs income = 5x = 500.Ratio & Proportion - Quant/Math - CAT 2008 Question 4 the day: February17, 2003The question for the day is from the topic of Ratio and Proportion. The proportion of milk and water in 3 samples is 2:1, 3:2 and 5:3. A mixture comprising of equal quantities of all 3 samples is made. The proportion of milk and water in the mixture is (1) 2:1 (2) 5:1 (3) 99:61 (4) 227:133Correct Answer - (4)Solution:Proportion of milk in 3 samples is 2/3, 3/5, 5/8.Proportion of water in 3 samples is 1/3, 2/5, 3/8.Since equal quantities are taken,Total proportion of milk is 2/3 + 3/5 + 5/8 = 227/120 World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from proportion of water is 1/3 + 2/5 + 3/8 = 133/120Proportion of milk and water in the solution is = 227:133So Choice (4) is the right answer.Ratio and Proportion - Quant/Math - CAT 2008Question 4 the day: March 26, 2003The question for the day is from the topic of Ratio and Proportion A group of workers can do a piece of work in 24 days. However as 7 of them were absent it took 30 days to complete the work. How many people actually worked on the job to complete it? (1) 35 (2) 30 (3) 28 (4) 42Correct Answer - (3)Solution:Let the original number of workers in the group be xTherefore, actual number of workers = x-7.We know that the number of manhours required to do the job is the same in boththe cases.Therefore, x (24) = (x-7).3024x = 30x - 2106x = 210x = 35.Therfore, the actual number of workers who worked to complete the job = x - 7 = 35-7 = 28.Ratio and Proportion - Quant/Math - CAT 2008Question 4 the day: March 31, 2003The question for the day is from the topic of Ratio and Proportion. A, B and C play cricket. As runs are to Bs runs and Bs runs are to Cs as 3:2. They get altogether 342 runs. How many runs did A make? May 16, 2003The question for the day is from the topic of Ratio and Proportion. A fort has provisions for 60 days. If after 15 days 500 men strengthen them and the food lasts 40 days longer, how many men are there in the fort? (1) 3500 (2) 4000 (3) 6000 (4) None of theseCorrect Answer - (2)Solution:Let there be x men in the beginning so that after 15 days the food for them is leftfor 45 days.After adding 500 men the food lasts for only 40 days.Now (x+500) men can have the same food for 40 days.Therefore by equating the amount of food we get,45 * x = (x + 500) * 4045x = (x+500) * 405x = 20,000x = 4,000Therefore there are 4,000 men in the fort.Ratio and Proportion - Quant/Math - CAT 2008Question 4 the day: May 26, 2003The question for the day is from the topic of Ratio and Proportion. The ratio of marks obtained by vinod and Basu is 6:5. If the combined average of their percentage is 68.75 and their sum of the marks is 275, find the total marks for which exam was conducted. (1) 150 (2) 200 (3) 400 (4) None of these.Correct Answer - (2) World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Vinod marks be 6x and Basus is 5x. Therefore, the sum of the marks = 6x + 5x= 11x.But the sum of the marks is given as 275 = 11x. We get x = 25 therefore, vinodmarks is 6x = 150 and Basu marks = 5x = 125.Therefore, the combined average of their marks = (150 + 125) / 2 = 137.5.If the total mark of the exam is 100 then their combined average of their percentageis 68.75Therefore, if their combined average of their percentage is 137.5 then the totalmarks would be (137.5 / 68.75)*100 = 200.Ratio and Proportion - Quant/Math - CAT 2008Question 4 the day: October 22, 2003The question for the day is from the topic of Ratio and Proportion. The present ages of A and B are as 6 : 4. Five years ago their ages were in the ratio 5 : 3. Find their present ages. (1) 42, 28 (2) 36, 24 (3) 30, 20 (4) 25, 15Correct Answer - (3)Solution:Go from the choicesChoice (3) 30 and 20 are in the ratio of 6: 4Five years ago their ages would be 25 and 15 which are in the ratio 5 : 3.Hence choice (3) is the right answer. November 11, 2003The question for the day is from the topic of Ratio and Proportion. A, B and C enter into a partnership by investing Rs.3600, Rs.4400 and Rs.2800. A is a working partner and gets a fourth of the profit for his services and the remaining profit is divided amongst the three in the rate of their investments. What is the amount of profit that B gets if A gets a total of Rs. 8000? (1) 4888.88 (2) 9333.33 (3) 4000 (4) 3666.66Correct Answer - (1)Solution:Let x be the profit.Their investment ratio = 3600: 4400: 2800 = 9 : 11 : 7As profit of Rs. 8000 = (1/4 * x) + 1/3(3/4*x) = 1/2 * xx = Rs. 16,000Therefore Bs profit = 11/27(3/4 * 16000) = Rs. 4888.88Ratio Proportion. Quant/Math - CAT 2008Question 4 the day: March 18, 2004The question for the day is from the topic ratio proportion. A, B and C, each of them working alone can complete a job in 6, 8 and 12 days respectively. If all three of them work together to complete a job and earn Rs.2340, what ill be C's share of the earnings? (1) Rs.520 (2) Rs.1080 (3) Rs.1170 (4) Rs.630Correct choice - (1) Correct Answer -(Rs.520)Solution:A, B and C will share the amount of Rs. 2340 in the ratio of the amounts of work World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from by them.As A takes 6 days to complete the job, if A works alone, A will be able to complete1/6 th of the work in a day.Similarly, B will complete 1/8 th and C will complete 1/12 th of the work.So, the ratio of the work done by A : B : C when they work together will be equal to1/6 : 1/8 : 1/12Multiplying the numerator of all 3 fractions by 24, the LCM of 6, 8 and 12 will notchange the relative values of the three values.We get24/6 : 24/8 : 24/12 = 4 : 3 : 2.i.e., the ratio in which A: B : C will share Rs.2340 will be 4 : 3 : 2.Hence, C's share will be 2/9 * 2340= Rs.520. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from & Allegations QuestionsMixtures and Allegations - Quant/Math - CAT 2008Question 4 the day: September 06, 2002The question for the day is from the topic of Mixtures and Allegations. How many litres of water should be added to a 30 litre mixture of milk and water containing milk and water in the ratio of 7 : 3 such that the resultant mixture has 40% water in it? (1) 7 litres (2) 10 litres (3) 5 litres (4) None of theseCorrect Answer - (3)Solution:30 litres of the mixture has milk and water in the ratio 7 : 3. i.e. the solution has 21litres of milk and 9 litres of water.When you add more water, the amount of milk in the mixture remains constant at 21litres. In the first case, before addition of further water, 21 litres of milk accounts for70% by volume. After water is added, the new mixture contains 60% milk and 40%water.Therefore, the 21 litres of milk accounts for 60% by volume.Hence, 100% volume = 21/0.6= 35 litres.We started with 30 litres and ended up with 35 litres. Therefore, 5 litres of water wasadded.Mixtures and Alligation - Quant/Math - CAT 2008Question 4 the day: October 11, 2002The question for the day is from the topic of Mixtures and Alligation. How many kgs of Basmati rice costing Rs.42/kg should a shopkeeper mix with 25 kgs of ordinary rice costing Rs.24 per kg so that he makes a profit of 25% on selling the mixture at Rs.40/kg? (1) 20 kgs (2) 12.5 kgs (3) 16 kgs (4) 200 kgsCorrect Answer - (1) World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from the amount of Basmati rice being mixed be x kgs. As the trader makes25% profit by selling the mixture at Rs.40/kg, his cost /kg of the mixture =Rs.32/kg.i.e. (x * 42) + (25 * 24) = 32 (x + 25)=> 42x + 600 = 32x + 800=> 10x = 200 or x = 20 kgs.Mixtures and Alligations - Quant/Math - CAT 2008Question 4 the day: June 16, 2003The question for the day is from the topic of Mixtures and Alligations. How many litres of a 12 litre mixture containing milk and water in the ratio of 2 : 3 be replaced with pure milk so that the resultant mixture contains milk and water in equal proportion? (1) 4 litres (2) 2 litres (3) 1 litre (4) 1.5 litresCorrect Answer - (2)Solution:The mixture contains 40% milk and 60% water in it. That is 4.8 litres of milk and 7.2litres of water.Now we are replacing the mixture with pure milk so that the amount of milk andwater in the mixture is 50% and 50%.That is we will end up with 6 litres of milk and6 litres of water.Water gets reduced by 1.2 litres.To remove 1.2 litres of water from the original mixture containing 60% water, weneed to remove 1.2 / 0.6 litres of the mixture = 2litres July 07, 2003The question for the day is from the topic Mixtures. A sample of x litres from a container having a 60 litre mixture of milk and water containing milk and water in the ratio of 2 : 3 is replaced with pure milk so that the container will have milk and water in equal proportions. What is the value of x? (1) 6 litres (2) 10 litres (3) 30 litres (4) None of theseCorrect Answer - (2)Solution:The best way to solve this problem is to go from the answer choices.The mixture of 60 litres has in it 24 litres of milk and 36 litres of water. (2 : 3 :: milk: water)When you remove x litres from it, you will remove 0.4 x litres of milk and 0.6 x litresof water from it.Take choice (2). According to this choice, x = 10.So, when one removes, 10 litres of the mixture, one is removing 4 litres of milk and6 litres of water.Therefore, there will be 20 litres of milk and 30 litres of water in the container.Now, when you add 10 litres of milk, you will have 30 litres of milk and 30 litres ofwater – i.e. milk and water are in equal proportion.Mixtures and Alligation - Quant/Math - CAT 2008Question 4 the day: August 06, 2003The question for the day is from the topic of Mixtures and Alligation. A zookeeper counted the heads of the animals in a zoo and found it to be 80. When he counted the legs of the animals he found it to be 260. If the zoo had either pigeons or horses, how many horses were there in the zoo? (1) 40 (2) 30 (3) 50 (4) 60 World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Answer - (3)Solution:Let the number of horses = xThen the number of pigeons = 80 – x.Each pigeon has 2 legs and each horse has 4 legs.Therefore, total number of legs = 4x + 2(80-x) = 260=>4x + 160 – 2x = 260=>2x = 100=>x = 50.Mixtures and Alligations - Quant/Math - CAT 2008Question 4 the day: July 4, 2002 In what ratio must a person mix three kinds of tea costing Rs.60/kg, Rs.75/kg and Rs.100 /kg so that the resultant mixture when sold at Rs.96/kg yields a profit of 20%? (1) 1 : 2 : 4 (2) 3 : 7 : 6 (3) 1 : 4 : 2 (4) None of theseCorrect Answer - (3)Solution:The resultant mixture is sold at a profit of 20% at Rs.96/kgi.e. 1.2 (cost) = Rs.96 => Cost = 96/1.2 = Rs.80 / kg.Let the three varities be A, B, and C costing Rs.60, Rs.75 and Rs.100 respectively.The mean price falls between B and C.Hence the following method should be used to find the ratio in which they should bemixed.Step 1. Find out the ratio of QA : QC using alligation rule Qa/Qc = 100-80 / 80-60 1/1Step 2. Find out the ratio of QB : QC using alligation rule Qb / Qc = 100-80/80-75= 4/1Step 3. QC, the resultant ratio of variety c can be found by adding the value of QC in World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from 1 and step 2 = 1 + 1 = 2.However, in CAT if you try and solve the problem using the above method, you willend up spending more than 2, and may be 3 minutes on this problem, which is acriminal mismanagement of time.The best way to solve a problem of this kind in CAT is to go from the answer choicesas shown below The resultant ratio QA : QB : QC :: 1 : 4 : 2. 1 kg of variety A at Rs.60 is mixed with 4 kgs of variety B at Rs.75 and 2 kgs of variety C at Rs.100. The total cost for the 7 kgs = 60 + (4 * 75) + (2 * 100) = 60 +300 + 200 = 560. Cost per kg of the mixture = 560/7 = 80 kgs.Even assuming that you hit upon the right answer as the last choice, you will still bebetter of going back from the answerMixtures and Alligations - Quant/Math - CAT 2008Question 4 the day: August 19, 2002The question for the day is from the topic of Mixtures and Alligations. A merchant mixes three varieties of rice costing Rs.20/kg, Rs.24/kg and Rs.30/kg and sells the mixture at a profit of 20% at Rs.30 / kg. How many kgs of the second variety will be in the mixture if 2 kgs of the third variety is there in the mixture? (1) 1 kg (2) 5 kgs (3) 3 kgs (4) 6 kgsCorrect Answer - (2)Solution:If the selling price of mixture is Rs.30/kg and the merchant makes a profit of 20%,then the cost price of the mixture = 30/1.2 = Rs.25/kg.We need to find out the ratio in which the three varieties are mixed to obtain amixture costing Rs.25 /kg.Let variety A cost Rs.20/kg, variety B cost Rs.24 / kg and variety C cost Rs.30/kg.The mean desired price falls between B and C.Step 1: Find out the ratio QA : QC using alligation rule. Qa / Qc = 30-25 / 25-20=1/1 World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from 2: Find out the ratio QB : QC using alligation rule. Qb / Qc = 30-25 / 25-24=5/1Step 3: QC is found by adding the value of QC in step 1 and step 2 = 1 + 1 = 2Therefore, the required ratio = 1 : 5 : 2If there are 2 kgs of the third variety in the mixture, then there will be 5 kgs of thesecond variety in the mixture.Note: This is a problem to be skipped, at least in the first go. If you were able tosolve at least 30 other problems in quant, then you should look at this problem.Quantitative General Questionbank - CAT 2007 Sample QuestionsQ. The annual salary of a person in the year2000 is 20% more than that of the year1999.His annual salary is increased by Rs.36,000 in the year 2001 over that of theyear 2,000.If the increase in the annual salary of the person in the year 2002 overthe year 2001 is 5 percentage points less than the increase in year 2001 over year2000,and his annual salary in the year 2002 is Rs.2,16,000,then what was his salaryin the year 1999?A. Rs.1,00,000B. Rs.1,20,000C. Rs.1,25,000D. Cannot be determined ans: DQ. Steve Warne captained his team in 120 one day cricket matches with a successrate of 75%.For the first m matches,his success rate was 70%,55% for the next nmatches and 90% for the last p matches.How many matches did he win in the firstm + n matches he captained?A. 30B. 36C. 45D. Cannot be determined ans: BQ. The volume of a cubiod increase by 40.4%.The total length of all the edges isincreased by 20% and the lateral surface area is increased by 13.4%.What is thepercentage increase/decrease in the height of the cubiod,if the ratio oflength,breadth and height is 3 : 2 : 1?A. 10% increaseB. 10% decreaseC. 20% increaseD. 20% decrease ans: BQ. Three times a number is 20% more than twice another number when increasedby 105.If twice the first number increased by 36 is 20% less than three times of thesecond number,then what is the first number? World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from 150B. 162C. 180D. None of these ans: BQ. In the year 2001,XYZ motors sold 50%,20%,30% of the total motorcycles sold inthat year in the first 3 months,next 4 months and last 5 months respectively.In theyear 2002,the increase in the number of motorcycles sold in the first 7 months is40% over the same period of the previous year and increase in the number ofmotorcycles sold in the last 9 months is 20% over the corresponding period of theprevious year.What is the minimum percentage increase in the number ofmotorcycles sold in year 2002 over 2001 if the increase in the number ofmotorcycles sold from April 2002 to July 2002 over the same 4 months of theprevious year is not more than 100%?A. 18%B. 38%C. 58%D. None of these ans: AQ. A,B and C contest an election from a particular constituency.A and B together got50% more votes than C.The vote share of A and C together is 30 percentage pointsmore than the vote share of B.Who won the election?A. AB. BC. CD. Cannot be determined ans: CQ. P,Q and R scored 36%,41% and 51% respectively in a test.R passed the test andQ failed the test.If one of them failed by 21 marks and R passed by 39 marks,thenwhat is the total marks in the test?A. 1,200B. 600C. 400D. Cannot be determined ans: DQ. A,B and C scored 28%,36% and 53% in an examination respectively.B and Cpassed the examination but A failed.One of them passed by 33 marks and A failed by17 marks.What is the pass mark in the examination?A. 192B. 73C. 175D. Cannot be determined ans: A World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from A man divides Rs.9,000 into 3 unequal parts and invests them at 5%,6% and 8%per annum.At the end of one year he receives an interest of Rs.580 on his totalinvestment.If he receives equal interest from two of his investments,how much didhe invest at 6%,which is more than the investment at 5%?A. Rs. 3,500B. Rs. 4,000C. Rs. 3,000D. Cannot be determined ans: BQ. Instead of increasing the salary of a salesman twice successively by 20%,theemployer has given a one-time 40% hike.What is the loss or gain for the employee ifhis original salary was Rs.1,000?A. Rs. 4 gainB. Rs. 40 lossC. Rs. 40 gainD. No loss,no gain ans: BQ. The salary of a salesman is first increased twice succesively by 15% and thendecreased twice successively by 15%.What is the approximate effective change inhis original salary?A. 10% increaseB. 5% decreaseC. 19% decreaseD. 25% increase ans: BQ. in a city,20% of the total population is the student community,which is notemployed.Of the remaining,56.25% are employed.If the number of non-studentswho are unemployed is 14,000,then find the population of the city.A. 56,000B. 70,000C. 40,000D. 60,000 ans: CQ. Three candidates A,B and C contest an assembly seat.A got as many more votesthan B as B got more than C.If A won the election by a majority of 28,000 votes andC got 52,000 votes.find the percentage of votes polled to A.9All the votes polled arevalid).A. 45%B. 36%C. 33.33%D. Data insufficient ans: AQ. Some articles were sold at a certain sellingprice.When the price of each article World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from increased by 20%,the revenue from the sales decreased by 10% and becameRs.2,160.If the new price of each article is Rs.36 then find the number of articlessold at the original price.A. 75B. 80C. 60D. 90 ans: BQ. Out of four numbers,the second and the third numbers respectively are 50% and150% more than the first number.If the fourth number is 25 more than the thirdnumber and five times the first number,then by what percentage is the scondnumber less than the fourth number?A. 38.46%B. 70%C. 60%D. 41.66% ans: B-------------------------------------------------------------------------------1. Some work is done by two people in 24 minutes. One of them can do this workalone in 40 minutes. How much time does the second person take to do the samework ?Ans. 60 minutes2. A car is filled with four and half gallons of fuel for a round trip.If the amount offuel taken while going is 1/4 more than the amount taken for coming, what is theamount of fuel consumed while coming back?Ans.2 gallons3. The lowest temperature in the night in a city A is 1/3 more than 1/2 the highestduring the day. Sum of the lowest temperature and the highest temperature is 100degrees. Then what is the low temp?Ans.40 degrees4. Javagal, who decided to go to weekened trip should not exceed 8 hours driving ina day. The average speed of forward journey is 40 miles/hr.Due to traffic onsundays, the return journeys average speed is 30 m/h. How far he can select apicnic spot? World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from 120 milesb) between 120 and 140 milesc) 160 milesAns. 120 miles5. A salesperson by mistake multiplied a number and got the answer as 3, instead ofdividing the number by 3.What is the answer he should have actually got?Ans. 36. A building with height D shadow upto G. What is the height of a neighbouringbuilding with a shadow of C feet.Ans. (C*D)/G7. A person was fined for exceeding the speed limit by 10 mph. Another person wasalso fined for exceeding the same speed limit by twice the same. If the secondperson was travelling at a speed of 35 mph, find the speed limit.Ans. 15 mph8. A bus started from bustand at 8.00am, and after staying for 30 minutes at adestination, it returned back to the busstand. The destination is 27 miles from thebusstand. The speed of the bus is 18mph. During the return journey bus travels with50% faster speed.At what time does it return to the busstand?Ans. 11.00am9. In a mixture, R is 2 parts and S is 1 part. In order to make S to 25% of themixture, how much of R is to be added?Ans.One part of R10. Wind flows 160 miles in 330 min, for travelling 80 miles how much time does itrequire?Ans. 2 hrs 45 mins11. With a 4/5 full tank a vehicle can travel 12 miles, how far can it travel with a 1/3full tank World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from 5 miles12. There are two trees in a lawn. One grows at a rate 3/5 of the other in 4 years. Ifthe total growth of trees is 8 ft. What is the height of the smaller tree after 2 yearsAns. 1 1/2 feet13. Refer to the figure below.A ship started from P and moves at a speed of I milesper hour and another ship starts from L and moving with H miles per hoursimultaneously.Where do the two ships meet?||---g---||---h---||---i---||---j---||---k---||---l---||PG H I J K L are the various stops in between denoted by || . The values g, h, i, j, k,l denote the distance between the ports.Ans. Between I and J, closer to J14. If A is travelling at 72 km per hour on a highway. B is travelling at a speed of 25meters per second on a highway. What is the difference in their speeds in m/sec.Ans. 1 m/sec15. What is the percentage represented by 0.03 * 0.05 ?(a)0.0015(b)0.000015(c)0.15(d)15Ans.B16. (x-a)(x-b)(x-c)....(x-z) = ?(a) 1(b) -1(c) 0(d) Cant be determinedAns. C17. If a = 1, b = 2, c = 3.......z = 26 what is the value of p+q+r ? World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from B18. A is 8 miles east of B. C is 10 miles north of B. D is 13 miles east of C and E is 2 miles north of D. Find shortest distance between A and E.(a) 5 miles(b) 6miles(c) 13 miles(d) 18 milesAns. C19. If z = 1, y = 2.......a = 26. Find the value of z + y + x + .......+a.(a) 351(b) 221(c) 400(d) 200Ans. A20. There are 30 socks in a bag. Out of these 60 % are green and the rest are blue. What is the maximum number of times that socks have to be taken out so thatatleast 1 blue pair is found.(a) 21(b) 2(c)18(d) 20Ans. D21. How many two digit numbers have their square ending with 8.(a) 13(b) 12 World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from 0(d) 11Ans. C22. How many numbers are there between 100 and 300 with 2 in the end and 2 inthe beginning.(a) 10(b) 9(c) 11(d) none of theseAns. A23. 0.000006 * 0.0000007 = ?(a) 0.0000000042(b) 0.000000000042(c) 0.0000000000042(d) 0.00000000000042Ans. B24. You have Rs 1000 with 8% p.a compounded every 6 months. What is the total interest you get after 1 year.(a) Rs.116.40(b) Rs.345.60(c) Rs.224.50(d) Rs.160Ans. A25. If x + y =12, x-y=2 Find x + 2y.(a) 12(b) 17(c) 14(d) none of these World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from B26. With one gallon of petrol a person moves at a speed of 50 mph and covers 16miles. 3/4th of the distance is covered while moving at 60 mph. How many gallons does he need to cover 120 miles in 60 mph.27. A tap drains at x speed while tap B is closed. When both taps are open they drain at y speed. What is the speed of draining when only tap B is open(a) x - y(b) y-x(c) x(d) cant be determinedAns. B28. What is twenty percent of 25 % of 20.(a) 2(b)1(c) 5(d) 4Ans. B29. A rectangle has the dimensions 6ft * 4ft. How many squares of 0.5 inches will it need to completely fill it.(a) 32000(b) 12824(c) 13824(d) 18324Ans. CVEDIC MATHEMATICS World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from Mathematics is the name given to the ancient system of Mathematics whichwas discovered from the Vedas between 1911 and 1918 by Sri Bharati Krsna Tirthaji(1884-1960). According to his research all of mathematics is based on sixteen Sutrasor word-formulae. For example, Vertically and Crosswise` is one of these Sutras.These formulae describe the way the mind naturally works and are therefore a greathelp in directing the student to the appropriate method of solution.Perhaps the most striking feature of the Vedic system is its coherence. Instead of ahotch-potch of unrelated techniques the whole system is beautifully interrelated andunified: the general multiplication method, for example, is easily reversed to allowone-line divisions and the simple squaring method can be reversed to give one-linesquare roots. And these are all easily understood. This unifying quality is verysatisfying, it makes mathematics easy and enjoyable and encourages innovation.In the Vedic system difficult problems or huge sums can often be solvedimmediately by the Vedic method. These striking and beautiful methods are just apart of a complete system of mathematics which is far more systematic than themodern system. Vedic Mathematics manifests the coherent and unified structure ofmathematics and the methods are complementary, direct and easy.The simplicity of Vedic Mathematics means that calculations can be carried outmentally (though the methods can also be written down). There are manyadvantages in using a flexible, mental system. Pupils can invent their own methods,they are not limited to the one correct method. This leads to more creative,interested and intelligent pupils.Interest in the Vedic system is growing in education where mathematics teachers arelooking for something better and finding the Vedic system is the answer. Research isbeing carried out in many areas including the effects of learning Vedic Maths onchildren; developing new, powerful but easy applications of the Vedic Sutras ingeometry, calculus, computing etc.But the real beauty and effectiveness of Vedic Mathematics cannot be fullyappreciated without actually practising the system. One can then see that it isperhaps the most refined and efficient mathematical system possible.Base MethodThis is very suitable when numbers are close to a base like 10, 100, 1000 or so on.Lets take an example:106 × 108Here the base is 100 and the surplus is 6 and 8 for the two numbers. The answerwill be found in two parts, the right-hand should have only two digits (because baseis 100) and will be the product of the surpluses. Thus, the right-hand part will be 6× 8, i.e. 48. The left-hand part will be one multiplicand plus the surplus of the othermultiplicand. The left part of the answer in this case will be 106 + 8 or for thatmatter 108 + 6 i.e. 114. The answer is 11448.12 X 14. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from would the most suitable base. In the current example, the surplus numbers are+2 and +4.If 8x7 were to be performed and base of 10 were chosen, then -2 and -3 wouldhave been the deficit numbers.Try the following numbers(a) 13 X 16 (b) 16 X 18 (c) 18 X 19 (d) 22 X 24Once you get comfortable, do not use any paper or pen.27 X 28 322 #9; #9; 23 X 18 46 X 48 5255 58253 X 57 622 382 42 X 46 #9; #9; 9698 92 X 9399 X 99 #9; #9; 102 X 105 98 X 107 112X113 1082 123 X 127USING OTHER BASESIn 46 X 48, the base chosen is 50 and multiplication of 44 by 50 is better done likethis: take the half of 44 and put two zeros at the end, because 50 is same as 100/2.Therefore, product will be 2200. It would be lengthy to multiply 44 by 5 and put azero at the end. In general, whenever we want to multiply anything by 5, simplyhalve it and put a zero.Multiply 32 by 25. Most of the students would take 30 as the base. The method iscorrect but nonetheless lengthier. Better technique is to understand that 25 is sameas one-fourth. Therefore, one-fourth of 32 is 8 and hence the answer is 800.An application of Base Method to learn multiplications of the type 3238, where unitsdigit summation is 10 and digits other than units digit are same in both thenumbers. In the above example, 2 + 8 = 10 and 3 in 32 is same as 3 in 38.Therefore method can be applied. The method is simple to apply. The group ofdigits other than units digit, in this case 3, is multiplied by the number next toitself. Therefore, 3 is multiplied by 4 to obtain 12, which will form the left part of theanswer. The units digits are multiplied to obtain 16 (in this case), which will formthe right part of the answer. Therefore, the answer is 1216.Try these now53 X 57 91 X 99 106 X 104 123 X 127The rule for squares of numbers ending with 5. e.g., 652. This is same as 65 X 65and since this multiplication satisfies the criteria that units digit summation is 10and rest of the numbers are same, we can apply the method. Therefore, the answeris 42 / 25 = 4225. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from these:352 952 1252 2052CUBINGFinding the cubes of numbers close to the powers of 10. e.g., cubes of 998, 1004,100012, 10007, 996, 9988, etc. Some of the numbers are in surplus and others arein deficit. Explain the method as given below.Find (10004)3Step (I) : Base is 10000. Provide three spaces in the answer.The base contains 4zeros. Hence, the second and third space must contain exactly 4 digits. 1 0 0 0 4 = —/ —/ —Step (II) : The surplus is (+4). If surplus is written as a, perform the operation3a and add to the base 10000 to get 10012. Put this in the 1st space.1 0 0 0 4 = 1 0 0 1 2 /—/—Step (III) : The new surplus is (+12). Multiply the new surplus by the old surplus,i.e. (+4)(+12) = (+48). According to the rule written in the step (I), 48 is writtenas 0048.1 0 0 0 4 = 1 0 0 1 2 / 0 0 4 8 /—Step (IV) : The last space will be filled by the cube of the old surplus (+4).Therefore, 43 = 64, which is written as 0064.10004=10012/0048/0064Therefore, the answer is 1001200480064.Find (998)3Step (I) : Base = 1000. Hence, exactly 3 digits must be there in the 2nd and 3rdspace.The deficit = (+2) 9 9 8 = —/—/—Step (II) : Multiply the deficit by 3 and subtract (because this is the case of deficit)from the base. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from words which have been removed in this reproduction. Also the passagesappearing in the actual paper are much lengthier.Directions: Answer the questions given below the passage or statement as true, falseor cant say.PASSAGE A: My father has no brothers. He has three sisters who has two childseach.Answer 1-5 based on the passage A1.My grandfather has two sons .2. Three of my aunts have two sons3. My father is only child to his father4. I have six cousins from my mother side5. I have one unclePASSAGE B: Ether injected into gallablader to dissolve colestrol based gallstones.This type one day treatment is enough for gallstones not for calcium stones. Thismethod is alternative to surgery for millions of people who are suffering from thisdisease.Answer questions 6-9 based on passage B6.Calcium stones can be cured in oneday7. Hundreds of people contains calcium stones8. Surgery is the only treatment to calcium stones9. Ether will be injected into the gallbleder to cure the cholestrol based gall stonesPASSAGE C: Hacking is illegal entry into another computer. This happens mostlybecause of lack of knowledge of computer networking. With networks one machinecan access to another machine.Hacking go about without knowing that each networkis accredited to use network facility.Answer questions 10-12 based on passage B10. Hackers never break the code of the company which they work for11. Hacking is the only vulnerability of the computers for the usage of the data World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from is done mostly due to the lack of computer knowledgePASSAGE C:Alphine tunnels are closed tunnels.In the past 30 yrs not even a single accident hasbeen recorded for there is one accident in the rail road system. Even in case of a fireaccident it is possible to shift the passengers into adjacent wagons and even the livefire can be detected and extinguished with in the duration of 30 min.Answer questions 13-16 based on passage C13. No accident can occur in the closed tunnels14. Fire is allowed to live for 30 min16. All the care that travel in the tunnels will be carried by rail shutters.PASSAGE D:In the past helicopters were forced to ground or crash because of the formation ofthe ice on the rotors and engines. A new electronic device has been developed whichcan detect the watercontent in the atmosphere and warns the pilot if thetemperature is below freezing temperature about the formation of the ice on therotors and wings.Answer questions 17-20 based on passage D17.The electronic device can avoid formation of the ice on the wings18. There will be the malfunction of rotor & engine because of formation of ice19. The helicopters were to be crashed or grounded20. There is only one device that warn about the formation of icePASSAGE E:In the survey conducted in mumbai out of 63 newly married house wives not a singlehouse wife felt that the husbands should take equal part in the household work asthey felt they loose their power over their husbands. Inspite of their careers they optto do the kitchen work themselves after coming back to home. the wives get half asmuch leisure time as the husbands get at the week ends.Answer questions 21-23 based on passage E21.Housewives want the husbands to take part equally in the household22. Wives have half as much leisure time as the husbands have World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from 39% of the men will work equally in the house in cleaning and washingPASSAGE F:Copernicus is the intelligent. In the days of copernicus the transport and technologydevelopment was less & it took place weeks to comunicate a message at thattime,wherein we can send it through satellite with in no time.Even with this fastdevelopments it has become difficult to understand each other.Answer questions 24-27 based on passage F24. People were not intelligent during Copernicus days25. Transport facilities are very much improved in noe a days26. Even with the fast developments of the technology we cant live happily.27. We can understand the people very much with the development ofcommunication PASSAGE G:Senior managers warned the workers that because of the intfoductorsof japanese industry in the car market. There is the threat to the workers.They alsosaid that there will be the reduction in the purchase of the sales of car in public.theinterest rates of the car will be increased with the loss in demand.Answer questions 28-31 based on passage G28. Japanese workers are taking over the jobs of indian industry.29. Managers said car interests will go down after seeing the raise in interest rates.30. Japanese investments are ceasing to end in the car industry.31. People are very interested to buy the cars.PASSAGE H:In the totalitariturican days,the words have very much devalued.In thepresent day,they are becoming domestic that is the words will be much moredevalued. In that days, the words will be very much effected in political area.but atpresent,the words came very cheap .We can say they come free at cost.Answer questions 32-34 based on passage H32.Totalitarian society words are devalued.33. Totalitarians will have to come much about words34. The art totalitatian society the words are used for the political speeches. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from I:There should be copyright for all arts. The reele has came that all thearts has come under one copy right society,they were use the money that come fromthe arts for the developments . There may be a lot of money will come from theTagore works. We have to ask the benifiters from Tagore work to help for thedevelopment of his works.Answer questions 35-39 based on passage I35. Tagore works are came under this copy right rule.36. People are free to go to the public because of the copy right rule.38. People gives to theater and collect the money for development.39. We have ask the Tagore residents to help for the developments of art.Directions for questions 40-45: In each question, a series of letters satisfying acertain pattern are given. Identify the pattern and then find the letter/letters that willcome in place of the blank/blanks.40. a, c, e, g, _(a) h(b) i(c) d(d) j41. a, e, i, m, q, u, _, _(a) y, c(b) b, f(c) g, i(d) none42. ay , bz , cw , dx ,__(a) gu(b) ev(c) fv(d) eu43. 1, 2, 3, 5, 7, 11, __(a) 15(b) 9(c) 13(d) 12 World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at
Quantitative Aptitude eBook for CAT from kp , lo , mn , __(a) nm(b) np(c) op(d) pq45. abc , zyx , def , wvu , ___(a) ghi(b) tsr(c) ihg(d) strDirections for questions 46 to 51: Select the alternative that logically follows formthe two given statements.46. All books are pages. All pages are boxes.(a) All boxes are books(b) All books are boxes(c) No books are boxes(d) Both (a) and (b) are correct47. No apple is an orange. All bananas are oranges.(a) All apples are oranges(b) Some apples are oranges(c) No apple is a banana(d) None of the above48. All pens are elephants. Some elephants are cats.(a) Some pens are cats(b) No pens are cats(c) All pens are cats(d) None of the above49. All shares are debentures.No debentures are deposits.(a) All shares are deposits(b) Some shares are deposits(c) No shares are deposits(d) None of the above50. Many fathers are brothers. All brothers are priests. World's Largest Portal on MBA Information & Job Join MBA Community at Over 5,000 Testing Interview Questions at |
PREFACE
In recognition of the fact that people preparing for the GMAT have widely
varying backgrounds in mathematics, this book provides an orientation
to the math content of the test, an introduction to the formats of the math
test questions, and practice with GMAT-style math questions. There is also
a complete description of the recently added Integrated Reasoning test
section, as well as practice GMAT-style Integrated Reasoning questions.
The mathematics on the GMAT is no more advanced than the mathematics taught in high school. The math review materials in this book
are structured so that you may select the topics you wish to review. Four
review chapters provide explanations, examples, and practice problems
covering number properties, arithmetic, algebra, and geometry. The
topics are explained in detail, and several examples of each concept are
provided. Throughout the chapters, practice problems give you a chance
to sharpen your skills. Each chapter ends with a test covering the concepts taught in that chapter. Following each unit test there are also GMAT
Solved Problems and GMAT Practice Problems. These provide practice
with GMAT-style math questions covering the content of each chapter.
Finally, at the end of the book there are also two tests modeled after the
GMAT mathematics section, with the same number of questions and the
same time limit. You can use these tests to assess your readiness to take
the actual GMAT math section.
GMAT Integrated Reasoning is related to mathematics in that you
are required to use your math skills to interpret and manipulate numerical and statistical data and arrive at logical conclusions. The Integrated
Reasoning chapter in this book will explain this process, give you some
tips for solving problems of this type, and provide practice with sample GMAT-style Integrated Reasoning questions. At the end of the book
there is also a set of practice GMAT Integrated Reasoning questions
that you can use to test your mastery of this question type.
Using this book to review your math knowledge, to learn about
GMAT math and Integrated Reasoning question formats, and to practice your skills with both question types will boost your test-taking confidence and make you better prepared for test day.
Robert E. Moyer, Ph.D.
Associate Professor of Mathematics
Southwest Minnesota State University
vii
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ABOUT THE AUTHOR
Dr. Robert E. Moyer has been teaching mathematics and mathematics education at Southwest Minnesota State University in Marshall,
Minnesota, since 2002. Before coming to SMSU, he taught at Fort
Valley State University in Fort Valley, Georgia, from 1985 to 2000,
serving as head of the Department of Mathematics and Physics from
1992 to 1994.
Prior to teaching at the university level, Dr. Moyer spent 7 years
as the mathematics consultant for a five-county Regional Educational
Service Agency in central Georgia and 12 years as a high school mathematics teacher in Illinois. He has developed and taught numerous
in-service courses for mathematics teachers.
He received his Doctor of Philosophy in Mathematics Education
from the University of Illinois (Urbana-Champaign) in 1974. He received
his Master of Science in 1967 and his Bachelor of Science in 1964,
both in Mathematics Education from Southern Illinois University
(Carbondale).
ACKNOWLEDGMENT
The writing of this book has been greatly aided and assisted by my
daughter, Michelle Moyer. She did research on the tests and the
mathematics content on them, created the graphics used in the manuscript, and edited the manuscript. Her work also aided in the consistency
of style, chapter format, and overall structure. I owe her a great deal of
thanks and appreciation for all the support she lent to the completion of
the manuscript.
ix
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SECTION I
INTRODUCTION
Graduate business schools consider a variety of factors when making
decisions about which applicants to admit to their programs. These
factors include educational background, work experience, recommendations, personal essays, and interviews. One factor often considered in
admissions decisions is the applicant's performance on a standardized
examination. The most common graduate business school admissions
test is the Graduate Management Admission Test, generally called the
GMAT®.
The Graduate Management Admission Council oversees the GMAT.
The GMAT is developed by ACT, Inc., and is delivered by Pearson VUE.
The GMAT is designed to help graduate schools assess the qualifications of applicants for advanced study in business and management.
The test is intended to be only one predictor of academic performance
in the core curriculum of a graduate management program. The GMAT
does not assume that test takers have specific knowledge of business or
any other content areas.
As of mid–2012, the GMAT consists of 4 sections: the Analytical
Writing Assessment, Quantitative, Verbal, and Integrated Reasoning.
This book focuses on the Quantitative and Integrated Reasoning sections. Both sections measure your ability to solve problems, to reason
mathematically, and to interpret data.
The GMAT uses a computer-adaptive format to deliver the Quantitative and Verbal questions. The computer selects a question based on
whether the previous question was answered correctly. If the previous
question was answered correctly, the difficulty level of the new question will be greater than that of the previous question; if the previous
question was answered incorrectly, the next question will be easier. The
content area of the question is the same whether a more difficult or an
easier question was selected. Your score on the test is based on both the
number of questions answered correctly and the level of difficulty of
those questions.
The computer-adaptive format imposes some very important conditions on the testing situation. First, you may not go back to a question,
so you must answer each question as you get to it. If you are not sure
of the answer, eliminate as many answer choices as you can and then
select the best choice from the smaller list. Second, you need to answer
all questions, or there will be a penalty for not completing the section.
In the Quantitative (mathematics) section, you are given 75 minutes
to answer 37 questions, or about 2 minutes per question. You need to
keep your eye on the time left and the number of questions remaining.
You will do much better if you pace yourself rather than rush through
1
2
SECTION I
the last few items. Missing several questions in a row, as you may if you
rush, will hurt your score in two ways: your number of correct answers
will be lower, of course, but also the questions you answer correctly
after that point will affect your score less because the difficulty level
will be lower.
When you prepare for the test, try to do three things: make sure
you know the mathematics content of the test, familiarize yourself with
the format of the test and questions, and practice the procedures so
that you are able to complete the test in the allotted time. This book
is designed to help you meet these three goals as you prepare for the
Quantitative and Integrated Reasoning Sections of the GMAT; the practice tests will let you know if you have accomplished these goals.
For general information about registering for and taking the GMAT,
visit the GMAT website at
CHAPTER 1
THE GMAT MATHEMATICS
SECTION
The GMAT Quantitative (mathematics) section is given as a computeradaptive test. It is a set of multiple-choice questions with five answer
choices each.
The computer presents you with one question at a time. The computer
then scores the current question and uses that information to select
the next question. If the question is answered correctly, the next question selected from the list of questions for the content area is slightly
more difficult than the question answered correctly. If the previous
question was answered incorrectly, the question selected is less difficult than the one just missed. Because the computer scores each
question before presenting the next one, you must complete one
question before you can go on to the next. Since you must answer
a question before proceeding to the next question in a computeradaptive test, you are asked to confirm your answer before going on to
the next question.
Time management is important. The computer will show an onscreen clock that counts down the time remaining on the section. The
clock can be hidden, but unless the clock is a distraction, leaving it
visible is generally helpful in managing your time. Whether or not you
hide the clock, it will alert you when there are 5 minutes left to work on
the current section.
The GMAT Quantitative section contains 37 questions with a
75-minute time limit. To complete the section in the time allotted, you
need to answer each question in an average time of about 2 minutes.
Not completing the section will result in a penalty and could significantly lower your Quantitative score. Failing to answer a question
has a greater negative impact on your score than answering the
question incorrectly. A steady pace is the best way to achieve your
highest possible score because rushing at the end means you may miss
questions covering content that you know very well.
The GMAT measures mathematics skills that are acquired over a
period of many years. Many of the skills are developed through the
curriculum of the average high school. The purpose of the Quantitative section is to determine whether you have the knowledge and skills
needed in a graduate business program. You have previously learned
the mathematics needed for the test, and you only need to review it to
be prepared for the Quantitative section.
The questions come in two basic formats: problem solving and
data sufficiency. Problem-solving questions should be familiar to you;
a question with five answer choices is presented, and you choose the
correct answer. This format is used on most standardized tests. The
3
4
CONQUERING GMAT MATH AND INTEGRATED REASONING
data-sufficiency format is unique to the GMAT. In this format, you
are given two statements and a question. You must decide if each of
the statements is sufficient to answer the question alone, if the two
statements taken together are sufficient to answer the question, or if
the statements, even taken together, are not sufficient to answer the
question.
CHAPTER 2
THE MATHEMATICS YOU
NEED TO REVIEW
Since the GMAT is taken by people with a wide variety of educational backgrounds, the test uses mathematical skills and concepts that are assumed
to be common for all test takers. The test questions use arithmetic, algebra,
geometry, and basic statistics. You will be expected to apply basic mathematical skills, understand elementary mathematical concepts, reason quantitatively, recognize information relevant to the problem, and determine if
there is sufficient information to solve a problem.
You will not be expected to know advanced statistics, trigonometry, or calculus, or to write a proof. The GMAT does not test specialized or advanced
knowledge of mathematics. In general, the mathematical knowledge and
skills needed do not extend beyond what is usually covered in the curriculum
of the average high school.
You will be expected to recognize standard symbols such as = (equal to),
= (not equal to), < (less than), > (greater than), || (parallel to), and ⊥ (perpendicular to). All numbers used will be real numbers. Fractions, decimals,
and percentages may be used. The broad areas of mathematical knowledge needed for success on the GMAT are number properties, arithmetic
computation, algebra, geometry, and some basic statistics.
Number properties include such concepts as even and odd numbers,
prime numbers, divisibility, rounding, and signed (positive and negative)
numbers.
Arithmetic computation includes the order of operations, fractions
(including computation with fractions), decimals, and averages. You may
also be asked to solve word problems using arithmetic concepts.
The algebra needed on the GMAT includes linear equations, operations
with algebraic expressions, powers and roots, standard deviation, inequalities, quadratic equations, systems of equations, and radicals. Again, algebra
concepts may be part of a word problem you are asked to solve.
Geometry topics include the properties of points, lines, planes, and polygons; you may be asked to calculate area, perimeter, and volume, or to
explore coordinate geometry.
When units of measure are used, they may be in English (U.S. Customary
System) or metric units. If you need to convert between units of measure,
the conversion relationship will be given, except for common ones such as
converting minutes to hours, inches to feet, or centimeters to meters.
Although simple graphs or tables may be used in a question, you will not
be asked to construct the graph or table; you will only need to interpret the
data in a given graph or table. Since constructing graphs is not part of the
GMAT, those procedures are not included in the mathematics review.
When answering any question on the GMAT, you first need to read the
question carefully to see what is being asked. Then recall the mathematical
concepts needed to relate the information you are given in a way that will
enable you to solve the problem.
5
6
CONQUERING GMAT MATH AND INTEGRATED REASONING
If you have completed an average high school mathematics program, you
have previously been taught the mathematics you need for the GMAT. The
review of arithmetic, algebra, and geometry provided in this book will help
you to refresh your memory of the mathematical skills and knowledge you
previously learned.
If you were not satisfied with your previous level of mathematical knowledge in a given area, then review the material provided on that topic in
greater detail, making sure you fully understand each section before going
on to the next one.
CHAPTER 3
THE GMAT INTEGRATED
REASONING SECTION
As of mid-2012, the GMAT includes a separately scored section called
Integrated Reasoning. This section tests your ability to use information
to solve complex problems. The GMAT Integrated Reasoning section
may seem intimidating because these types of questions have never
appeared on a standardized test before. However, the questions really
just test skills you've always used in school and when taking other tests.
The difference is that the skills must be combined to answer questions
correctly.
The Integrated Reasoning section is intended to provide business
schools with additional information to help evaluate admissions candidates. The decision-making skills that candidates display in answering the questions can help schools identify which candidates are most
likely to be successful within the classroom and in their careers.
The Integrated Reasoning section has a 30-minute time limit.
According to the test makers, the section includes 12 questions, some
of which may have multiple parts. A special online calculator is available to use for this section of the test only. You may not bring your own
calculator, and you cannot use the online calculator for any other section of the test.
Integrated Reasoning questions test your ability to solve complicated problems using information from multiple sources. They test
your logic and reasoning abilities, your skills at analyzing and synthesizing information, and your math and computation skills. They
also test your ability to convert between graphical and verbal representations of ideas. Several different skills may be tested by a single
question.
Integrated Reasoning questions do not test your business knowledge, but they do test the types of real world skills you would use in the
classroom or on the job. While you might never need to measure the
hypotenuse of a right triangle over the course of your career, you will
likely be required to read text, tables, and charts and to make decisions
based on complex information.
Based on samples provided by the test makers, there are four types
of GMAT Integrated Reasoning questions:
1.
2.
3.
4.
Table Analysis
Graphics Interpretation
Multi-Source Reasoning
Two-Part Analysis
For more information about each question type, see Chapter 6, "GMAT
Integrated Reasoning Questions."
7
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SECTION II
ITEM FORMATS
The GMAT Quantitative section has only multiple-choice questions.
There are 37 questions in the section. They are divided into two formats: problem solving, with approximately 22 questions, and data sufficiency, with approximately 15 questions. Each one has five answer
choices.
The problem-solving questions may be word problems or computations. The data-sufficiency questions measure your ability to determine
how much information is needed to solve a problem. For these questions, you must decide if enough data is given to enable you to arrive at
an answer; you do not need to actually find that answer.
Because the time limit for the Quantitative section is 75 minutes,
you need to complete each item in 2 minutes or less. Because of the
computer-adaptive format, you need to start each question knowing
that you must answer it in order to go on to the next one. Also, you
cannot go back to a question later and change your answer. You have
just one chance to answer each question. Read the question, consider
the relevant mathematics you know, and apply logical reasoning to the
situation. This should allow you to answer the question or to eliminate
some of the answer choices so that you can take an educated guess.
The 30-minute GMAT Integrated Reasoning section, based on the
samples released by the test makers, has 12 questions in at least four
formats: Table Analysis, Graphics Interpretation, Multi-Source Reasoning, and Two-Part Analysis. Many questions include a chart, a graph, or
another graphic. Some may have more than one part. Some are multiple choice, but others may ask you to pick true or false, or to answer yes
or no. Again, read the question, consider the relevant mathematics you
know, and apply logical reasoning. Most likely you will need to solve a
complex problem by analyzing information from multiple sources.
9
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CHAPTER
GMAT PROBLEM-SOLVING
QUESTIONS
ITEM FORMATS
About two-thirds of the 37 questions on the Quantitative section of the GMAT
are of the general problem-solving type. Each question has five answer
choices. The questions focus on the given information and reasoning that
you supply to select the best answer. A good strategy is to eliminate at least
two answers and, if you cannot eliminate any more, to select the best answer
from the remaining choices.
Any number in the problems will be a real number unless there is a further
restriction on the variables. Operations among real numbers are assumed.
Figures show general relationships such as straight lines, collinear points,
and adjacent angles. In general, you cannot determine measures of angles or
line segments based on a figure alone. In a few cases, you will be told that a
figure has been drawn to scale. When a figure has been drawn to scale, you
may use the lengths in the drawing to help you solve the problem. Similarly,
angle measures can be estimated from figures drawn to scale.
Example 1
b
7
a
If
=
, then what does equal?
a+b
12
b
A.
5
12
B.
5
7
C.
7
5
D.
7
19
E.
19
12
Solution
b
7
=
is a proportion, you can use two properties to
a+b
12
12
a+b
=
; then use
transform it. First, use the reciprocal property to get
b
7
12 − 7
a
5
a+b−b
=
. So = , and answer B
the subtraction property to get
b
7
b
7
is correct.
Because
11
12
CONQUERING GMAT MATH AND INTEGRATED REASONING
Example 2
In circle P, the two chords intersect
at point X, with the lengths as indicated in the figure. Which could not
be the sum of lengths a and b, if a
and b are integers?
A.
B.
C.
D.
E.
49
30
26
16
14
A
D
X
B
C
P
Solution
When two chords intersect within a circle, the product of the segments on
one chord is equal to the product of the segments on the other chord. Since
the segments of the first chord are 6 and 8, the product of the lengths is 48.
Thus, the product of the lengths a and b must be 48, and possible lengths
are 48 and 1, 24 and 2, 12 and 4, and 8 and 6. So 49, 26, 16, and 14 are
possible values for a + b. The correct answer is B, since 30 is not the sum of
two integer factors of 48.
Example 3
In one can of mixed nuts, 30% is peanuts. In another can of mixed nuts
that is one-half the size of the first one, 40% is peanuts. If both cans are
emptied into the same bowl, what percentage of the mixed nuts in the bowl
is peanuts?
2
A. 16 %
3
B. 20%
C. 25%
1
D. 33 %
3
E. 35%
Solution
Let the first can contain 16 ounces of nuts, so the second can contains 8
ounces of nuts. Thirty percent of 16 ounces is 4.8 ounces of peanuts, and
40% of 8 ounces is 4.2 ounces of peanuts. In the bowl there is (4.8 + 3.2)
1
1
1
8
= = 33 %. So 33 %
ounces of the (16 + 8) ounces in the bowl, and
24
3
3
3
of the nuts are peanuts, and D is the correct answer.
Example 4
What is the sum of the prime numbers between
A.
B.
C.
D.
E.
15
16
17
18
25
1
1
and 9 ?
2
5
CHAPTER 4 / GMAT PROBLEM-SOLVING QUESTIONS
13
Solution
1
1
and 9 are 2, 3, 5, and 7. The sum of these
2
5
prime numbers is 17, so the answer is C.
The prime numbers between
Example 5
2
3
pint of red paint and pint of white paint to make a
4
3
new paint color called Perfect Pink. How many pints of red paint would be
needed to make 34 pints of Perfect Pink paint?
A paint store mixes
A. 9
B. 16
C. 18
D. 25
1
3
E. 28
1
2
Solution
First, determine how much paint the recipe for Perfect Pink will make.
3
2
9
8
17
5
pint +
pint =
pint +
pint =
pints, or 1
pints. The ratio
4
3
12
12
12
12
of red paint in the recipe is the same as it will be in the 34 pints of paint. Let
N be the number of pints of red paint needed.
N
3/4
=
1(5/12)
34
3
5
(34) = 1 (N)
4
12
102
17
=
N
4
12
102 17
÷
=N
4
12
18 = N
Thus, 18 pints of red paint are needed, and the answer is C.
SOLUTION STRATEGIES
1. Apply a general rule or formula to answer the question.
In Example 2, you can apply a property from geometry that says that
when two chords intersect inside a circle, the segments formed have
lengths such that the product of the segment lengths is the same for each
chord.
14
CONQUERING GMAT MATH AND INTEGRATED REASONING
2. Apply basic properties of numbers.
In Example 4, you can use the definition of a prime number so that you
do not include 1, but do include 2.
3. Eliminate as many answers as possible so that you can select from
a smaller set of answer choices.
In Example 3, you can eliminate some of the answers by noting that since
each can of mixed nuts is at least 30% peanuts, the mixture of the two cans
will be least 30% peanuts. Thus, before doing any computation, you could
eliminate answers A, B, and C. Therefore, if you need to guess, you only
have two answer choices left and have increased your odds of guessing
correctly.
4. Substitute answers into the given question to see which one produces the correct result.
7
a
b
=
, and you want the value of . You
In Example 1, you are given
a+b
12
b
1
b
.
by b to get a
can divide the numerator and denominator of
a+b
+1
b
Now you can substitute the answer choices into the expression to see
7
12
which answer produces a value of
. Answer A produces
, so it is
12
17
7
, so it is correct. Since this type of question
wrong. Answer B produces
12
has only one correct answer, you know the correct answer is B. You do
not have to test the rest of the answer choices.
This strategy cannot be employed on the majority of questions, but
you can use it when you can see a way to quickly test the answer choices.
5. Break down the situation into individual steps.
In Example 5, you have an everyday situation of mixing paint. Break the
problem down into steps. First, find the total amount of paint the formula makes. Then set up a proportion to find the increased amount of
red paint. Taking word problems one step at a time makes them more
manageable.
EXERCISES
1. If a jewelry store wants to sell a necklace for
$179.95 next week at a 60% off sale, how much
is the price of the necklace this week?
A. $71.98
B. $251.93
C. $287.92
D. $399.92
E. $449.88
2. Find the median for this set of data: 9, 2, 5, 7,
10, 9, 2, 8, 11, 10.
A. 8
B. 8.5
C. 9
D. 9.5
E. 10
3. Which number is divisible by 3, 4, 5, and 6?
A. 30
B. 48
C. 75
D. 120
E. 160
4. The length of a rectangle is 4 centimeters
longer than the width, and the perimeter is 96
centimeters. How many square centimeters are
there in the area of the rectangle?
A. 48
B. 396
C. 572
D. 1,760
E. 2,288
CHAPTER 4 / GMAT PROBLEM-SOLVING QUESTIONS
15
5. Which quadratic equation has roots of 4 and
1
?
2
A. 4x2 − 9x + 2 = 0
B. 2x2 − 9x + 4 = 0
C. 2x3 − 9x2 + 4x = 0
D. 2x2 + 9x + 4 = 0
E. x2 − 2x +
1
=0
2
SOLUTIONS
1. E First, eliminate as many answers as possible so that you can select from a smaller
set of numbers. If 60% off the original price
leaves $179.95, then the original price is more
than twice the sale price—greater than $360.
Eliminate choices A, B, and C. Then substitute answers into the question to see which
one produces the correct result. Try choice E.
Now 60% of $449.88 is $269.93, and $449.88
− $269.93 = $179.95, so choice E is correct.
2. B Apply basic properties of numbers. In this
case, apply the definition of the median as
the middle value in the ordered sequence
of values. To find the median, you need to
arrange the data in order from lowest to highest: 2, 2, 5, 7, 8, 9, 9, 10, 10, 11. Since there is
an even number of values, you average the two
middle values to get the median Md.
Md = (8 + 9) ÷ 2
Md = 8.5
3. D Apply basic properties of numbers. In this
case, apply the divisibility rules for 3, 4, and 5.
Since any number divisible by 3 and 4 is divisible by 6, there is no need to check separately
for divisibility by 6. When a number is divisible by 5, its units digit must be either 0 or 5. If
a number is divisible by 3, then the sum of the
digits must be divisible by 3. To be divisible
by 4, the last two digits must form a number
divisible by 4. Since you want an answer that
ends in 0 or 5, answer B can be eliminated.
The sum of the digits must be divisible by 3,
so answer E can be eliminated. Finally, the
last two digits of the number must be divisible
by 4, so answers A and C can be eliminated.
The correct answer is D.
4. C Break the situation down into individual
steps. Apply the formulas for perimeter and
area of a rectangle. The perimeter of a rectangle is given by the formula P = 2l +2w, and the
area is given by the formula A = lw. First, find
the length and width. Let w equal the width
of the given rectangle. The length can then be
represented as w + 4.
2w + 2(w + 4) = 96
2w + 2w + 8 = 96
4w + 8 = 96
4w = 88
w = 22
Now find the length by adding 4.
l = 26
Now apply the formula for area.
A = lw = 26(22) = 572
5. B Apply a general rule or formula to answer
the question. In this case, apply the factoring
procedure, and then find the solution for each
factor. Note that answer C is not a quadratic
equation and can be eliminated immediately.
If the roots of a quadratic equation are 4 and
1
1
, then x = 4 and x = will yield x−4 = 0 and
2
2
2x − 1 = 0. The quadratic equation is therefore
(x − 4)(2x − 1) = 0, which is 2x2 − 9x + 4 = 0.
The correct answer is B.
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CHAPTER
GMAT DATA-SUFFICIENCY
QUESTIONS
ITEM FORMATS
GMAT data-sufficiency questions have as their focus not finding the solution
to the problem, but determining whether or not there is enough information
to solve it. In each item, you are given a situation and then two statements,
and you are asked to determine whether one of the statements, both of the
statements, or neither of the statements provides enough information to
solve the problem.
Data-sufficiency questions occur only on the GMAT. About one-third of
the 37 mathematics questions are of this type.
Once you practice a few of this type of question, you will see that they
often take much less time than do the problem-solving questions. You do
not have to solve the problem, just decide whether it can be solved.
The answer choices are the same for each question of this type:
A Statement (1) ALONE is sufficient, but statement (2) is not sufficient.
B Statement (2) ALONE is sufficient, but statement (1) is not sufficient.
C BOTH statements TOGETHER are sufficient, but NEITHER statement
ALONE is sufficient.
D EACH statement ALONE is sufficient.
E Statements (1) and (2) TOGETHER are NOT sufficient.
Important points to remember when answering this type of question
are that all numbers in the problem are real numbers; figures are always
consistent with the given information but may conflict with either or both
statements; all lines in the figures are straight lines; the position of points,
line segments, and angles in a figure exist in the order shown; and angle
measures are all greater than zero.
In a data-sufficiency question that asks for a numerical answer, it must
be possible to determine that quantity's value exactly for a statement to be
sufficient. You do not have to find the value, just know that with the given
information and one or both statements, the value can be determined.
Study the following examples to see how the directions below apply to
each item. Statement 1 ALONE is sufficient, but statement 2 alone is not sufficient.
B Statement 2 ALONE is sufficient, but statement 1 alone is not sufficient.
17
18
CONQUERING GMAT MATH AND INTEGRATED REASONING
C BOTH statements TOGETHER are sufficient, but NEITHER statement
ALONE is sufficient.
D EACH statement ALONE is sufficient.
E Statements 1 and 2 TOGETHER are NOT sufficient.
Example 1
A rectangle has a perimeter of 96 centimeters. What are the dimensions of
the rectangle?
1. The area is 572 square centimeters (cm2 ).
2. The width is 4 centimeters shorter than the length.
A.
B.
C.
D.
E.
Solution
For statement 1, if the area is 572 cm2 , then you have lw = 572 and 2l + 2w =
96. If you have two equations with the same two variables, you can solve. For
example, solving the system, you get l = 26 cm and w = 22 cm or l = 22 cm
and w = 26 cm. The dimensions are 22 and 26 cm; thus, statement 1 is
sufficient. Note that you do not need to actually solve it, just know that you
can solve it.
For statement 2, if the width is 4 less than the length, you have w = l − 4
and 2l + 2w = 96. Again, you can solve. For example, solving the system, you
get l = 26 cm and w = 22 cm. The dimensions of the rectangle are 22 and
26 cm; thus, statement 2 is sufficient.
Since each statement is sufficient alone, the answer is D.
Example 2
Here G, H, I, J, and K are consecutive whole numbers. When is G× H× I >
12?
1. G ≥ 2
2. G is odd.
A.
B.
C.
D.
E.
Solution
Using statement 1, if G ≥ 2, then G × H × I = 2 × 3 × 4 = 24, at least, and
24 > 12. Statement 1 is sufficient.
Using statement 2, if G is odd, then G can be 1. If G = 1, then G × H × I =
1×2×3 = 6, and 6 < 12. Also G could be 3, and then G×H×I = 3×4×5 = 60,
and 60 > 12. Statement 2 is not sufficient.
Thus, the correct answer is A.
Example 3
Lisa bought $50 worth of gas for her truck. How far can Lisa travel in her
truck using this amount of gas?
1. The gas Lisa bought cost $2.75 a gallon.
2. Lisa's truck gets 24 miles per gallon of gas.
A.
B.
C.
D.
E.
CHAPTER 5 / GMAT DATA-SUFFICIENTLY QUESTIONS
19
Solution
To determine how far Lisa can travel on $50 worth of gas, you need to know
the number of gallons of gas that she purchased and the number of miles
per gallon (mpg) Lisa's truck gets. (miles per gallon) × (number of gallons
of gas) = distance traveled.
Using statement 1, knowing that gas cost $2.75 a gallon, you can determine the number of gallons of gas purchased: $50 ÷ $2.75, which is about
18.18 gallons. This is not sufficient to determine how far Lisa can travel.
Using statement 2, knowing that her truck gets 24 miles per gallon is not
sufficient to determine how far Lisa can travel.
However, knowing the cost per gallon of gas yields the number of gallons
of gas Lisa purchased (about 18.18), and knowing the mpg for her truck is
24, you can determine the distance that Lisa can travel: 24 ×18.18 = 436.32
miles.
Since it takes both statements together to be able to determine the
distance, the answer is C.
Example 4
If x3 + y3 , what is the value of x?
1. y = 3
2. x < 0
A.
B.
C.
D.
E.
Solution
For statement 1, knowing y = 3 just yields x3 + 27, which is not an equation;
you cannot determine x. Thus, statement 1 is not sufficient.
For statement 2, knowing x < 0 does not yield a value for x. Thus, statement 2 is not sufficient.
Combining y = 3 and x < 0 still does not let you determine the value of
x, so together, statements 1 and 2 are not sufficient. Thus, the answer must
be E.
Example 5
David has 26 coins in a jar, and all the coins are dimes and nickels. How
many nickels are in the jar?
1. David spent part of the money on a soft drink.
2. The value of the coins is $1.85.
A.
B.
C.
D.
E.
Solution
For statement 1, the fact that David spent some of the money does not
allow you to determine the number of nickels he has, so statement 1 is not
sufficient.
For statement 2, if the value of the coins is $1.85, then if you let n be the
number of nickels and d be the number of dimes, 0.05n + 0.10d = 1.85, and
n+ d = 26. Solving the system of equations tells you that there are 15 nickels
and 11 dimes, so statement 2 is sufficient to solve the problem.
Since statement 2 is sufficient and statement 1 is not, the answer is B.
20
CONQUERING GMAT MATH AND INTEGRATED REASONING
SOLUTION STRATEGIES
1. Work only as far as you must to be sure the question has an answer.
In Example 3, once you are sure you can determine the number of gallons
of gas purchased and the number of miles per gallon, you know the
answer is C because it takes both statements to get an answer.
2. Be sure to try each statement separately to get an answer to the
question.
In Example 1, after statement 1 yields an answer, there is a tendency to
stop and say the answer is A. However, the answer could also be D, if
statement 2 also yields an answer. To be sure whether the answer is A or
D, you must try each statement separately.
3. If statement 1 does not yield an answer to the question, check
statement 2 to see if it will yield an answer to the question.
The tendency is to mark answer B, but when statement 1 does not yield
a solution, the answer could be B, C, or E. When you try statement 2 and
it yields an answer to the question, the correct answer is B as in Example
5. If it does not yield an answer, then the answer choice could be C or E
as in Example 3 and Example 4, respectively.
4. If neither statement alone yields an answer to the question, be sure
to consider them together.
In Example 3, it takes both statements together to get an answer to the
question.
5. Remember that when both statements fail to yield an answer individually, they can still fail to yield an answer when taken together.
In Example 4, the statements individually do not yield an answer to the
question, and when taken together, they still don't yield an answer to the
question. Often people assume that when both statements fail individually to produce an answer, the answer is always C or is always E. Neither
assumption is valid. You must check the two statements taken together to
determine which is actually the correct answer. If, when taken together,
they produce an answer to the question, the correct answer is C. If, when
taken together, they fail to produce an answer to the question, the correct
answer is E.
EXERCISES. Statement 1 ALONE is sufficient, but statement 2 alone is not sufficient.
B. Statement 2 ALONE is sufficient, but statement 1 alone22
CONQUERING GMAT MATH AND INTEGRATED REASONING
sufficient. Using statement 2, knowing that AB = 9, BC = 12, and CA =
15 allows you to determine that triangle ABC is a right triangle. Since
2
AB2 + BC 2 = 92 + 122 = 152 = CA , triangle ABC is a right triangle.
The legs of the right triangle, AB and BC, can be the base and altitude of
the triangle, so the area of triangle ABC is 1/2bh = 1/2 · 9 · 12 = 54. Thus,
statement 2 is sufficient alone. As indicated by strategy 3, you must check
statement 2 alone. Since it is sufficient, the answer is B.
3. D For statement 1, if 1/2(x + y) = w, then x + y = 2w and w + x + y = 3w,
so the average of w, x, and y is 3w ÷ 3 = w and z = w. Thus, statement
1 is sufficient. Using strategy 2, you need to check statement 2 alone to
determine if the answer is A or D. For statement 2, if w = x = y, then
w + x + y = 3w and (w + x + y) ÷ 3 = 3w ÷ 3 = w. Thus, w is the average
and w = z, so statement 2 is sufficient. Since each statement is sufficient
alone, the answer is D.
4. D For statement 1, if the perimeter of the square is 28, then 4s = 28
and s = 7. The area of a square is s2 , so s2 = 72 = 49. Thus, statement
1 is sufficient. Using strategy 2, you need to see if statement 2 alone is
sufficient. The diagonal forms an isosceles right triangle with two of the
sides of the square. If the diagonal is7 2, then s2 + s2 = (7 2)2 and
2s2 = 98. So s2 = 49. Thus, statement 2 is sufficient. Since each statement
alone is sufficient, the answer is D.
5. A As stated in the problem, if the ratio of x : y : z = 1 : 2 : 3, then y = 2x
and z = 3x. Thus, x + y + z = 6x. For statement 1, if x + z = 16, then
x + 3x = 16, and x = 4. So x + y + z = 6x = 6(4) = 24. Thus, statement 1 is
sufficient. From strategy 2, you know to check each statement. Statement
2 does not provide any additional information. Since the ratio is 1 : 2 :
3, you can already conclude that x < y < z. Thus, statement 2 is not
sufficient alone. Since only statement 1 yielded a result for the question,
A is the correct answer for this question.
Note: The exercises in this chapter were worked out to the point where a solution to the problem was found (if possible), which is not needed to answer
a data-sufficiency question on the GMAT. It was done here to make sure the
justification for the answer was clear.
In general, you only have to work a data-sufficiency problem to the point at
which you know there will be (or definitely will not be) a meaningful answer.
When an answer is found that is not acceptable for the problem, you do not
have sufficient data to solve the problem.
CHAPTER 6
GMAT INTEGRATED
REASONING QUESTIONS
According to the samples provided by the test makers, the GMAT Integrated Reasoning section uses at least four question formats: Table Analysis, Graphics Interpretation, Multi-Source Reasoning, and Two-Part
Analysis. Each format requires you to solve complex problems using
information from multiple sources. Understanding these formats will
help you to know what to expect and how to approach each question.
TABLE ANALYSIS
Table Analysis questions require you to analyze data in spreadsheets or
tables. The questions may contain spreadsheets or tables that can be
sorted. You will have to sort the data to determine the accuracy of given
answer statements.
Example
The table below gives information on total deliveries (total delivery
trips made to addresses on record) and total items delivered (packages and letters) in 2010 by a private company for a one-year period to
21 zip codes throughout the country. The 21 zip codes fall among the
top 35 for this annual period in terms of both total deliveries and total
items delivered by the company. In addition to providing the numbers
of total deliveries and total items delivered for each route, the table
also gives the percent of increase or decrease over the numbers for
2009 and the rank of the route for total deliveries and total pieces
delivered.
[NOTE: On the real exam, you will have the ability to sort the table by any
of its columns. Columns can be sorted in ascending order only. The table
is shown below sorted in different ways.]
Sorted by Percent Change in Deliveries (Column 5)
Delivery Route
City
State Zip Code
Washington
DC
20011
Bellevue
WA
98004
Pensacola
FL
32506
Little Rock
AR
72203
Miami
FL
33124
Cambridge
MA
02138
Baltimore
MD
21201
Deliveries
Number % Change
73,997
–8.3
41,653
–1.9
69,472
–1.2
76,247
–0.9
69,804
–0.9
47,181
–0.8
57,632
–0.3
Rank
5
22
9
4
8
21
16
Items Delivered
Number % Change Rank
126,048
6.2
21
125,297
4.1
22
290,771
–1.7
1
204,956
–0.6
10
127,793
–7.0
20
160,032
9.9
17
210,955
–1.7
8
(Continued)
23
26
CONQUERING GMAT MATH AND INTEGRATED REASONING
Review each of the statements below. Based on information provided in
the tables, indicate whether the statement is true or false.
True False
The delivery route with the median rank based on total
number of deliveries is the same as the route with the
median rank based on total number of items delivered.
The total number of deliveries to Cambridge, MA, in 2009
was made than
delivery routes experiencing a percent decrease in the
number of items delivered.
Solution
True False
The delivery route with the median rank based on total
number of deliveries is the same as the route with the
median rank based on total number of items delivered.
The total number of items delivered in Cambridge, MA, in
2009 was than delivery
routes experiencing a percent decrease in the number of
items delivered.
The delivery route with the median rank based on total number of
deliveries is Atlanta, GA (30305). The delivery route with the median
rank based on total number of items delivered is St. Louis, MO
(63101).
In 2010, there were approximately 160,000 items delivered in Cambridge. This represented an increase of 10 percent over the previous
year. In 2009, therefore, approximately 16,000 fewer items were delivered to this route. The 2009 items totaled about 145,000.
The delivery route experiencing the greatest percent increase in
total deliveries from 2009 to 2010 is Atlanta, GA (30302). The delivery
route that saw the greatest increase in the percent of items delivered is
Cambridge, MA (02138).
In 2010, 7 delivery routes experienced a percent decrease in the
number of deliveries. There were 10 delivery routes that experienced a
percent decrease in the number of items delivered.
CHAPTER 6 / GMAT INTEGRATED REASONING QUESTIONS
27
Solution Strategies
Here are some helpful strategies for approaching Table Analysis
questions:
1. Tables are organized in columns and rows. The columns go up and
down the table vertically from top to bottom, and the rows go across
the table horizontally from left to right. Each unit of data is presented in a cell.
2. The first row of a table is called the header row. The header row contains category names that identify the data in each column.
3. To sort the information in a table, click on the header cell of a column. The data in the entire table will be reorganized according
to that column. Information can be sorted from lowest to highest
(1–100 or A–Z) only.
4. Use estimates where possible to calculate answers quickly.
GRAPHICS INTERPRETATION
Graphics Interpretation questions contain graphs, images, or charts.
You will be required to review the image and interpret it to answer
a question. Graphics Interpretation questions contain fill-in-the-blank
answer statements. To select the correct answer, you must choose from
several options on a drop-down list.
Example
The graph above is a scatter plot with 60 points, each representing the
number of daily hours of sunlight to which 60 plants were exposed,
and the corresponding height, measured in centimeters, that each plant
attained. The plant heights were measured after six weeks of consistent
sun exposure. The solid line is the regression line, and the dashed line
is the line through the points (1, 4) and (7, 8). Select the best answer
to fill in the blanks in each of the statements below based on the data
shown in the graph.
28
CONQUERING GMAT MATH AND INTEGRATED REASONING
The relationship between the hours of sun exposure and plant height
is __________.
A. zero
B. negative
C. positive
The slope of the dashed line is __________ the slope of the regression
line.
A. greater than
B. less than
C. equal to
The number of plants that received more than six hours of daily sun
exposure is closest to ________ percent of 60.
A. 0
B. 5
C. 10
D. 20
E. 40
Solution
The relationship between the hours of sun exposure and plant height is
positive. As sun exposure increases, so does plant height. The slope of
the dashed line is less than that of the solid line. The solid line is steeper
than the dashed line, so its slope is the larger of the two. The number of
plants that received more than six hours of daily sun exposure is closest
to 5 percent of 60. Exactly three plants received more than six hours
of daily sun exposure, so 5 percent of the plants received this amount.
Solution Strategies
For these questions, you'll need to use data analysis, percentages, and
coordinate geometry skills. Here are some key points:
1. Two factors are positively related if one increases as the other does.
If one increases and the other decreases, the relationship between
the two is negative.
2. On a graph, the slope of a line is a measure of its steepness. The
steeper the line, the greater the slope.
3. If a line slants upward from left to right, its slope is positive. If the
line slants downward from left to right, its slope is negative.
4. To calculate the percentage represented by part of a whole, divide
the part by the total. In this example, to calculate the percentage
represented by three plants, divide 3 by 60. The percentage is 0.05,
or 5 percent.
5. Where possible, use estimation to answer Graphics Interpretation
questions. You don't have to calculate the slopes of the lines, for
instance, if you know that steeper lines have greater slopes.
CHAPTER 6 / GMAT INTEGRATED REASONING QUESTIONS
29
MULTI-SOURCE REASONING
Multi-Source Reasoning questions require you to examine multiple
sources and calculate the correct answers to problems. Two to three
sources of information will be provided: these may include text, graphs,
charts, tables, or spreadsheets. You will have to consult more than one
source to answer each question.
Example
E-mail 1—E-mail from division director to donations coordinator
August 10, 9:37 a.m.
Yesterday I spoke with the computer training lab administrator to update
him on the status of donations for the school district's computer donations drive. He extended the donations deadline for another week, until
next Tuesday. Are we on track to receive enough donations from students'
families to meet our goal of computers for the new training lab? Do we
need to extend our request to local businesses too?
E-mail 2—E-mail from donations coordinator in response to the division
director's August 10, 9:37 a.m. message
August 10, 10:04 a.m.
To date we have received 40 computers. We need 100 computers
donated to meet our goal for the new training lab. We have requested
help from all of the students' families, so we should invite local businesses as well. In all of our past drives, including this one so far, we
have received donations from about 20 percent of those who received
requests. (Of course, we might always receive more or less than that
average, so we should consider the possibilities of not meeting the goal
or overspending the budget for the thank-you event.) Each individual
or organization donating a computer will receive two invitations to our
thank-you event to celebrate the opening of the lab. Refreshments and
supplies for the event are expected to run $20 per person. What is the
total budget for the
allow us to accommodate 2 attendees for each of the 100 computers
donated. The budget is firm, so we should take care to ensure that the event
costs stay within this amount. Although we do not have resources to extend
the budget, if necessary we could determine ways to reduce the cost per
person if we receive more donations than the original goal amount.
Consider each of the following statements. Does the information in the
three e-mails support the inference as stated?
30
CONQUERING GMAT MATH AND INTEGRATED REASONINGSolutionThe donations coordinator states in E-mail 2 that local businesses
should be invited to contribute to the drive.
The e-mails do not suggest that the donation coordinator does not
believe the goals of the drive can be met. This inference is not supported by the information provided.
The division director states in E-mail 3 that the budget for the
thank-you event is firm and cannot be extended.
The e-mails do not suggest a disagreement between the two over the
per-person budget for the thank-you event. The donations coordinator
mentions this amount in E-mail 2, and the division director proposes
that attempts might be made to reduce the cost per person if necessary.
Solution Strategies
1.
2.
3.
Multi-Source Reasoning questions are presented in "tabbed" format. To see the different sources, click on the tabs at the top of the
screen. You can view only one source at a time.
These questions give you more information than needed to arrive
at the answer. Sort through the information to determine what is
relevant before answering.
When determining whether an inference is supported, consider
the source materials carefully. A topic might be mentioned in the
sources without necessarily supporting an inference about it.
CHAPTER 6 / GMAT INTEGRATED REASONING QUESTIONS
31
TWO-PART ANALYSIS
The answers to these questions will have two components. Components are presented in table format, with one component per column.
To answer the questions, you must analyze different combinations of
the two components.
Example
Acme Company currently produces 7,500 circuit board units per year.
Brown Company currently produces 8,000 circuit board units per year.
The numbers of units produced by both companies are increasing each
year at a constant rate. If each of these companies continues to produce
an increased number of units annually at its constant rate, in 10 years
both companies will produce the same number of units for the first
time. After the 10-year mark, Acme Company will produce more units
per year than Brown Company.
In the table below, identify the rates of increase, in annual units produced, for each company that together meet the performance projections
given above. Select only one option in each column.
Acme Company
Brown Company
Rate of increase
(units per year)
15
25
50
100
125
140
Solution
The correct answer is 100 units per year for Acme Company and 50
units per year for Brown Company.
Acme Company
Brown Company
Rate of increase
(units per year)
15
25
50
100
125
140
If Acme Company increases its production by 100 units per year, in 10
years it will produce 8,500 units per year. Brown Company would reach
the 8,500-unit mark at the same time by producing 50 more units per
year. Every year after the first 10, Acme will produce more units than
Brown Company.
32
CONQUERING GMAT MATH AND INTEGRATED REASONING
Solution Strategies
1.
To answer Two-Part Analysis questions, you must consider both
components. In the example above, consider the increases for both
Acme Company and Brown Company.
2.
Two-Part Analysis questions may involve more than one outcome.
In the example above, two outcomes are projected: the companies
produce the same number of units in 10 years; and after 10 years,
Acme produces more units than Brown.
3.
This problem can be solved by working backwards. Start with a
number in the middle for Acme—say, 50 units per year. If Acme
increases its production by 50 units per year, in 10 years it will produce 8,000 units. If Brown increases its production by 50 units per
year, in 10 years it will produce 8,500 units.
The number of 50 for Acme is too small. Try the next larger number.
If Acme increases its production by 100 units per year, in 10 years it
will produce 8,500 units. This would match Brown's production at
an increase of 50 units per year.
4.
You can also use algebra to help find the answer. Set up an equation
for the first outcome:
7,500 ϩ 10x ϭ 8,000 ϩ 10y
Solve for x in terms of y:
7,500 ϩ 10x ϭ 8,000 ϩ 10y
10x ϭ 8,000 ϩ 10y – 7,500
10x ϭ 500 ϩ 10y
10 x
10
ϭ
500 ϩ10 y
10
x ϭ 50 ϩ y
This tells you that Acme's production (x) is 50 units more than Brown's
production (y). The only options that fit from the table are 100 units for
Acme and 50 units for Brown.
SECTION III
BASIC MATHEMATICS
REVIEW
The mathematics section on the GMAT requires a knowledge of mathematics
that is acquired over a period of years. This section will review topics in arithmetic, algebra, and geometry that could form the content of the mathematics
questions on the GMAT.
Each chapter includes definitions of key concepts, worked-out examples,
and practice exercises with explanations. Chapters conclude with a chapter
test in multiple-choice format. The examples let you review the concepts and
recall information you had learned previously. The practice exercises enable
you to demonstrate your understanding of the concepts. Finally, you apply
your knowledge to questions in a format similar to test questions.
Chapter , Number Properties, reviews the number line, the types of
real numbers, rounding numbers, computing with signed numbers, and the
properties of numbers. The properties of the special numbers 0, 1, and −1 are
reviewed, as well as those of even and odd numbers. Number theory properties such as prime and composite numbers, multiples, factors, and divisors
are covered as well as least common multiple, greatest common divisor, and
prime factorization.
In Chapter , Arithmetic Computation, the order of operations, properties
of operations, fractions, decimals, ratios, proportions, percents, averaging,
powers, and roots are discussed. The associative, commutative, and distributive properties are reviewed and applied. Converting among fractions,
decimals, and percents shows the relationships among the various forms of
rational numbers. Word problems will allow you to apply information you
know about numbers.
In Chapter , Algebra, you can review the basic concepts of algebra. You
will evaluate expressions, solve equations, and solve inequalities. Computation with algebraic expressions will include monomials, binomials, and
polynomials. Graphs of points and linear equations will be discussed. The
solution of quadratic equations by factoring and by using the quadratic
formula will be demonstrated. Algebraic word problems will include consecutive integer problems, age problems, mixture problems, and motion
problems.
Chapter , Geometry, will review the fundamental concepts of geometry.
Proofs will not be part of the GMAT, so they will not be considered. Angles,
angle relationships, the relationship between lines, and types of polygons
will be reviewed. Properties of triangles, parallelograms, rectangles, squares,
and circles will be discussed. Formulas for area, perimeter, and volume of
common geometric figures will be used in problems.
33
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CHAPTER
NUMBER PROPERTIES
THE NUMBER LINE
A number line is a line with a scale for locating numbers indicated. The scale
is determined when two numbers have their location indicated on the line.
Usually, 0 and 1 are located first, but that is not a requirement.
The origin of a number line is the location of zero on that number line.
The coordinate of a point on a number line is the number associated with
that point.
Figure .1 is an example of a number line. The space between two consecutive numbers is equal everywhere on the number line. On this number
line, the integers from negative 5 to positive 5 are shown, but all integers can
be located by using the scale shown.
Figure .1
Numbers not shown on the number line can be represented by estimating
their location between two consecutive numbers.
Example 1
Locate A =
1
, B=
2
2, and C = 3.1 on a number line, shown in Figure 7.2.
A
B
C
Figure .2
Solution
Since A = 1/2, it is located midway between 0 and 1. Since B = 2, you
have to approximate 2 to get 2 ≈ 1.414. You locate B at about 1.4, which
is a little less than halfway between 1 and 2. For C = 3.1, you locate C just
to the right of 3 on the number line.
In working out the above example, you used a common practice of having
the numbers increase as you go from left to right. Unless you are told otherwise, you may assume this is true for all number lines you encounter in this
book and on the GMAT. Also, there are an infinite number of points between
any two of the indicated coordinates. In general, you attempt to locate the
number by thinking of the interval between two numbers as divided into
halves. Then in each half, you locate the number by deciding if it is closer
to the left end of the space, in the middle of the space, or closer to the right
end of the space.
35
36
CONQUERING GMAT MATH AND INTEGRATED REASONING
If greater accuracy is needed on the number line, the space between two
numbers can be enlarged and subdivided into tenths of a unit. See Figure 7.3.
Figure 7.3
For even greater accuracy in locating points, the space between 1.2 and 1.3
could be enlarged and subdivided into 10 parts, or hundredths of a unit.
Number Line Exercises
1
1
2
A. Graph the numbers 3, , −5, 8, 1 , and − on a number line.
2
4
3
B. What are the coordinates of the points A, B, C, D, and E on the number
line in Figure 7.4?
B
E
A
D
C
Figure 7.4
Solutions
A.
1
1
B. A = 3, B = −1 , C = 7, D = 4 , E = 1
2
2
THE REAL NUMBERS
A real number is any number that can be the coordinates of a point on
5
2
1
a number line. The numbers 3, −4, 2.3, 5 , −7.21, − , 2, 5, and are all
3
3
7
examples of real numbers.
The set of counting numbers is the set of numbers 1, 2, 3, 4, . . . . Counting
numbers are evenly spaced on the number line. Each number is 1 more than
the previous number, except for 1, which is the smallest counting number.
The counting numbers are also known as the natural numbers.
The whole numbers are the counting numbers plus zero.
The integers are made up of the counting numbers, zero, and the
negatives of the counting numbers. The integers are { . . . , − 4, −3, −2,
−1, 0, 1, 2, 3, 4, . . . }.
The rational numbers are all the numbers that can be written as the ratio
of two integers a and b, when b is not zero. The rational numbers include
all the integers, since you can let a be any integer and b be 1. You also get
1 6
11
2
fractions such as , − , , and − . These fractions can also be written
3
4 5
7
1
1
= 1 ÷ 4 = 0.25;
= 1 ÷ 3 = 0.333 . . . .
as decimals through division:
4
3
Thus, you have rational numbers that yield finite or terminating decimals
CHAPTER 7 / NUMBER PROPERTIES
37
and rational numbers that yield infinite repeating decimals. Every rational
number can be written in decimal form.
The irrational numbers are the real numbers that are not rational. The
irrational numbers you are most familiar with include 2, 3, π , and e.
Irrational numbers can be written as infinite, nonrepeating decimals. There
are decimal approximations for 2 ≈ 1.41421 · · · , 3 ≈ 1.73205 · · · , π ≈
3.14159 · · · , and e ≈ 2.71828 · · · . However, these approximations do not
clearly show that the decimals are nonrepeating. To show a decimal that is
both infinite and nonrepeating, you need a pattern that is not based on repetition. One such pattern is to start with 1, then 01, then 001, then 0001, with
each step in the pattern adding another zero. You get 0.101001000100001 · · · .
Another pattern yields 0.1121231234 · · · .
The real numbers are made up of the combination of all rational numbers
and all irrational numbers. Thus, the set of real numbers is the set of all
decimals, finite and infinite.
Number-Type Exercises
1. Change each number to its decimal form.
A.
4
5
B.
3
4
C.
5
6
D.
17
10
E.
11
20
2. Classify each of these numbers. Include all number types the number
belongs to.
A.
5
B. −3
C. 0
D.
2
3
E. 8
Solutions
4
1. A.
= 4 ÷ 5 = 0.8
5
3
= 3 ÷ 4 = 0.75
4
5
C.
= 5 ÷ 6 = 0.8333 · · ·
6
17
D.
= 17 ÷ 10 = 1.7
10
11
E.
= 11 ÷ 20 = 0.55
20
B.
2. A.
5 is irrational, real
B. −3 is integer, rational, real
C. 0 is whole number, integer, rational, real
2
is rational, real
D.
3
E. 8 is counting number, whole number, integer, rational, real
ROUNDING NUMBERS
When numbers are approximated, the results need to be rounded to maintain the accuracy of the data. For example, frequently, when you work on
problems with money, you get results that contain a fractional part of a
38
CONQUERING GMAT MATH AND INTEGRATED REASONING
cent. In these cases, you round the result to the nearest cent. In general,
the accuracy is specified in the problem.
To round a number, first locate the digit that has the accuracy wanted;
then examine the digit to the right. If the digit to the right is 5 or more,
round up by increasing the accuracy digit by 1 and dropping all digits to the
right of it. If the digit to the right of the accuracy digit is 4 or less, round
down by leaving the accuracy digit the same and dropping all digits to the
right of it.
If the digits dropped are to the left of the decimal point, they are replaced
with zeros.
Example 2
Consider the number 2,643.718.
A.
B.
C.
D.
E.
Round the number to the nearest tenth.
Round the number to the nearest hundredth.
Round the number to the nearest thousand.
Round the number to the nearest unit.
Round the number to the nearest hundred.
Solution
A. There is a 7 in the tenths digit, and the digit to the right of the tenths digit
is a 1. Thus the tenth digit stays the same, and the digits to the right of
it are dropped. When 2,643.718 is rounded to the nearest tenth, you get
2,643.7.
B. There is a 1 in the hundredths digit and an 8 in the next place to the
right. Thus, you increase the 1 by 1 and drop all digits to the right. When
rounded to the nearest hundredth, 2,643.718 becomes 2,643.72.
C. There is a 2 in the thousands place, and the digit to the right is a 6, so
you add 1 to the 2 and drop all digits to the right. Since the thousands
place is to the left of the decimal point, you have to fill in zeros for the
dropped digits between the thousands digit and the decimal point. Thus,
when you round 2,643.718 to the nearest thousand, you get 3,000.
D. When rounding 2,643.718 to the nearest unit, you have a 3 in the units
place and a 7 in the place to the right. The result of the rounding is 2,644.
E. When rounding to the nearest hundred, you note that there is a 6 in the
hundreds place and a 4 in the place to the right. Thus, you leave the 6
unchanged, and you drop the digits to the right of the hundreds place.
Since the hundreds place is to the left of the decimal point, you must
fill in zeros for the dropped digits between the hundreds place and the
decimal point. The answer after rounding to hundreds is 2,600.
Rounding Exercises
A.
B.
C.
D.
E.
F.
G.
Round 4.536 to the nearest tenth.
Round 5.8165 to the nearest thousandth.
Round 76,472 to the nearest hundred.
Round 268.463 to the nearest ten.
Round $486.238 to the nearest cent.
Round 4,563.75 to the nearest unit.
Round 1,436.3 to the nearest whole number.
44
CONQUERING GMAT MATH AND INTEGRATED REASONING
Zero is the only number that is neither positive nor negative. Preceding
zero with a "+" sign or a "−" sign does not make it positive or negative.
The absolute value of a number is the distance a number is from zero on
a number line.
The absolute value of a is denoted by |a|.
Example 12
Find the absolute value of each number.
A. |+5|
B. |−3|
C. |0|
D. −
2
3
E. +
1
2
Solution
A. When you look at the number line in Figure 7.7 and locate +5, you find
that the distance from +5 to 0 on the number line is 5. Thus |+5| = 5.
B. When you look at Figure 7.7 and locate −3, you see that the distance
from −3 to 0 is 3 units. Thus |−3| = 3.
C. When you look at Figure 7.7 and locate 0, you see that the distance from
0 is 0 units. Thus, |0| = 0.
2
2
D. − =
3
3
1
1
=
E.
2
2
Adding Signed Numbers
To add two numbers with like signs, you add the absolute value of the
numbers and precede the value with the common sign.
Example 13
Add these numbers.
A. +5
+2
B. +7
+6
C. −3
−2
D. −5
−4
E. −8
−3
Solution
A. |+5| = 5, |+2| = 2, and 5 + 2 = 7. The answer is positive.
(+5) + (+2) = +7
B. |+7| = 7, |+6| = 6, and 7 + 6 = 13. The answer is positive.
(+7) + (+6) = +13
C. |−3| = 3, |−2| = 2, and 3 + 2 = 5. The answer is negative.
(−3) + (−2) = −5
D. |−5| = 5, |−4| = 4, and 5 + 4 = 9. The answer is negative.
(−5) + (−4) = −9
E. |−8| = 8, |−3| = 3, and 8 + 3 = 11. The answer is negative.
(−8) + (−3) = −11
To add two numbers with unlike signs, you find the absolute value of
the two numbers and subtract the smaller absolute value from the greater
absolute value. The sign of the answer is the sign of the number with the
greater absolute value.
52
CONQUERING GMAT MATH AND INTEGRATED REASONING
H. Dividing by zero is undefined. 4 ÷ 0 =
undefined
I. Dividing a number by −1 gives its opposite.
(−3) ÷ (−1) = +3
J. Multiplying by +1 leaves the number
unchanged. (−17)(+1) = −17
K. This is the sum of opposites, so the result
is 0. (−13) + (+13) = 0
L. A number is divided by its opposite, so the
answer is −1. 16 ÷ (−16) = −1
M. Dividing by +1 leaves the number
unchanged. (−7) ÷ (+1) = −7
N. Dividing a number by −1 gives its opposite.
8 ÷ (−1) = −8
O. Dividing a nonzero number by itself gives
1 as the answer. 5 ÷ 5 = 1
P. Adding 0 to a number leaves the number
unchanged. 0 + (−8) = −8
ODD AND EVEN NUMBERS
An even number is an integer that is divisible by 2.
An odd number is an integer that has a remainder of 1 when divided
by 2.
An even number can be written in the form 2n, where n is an integer.
The even numbers are . . . , − 6, −4, −2, 0, 2, 4, 6, . . . . To get the positive even
numbers, restrict n to the counting numbers.
An odd number can be written in the form 2k + 1, where k is an integer.
The odd numbers are . . . , − 5, −3, −1, 1, 3, 5, . . . .
Every integer is either even or odd. No integer can be both even and odd.
Zero is an even number. The smallest positive odd number is 1. The smallest
positive even number is 2.
An even number added to another even number always yields a third even
number. 2 + 6 = 8, 0 + (−4) = −4, 16 + (−4) = 12.
An even number subtracted from another even number always yields a
third even number. 4 − 12 = −8, 16 − 2 = 14, 0 − 8 = −8, 10 − (−8) = 18.
When an even number is multiplied by an even number, the result is an
even number. 4 × 6 = 24, 8 × 0 = 0, −4 × 2 = −8, −6 × −10 = 60.
However, when you divide an even number by another even number, you
do not always get an even number, and you may not get an integer at all.
1
18 ÷ (−2) = −9, 24 ÷ 6 = 4, 6 ÷ 2 = 3, 4 ÷ 8 = , 6 ÷ 0 = undefined.
2
Summary:
even + even = even
even − even = even
even × even = even
An odd number added to an odd number is an even number. 3 + 7 = 10,
−5 + 11 = 6, −3 + 3 = 0, 1 + 1 = 2.
When an odd number is subtracted from an odd number, the answer is an
even number. 7 − 13 = −6, 5 − 3 = 2, 7 − 7 = 0, 11 − 3 = 8.
When two odd numbers are multiplied, the product will be an odd number.
3 × 5 = 15, 7 × −5 = −35, 7 × (−9) = −63.
When two odd numbers are divided, the result is always defined, but may
not be an integer. It may be odd or not an integer at all. 21 ÷ 3 = 7, 15 ÷ (−5) =
5
11
, 5÷7= .
−3, 11 ÷ 5 =
5
7
Summary:
odd + odd = even
odd − odd = even
odd × odd = odd
Adding an even number and an odd number always gives an odd number for
the answer. 3 + 4 = 7, 8 + (−5) = 3, 11 + 12 = 23, 15 + 0 = 15, 8 + 5 = 13.
CHAPTER 7 / NUMBER PROPERTIES
55
B. The multiples of 30 are 30, 60, 90, 120, 150, 180, 210, . . . .
The multiples of 24 are 24, 48, 72, 96, 120, 144, 168, 192, . . . .
120 is the only number common to both lists. LCM (30, 24) = 120.
C. The multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112,
120, 128, . . . .
The multiples of 15 are 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, . . ..
The only number in both lists is 120. LCM (8, 15) = 120.
Notes:
• If you do not have a multiple in common in your lists, extend your list of
multiples until one is reached.
• For counting numbers a and b, LCM (a, b) ≤ a × b. This means that the least
common multiple of two numbers is less than or equal to the product of the
two numbers.
Example 25
Find the least common multiple for these sets of three numbers.
A. 5, 12, 8
B. 10, 25, 15
Solution
A. The multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75,
80, 85, 90, 95, 100, 105, 110, 115, 120, 125, . . . .
The multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, . . . .
The multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112,
120, 128, 136, . . . .
The number common to all three lists is 120. LCM (5, 12, 8) = 120.
B. The multiples of 10 are 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130,
140, 150, 160, 170, 180, . . . .
The multiples of 25 are 25, 50, 75, 100, 125, 150, 175, 200, . . . .
The multiples of 15 are 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165,
180, . . . .
The number common to all three lists is 150. LCM (10, 25, 15) = 150.
Factors and Greatest Common Divisors
A counting number a is a divisor of a counting number b if there is a counting
number c such that a × c = b.
Since a × c = b, we also say that a is a factor of b.
In most cases the terms factor and divisor are used interchangeably; however,
there are times when they have a slight difference. Since you are using counting
numbers, this difference is not relevant on the GMAT.
The greatest common divisor (GCD) for two or more counting numbers is
the largest counting number that is a divisor of each of the counting numbers.
To find the divisors of a counting number such as 12, write down the factors,
which come in pairs. However, when both factors are the same, you list it just
once. The divisors of 12 are 1 and 12, 2 and 6, and 3 and 4. You usually list the
divisors in numerical order. Thus, you say that the divisors of 12 are 1, 2, 3, 4,
6, and 12. Since 12 is equal to the number itself, you say that 12 is an improper
divisor of 12, and 1, 2, 3, 4, and 6 are the proper divisors of 12.
Example 26
List all the divisors of the given numbers.
A. 16
B. 20
C. 21
D. 23
E. 36
CHAPTER 7 / NUMBER PROPERTIES
57
B. 1 × 39 = 39, 3 × 13 = 39
The divisors of 39 are 1, 3, 13, 39.
1 × 65 = 65, 5 × 13 = 65
The divisors of 65 are 1, 5, 13, 65.
1 × 91 = 91, 7 × 13 = 91
The divisors of 91 are 1, 7, 13, 91.
The common divisors of 39, 65, and 91 are 1 and 13. GCD (39, 65, 91) = 13.
C. 1 × 250 = 250, 2 × 125 = 250, 5 × 50 = 250, 10 × 25 = 250
The divisors
of 250 are 1, 2, 5, 10, 25, 50, 125, 250.
1 × 375 = 375, 3 × 125 = 375, 5 × 75 = 375, 15 × 25 = 375
The divisors
of 375 are 1, 3, 5, 15, 25, 75, 125, 375.
1 × 625 = 625, 5 × 125 = 625, 25 × 25 = 625
The divisors of 625 are 1,
5, 25, 125, 625.
The common divisors of 250, 375, and 625 are 1, 5, 25, and 125.
GCD (250, 375, 625) = 125.
When a number is a divisor of a second number, it means that there is a
counting number such that the product of it and the divisor equals the second
number. However, when the first number is not a divisor of the second, you
cannot find a counting number that will yield the second. In this case, you try
to get as close as you can and stay less than the second number. To make them
equal, you add the missing amount, called the remainder.
If a is a divisor of b, then there is a counting number c such that a × c = b.
If a is not a divisor of b, then there exist counting numbers q and r such that
a × q + r = b, where 0 < r < a.
Prime Numbers
A prime number is a counting number greater than 1 such that its only counting
number divisors are 1 and itself. Since a prime number is greater than 1, clearly
0 and 1 are not prime numbers.
A composite number is a counting number greater than 1 such that it has
at least three divisors. A counting number greater than 1 is either prime or
composite. No counting number is both prime and composite.
The prime factorization of a number is a set of prime numbers such that the
product of these prime factors will yield the given number. A prime factor may
be used more than once in the product. The prime factorization of 6 is 6 = 2 × 3,
and the prime factorization of 8 is 8 = 2 × 2 × 2, or 8 = 23 .
Exercise
Find the prime factorization of each number.
A. 24
B. 50
C. 72
D. 23
E. 39
Solution
A.
B.
C.
D.
E.
24 = 2 × 12 = 2 × 2 × 6 = 2 × 2 × 2 × 3 = 23 × 3
50 = 2 × 25 = 2 × 5 × 5 = 2 × 52
72 = 2 × 36 = 2 × 2 × 18 = 2 × 2 × 2 × 9 = 2 × 2 × 2 × 3 × 3 = 23 × 32
23 = 23
39 = 3 × 13
In finding the prime factorization of a counting number, look for divisors (or
factors) of the number that are prime. Thus, it is helpful to know some prime
numbers. The prime numbers less than 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29,
31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97. It is important to
notice that 2 is the only even number that is prime, since for a number to be
even means that it is divisible by 2.
58
CONQUERING GMAT MATH AND INTEGRATED REASONING
Example 29
Find the prime factorization of each number.
A. 54
B. 150
C. 168
D. 500
E. 144
Solution
A.
B.
C.
D.
E.
54 = 2 × 27 = 2 × 3 × 9 = 2 × 3 × 3 × 3 = 2 × 33
150 = 2 × 75 = 2 × 3 × 25 = 2 × 3 × 5 × 5 = 2 × 3 × 52
168 = 2 × 84 = 2 × 2 × 42 = 2 × 2 × 2 × 21 = 2 × 2 × 2 × 3 × 7 = 23 × 3 × 7
500 = 2 × 250 = 2 × 2 × 125 = 2 × 2 × 5 × 25 = 2 × 2 × 5 × 5 × 5 = 22 × 53
144 = 2 × 72 = 2 × 2 × 36 = 2 × 2 × 2 × 18 = 2 × 2 × 2 × 2 × 9 = 2 × 2 × 2×
2 × 3 × 3 = 24 × 32
Finding the prime factorization of a number requires you to divide the number by primes until you find one that is a divisor; then you use that same prime to
see if it is a divisor of the second factor. You continue to work on the factor that
is not known to be a prime until you have reduced it to a prime. You check your
work by multiplying all the prime factors together to get the original number.
DIVISIBILITY TESTS
One thing that will help you to find the prime factorization of a number is a
quick way to tell if a number is divisible by smaller numbers.
• A counting number is divisible by 2 if its units digit is an even number. Thus,
if the rightmost digit of a counting number is 0, 2, 4, 6, or 8, the number is
divisible by 2.
• A counting number is divisible by 3 if the sum of its digits is divisible by 3.
• A counting number is divisible by 5 if its units digit is 0 or 5.
• A counting number is divisible by 4 if the two-digit number formed by the tens
and units digits is divisible by 4.
• A counting number is divisible by 6 if it is divisible by 2 and also by 3. Thus,
if the units digit is even and the sum of all digits is divisible by 3, then the
number is divisible by 6.
• A counting number is divisible by 9 if the sum of its digits is divisible by 9.
• A counting number is divisible by 10 if the units digit is 0.
Exercise
Which of these numbers are divisible by 2 or 4?
A. 68,532
B. 384,670
C. 165,501
D. 483,000
E. 759,258
Solution
A. Since the units digit is 2, which is even, 68,532 is divisible by 2.
Since the last two digits are 32, which is divisible by 4, the number 68,532 is
divisible by 4.
B. Since the units digits is 0, which is even, 384,670 is divisible by 2.
Since the last two digits are 70, which is not divisible by 4, the number
384,670 is not divisible by 4.
C. Since the units digit is 1, which is odd, 165,501 is not divisible by 2 or by 4.
D. Since the units digit is 0, which is even, 483,000 is divisible by 2.
Since the last two digits are 00, which is divisible by 4, the number 483,000
is divisible by 4.
E. Since the units digit is 8, which is even, 759,258 is divisible by 2.
Since the last two digits are 58, which is not divisible by 4, the number
759,258 is not divisible by 4.60
CONQUERING GMAT MATH AND INTEGRATED REASONING
Solution
A. Since the units digit is 5, the number 68,535 is divisible by 5.
Since the units digit is not 0, the number 68,535 is not divisible by 10.
B. Since the units digit is 0, the number 384,670 is divisible by 5.
Since the units digit is 0, the number 384,670 is divisible by 10.
C. Since the units digit is 1, the number 561,501 is not divisible by 5 or by 10.
D. Since the units digit is 0, the number 483,000 is divisible by 5.
Since the units digit is 0, the number 483,000 is divisible by 10.
E. Since the units digit is 5, the number 759,205 is divisible by 5.
Since the units digit is not 0, the number 759,205 is not divisible by 10.
A number that has a units digit of 5 is divisible by 5, but not by 10. Also, when
the units digit of a number is 0, it will always be divisible by both 5 and 10.
GCD AND LCM REVISITED
A second way to find the GCD and LCM of a number is to use prime factorization. When you are finding the divisors of numbers with many factors, it is easy
to overlook a factor that may be a common divisor. Similarly, in listing the multiples of a number, any error in finding one could cause other numbers to be
wrong. It could also take a long time to find the common multiple if it is large.
To find the least common multiple (LCM) by using the prime factorization,
you must first find the prime factorization of each number. Then you need to find
the greatest power each different factor of the number has. The least common
multiple will be the product of each different factor to the greatest power at
which it occurs.
Exercise
Find the least common multiple of these sets of numbers.
A. 150 and 225
B. 63 and 84
C. 24, 60, and 96
Solution
A. 150 = 2 · 75 = 2 · 3 · 25 = 2 · 3 · 5 · 5 = 2 · 3 · 52
225 = 3 · 75 = 3 · 3 · 25 = 3 · 3 · 5 · 5 = 32 · 52
The different prime factors are 2, 3, and 5. The greatest power of 2 is 1, the
greatest power of 3 is 2, and the greatest power of 5 is 2.
LCM (150, 225) = 21 · 32 · 52 = 2 · 9 · 25 = 450
B. 63 = 3 · 21 = 3 · 3 · 7 = 32 · 7
84 = 2 · 42 = 2 · 2 · 21 = 2 · 2 · 3 · 7 = 22 · 3 · 7
The greatest power of 2 is 2, the greatest power of 3 is 2, and the greatest
power for 7 is 1.
LCM (63, 84) = 22 · 32 · 7 = 4 · 9 · 7 = 252
C. 24 = 2 · 12 = 2 · 2 · 6 = 2 · 2 · 2 · 3 = 23 · 3
60 = 2 · 30 = 2 · 2 · 15 = 2 · 2 · 3 · 5 = 22 · 3 · 5
96 = 2 · 48 = 2 · 2 · 24 = 2 · 2 · 2 · 12 = 2 · 2 · 2 · 2 · 6 = 2 · 2 · 2 · 2 · 2 · 3 = 25 · 3
The greatest occurring power of 2 is 5; the greatest power of 3 is 1, and the
greatest power of 5 is 1.
LCM (24, 60, 96) = 25 · 3 · 5 = 32 · 3 · 5 = 480
To find the greatest common divisor (GCD) of a set of numbers, first find the
prime factorization of each number. Next identify the primes that are common
to all the numbers and then the greatest common power that these primes have.
The greatest common divisor of the numbers is the product of these common
divisors to their greatest common powers.
07_Moyer.indd 60
15/11/11 12:46 PM72
CONQUERING GMAT MATH AND INTEGRATED REASONING
ANSWER KEY
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
B
B
A
E
B
E
E
D
C
A
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
E
E
C
D
C
E
D
A
A
E
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
D
C
A
B
C
E
A
A
E
B
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
C
A
C
D
D
A
B
D
A
C
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
C
D
C
A
E
D
C
A
C
D
SOLUTIONS
3
4
The point P is between −1 and −2. The units are divided into quarter units,
3
so P is at −1 .
4
1. B −1
2. B Point H
The unit between 3 and 4 is divided into twentieths, with darker lines at
2
each fifth. Thus H is located at 3 .
5
3. A Point R
The point R is farthest left on the number line, so it represents the smallest
value of the labeled points.
4. E 5
The counting numbers are 1, 2, 3, 4, . . . . The only counting number listed
is 5.
5. B 0
The whole numbers are 0, 1, 2, 3, . . . . The only whole number listed is 0.
6. E π
2 1
Since 3, − , 1 , and 0 are all rational numbers, the only irrational number
3 2
is π .
7. E An integer is an irrational number.
Since the integers are rational numbers, they cannot also be irrational
numbers. The false statement is that "An integer is an irrational number."
8. D 3.2
There is a 1 in the tenths place, and the digit to the right is 5. You increase
the tenths digit by 1 to get 2, and you drop the digits to the right of the tenths
place.
9. C 5.6351
To round up, the digit to the right of the hundredths place must be 5 or
greater, and this is only true of 5.6351.
CHAPTER 7 / NUMBER PROPERTIES
73
10. A 2.34647
The thousandths digit will not round up if the digit to the right of the
thousandths place is 4 or less. This is only true of 2.34647.
11. E 1,700
The hundreds digit is 6, and the digit to the right is an 8, so you add 1 to 6 to
get 7. Drop the digits to the right of the hundreds, but to maintain the place
value of the thousands and hundreds digits, you must add zeros in the tens
place and the ones place.
12. E 290,000
The digit in the ten-thousands place is an 8, and the digit to the right is a 6.
Add 1 to the 8 to get 9 and drop the digits to the right of the ten-thousands.
You must replace the dropped digits between the ten-thousands digit and
the decimal point with zeros.
13. C 4,316.47
There is a 7 in the hundredths place, and to the right of the hundredths place
the digit is 4 or less. You drop all the digits to the right of the hundredths
place.
14. D 4 × 101 + 3 × 100 + 5 × 10−1
43.5 = 40 + 3 +
5
= 4 × 101 + 3 × 100 + 5 × 10−1
10
15. C 2 × 103 + 6 × 101 + 5 × 100
2,065 = 2,000 + 60 + 5 = 2 × 103 + 6 × 101 + 5 × 100
16. E 305,070
3 × 105 + 5 × 103 + 7 × 101 = 3 × 100,000 + 5 × 1,000 + 7 × 10
= 300,000 + 5,000 + 70 = 305,070
17. D 2 × 10−1 + 8 × 10−2 + 4 × 10−3 + 1 × 10−4
0.2841 = 0.2 + 0.08 + 0.004 + 0.0001
= 2 × 10−1 + 8 × 10−2 + 4 × 10−3 + 1 × 10−4
18. A −5
Since negative numbers are less than any positive numbers, −5 and −2 are
the lesser numbers. −5 is to the left of −2 on the number line, so −5 is the
smallest number.
19. A |−10|
|−10| = 10, |−5| = 5, |0| = 0, so 10 is the greatest number.
20. E −10
(−7) + (−3) is the sum of two numbers with the same sign. |−7| = 7, |−3| =
3. 7 + 3 = 10, and the sign is negative, so the answer is −10.
21. D −1
(−6) + (+5) is the sum of two numbers with unlike signs. |−6| = 6, |+5| =
5. 6 − 5 = 1, and 6 > 5, so the sign is negative. The answer is −1.
74
CONQUERING GMAT MATH AND INTEGRATED REASONING
22. C +3
(+8) + (−5) is the sum of two numbers with unlike signs. |+8| = 8, |−5| =
5. 8 − 5 = 3, and 8 > 5, so the sign is positive. The answer is +3.
23. A −11
(−7) − (+4) is the subtraction of +4 from −7. You add the opposite of
+4, which is −4, to −7. This is the sum of two numbers with like signs.
|−7| = 7, |−4| = 4. 7 + 4 = 11, and the sign is negative. The answer is −11.
24. B +1
(+6) − (+5) is the subtraction of +5 from +6. We add the opposite of +5,
which is −5, to +6. This is the sum of two numbers with unlike signs. |+6| =
6, |−5| = 5. 6 − 5 = 1, and 6 > 5, so the sign is positive. The answer is +1.
25. C 0
Since you are subtracting a number from itself, the answer is zero.
26. E +10
Since you are multiplying two numbers with like signs, the answer is
positive. |−5| = 5, |−2| = 2, and 5 × 2 = 10. The answer is +10.
27. A −24
Since you are multiplying two numbers with unlike signs, the answer is
negative. |−6| = 6, |+4| = 4, and 6 × 4 = 24. The answer is −24.
28. A −36
Since you are multiplying two numbers with unlike signs, the answer is
negative. |−6| = 6, |+6| = 6, and 6 × 6 = 36. The answer is −36.
29. E +2
Since you are dividing two numbers with like signs, the answer is positive.
|−6| = 6, |−3| = 3, and 6 ÷ 3 = 2. The answer is +2.
30. B −4
Since you are dividing two numbers with unlike signs, the answer is
negative. |−28| = 28, |+7| = 7, and 28 ÷ 7 = 4. The answer is −4.
31. C −1
(+5) + (−3) + (−8) + (+4) + (+1) = [(+5) + (+4) + (+1)] + [(−3) + (−8)]
= (+10) + (−11) = −(11 − 10) = −1
32. A −36
(−3) × (−1) × (+2) × (+3) × (−2) is a product of more than two signed
numbers. Since there are an odd number of negative factors, the product
is negative. −(3 × 1 × 2 × 3 × 2) = −36
33. C 0
Since one factor in the product is 0, the product is 0.
34. D odd × odd
even + even = even, odd + odd = even, even × odd = even, odd × odd
= odd, odd − odd = even
76
CONQUERING GMAT MATH AND INTEGRATED REASONING
46. D 22 · 32 · 5
180 = 2 · 90 = 2 · 2 · 45 = 2 · 2 · 3 · 15 = 2 · 2 · 3 · 3 · 5 = 22 · 32 · 5
47. C 22 · 3 · 72
588 = 2 · 294 = 2 · 2 · 147 = 2 · 2 · 3 · 49 = 2 · 2 · 3 · 7 · 7 = 22 · 3 · 72
48. A 2,372
Only 2,372, 5,622, and 9,418 have units digits that are even, so they are the
only choices divisible by 2.
Looking at the last two digits of each of these numbers, you see that only
2,372 has the last two digits making a number divisible by 4.
2,372 is divisible by 2 and 4.
49. C 4,865
To be divisible by 5, the units digit must be 5 or 0, so 6,450, 3,790, and 4,865
are divisible by 5.
To be divisible by 10, the units digit must be 0. Thus, only 4,865 is divisible
by 5 and NOT divisible by 10.
50. D 6,201
To be divisible by 6, the units digit must be even, so 1,743 and 6,201 are NOT
divisible by 6.
1 + 7 + 4 + 3 = 15, which is divisible by 3, but not divisible by 9.
6 + 2 + 0 + 1 = 9, which is divisible by 3 and by 9. So 6,201 is the only
answer choice that is divisible by 3 and by 9, but not divisible by 6.
80
CONQUERING GMAT MATH AND INTEGRATED REASONING
8. B From statement 1, you have x = y, but this just lets you say that x + y =
2x. You still do not know the value of x + y. So statement 1 is not sufficient.
From statement 2, you have y = 3 − x, which can be written as x + y = 3.
Thus, statement 2 is sufficient. Since statement 2 is sufficient and statement
1 is not sufficient, the answer is B.
9. D According to the graph, Cola T is selected by 20% of the people responding. If there were 10,000 people involved in the Cola Choice experiment,
then Cola T was chosen by 20% of 10,000 people, or 0.20 ×10, 000 = 2, 000
people who chose Cola T. Thus, the answer is D.
10. E Cola S was chosen by 30% of the participants, and Cola U was chosen by
40%. 30% + 40% = 70%. Thus, the answer is E.
GMAT PRACTICE PROBLEMS
For each question, select the best answer.
1. Which number is the greatest common divisor (GCD) of 45 and 75?
A.
B.
C.
D.
E.
3,375
225
15
5
3
2. Which number is the same as 5 tenths, 4 hundreds, 3 tens, and 8
ones?
A.
B.
C.
D.
E.
43.85
45.83
438.5
458.3
5,438
3. Which number is a rational number that is NOT a whole number?
A.
B.
C.
D.
E.
−12
−3
2 1 1
2 − −
3 2 6
0
9
5(−2)
4. Which is the quotient for 281.68 ÷ 28?
A.
B.
C.
D.
E.
1.6
16
10.06
10.6
100.6
CHAPTER 7 / NUMBER PROPERTIES
81
5. Which number is divisible by 4 but not divisible by 8?
A.
B.
C.
D.
E.
256
201
122
104
68
6. Which is the smallest of the prime factors of 161?
A.
B.
C.
D.
E.
3
5
7
11
37
7. Is n less than 2.6?
1. n < 2.7
2. n < 2.58. If n is a member of the set {30, 32, 35, 38, 39, 40} what is the value
of n?
1. n is odd
2. n is a multiple of 5CHAPTER 8
ARITHMETIC COMPUTATION
SYMBOLS
=
>
<
≥
≤
=
equals
is greater than
is less than
is greater than or equal to
is less than or equal to
is not equal to
x2
x squared
x3
x cubed
square root of x
absolute value of x
addition, plus
subtraction, minus
a times b, multiplication
a times b, multiplication
a times b, multiplication
a times b, multiplication
a divided by b, division
√
x
|x|
+
−
a×b
a · b
a∗b
ab
a÷b
a
b
a:b
%
a divided by b, division
the ratio of a to b
percent
ORDER OF OPERATIONS
When there is more than one operation in an expression, you have to use
the standard order of operations to do the computation in order to get
a consistent, correct, result. You use the order PEMDAS, which stands
for parentheses, exponents, multiplication and division, and addition and
subtraction.
The first step is to work within each set of parentheses or brackets.
A fraction bar means that the numerator and denominator are worked
separately as though each were enclosed by parentheses. The work inside
parentheses follows these rules also.
The second level is exponents, which means that you simplify all powers
and roots before trying to use them in computing.
The next level is multiplication and division. These operations are done
in the order that they occur from left to right in the problem. In 6 × 5 ÷ 2 × 4,
you multiply 6 × 5 first to get 30 ÷ 2 × 4, then you divide 30 by 2 and get
15 × 4. Finally, you multiply 15 by 4 to get 60. Multiplication and division
are of equal rank in the order of operations.
The last step is to do the addition and subtraction in the order in
which they occur from left to right. In 15 − 7 + 4, you subtract first, then
85
CHAPTER 8 / ARITHMETIC COMPUTATION
3. A. Because 6 + (−6) = 0 and −6 + 6 = 0, −6 is
the addition inverse of 6.
B. Because −7 + 7 = 0 and 7 + (−7) = 0, 7 is
the addition inverse of −7.
C. Because 0 + 0 = 0, 0 is the addition inverse
of 0.
D. Because −3 + (+3) = 0 and +3 + (−3) = 0,
+3 is the addition inverse of −3.
1
1
1
1
= 0 and − + = 0,
E. Because + −
2
2
2
2
1
1
− is the addition inverse of .
2
2
1
1
1
= 1 and ∗ 5 = 1, is the
5
5
5
multiplication inverse of 5.
1
1
= 1 and − ∗(−3) = 1,
Because −3∗ −
3
3
1
− is the multiplication inverse of −3.
3
1
1
Because ∗ 2 = 1 and 2 ∗ = 1, 2 is the
2
2
1
multiplication inverse of .
2
3
3
2
2
= 1 and − ∗ −
=
Because − ∗ −
3
2
2
3
2
3
1, − is the multiplication inverse of − .
2
3
Because 1 ∗ 1 = 1, 1 is the multiplication
inverse of 1.
Because −1 ∗ (−1) = 1, −1 is the multiplication inverse of −1.
4. A. Because 5 ∗
B.
C.
D.
E.
F.
89
5. A. Because 0 was added to (9 + 3), this shows
the addition identity property.
B. Because all that we changed was the grouping, this shows the associative property for
multiplication.
C. Because the order of the factors 5 and 3 +
8 is all that has been changed, the property
is commutative for multiplication.
D. The common factor of 5 has been taken out
of the two terms, so it is the distributive
property.
E. Opposites are being added, so this is the
addition inverse property.
F. The grouping is changed, so this is the
associative property for addition.
G. Since the −2 has been distributed over the
sum, this is the distributive property.
H. You are multiplying reciprocals, so this is
the multiplication inverse.
I. You are multiplying by 1, so this is the
multiplication identity.
J. The order of the factors has changed, so
this is the commutative property of multiplication.
K. The order of the addends has changed,
so this is the commutative property of
addition.
FRACTIONS
A fraction is made up of three parts: the numerator, the fraction bar, and
the denominator. The numerator tells you how many parts you have, and
the denominator tells you how many parts the whole was divided into. The
fraction bar is read as "out of." Otherwise it is used as a grouping indicator
to remind you to simplify the numerator and denominator separately before
doing the division.
A fraction can tell you what part of a whole unit you have, or even that
you have more parts than are needed for one whole.
3
means that you have 3 of the 4 parts that the whole is
The fraction
4
5
divided into, while means that you have all 4 parts of the one whole, plus
4
1 part of another whole that is divided into 4 equal parts.
a
a
In a fraction where a and b are whole numbers and b is not zero, is a
b
b
proper fraction when a < b. If a = b or a > b, it is an improper fraction.
You can use fractions to express the ratio of two quantities. If there are
units with the numbers, then they need to be the same units. When the units
are different, you convert to a common unit.
90
CONQUERING GMAT MATH AND INTEGRATED REASONING
a
. This use of fractions
b
is especially helpful when you are dividing a smaller counting number by a
larger one.
11
4 7
, you can change
When you have an improper fraction such as , , or
4 5
2
4
them so that the number of whole units involved is clear. For , you have all
4
7
4
4 parts for 1 unit, and = 1. With , you have 2 more parts than needed to
4
5
2
2
2
7
make a whole unit, and you say that = 1 . It is understood that 1 = 1+ .
5
5
5
5
11
, you see that 11 ÷ 2 yields 5 with a remainder of 1. The remainder is
With
2
11
1
1
=5 .
1 of 2 parts needed to make the next whole, so it represents . Thus,
2
2
2
It is very helpful to be able to convert between improper fractions and mixed
numbers.
A fraction can be used to indicate division; a ÷b =
Example 4
Convert these improper fractions to mixed numbers.
A.
7
3
B.
5
2
C.
8
5
D.
14
5
E.
13
8
Solution
A.
B.
C.
D.
E.
7
7
1
= 7 ÷ 3 = 2 with remainder 1
=2
3
3
3
5
5
1
= 5 ÷ 2 = 2 with remainder 1
=2
2
2
2
8
8
3
= 8 ÷ 5 = 1 with remainder 3
=1
5
5
5
14
14
4
= 14 ÷ 5 = 2 with remainder 4
=2
5
5
5
13
13
5
= 13 ÷ 8 = 1 with remainder 5
=1
8
8
8
Note that in each case, the remainder tells you how many parts of the
next unit you have, but the denominator is needed to show how many parts
are needed to make the whole.
A complex fraction has fractions in the numerator, the denominator, or
1 + (3/4)
1/2 3 2/3
,
,
, and
.
both. Examples of complex fractions are
5 5/6 1/5
3/10
One way to simplify complex fractions is to find the least common multiple of all the fractions in the numerator and denominator and then multiply
the numerator and denominator by this number.
CHAPTER 8 / ARITHMETIC COMPUTATION
93
D. GCD(18, 24) = 6
18
18 ÷ 6
3
=
=
24
24 ÷ 6
4
E. GCD(42, 60) = 6
42
42 ÷ 6
7
=
=
60
60 ÷ 6
10
F. GCD(40, 200) = 40
40
40 ÷ 40
1
=
=
200
200 ÷ 40
5
Comparing fractions allows you to determine whether the two fractions
are equal or whether one of the fractions is greater than the other. There
are many ways to make the comparisons, and you need to decide which
procedure works best for you.
When you compare fractions with equal denominators, the fraction with
the greater numerator is the greater fraction.
15
15
11
11
and
, for example, since 15 > 11,
>
.
To compare
25
25
25
25
4
3
This procedure can be used to compare and if you change them to
8
7
equivalent fractions. The easiest denominator to use is the product of the
given denominators (even though the least common multiple of them might
be much smaller). 7 × 8 = 56, so change each fraction to an equivalent
3∗7
21
4
4∗8
32
3
=
, and =
=
. Since
one with a denominator of 56. =
8
8∗7
56
7
7∗8
56
21
4
3
32
>
and > .
32 > 21,
56
56
7
8
Occasionally, the fractions you want to compare will have equal numerators. In this case, the fraction with the smallest denominator will be
11
11
and
. Since with sevenths the whole is
the greater fraction. Compare
7
5
divided into 7 parts and with fifths the whole is divided into 5 parts, the fifths
are larger pieces than the sevenths, so 11 fifths is greater than 11 sevenths,
11
11
>
.
or
5
7
Another procedure that can be helpful is to make a mental comparison
1
3
1
1
to another number. For example, is less than , and is more than , so
3
2
5
2
1
3
> . The comparison value needs to be such that one fraction is greater
5
2
than this value and the other is less than it.
The cross-multiplication procedure is a relatively easy method to use.
c
a
c
a
To compare and , cross-multiply to get ad and bc. If ad > bc, then > .
b
d
b
d
c
a
c
3
4
a
If ad = bc, then = , and if ad < bc, then < . To compare and ,
b
d
b
d
8
7
3
4
cross-multiply to get 21 and 32. 21 < 32, so < .
8
7
106
CONQUERING GMAT MATH AND INTEGRATED REASONING
10. A.
1 2 1 1 1
1 2
1
6
1
+ − × ÷ = + −
× = +
2 3 4 3 6
2 3 12 1
2
D.
2
2 1
− =
3 2
3
B.
12
3
9
3
=
−
=
10
10 10
10
20 18 15 20 25
2 3 1 2 5
+ − + − =
+ − + −
=
3 5 2 3 6
30 30 30 30 30
18
3
58 40
−
=
=
30 30
30
5
C.
3 8 4×3
3×8
3 5 4 3
÷ − × = × −
=
−
4 8 5 8
4 5 5×8
4×5
E.
5 3 2
3×5
3 2
3
× ÷ × =
÷ × =
10 9 4 3
10 × 9 4 3
4
2
8
4
1 4 2
× × =
× =
=
6 3 3
18 3
54
27
3 3 35
3 3 × 35
3 1 3 6
× + ÷
= + ×
= +
=
4 2 5 35
8 5 6
8 5×6
3 28
31
7
3 7
+ = +
=
=3
8 2
8
8
8
8
DECIMALS
A decimal is made up of a whole number part (which can be zero), a
decimal point, and a decimal fraction. For example, 2.1345, 720.864913,
0.4823, 0.222· · · , and 0.9090090009· · · are all examples of decimals.
In 0.631579824, the 6 is in the tenths place, 3 is in the hundredths
place, 1 is in the thousandths place, 5 is in the ten-thousandths place, 7
is in the hundred-thousandths place, 9 is in the millionths place, 8 is in
the ten-millionths place, 2 is in the hundred-millionths place, and 4 is in
the billionths place. You can continue naming decimal places indefinitely,
but the ones already named go far beyond what you will need for the
GMAT.
When you compare decimal fractions, start at the decimal and compare
the digits one by one until you find a difference in the digits. The one with
the greater number in this place is the greater number.
Example 18
Which decimal is greater?
A. 0.7 or 0.074
B. 0.9086 or 0.908
C. 0.608 or 0.64
Solution
A. 0.7 has a 7 in the tenths place, while 0.074 has a 0 in the tenths place, so
0.7 > 0.074.
B. 0.9086 has 9 in the tenths place, and 0.908 has a 9 in the tenths place.
0.9086 has a 0 in the hundredths place, and 0.908 has a 0 in the hundredths place. 0.9086 has an 8 in the thousandths place, and 0.908 has
an 8 in the thousandths place. 0.9086 has an additional digit in the
ten-thousandths place, so 0.9086 > 0.908.
C. Both numbers have a 6 in the tenths place. 0.608 has a 0 in the hundredths
place, while 0.64 has a 4 in the hundredths place. 0.64 > 0.608.
When you convert decimals to fractions, look at two different situations:
one in which the decimal is finite and one in which the decimal is infinite
and repeating.
CHAPTER 8 / ARITHMETIC COMPUTATION
107
Example 19
Write each decimal as a fraction.
A. 0.265
B. 0.41
C. 0.25
Solution
A. 0.265 is 265 thousandths, so 0.265 =
D. 0.65
E. 0.104
265
53
=
.
1,000
200
41
.
100
1
25
= .
C. 0.25 is 25 hundredths, so 0.25 =
100
4
13
65
=
.
D. 0.65 is 65 hundredths, so 0.65 =
100
20
13
104
=
.
E. 0.104 is 104 thousandths, so 0.104 =
1,000
125
B. 0.41 is 41 hundredths, so 0.41 =
Note that you write the digits in the decimal over a 1 followed by zeros.
The number of zeros is equal to the number of digits in the decimal.
When a decimal has all its digits in repeating groups, you can write it as a
fraction by writing one repeating group over the same number of 9s as there
are digits in the repeating group.
Example 20
Change these repeating decimals to fractions.
A. 0.333· · ·
B. 0.252525· · ·
C. 0.090909· · ·
D. 0.123123123· · ·
Solution
A. 0.333· · · has one digit that repeats, 3. You write this repeating value over
3
1
a 9 since only one digit repeats. So 0.333 · · · = = .
9
3
B. 0.252525· · · has two digits that repeat, 25. You write the repeating value
25
.
over two 9s, since two digits repeat. So 0.252525 · · · =
99
9
09
=
=
C. 0.090909· · · has two digits that repeat, 09. So 0.090909 · · · =
99
99
1
.
11
D. 0.123123123· · · has three digits that repeat, 123. So 0.123123123 · · · =
123
41
=
.
999
333
If a decimal has some digits between the decimal point and the start of
the first repeating group, you have to modify the procedure above. In this
case, you write the nonrepeating digits and one repeating group minus the
nonrepeating digits over a group of 9s followed by a group of 0s. The number of 9s is the same as the number of digits in the repeating group, and
the number of 0s is equal to the number of digits that do not repeat. Thus,
12,441
4,147
12,453 − 12
=
=
.
0.12453453453 · · · =
99,900
99,900
33,300
108
CONQUERING GMAT MATH AND INTEGRATED REASONING
Example 21
Change these decimals to fractions.
A. 0.4353535· · ·
B. 0.142333· · ·
C. 0.12373737· · ·
Solution
435 − 4
431
=
990
990
1,281
427
1,423 − 142
=
=
B. 0.142333 · · · =
9,000
9,000
3,000
1,225
49
1,237 − 12
=
=
C. 0.12373737 · · · =
9,900
9,900
396
A. 0.4353535 · · · =
You can also change fractions to decimals. The easiest fractions to change
to decimals are those whose denominators are powers of 10, such as 10, 100,
263
9
17
= 0.17,
= 0.263,
= 0.09.
or 1,000.
100
1,000
100
In each case, the numerator is written as a whole number so the decimal
point is at the right, and then you count to the left of that decimal point the
same number of places as there are 0s in the power of 10.
For all other fractions, you do the indicated division until the division
terminates with a remainder of zero, or until it repeats a remainder. If the
division terminates, you have a finite decimal with the decimal in the quotient as the answer. If you get a repeated remainder, you get a repeating
decimal. The repeating group is the part of the quotient found from the first
time the remainder occurred to the second time it occurred.
Note that a fraction in lowest terms will terminate only when 2 and 5
are the only prime divisors of the denominator. If the denominator has
any prime divisor other than 2 or 5, that fraction will yield a repeating
decimal.
Example 22
Change these fractions to decimals.
A.
19
100
B.
7
1,000
C.
3
10
Solution
A.
B.
C.
D.
E.
19
19
is 19 hundredths, so
= 0.19.
100
100
7
7
is 7 thousandths, so
= 0.007.
1,000
1,000
3
3
is 3 tenths, so
= 0.3.
10
10
7
7
is 7 hundredths, so
= 0.07.
100
100
263
263
is 263 thousandths, so
= 0.263.
1,000
1,000
D.
7
100
E.
263
1,000
122
CONQUERING GMAT MATH AND INTEGRATED REASONING
Operation
Addition
Subtraction
Multiplication
Division
Keywords
Sum, total, all together, and, added to, combined, exceeds, increased by,
more than, greater than, plus
Minus, difference, decreased by, less, diminished by, reduced by,
subtracted from, deducted
Times, twice, doubled, tripled, product, multiplied by, halved
Quotient, divided by, divide
PRACTICE PROBLEMS
1. In a class, 6 students received A's, 12 received
B's, 10 C's, 8 D's, and 4 F's. What part of the
class received A's?
6. A book that regularly sold for $19.49 is sold
after giving it a $2.98 price reduction. How
much did the book sell for?
2. In 36 times at bat, Jones got 8 singles, 3 doubles, 1 triple, and 3 home runs. What part of
the time did Jones get a hit?
7. A sales clerk received $218.40 regular pay,
$28.50 overtime pay, $36.14 in commissions,
and a $25 bonus on a payday. How much
money did the clerk receive on that payday?
3. The slope of a roof is the ratio of the rise of
the roof to the run of the roof. What is the
slope of a roof that rises 6 feet in a run of 24
feet?
4. If four-ninths of a class of 36 students is girls,
how many girls are there in the class?
5. If a hydrochloric acid solution has two-thirds
water, how much water is used to make 24
ounces of this solution?
8. A finance company made a loan of $100 for
six monthly payments of $21.55. How much
money was repaid to the loan company?
9. Jane drove 331 miles in 6 hours. To the nearest tenth of a mile, what was the average
number of miles Jane drove per hour?
10. Harry bought a suit for $194.95 and a shirt
for $18.45. How much more than the shirt did
the suit cost?
SOLUTIONS
1. Read: You know how many students received
each grade, so you can add them to find how
many students are in the class. 6 + 12 + 10 +
8 + 4 = 40 students in the class.
Plan: Write the fraction for the number of
students receiving A's, 6, divided by the number of students in the class, 40. 6/40
Solve: Simplify the fraction. 6/40 = 3/20
Answer: Did you find the fraction of students
getting A's as a part of the class? Yes, the 6 for
the students getting A's is compared to the 40
for the students in the class.
2. 8 singles, 3 doubles, 1 triple, 3 home runs. So
8 + 3 + 1 + 3 = 15 hits.
15
=
The part of the time Jones got a hit is
36
5
.
12
3. Slope =
4.
5.
rise
6 ft
1
=
=
run
24 ft
4
4 36
4 × 36
4
girls out of 36 students. ×
=
=
9
9
1
9×1
4×4
= 16 girls
1
2 24
2 × 24
2
water out of 24 ounces. ×
=
=
3
3
1
3×1
2×8
= 16 ounces
1
6.
$19.49
− $2.98
16.51
regular price
price reduction
selling price
7.
$218.40
28.50
36.14
+ 25
$308.04
regular pay
overtime pay
commissions
bonus
total received
CHAPTER 8 / ARITHMETIC COMPUTATION
8.
$21.55
×
6
$129.30
123
10. $194.95 suit
− 18.45 shirt
$176.50 difference
each payment
payments made
total repaid
9. 331 miles driven in 6 hours
55.16
6 |331.00
30
31
30
10
6
40
36
4
55.16 is 55.2 rounded to the nearest tenth.
Jane averaged 55.2 miles per hour.
RATIO AND PROPORTIONS
A ratio is a quotient of two quantities. The ratio of a to b is written as a to
a
b, , or a : b. When ratios are used to compare numbers that have units of
b
measure, the units need to be the same for both numbers. You can find the
ratio of one part to another, such as the ratio of boys to girls in a class, or of
the parts to the whole, such as ratio of boys or girls to the whole class.
If there are 12 boys and 18 girls in a class, then there are 12 + 18 = 30
students in the class. The ratio of boys to girls is 12 : 18, which simplifies to
2 : 3. The ratio of girls to the class is 18 : 30, or 3 : 5.
If the ratio of yes votes to no votes is 175 : 350, then the ratio of yes votes
to total votes is 175 : (175 + 350) = 175 : 525 = 1 : 3, and the ratio of no
votes to total votes is 350 : (175 + 350) = 350 : 525 = 2 : 3.
When the ratio is known, you do not necessarily need to know the actual
4
6
2
amounts of the quantities compared. The ratio 2 to 3 = 2 : 3 = = = =
3
6
9
2n
= 2n : 3n, for all counting numbers n.
3n
Example 31
State the ratios requested.
A.
B.
C.
D.
The ratio of a nickel to a quarter
The ratio of 3 hours to a day
The ratio of boys to students in a class of 40 students with 22 girls
The ratio of the short part to the long part of a 10-foot board when the
shorter of the two parts is 4 feet
Solution
A. Nickel = 5 cents
quarter = 25 cents
5
1
nickel
=
=
quarter
25
5
B. 3 hours
1 day = 24 hours
3
1
3 hours
=
=
1 day
24
8
124
CONQUERING GMAT MATH AND INTEGRATED REASONING
C. Girls = 22
class = 40
boys = 40 − 22 = 18
boys
18
9
=
=
class
40
20
D. Shorter part = 4 ft
total = 10 ft
longer part = 10 ft − 4 ft = 6 ft
shorter part
4
2
= =
longer part
6
3
An extended ratio can have more than two parts, such as a : b : c or a : b
: c : d. If a board is 12 feet long and cut into three parts that have a ratio 1 :
2 : 3, the lengths are 1n : 2n : 3n. Since the total length is 12 feet, the lengths
are 2 feet, 4 feet, and 6 feet because 2 ft + 4 ft + 6 ft = 12 ft. You can work
this out from the ratio 1n : 2n : 3n by using the equation 1n + 2n + 3n = 12.
Thus, 6n = 12 and n = 2. So the parts are 1n = 1(2) = 2 ft, 2n = 2(2) = 4 ft,
and 3n = 3(2) = 6 ft.
Example 32
Find the parts for each ratio.
A. A length of cloth is 40 feet long. It is cut into lengths that have the ratio
2 : 3 : 5. What are the lengths of the parts?
B. A banner is 18 feet long and is cut into three parts with the ratio 1 : 3 : 5.
What are the lengths of the parts?
C. A board is 48 inches long and is cut into four parts with the ratio 1 : 1 : 3
: 5. Find the lengths of the parts.
Solution
A. 2 : 3 : 5 ratio and the total length is 40 feet
2n + 3n + 5n = 40;
10n = 40;
n= 4
2n = 2(4) = 8 feet; 3n = 3(4) = 12 feet; 5n = 5(4) = 20 feet
The parts are 8, 12, and 20 feet.
B. 1 : 3 : 5 ratio and the total length is 18 feet
1n + 3n + 5n = 18;
9n = 18;
n= 2
1n = 1(2) = 2 feet; 3n = 3(2) = 6 feet; 5n = 5(2) = 10 feet
The parts are 2, 6, and 10 feet.
C. 1 : 1 : 3 : 5 ratio and the total length is 48 inches
n + n + 3n + 5n = 48;
10n = 48;
n = 4.8
n = 4.8 inches; n = 4.8 inches; 3n = 3(4.8) = 14.4 inches; 5n = 5(4.8) =
24 inches
The parts are 4.8, 4.8, 14.4, and 24 inches.
A proportion is a statement that two ratios are equal. A proportion can
c
a
be stated as "a is to b as c is to d" and written as a : b : : c : d or = . In
b
d
the proportion, a and d are called the extremes and b and c are the means.
When you want to solve a proportion, you cross-multiply to get the product
of the means equal to the product of the extremes.
Example 33
Write each as a proportion.
A.
B.
C.
D.
26 is to 13 is the same as 6 is to 3.
4 is to 12 as 5 is to 15.
45 is to 80 is the same as 18 is to 32.
150 is to 100 as 54 is to 36.
126
CONQUERING GMAT MATH AND INTEGRATED REASONING
Two quantities are directly proportional when a constant k times one of
them gives the other. The distance d that a car travels is directly proportional
to the time t that it travels, so d = kt. The amount of sales tax t that you pay
is proportional to the cost c of the items purchased, so t = kc.
Example 35
Write the proportion for each problem.
A. The number of commercials c in a television program is directly proportional to the length of the program m in minutes.
B. The number of miles m that you run in a given time is directly proportional to your rate r of speed.
C. The number of books b that you read in a given time is directly proportional to the number of hours h that you spend reading.
Solution
A. c = km
B. m = kr
C. b = kh
When the product of two quantities is a constant, the quantities are said
to be inversely proportional. The number of tickets to a play t that you get
for $100 is inversely proportional to the cost c of a ticket to the play. So
tc = $100. The number of workers w needed to finish a job in 6 days is
inversely proportional to the rate r of work done per day by a worker. So
wr = 6.
Example 36
Write the proportion for each problem.
A. The number n of cookies in a box with volume 100 in3 is inversely
proportional to the size s of the cookies.
B. The number n of mowers it takes to mow a field in 10 hours is inversely
proportional to the area a that each mower cuts per hour.
C. The number of movie tickets t that you can get for $50.00 is inversely
proportional to the price p of a ticket.
Solution
A. ns = 100
B. na = 10
C. tp = 50
Properties of Proportions
For a proportion
1. ad = bc
4.
c+d
a+b
=
b
d
a
c
= :
b
d
d
b
=
a
c
a−b
c−d
5.
=
b
d
2.
a
b
=
c
d
a+b
c+d
6.
=
a−b
c−d
3.
Each of these properties is a transformation of the given proportion.
CHAPTER 8 / ARITHMETIC COMPUTATION
127
Even if you do not know the amount of each quantity, sometimes you
1
can still find the ratio of the quantities. If you know that of the students
3
1
like carrots and of the students like turnips, the ratio of students who like
8
1
1
carrots C to students who like turnips T is to .
3
8
1/3
1 1
1 8
8
C
=
= ÷ = × =
T
1/8
3 8
3 1
3
Thus, there are 8 students who like carrots for every 3 who like turnips.
PRACTICE PROBLEMS
1. Write the ratio requested.
A. Ratio of 5 days to a week
B. Ratio of girls to boys in a class of 35
students with 18 boys
C. Ratio of a dime to a quarter
D. Ratio of a foot to a yard
2. Find the values requested.
A. If a foot-long hotdog was cut into two parts
that have a ratio of 1 : 2, how long is each
part?
B. If a collection of 54 coins, only nickels and
dimes, has a ratio of nickels to dimes of
4 : 5, how many of each coin are there?
C. If a collection of 1,000 nickels, dimes, and
quarters has a ratio of nickels to dimes to
quarters of 5 : 3 : 2, how many of each coin
are in the collection?
3. Write each as a proportion.
A. 50 is to 12 as 150 is to 36.
B. 14 to 12 is the same as 42 to 36.
C. 45 is to 90 as 225 is to 450.
4. Solve these proportions.
A. A recipe calls for 1 cup of flour to 2 cups of
sugar. If the next batch uses 3 cups of flour,
how much sugar is needed?
B. A chili recipe calls for 4 cups of beans to
2 cups of tomatoes. If you use 5 cups of
tomatoes, how many cups of beans are
needed?
C. If 5 stamps cost $1.95, how much would
20 stamps cost?
D. If you get 3 oranges for 70 cents, how many
oranges can you get for $3.50?
5. Write the proportion.
A. The number of dollars d in your pay is
directly proportional to the number of
hours h that you work.
B. The number of meals m that you eat on a
trip is directly proportional to the number
of days d in the trip.
C. The number of pages p copied is directly
proportional to the number of minutes m
that the copier runs.
D. The number of lunches l that you get for
$50 is inversely proportional to the cost c
of a meal.
E. The number of candy bars c that you can
get for $20 is inversely proportional to the
price p of the candy bar.
6. Find the ratio of a to b.
A. In a sentence,
2
of the letters are a's and
5
1
of the letters are b's.
20
B.
17
a−b
=
b
2
C.
3
b
=
a
10
D.
5
3
of a class are girls and of the class are
8
8
girls
a
boys and =
b
boys
CHAPTER 8 / ARITHMETIC COMPUTATION
129
MOTION AND WORK PROBLEMS
The motion formula d = r t states that the distance traveled is equal to the
rate of travel multiplied by time traveled. The rate is distance per unit of
time and the unit of time used to express the rate must match the unit used
to express the time traveled.
d
d
time: t =
t
r
If a bicycle travels at 8 miles per hour for 3 hours, then the bicycle has
traveled d = 8 miles/hour times 3 hours = 24 miles.
However, if a car traveled 50 miles per hour for 30 minutes, it did NOT
travel 50 × 30 miles since 50 is in miles per hour and the time is in minutes. The correct result is found by changing 30 minutes to 0.5 hour. d = 50
miles/hour × 0.5 hour = 25 miles.
Distance: d = r t
rate: r =
Example 37
Solve each of these motion problems.
A. A driver drove for 10 hours at the average rate of 53 miles per hour (mph) to
get from Memphis to Chicago. How far was it from Memphis to Chicago?
B. A plane traveled the 328 miles from Limestone to Lincoln in 2 hours.
What was the plane's average rate of speed?
C. A boat traveled 120 miles from Wabash to Wilson at the average rate of
60 miles per hour. How long did the boat trip take?
Solution
A. d = r t
d = 53(10)
d = 530 miles
It was 530 miles from Memphis to Chicago.
d
B. r =
t
r = 328 ÷ 2
r = 164 mph
The average rate of the plane was 164 mph.
d
C. t =
r
t = 120 ÷ 60
t = 2 hours
The boat trip took 2 hours.
On the GMAT, you need to be able to convert one unit of measure to
another. The units will be in the same measurement system only. Also, only
common conversions will be used.
Time
60 minutes = 1 hour
60 seconds = 1 minute
24 hours = 1 day
7 days = 1 week
52 weeks = 1 year
12 months = 1 year
U.S. Customary System
or English Units
12 inches = 1 foot
3 feet = 1 yard
5,280 feet = 1 mile
1,760 yards = 1 mile
16 ounces = 1 pound
Metric Units
10 millimeters = 1 centimeter
100 centimeters = 1 meter
1,000 meters = 1 kilometer
1,000 milligrams = 1 gram
1,000 grams = 1 kilogram
1,000 milliliters = 1 liter
130
CONQUERING GMAT MATH AND INTEGRATED REASONING
Work problems involve individuals or machines working together to
accomplish a task when you know how long it takes each to do the work
1
individually. If a person can do a job in 5 hours, then that person can do
5
3
2
of the work in 1 hour. Thus, the person can do of the job in 2 hours, in
5
5
4
3 hours, and in 4 hours. There is an assumption that a person always works
5
at a constant rate.
The principle behind work problems is that the time worked t times the
rate of work r gives the amount of work w done: w = tr . For each person
working, calculate the part of the work done and add all the parts together,
to get one unit of work completed.
Example 38
Solve these work problems.
A. Sue can paint a room in 4 hours, and Sam can paint the same room in
5 hours. How long would it take them to paint the room together?
B. Abe and Ben, working together, can build a cabinet in 6 days. Abe works
twice as fast as Ben. How long would it take each of them to build an
identical cabinet on his own?
C. A holding tank can be filled by three pipes (A, B, and C) in 1 hour, 1.5
hours, and 3 hours, respectively. In how many minutes can the tank be
filled by the three pipes together?
D. A copy machine makes 40 copies per minute. A second copy machine can
make 30 copies per minute. If the two machines work together, how long
would it take them to produce 1,400 copies?
Solution
1
of the room per hour.
4
1
Sam paints the room in 5 hours, so she does of the room per hour.
5
They work together for t hours.
t
t
+ =1
4 5
A. Sue paints the room in 4 hours, so she does
5t + 4t = 20
9t = 20
2
hours
9
2
It takes Sue and Sam 2 hours to paint the room together.
9
B. t = time for Abe and 2t = time for Ben.
6
6
+
=1
t
2t
t=2
12 + 6 = 2t
18 = 2t
9=t
It would take Abe 9 hours working alone, and it would take Ben 18 hours
working alone.
CHAPTER 8 / ARITHMETIC COMPUTATION
131
C. t = the time, in minutes, it would take the pipes together to fill the tank.
t
t
t
+
+
=1
60 90 180
3t + 2t + t = 180
6t = 180
t = 30
It takes the three pipes 30 minutes to fill the tank together.
D. t = the number of minutes together
40t + 30t = 1,400
70t = 1,400
t = 20
It takes the two copiers 20 minutes to produce 1,400 copies.
When it takes person A x hours to do a task alone and it takes person B
y hours to do the same task alone, then the time it will take them together
will be more than one-half of the faster time but less than one-half of the
slower time. When you have a multiple-choice question that has two people
or machines working together, use the fact that one-half of the faster time <
together time < one-half of the slower time to eliminate some of the given
answer choices. This is a quick check on the answers to parts A, B, and D in
the example above.
PRACTICE PROBLEMS
1. A train left Los Angeles at 8 a.m. headed
toward San Francisco. At the same time,
another train left San Francisco headed
toward Los Angeles. The first train travels at
60 mph and the second train travels at 50 mph.
If the distance between Los Angeles and San
Francisco is 440 miles, how long will it take
the trains to meet?
4. Carlos can run a mile in 6 minutes, and Kevin
can run a mile in 8 minutes. If Carlos gives
Kevin a 1-minute head start, how far will
Kevin run before Carlos passes him?
2. Two planes leave Chicago at the same time,
one headed for New York and the other headed
in the opposite direction for Denver. The plane
headed for New York traveled at 600 mph,
while the plane headed for Denver traveled at
150 mph. How long did each plane fly before
they were 900 miles apart?
6. David can type a paper in 5 hours, but when
he works with Jim, it takes only 2 hours. How
long would it take Jim to type the paper alone?
3. Two cars leave Sioux Falls headed toward
Fargo. The first car left at 7 a.m., traveling at
60 mph. The second car left at 8 a.m., traveling
at 70 mph. How long would it take the second
car to overtake the first car?
5. Bev can dig a ditch in 4 hours, and Jan can dig
a ditch in 3 hours. If Bev and Jan dig a ditch
together, how long would it take?
7. Kim can paint a barn in 5 days, and Alyssa can
paint it in 8 days. Kim and Alyssa start painting a barn together, but after 2 days Alyssa
gets sick and Kim finishes the work alone.
How long did it take Kim to finish painting the
barn?
8. Jay can prune an orchard in 50 hours, and Ray
can prune the orchard in 40 hours. How long
will it take them, working together, to prune
the orchard?
138
CONQUERING GMAT MATH AND INTEGRATED REASONING
Example 47
Solve these percent problems.
A. A car has a sticker price of $26,410. The customer bargained for 5.5% off.
How much did the car sell for?
B. A lamp was usually sold for $60. It was discounted 20%. What was the
sale price of the lamp?
C. A table was sold at a 40% discount sale for $150. What was the original
price of the table?
Solution
A. Original price − discount = sale price
$26,410 − 0.055($26,410) = S
$26,410 − $1, 452.55 = S
$24,957.45 = S
The selling price of the car was $24,957.45.
B. $60 − 0.2($60) = S
$60 − $12 = S
$48 = S
The sale price of the lamp was $48.
C. Original price = 100% of P
Discount = 40% of P
Sale price = 1.00P − 0.40P
0.6P = $150
P = $150 ÷ 0.6
P = $250
The original price of the table was $250.
In some problems, more than one percentage is used. These can be multiple discounts, sales tax and a tip, or a markup and then a discount, among
many other possibilities.
Example 48
Solve these percent problems.
A. A vase was purchased at a wholesale warehouse for $120. The vase was
marked up 40% in a retail store. The store owner sold the vase at a 25%
discount. What was the selling price of the vase?
B. In a restaurant, a meal costs $32.75. There is a 6% tax to be added before
you get the bill. If you leave the wait staff a 20% tip, what is the total cost
of the meal?
C. The workers at a company took a 15% cut in pay in 1999 and received a
15% pay raise in 2000. If a worker made $19.40 per hour before the 1999
pay cut, how much would this worker be earning after the 2000 pay raise?
Solution
A. Cost = $120
Retail price = $120 + 0.4($120) = $120 + $48 = $168
Sale price = $168 − 0.25($168) = $168 − $42 = $126
The selling price of the vase was $126.
B. Cost of meal = $32.75
Meal with tax = $32.75 + 0.06($32.75) = $32.75 + $1.965 = $34.715 =
$34.72
Meal with tax and tip: = $34.72 + 0.2($34.72) = $34.72 + $6.944 =
$41.664 = $41.66
The total cost of the meal is $41.66.
CHAPTER 8 / ARITHMETIC COMPUTATION
139
C. Original pay rate = $19.40
Pay after 1999 cut = $19.40 − 0.15($19.40) = $19.40 − $2.91 = $16.49
Pay after 2000 raise = $16.49 + 0.15($16.49) = $16.49 + $2.4735 =
$18.9635 = $18.96
The worker's pay after the raise in 2000 would be $18.96.
PRACTICE PROBLEMS
1. Solve these percentage problems.
A. What number is 32% of 65?
B. What percent of 90 is 27%?
C. 16 is 40% of what number?
D. What percent of 1,032 is 645?
E. What number is 8% of 247?
F. What number is 120% of 216?
G. 245% of what number is 98?
H. 0.4% of what number is 2?
I. 28 is what percent of 25?
J. 27 is what percent of 60?
2. A. What would Sharon's salary be after a 4.5%
pay raise, if she makes $25,000 per year
now?
B. Jack weighed 65 kg in January and by July
he had lost 13 kg. What percent of his body
weight did he lose?
C. In the freshman class at City College, 396
students are from Iowa. If 18% of the freshman class is from Iowa, how large is the
freshman class at City College?
D. A real estate agent receives a 6% commission on the sale price of the property. How
much will the real estate agent earn on the
sale of a lot for $38,000?
E. Thrifty Mart marks up the price of greeting cards by 42%. If a card costs Thrifty
Mart $2, what will the selling price of the
card be?
3. A. Deuce Hardware plans a Fourth of July sale
in which all garden tools will be marked
down 22%. What will be the sale price of
a wheelbarrow that regularly sells for $50?
B. A paring knife has a wholesale price of
$1.20. If it is marked up 30%, what will be
the retail price for the knife?
C. A suit sold for $150 at a 25% off sale. What
was the original price of the suit?
D. You select items with a total price of $38.90.
If the sales tax is 5%, what is the total cost
of these items?
E. A table has a list price of $1,200. The store
discounted the table by 30%, but it did not
sell. At a clearance sale, the store took 20%
off the discounted price. What is the price
of the table at the clearance sale?
F. You order a meal with a price of $35. There
is a tax of 6% on the meal. If you leave a
20% tip (after tax), what is the total cost of
the meal?
G. Jerry Smith was hired at an annual salary
of $26,500. He receives a raise of 11% in
January, then another 6% raise in June.
What is his salary after the raise in June?
SOLUTIONS
1. A.
B.
C.
D.
E.
F.
G.
H.
I.
J.
32% of 65 = 0.32(65) = 20.8
W = 20.8
27 ÷ 90 = 0.3
0.3 = 30%
P = 30%
40% = 0.4
16 ÷ 0.4 = 40
N = 40
645 ÷ 1,032 = 0.625,
0.625 = 62.5%
P = 62.5%
0.08(247) = 19.76
W = 19.76
1.20(216) = 259.2
W = 259.2
98 ÷ 2.45 = 40
N = 40
2 ÷ 0.004 = 500
N = 500
28 ÷ 25 = 1.12
1.12 = 112%
P = 112%
27 ÷ 60 = 0.45
0.45 = 45%
P = 45%
2. A. $25,000 + 0.45($25,000) = S
$25,000 + $1,125 = S
$26,125 = S
Sharon's salary after the raise would be
$26,125.
B. 13 ÷ 65 = 20%
Jack lost 20% of his body weight.
C. 396 ÷ 0.18 = 22
There are 2,200 freshmen at City College.
D. 0.06($38,000) = $2,280
The real estate agent will earn a commission of $2,280.
E. $2 + 0.42($2) = S
$2 + $0.84 = S
$2.84 = S
Thrifty Mart will sell the card for $2.84.
CHAPTER 8 / ARITHMETIC COMPUTATION
141
C. SUM = 4 + 2 + 24 + 14 + 34 + 8 + 10 + 20 = 116
N=8
AVE = SUM ÷ N = 116 ÷ 8 = 14.5
D. SUM = 5 + 1 + 7 + 3 + 1 = 17
N=5
AVE = SUM ÷ N = 17 ÷ 5 = 3.4
E. SUM = 25+2+5+6+5+23+22+7+10+15+21+23 = 164
N = 12
AVE = SUM ÷ N = 164 ÷ 12 = 13.666 · · · = 13.7
If all values but one are known and the average is also known, then the
last value can be determined. Suppose that 8, 16, 4, and 10 are four of five
values that have an average of 10. If five values have a mean of 10, then the
sum of the values must be 5(10) = 50. The sum of the four known values is
8 + 16 + 4 + 10 = 38. The missing value is 50 − 38 = 12.
Example 50
Find the missing value.
A. If the average of six values is 28 and five of the six values are 29, 19, 23,
20, and 43, what is the sixth value?
B. If the average of five numbers is 41 and the numbers include 46, 35, 38,
and 41, what is the missing number?
C. If Sara scored 77, 89, 98, 97, 99, and 91 on the first six tests, what does
she need to score on the seventh test to get an average of 93 for the seven
tests?
D. If Joe scored 77, 80, and 81 on three quizzes, what does he need to score
on the next quiz to have an 82 average?
Solution
A. 6(28) = 168
29 + 19 + 23 + 20 + 43 = 134
168 − 134 = 34
The sixth number is 34.
B. 5(41) = 205
46 + 35 + 38 + 41 = 160
205 − 160 = 45
The missing number is 45.
C. 7(93) = 651
77 + 89 + 98 + 97 + 99 + 91 = 551
651 − 551 = 100
She needs to score 100 on the seventh test.
D. 4(82) = 328
77 + 80 + 81 = 238
328 − 238 = 90
He needs to score 90 on the fourth quiz.
To find the average of two or more averages, weight each average with the
number of values in the average and then divide the sum of the weighted
averages by the sum of the numbers for each of the averages. If A is the
average of h values, B is the average of i values, and C is the average of
j values, then the combined average is the sum of the weighted averages
divided by the sum of the weights. SUM = hA + iB + jC and N = h + i + j.
AVE = (hA + iB + jC) ÷ (h + i + j).
142
CONQUERING GMAT MATH AND INTEGRATED REASONING
Example 51
Find the weighted averages.
A. The average score on a test for 400 students in Crawford County is 650.
The average score on the same test for 600 students in Edwards County
is 680. What is the combined average for the students in Crawford and
Edwards counties?
B. Miss Jackson has 30 students in first period who averaged 85 on the state
algebra test, 24 students in second period averaged a 90 on the algebra
test, and 22 students in fifth period averaged an 82 on the algebra test.
What was the average for these three classes?
C. The average traffic fine was $90 on Monday for the 18 people ticketed,
the average fine was $76 on Tuesday for the 20 people ticketed, and the
average fine was $80 on Wednesday for the 24 people ticketed. What was
the average fine for the three days?
Solution
A. SUM = 400(650) + 600(680) = 260,000 + 408,000 = 668,000
N = 400 + 600 = 1,000
AVE = 668,000 ÷ 1,000 = 668
B. SUM = 30(85) + 24(90) + 22(82) = 2,550 + 2,160 + 1,804 = 6,514
N = 30 + 24 + 22 = 76
AVE = 6,514 ÷ 76 = 85.7105 = 85.7
C. SUM = 18($90) + 20($76) + 24($80) = $1,620 + $1,520 + $1,920 = $5,060
N = 18 + 20 + 24 = 62
AVE = $5,060 ÷ 62 = $81.6129 = $81.61
When consecutive integers are averaged, the average will always equal
the average of the first and last numbers in the sequence. The average of 5,
6, 7, 8, and 9 is 35 ÷ 5 = 7 in the traditional way and (5 + 9) ÷ 2 = 7 in the
way for consecutive integers.
Example 52
Find the average of these sequences of consecutive integers.
A. 1, 2, 3, 4, 5, 6, 7, 8
B. 7, 8, 9, 10, 11, 12
C. 25, 26, 27, 28, 29
Solution
A. AVE = (1 + 8) ÷ 2 = 9 ÷ 2 = 4.5
B. AVE = (7 + 12) ÷ 2 = 19 ÷ 2 = 9.5
C. AVE = (25 + 29) ÷ 2 = 54 ÷ 2 = 27
The mode of a set of values is the value that occurs most often. If
all values occur the same number of times, there is no mode. If two or
more values occur with the greatest frequency, then each of the values is a
mode.
CHAPTER 8 / ARITHMETIC COMPUTATION
143
Example 53
Find the mode for the set of values.
A.
B.
C.
D.
2, 8, 7, 6, 4, 3, 8, 5, 1
3, 7, 2, 6, 4, 1, 6, 5, 7, 9
8, 7, 2, 5, 9, 4, 1, 3
8, 7, 6, 9, 4, 18, 6, 3, 2, 6, 7
Solution
A.
B.
C.
D.
Since 8 occurs twice and all other values occur once, 8 is the mode.
Both 6 and 7 occur twice, so 6 and 7 are both modes.
All the values occur once, so there is no mode.
Since 6 occurs three times and no other value occurs more than twice, 6
is the mode.
The median is the middle value of a set, or the average of the two middle
values, when the values are arranged in order from least to greatest. The
median of the values 1, 3, 4, 7, 10 is 4 since it is the middle value in the
ordered values. The median of the values 1, 3, 4, 7, 10, 20 is the average of
the two middle values 4 and 7, or (4 + 7) ÷ 2 = 5.5.
Example 54
Find the median for each set of values.
A.
B.
C.
D.
E.
24, 6, 7, 23, 13, 12, 18
17, 15, 9, 13, 21, 32, 41, 7, 12
147, 159, 132, 181, 174, 253
74, 81, 39, 74, 82, 74, 80, 100, 74, 42
11, 38, 73, 91, 16, 51, 39
Solution
A. The ordered values are 6, 7, 12, 13, 18, 23, 24.
The middle value of the seven values is 13.
The median is 13.
B. The ordered values are 7, 9, 12, 13, 15, 17, 21, 32, 41.
The middle value of the nine values is 15.
The median is 15.
C. The ordered values are 132, 147, 159, 174, 181, 253.
The middle two values of the six values are 159 and 174.
The average of the two middle values is (159 + 174) ÷ 2 = 166.5.
The median is 166.5.
D. The ordered values are 39, 42, 74, 74, 74, 74, 80, 81, 82, 100.
The middle two values of the 10 values are 74 and 74.
The median is 74 since the average of 74 and 74 is 74.
E. The ordered values are 11, 16, 38, 39, 51, 73, 91.
The middle value of the seven values is 39.
The median is 39.
The range is the easiest way to describe how a set of data spreads out. To
compute the range R, subtract the smallest value, the minimum, from the
greatest value, the maximum. Thus, R = Max − Min.
170
CONQUERING GMAT MATH AND INTEGRATED REASONING
5. Which number is equal to 1.68 × 4.5?
A.
B.
C.
D.
E.
0.756
7.56
75.6
756
756.000
6. Elena's bread recipe calls for 3 ounces of butter for each 4 cups
of flour used. She needs to make 4 times the original recipe. If 12
ounces of butter is used, then how many cups of flour are needed?
A.
B.
C.
D.
E.
1
4
9
13
16
7. A room is to be painted by Jane and Sally working together. How
long will it take them to paint the room together?
1. Jane can paint the room in 16 hours, working alone.
2. Sally can paint the room in 12 hours, working alone table was purchased for $100 by the owner of Bella Home
Furnishings. What is the selling price of the table?
1. Bella Home Furnishings marks up merchandise by 40%.
2. Bella Home Furnishings has a 25% discount sale every
TuesdayNumber of Students
Scores on a 5-Point Quiz
20
18
16
14
12
10
8
6
4
2
0
0
1
2
3
Scores
4
5
6
CHAPTER 8 / ARITHMETIC COMPUTATION
171
9. What was the median for the scores on the quiz?
A.
B.
C.
D.
E.
1
2
3
4
5
10. What was the mode for the scores on the quiz?
A.
B.
C.
D.
E.
1
2
3
4
5
SOLUTIONS
1. D The parentheses still contain the same numbers, so it is not an example of an associative property. Since the order of the two numbers inside
the parentheses has been changed, it is an example of a commutative property, and the operation between the numbers is multiplication.
Thus, the answer is D.
2. C
1
1 1
3
4
7
7
1
+
=5+
=5
3 +2 =3+2+ + =5+
4
3
4 3
12 12
12
12
Thus, the answer is C.
3. D
5 3
5 8
40
20
÷ = · =
=
6 8
6 3
18
9
Thus, the answer is D.
4. E 0.3282828· · · has one digit in the decimal part that does not repeat
and two digits in the repeating group. To write the fraction for the given
decimal, the numerator will be the digits from the decimal part to the
end of one repeating group, 328, minus the nonrepeating digit 3 or
328 − 3 = 325. The denominator will be two 9s followed by one 0, since
there are two digits in the repeating group and one nonrepeating digit, so
328
, and the answer
the denominator is 990. Therefore, 0.3282828· · · =
990
is E.
5. B 168 ×45 = 7, 560. In 1.68 × 4.5, there are two decimal places in 1.68
and one decimal place in 4.5, so the product should have three decimal
places. 1.68 ×4.5 = 7.560 = 7.56. Thus, the answer is B.
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CONQUERING GMAT MATH AND INTEGRATED REASONING
6. E The recipe calls for 3 ounces of butter for each 4 cups of flour. Elena
is using 12 ounces of butter, so let n represent the number of cups of
12
3
=
. From this proportion,
flour she needs. Write the proportion
4
n
3n = 48, and n = 16. Thus, the answer is E.
7. C Neither statement 1 nor statement 2 provides any information about
them working together. Thus, you need to consider the statements
together. Let n be the number of hours for them to do the job together,
n
n
+
= 1, which has a unique solution.
and write the equation
16 12
Both statements taken together are sufficient to answer the question,
but neither is sufficient alone. The answer is C.
8. A Statement 1 says that the table will be marked up 40% for a price of
$140. Thus, it is sufficient to answer the question.
Statement 2 says that on Tuesdays there is a sale, but does not provide
any information about the selling price.
Only statement 1 is sufficient, so the answer is A.
9. B The median is the value for which one-half the data is below and onehalf is above.
Score
Students
Cumulative
0
8
8
1
12
20
2
16
36
3
10
46
4
18
64
5
4
68
There are 68 values, so the middle value would be the average of the
34th and 35th values, both of which are 2 since the 21st through the
36th values are all 2. The median is 2, and the answer is B.
10. D The mode is the most frequent value. Since 4 was scored 18 times, it
is the mode. The answer is D.
GMAT PRACTICE PROBLEMS
For each question, select the best answer.
5
1. Which number is equivalent to 2 ?
8
A.
10
16
B.
7
8
C.
10
8
174
CONQUERING GMAT MATH AND INTEGRATED REASONING
6. In Mrs. Webster's algebra class, 8 students received A's, 14 received
B's, 12 received C's, 10 received D's, and 6 received F's. Which
number represents the ratio of B's to D's for the class?
A.
B.
C.
D.
E.
7 : 25
2:3
5:7
7:5
3:2
7. How much money was donated to The Hanks Foundation?
1. Individuals donated 35% of all money The Hanks Foundation
received.
2. Community groups donated 25% of all the money The Hanks
Foundation received number w with two digits to the right of the decimal point is
multiplied by the number n. How many decimal places are there
in the product w · n?
1. The number n has three digits to the right of the decimal
point.
2. Neither the hundredths digit in w nor the thousandths digit
in n is a multiple of 10Class Grade Distribution
F
8%
A
14%
D
12%
B
24%
C
42%
CHAPTER 8 / ARITHMETIC COMPUTATION
175
9. If there were 150 students in the class, how many students received
an A?
A.
B.
C.
D.
E.
35
28
21
14
7
10. What percentage of the class received a grade lower than a C?
A.
B.
C.
D.
E.
24%
20%
16%
12%
8%
228
CONQUERING GMAT MATH AND INTEGRATED REASONING
Solving Word Problems
1. Read the problem carefully, looking for key terms and concepts. Identify the question you must answer. A diagram may help you to interpret
the given information.
2. List all the unknown quantities in the problem and represent them in
terms of one variable if possible, such as x or x and y.
3. Use the information identified in step 1 to write algebraic relationships
among the quantities identified in step 2.
4. Combine the algebraic relationships into equations.
5. Solve the equation or system of equations.
6. Verify your results by checking against the facts in the problem.
Example 54
Solve these numerical word problems.
A. The sum of two numbers is 94, and the larger number is 5 less than twice
the smaller number. Find the numbers.
B. Two numbers have a sum of 18. Find the numbers if one number is 8
larger than the other.
C. Find three consecutive integers if their sum is 21.
D. Find three consecutive even integers such that the first plus twice the
second plus 4 times the third equals 174.
Solution
A. x = smaller number
Represent the smaller number by
using x.
2x − 5 = larger number
Represent the larger number.
x + 2x − 5 = 94
Add the two numbers to get the sum.
3x − 5 = 94
Combine like terms to simplify.
x = 33
Solve the equation.
2x − 5 = 2(33) − 5 = 66 − 5 = 61. Find the larger number.
The two numbers are 33 and 61.
Write an answer to the question asked.
B. x = smaller number
y = larger number
x + y = 18
x+8= y
x + y = 18
x + y = 18
+ x − y = −8
5 + y = 18
2x
= 10
y = 13
x
= 5
The two numbers are 5 and 13.
C. x = first integer
x + 1 = second integer
x + 2 = third integer
x + x + 1 + x + 2 = 21
3x + 3 = 21
3x = 18
x=6
x+1=7
x+2=8
The three consecutive integers are 6, 7, and 8.
CHAPTER 9 / ALGEBRA
229
D. x = first even integer
x + 2 = second even integer
x + 4 = third even integer
x + 2(x + 2) + 4(x + 4) = 174
x + 2x + 4 + 4x + 16 = 174
7x + 20 = 174
7x = 154
x = 22
x + 2 = 24
x + 4 = 26
The three consecutive even integers are 22, 24, and 26.
Example 55
Solve these age word problems.
A. Carlos is 3 years older than his brother Jose. In 4 years from now, the
sum of their ages will be 33 years. How old is each now?
B. Kia is 5 years younger than her sister Yvette. Three years ago, the sum of
their ages was 23. How old is each now?
Solution
A. x = age of Jose now
Represent the younger person's age as x.
x + 3 = age of Carlos now
Using x, represent the older person's age.
x + 4 + x + 3 + 4 = 33
Add their ages in 4 years to get that sum.
2x + 11 = 33
Combine like terms to simplify.
2x = 22
Solve the equation.
x = 11
Find the second person's age.
x + 3 = 14
Write an answer to the question asked.
Jose is 11 years old now and Carlos is 14 years old.
B. x = age of Yvette now
x − 5 = age of Kia now
x − 3 + x − 5 − 3 = 23
2x − 11 = 23
2x = 34
x = 17
x − 5 = 12
Kia is 12 years old now and Yvette is 17 years old.
Example 56
Solve these statistical word problems.
A. Ken's percent grades on five tests were 84, 72, 91, 64, and 83. Find the
average (arithmetic mean) grade.
B. Von had an average percent score on the first four tests of 84, and his average percent score on the next six tests was 92. What was Von's average
score for all the tests?
C. Marie's monthly school expenses were $64, $82, $51, $90, $67, $71, $58,
$94, and $63. What is Marie's median monthly school expense?
D. Tim's monthly food costs were $412, $408, $410, $408, $401, $410, and
$408. What is the mode for the monthly food costs?
CHAPTER 9 / ALGEBRA
231
Example 58
Solve these rate word problems.
A. Tyrone rides his bicycle 5 miles from his home to the city bus stop at
the rate of 8 miles per hour. He arrives in time to catch the bus to work,
which travels at 25 miles per hour. If he spends 1.5 hours traveling from
home to work, how far does he travel on the bus?
B. Maria jogs 15 miles to her sister's house to get her bicycle and then bicycles home. The total trip takes 3 hours. If she bicycles twice as fast as she
jogs, how fast does she bicycle?
C. Two drivers, Brenda and Julie, are 300 miles apart. Brenda drives at 30
miles per hour, and Julie drives at 45 miles per hour. If they drive toward
each other, how far will each have traveled when they meet?
D. At what rate must Eden travel to overtake Rex, who is traveling at a rate
20 miles per hour slower than Eden, if Eden starts 2 hours after Rex and
wants to overtake him in 4 hours?
E. Two planes start from Minneapolis at the same time and fly in opposite
directions, one averaging 40 miles per hour faster than the other. If they
are 2,000 miles apart in 5 hours, what is the average speed for each plane?
Solution
A. d = r t where d is the distance traveled, r is the rate, and t is the time.
x = distance traveled on bus
Represent the quantities.
5
= time spent riding bicycle
8
x
= time spent riding bus
25
5
x
3
3
+
=
1.5 hours = hours Add the times to get total time.
8 25
2
2
5
x
3
+ 200
= 200
200
8
25
2
LCM(8, 25, 2) = 200
Clear equation of fractions.
25(5) + 8(x) = 100(3)
Simplify the equation.
125 + 8(x) = 300
8x = 175
Solve the equation.
x = 21.875
Tyrone traveled 21.875 miles on
the bus.
Write the answer to the question.
B. x = miles per hour for jogging
2x = miles per hour for bicycling
15 15
+
=3
x
2x
30 + 15 = 6x
45 = 6x
7.5 = x
15 = 2x
Maria bicycles at the rate of 15 miles per hour.
232
CONQUERING GMAT MATH AND INTEGRATED REASONING
C. x = time they traveled
30x + 45x = 300
75x = 300
x=4
30x = 120
45x = 180
Brenda traveled 120 miles and Julie traveled 180 miles.
D. x = miles per hour for Eden
x − 20 = miles per hour for Rex
6(x − 20) = 4x
6x − 120 = 4x
2x − 120 = 0
2x = 120
x = 60
Eden needs to travel at 60 miles per hour.
E. x = rate of faster plane
x + 40 = rate of slower plane
5x + 5(x + 40) = 2,000
5x + 5x + 200 = 2,000
10x = 1,800
x = 180
x + 40 = 220
The planes travel at 180 and 220 miles per hour.
Example 59
Solve these work word problems.
A. Jamal can fill the vending machine in 45 minutes. When his sister, Violet,
helps him, it takes them 20 minutes. How long would it take Violet to fill
the machine by herself?
B. One pipe can fill a pool in 18 hours. Another pipe can fill it in 24 hours.
The drainpipe can empty the tank in 12 hours. With all three pipes open,
how long will it take to fill the pool?
C. One work crew can do a job in 8 days. After the first crew worked 3 days,
a second crew joins them, and together, the two crews finish the job in 3
more days. How long would it take the second crew to do the job alone?
D. One machine can wrap 200 boxes per hour. A newer machine can wrap
250 boxes per hour. How long would it take the two machines working
together to wrap 4,950 boxes?
E. Barbara and Wesley can paint a room together in 6 hours. If Barbara can
paint the room alone in 10 hours, how long would it take Wesley working
alone to paint the room?
Solution
A. x = number of minutes
for Violet alone
20 20
+
=1
x
45
900 + 20x = 45x
Represent each person's time.
Add work done by each person.
Solve equation.
CHAPTER 9 / ALGEBRA
233
900 = 25x
36 = x
It would take Violet 36
minutes working alone.
Write an answer to the question.
B. x = number of hours together
x
x
x
+
−
=1
LCM(18, 24, 12) = 72
18 24 12
x
x
x
72
= 72(1)
+
−
18 24 12
4x + 3x − 6x = 72
x = 72
With all three pipes open, it would take 72 hours to fill the pool.
C. x = number of days for second crew
6 3
+ =1
8 x
6x + 24 = 8x
24 = 2x
12 = x
The second crew could do the job alone in 12 days.
D. x = number of hours together
200x + 250x = 4,950
450x = 4,950
x = 11
It would take the two machines together 11 hours to do the job.
E. x = number of hours for Wesley
6
6
+ =1
10 x
6x + 60 = 10x
60 = 4x
15 = x
Wesley can paint the room alone in 15 hours.
Sometimes you will need to use data from a survey to answer questions.
A problem with the data occurs when you allow an object to have multiple
attributes. For example, a person might own both a desktop computer and
a laptop computer, so that person would belong to two categories.
You use Venn diagrams to represent the data in distinct categories while
showing relationships among the categories. Overlapping circles allow you
to visually analyze the data. For each of the given groups, use one circle.
With two groups there are four regions: group 1 only, group 2 only, both
groups, and neither group.
Example 60
Solve these problems by using Venn diagrams.
A. One hundred people were asked about the computers that they owned.
Laptop computers were owned by 55 people, desktop computers were
owned by 68 people, and 28 people owned both a laptop and a desktop
234
CONQUERING GMAT MATH AND INTEGRATED REASONING
computer. How many people did not own a computer? How many people
owned just one computer?
B. At a wine-tasting party, 150 people were asked which of the three wines
served that they liked.
68 like Boone's Farm Strawberry Hill.
68 like Ripple.
80 like Thunderbird.
35 like Boone's and Thunderbird.
30 like Ripple and Thunderbird.
28 like Boone's and Ripple.
20 like all three.
How many people like exactly two of these wines?
How many people like exactly one of these wines?
Solution
A. Two groups: L = laptops and D = desktops.
Draw a rectangle enclosing two intersecting circles.
D
L
Figure 9.5
Since N = 100, the sum of the numbers in the regions must total 100.
Since 28 people owned both a laptop and a desktop, put 28 in the
overlap of the circles labeled D and L. There are 55 laptop owners, so
55−28 = 27 goes in the part of L not shared with D. There are 68 desktop
owners, so 68−28 = 40 goes in the part of D not in L. You have accounted
for 40 + 28 + 27 = 95 of the 100 people, so 5 are outside the circles for
D and L. Thus, there are 5 people who do not own a computer. There are
40 people with desktops only and 27 people with laptops only, so there
are 40 + 27 = 67 people who own one computer.
B. Three groups: B = Boone's, R = Ripple, T = Thunderbird.
Draw a rectangle enclosing three intersecting circles.
B
R
T
Figure 9.6
CHAPTER 9 / ALGEBRA
235
Since N = 150, the numbers put in the eight regions must total 150. Since
20 people like all three wines, place 20 in the region common to B, R, and
T. There are 28 − 20 = 8 in the overlap of B and R and not in all three,
30 − 20 = 10 in the overlap of R and T not in all three, and 35 − 20 = 15 in
the overlap of B and T and not in all three. There are 68 − 15 − 20 − 8 = 25
in B only, 68 − 8 − 20 − 10 = 30 in R only, and 80 − 15 − 20 − 10 = 35 in T
only. There are 25 + 15 + 20 + 8 + 30 + 10 + 35 = 143 accounted for by those
who like at least one of the wines. There are 150 − 143 = 7 people who did
not like any of the wines.
There are 15 who like Boone's and Thunderbird only, 10 who like Ripple
and Thunderbird only, and 8 who like Boone's and Ripple only, so 33 people
like exactly two of these wines. There are 25 who like Boone's only, 30 who
like Ripple only, and 35 who like Thunderbird only, so 90 people like exactly
one of these wines.
Probability is the likelihood that a given event will occur. It is the ratio of
the number of favorable outcomes to the total number of outcomes. There
1
are two ways a coin can come up when it is flipped. P(heads) = . There
2
are six outcomes for a die when it is rolled: 1, 2, 3, 4, 5, or 6. Thus P(1) =
1
P(2) = P(3) = P(4) = P(5) = P(6) = .
6
The probability of an event is 0 ≤ P(event) ≤ 1. There is a probability of
0 only when the event cannot occur, and a probability of 1 when the event is
certain to occur. If P(E) is the probability event E will occur, and P(not E) is
the probability event E will not occur, then P(not E) = 1 − P(E).
Example 61
Find the probability of the event.
A. Two coins are flipped once. What is the probability of getting two tails?
B. A jar contains 3 yellow marbles, 4 green marbles, 2 black marbles, and
1 white marble. One marble is selected without looking. What is the
probability that the marble will be green?
C. A box contains 20 light switches, and 3 of them are defective. If you select
one switch from the box at random, what is the probability that it will
not be defective?
D. What is the probability that a card selected at random from a standard
deck of playing cards is a 10?
E. What is the probability that a card selected at random from a standard
deck of playing cards will be a 7 or a club?
Solution
1
.
4
B. 3 + 4 + 2 + 1 = 10 possible outcomes, 4 outcomes green. P(green) =
2
4
= .
10
5
3
C. 20 light switches and 3 are defective. P(defective) =
, P(not defective)
20
= 1 − P(defective).
A. The possible outcomes are HH, HT, TH, and TT. P(two tails) =
P(not defective) = 1 −
17
3
=
20
20
236
CONQUERING GMAT MATH AND INTEGRATED REASONING
D. There are 52 cards in a standard deck. There are four 10s: 10 of hearts,
1
4
=
.
10 of clubs, 10 of spades, and 10 of diamonds. P(10) =
52
13
E. There are 52 cards in a deck. There are 13 each in hearts, clubs, spades,
and diamonds. There are four 7s, one each in hearts, clubs, spades, and
diamonds. Thus, the 7 of clubs is counted twice, once as a club and once
as a 7. There are 13 clubs plus three 7s that are not clubs. P(7 or club) =
4
16
=
.
52
13
In computing probability, you must find the number of outcomes for the
event of interest and the number of outcomes that are possible. In the examples above, you were able to list all the outcomes or reason out how many
outcomes there were. This is not always possible, but there are tools that can
help you determine the number of outcomes without listing them. The most
common of these tools are the multiplication principle, permutations, and
combinations.
If an activity consists of distinct parts, then the multiplication principle
says that the number of ways the activity can be done is the product of the
number of ways for all the parts of the activity. If an activity consists of parts
A, B, and C and part A can be done in a ways, part B can be done in b ways,
and part C can be done in c ways, then the activity can be done in a × b × c
ways.
Example 62
Find the number of ways each activity can be done.
A. Kim selects a shirt, a pair of slacks, and a jacket to wear to work. She
has 7 shirts, 4 pairs of slacks, and 3 jackets in her closet. If she does not
worry about which clothes match, how many different outfits can she
create?
B. Randy flips a coin and then rolls a six-sided die. How many different
outcomes are there?
C. Noah is packing his backpack with toys for a trip to Grandpa Bruce's
house. He gets to take 1 car, 1 book, and 1 ball. If Noah has 4 cars, 6
books, and 3 balls, how many different sets of toys can he take?
Solution
A. Kim selects 1 each of her shirts, slacks, and jackets. She has 7×4×3 = 84
choices. There are 84 different outfits she could create.
B. Randy flips a coin, then rolls a die, so he has 2 × 6 = 12 outcomes. There
are 12 total outcomes for the activity.
C. Noah packs 1 each of his cars, books, and balls, so he has 4 × 6 × 3 = 72
choices. There are 72 different sets of toys that he could take.
When you select from a collection of objects, the number of ways the
selection can be made depends on the number of objects in the collection,
the number of objects being selected, and whether the order of selection
matters. If the objects selected are all going to be treated exactly the same
way, order is not important. However, if the objects will be treated differently
based on the sequence of selection, then order is important.
CHAPTER 9 / ALGEBRA
237
When there are n objects in the collection and you select r, 0 ≤ r ≤ n,
objects, and order is important, it is a permutation. If order is not important, it is a combination. For example, if there are 10 people in a race and
the first person to finish gets a trophy, and the second person to finish gets a
medal, and the third person to finish gets a ribbon, then you can use permutation to determine in how many different ways the prizes could be awarded.
In another race of 10 people, if the first three people to finish go to the next
round of competition, then you use combinations to determine the number
of ways the group of three people who go on to the next round could turn
n!
n!
and nCr =
are for computing the
out. The formulas n Pr =
(n − r )!
(n − r )! · r !
number of permutations and combinations, respectively. And n! is defined
to be the product of the counting number n and all the counting numbers
less than n down to and including 1. Use 0! = 1 to define 0!. The ! in these
expressions is read as "factorial."
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5,040 and 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
n! = n × (n − 1)! so 7! = 7 × 6! = 7 × 720 = 5, 040
Example 63
Evaluate each expression.
B. 8 P3
C. 9 C 7
A. 9 P7
D. 8 C3
Solution
9 × 8 × 7 × 6 × 5 × 4 × 3 × 2!
9!
=
= 181, 440
− 7)!
2!
(9
8 × 7 × 6 × 5!
8!
B. 8 P3 =
=
= 336
5!
(8 − 3)!
9 × 8 × 7!
72
9!
C. 9 C7 =
=
=
= 36
2! × 7!
2
(9 − 7)! × 7!
8 × 7 × 6 × 5!
8!
336
=
=
= 56
D. 8 C3 =
5! × 3!
6
(8 − 3)! × 3!
A.
9 P7
=
Example 64
1. Caf e Laura offers a lunch consisting of one of 4 soups, one of 7 sand´
wiches, one of 5 chips, and one of 3 drinks. How many different lunches
can it serve?
2. A sorority has 10 members, and 4 of them are to attend the national
convention. In how many ways can the 4 people attending the national
convention be selected?
3. There are 20 members on the city council. One is to be selected chairperson, a second person will be selected vice-chairperson, and a third
person will be selected secretary. In how many ways can these offices be
filled?
Solution
1. Since the activity has four separate parts that can be done in multiple
ways, use the multiplication principle. Lunch = soup, sandwich, chips,
and drink: 4 × 7 × 5 × 3 = 420. There are 420 possible lunches that could
be served at Caf e Laura.
´
238
CONQUERING GMAT MATH AND INTEGRATED REASONING
2. Since all 4 members selected will go to the convention, order of selection
is not important. Use combinations.
10 C 4
=
10 × 9 × 8 × 7 × 6!
10!
10 × 9 × 8 × 7
=
=
= 210
6! × 4!
4×3×2×1
(10 − 4)! × 4!
3. Since the 3 people selected are given offices based on their order of
selection, use permutations.
20 P3
=
20!
20!
20 × 19 × 18 × 17!
=
=
= 20 × 19 × 18 = 6, 840
17!
17!
(20 − 3)!
PRACTICE PROBLEMS
1. Find two numbers such that twice the first
plus 5 times the second is 20, and 4 times the
first less 3 times the second is 14.
2. Three more than twice a certain number is
57. Find the number.
3. Wendy's mother is 3 times as old as Wendy.
In 14 years, she will be twice as old as Wendy
is then. How old is each now?
4. Joan is 3 years older than Susan. Eight years
ago, Joan was 4 times the age of Susan. How
old is each now?
5. Michelle scored 95, 91, 98, 90, 96, and 100.
What is her average test score?
6. David weighed himself each week. In May,
he weighed himself 5 times with an average
weight of 185 pounds. In June, he weighed
himself 4 times with an average weight of 180
pounds. What is his average weight?
7. John's cell phone bills were $87, $81, $88,
$87, $84, $87, $89, $80, $78, $79, $81, and
$82 for the past year. What was the median
for his bills? What is the mode for his bills?
8. An 80% acid solution is mixed with a 20%
acid solution to get 3 gallons of a solution that
1
acid. How much of each acid solution
is
3
was used?
9. A mixture of 40 pounds of mixed nuts worth
$1.80 a pound is to be made from peanuts
costing $1.35 a pound added to fancy mixed
nuts costing $2.55 a pound. How many
pounds of each kind of nut should be used?
10. In her motorboat, Ruth can go downstream
in 1 hour less time than she can go upstream.
If the current is 5 miles per hour, how fast can
she travel in still water if it takes her 2 hours
to travel upstream the given distance?
11. Two drivers started toward each other from
towns 255 miles apart. One driver traveled at
40 miles per hour, and the other traveled at
45 miles per hour. How long did the people
drive until they met?
12. Amanda can mow a lawn in 1 hour 20 minutes. Kim can mow the same lawn in 2
hours. How long would it take them, working
together, to mow the lawn?
13. One computer can do a payroll in 12 hours.
A second computer can do the payroll in 6
hours. How long will it take to do the payroll if both computers work on the payroll at
the same time?
14. A survey was taken of 52 students at Macon
High School. They were asked which amusement parks they would like to visit on a class
trip: Six Flags, Disney World, and Opryland.
The data is summarized as follows.
28 preferred Six Flags.
25 preferred Disney World.
26 preferred Opryland.
10 preferred Opryland and Disney World.
11 preferred Disney World and Six Flags.
14 preferred Six Flags and Opryland.
6 preferred all three.
How many students prefer both Disney
World and Six Flags and did not prefer Opryland? How many did not prefer any of these
three sites?
CHAPTER 9 / ALGEBRA
259
6. Which number is a solution for
A.
B.
C.
D.
E.
x
x+5
= + 3?
4
8
−14
−7
0
7
14
7. x, y, and z are numbers that have a sum of 93. Find the numbers.
1. The numbers are primes.
2. The numbers are consecutive integers Jack is 4 years older than Daniel. What are their ages now?
1. 16 years ago, Jack was twice as old as Daniel.
2. 18 years ago, Jack was 3 times as old as DanielMillions of Dollars
Annual Sales
80
70
60
50
40
30
20
10
0
A
B
C
Business
D
E
9. Which two businesses together earned exactly $80 million?
A.
B.
C.
D.
E.
A and B
B and C
C and D
D and E
A and E
CHAPTER 10
GEOMETRY
π
pi (about 3.14)
is parallel to
is perpendicular to
angle
triangle
a degrees
right angle
⊥
∠
a◦
POINTS, LINES, AND ANGLES
Points are represented with dots and are named by capital letters. A line
extends indefinitely in two directions and can be named by naming two
points on it. Because a line has infinitely many points, a line can have many
names.
A
B
C
Figure 10.1
The line in Figure 10.1 can be named line AB, line AC, or line BC.
• A line segment is a part of a line between two points. The two points are
the endpoints of the line segment.
In Figure 10.1, there is a line segment AB, a line segment AC, and a line
segment BC. The line segment AB and the line segment BA are two names
for the same segments, AB = BA.
• A ray is a part of a line with exactly one endpoint. Ray AB is a ray with
endpoint A going through point B. Ray BA is a ray with endpoint B going
through point A. Ray AB = ray BA.
• An angle is the figure formed by two rays with a common endpoint. The
common endpoint of the rays is called the vertex of the angle, and the rays
are called the sides of an angle.
The symbol for an angle is ∠. An angle can be named by stating only its
vertex if no other angle has the same vertex. In general, an angle is named
by naming a point on one side of the angle, then its vertex, and finally a
point on the other side of the angle. In Figure 10.2, the angle can be named
∠A, ∠BAC, ∠CAB, ∠BAD, ∠DAB, and ∠EAC, among many other names.
B
E
A
C
D
Figure 10.2
265
266
CONQUERING GMAT MATH AND INTEGRATED REASONING
Angles are measured in degrees. There are 360◦ about a point and 180◦
about the point staying on one side of a line through that point.
The measure of an angle is the number of degrees of rotation it takes to
get from one side of the angle to the other.
• An acute angle is an angle whose measure is greater than 0◦ but less
than 90◦ .
• A right angle is an angle whose measure is exactly 90◦ .
• An obtuse angle is an angle whose measure is greater than 90◦ but less
than 180◦ .
• A straight angle is an angle whose sides are a pair of opposite rays and
the measure is exactly 180◦ .
• A reflex angle is an angle whose measure is greater than 180◦ , but less
than 360◦ .
• Adjacent angles are two angles in the same plane that have a common
vertex and a common side that separates the two angles.
If the rotation from one side of the angle to the other is 0◦ , the rays lie on
top of each other. Also, if the rotation between the sides of the angle is 360◦ ,
the rays lie on top of each other. In both cases, the angle looks like a single
ray.
• Two angles are complementary if the sum of the measures is 90◦ .
• Two angles are supplementary if the sum of the measures is 180◦ .
To be complementary or supplementary, the two angles do not have to
be adjacent. The sum of the measures of the two angles is all that matters
in deciding if the pair of angles is either supplementary or complementary. A way to remember which pair has which sum is to put the terms in
alphabetical order, complementary then supplementary, and put the sums
in numerical order, 90◦ then 180◦ . The results match up in the orders shown.
Example 1
Find the complement of each angle.
A. 17◦
B. 24◦
C. 87◦
D. 60◦
C. 103◦
D. 135◦
Solution
A.
B.
C.
D.
90◦ − 17◦
90◦ − 24◦
90◦ − 87◦
90◦ − 60◦
= 73◦
= 66◦
= 3◦
= 30◦
Example 2
Find the supplement of each angle.
A. 156◦
Solution
A.
B.
C.
D.
180◦ − 156◦ = 24◦
180◦ − 95◦ = 85◦
180◦ − 103◦ = 77◦
180◦ − 135◦ = 45◦
B. 95◦
CHAPTER 10 / GEOMETRY
267
• Vertical angles are two nonadjacent angles formed when two lines intersect.
• Two lines are perpendicular when they intersect to form right angles.
This can be written as line a ⊥ line b.
• Two lines in same plane are parallel if they do not intersect. This can be
written as line a || line b.
When two lines intersect, the angles in each pair of vertical angles formed
have equal measures. When two parallel lines are intersected by a third line,
several special types of angles are formed. In Figure 10.3, lines a and b are
parallel; line t is called a transversal.
Figure 10.3
In Figure 10.3, ∠ 1 and ∠5 are a pair of corresponding angles. The other
pairs of corresponding angles are ∠2 and ∠6, ∠3 and ∠7, and ∠4 and ∠8.
Because a || b, ∠1 = ∠5, ∠2 = ∠6, ∠3 = ∠7, and ∠4 = ∠8. Another
pair of special angles is ∠3 and ∠5, which are alternate interior angles,
and so are ∠4 and ∠6. Lines a || b, ∠3 = ∠5, and ∠4 = ∠6. The alternate exterior angles are ∠1 and ∠7, and ∠2 and ∠8, which are equal
since the lines are parallel. Angles 4 and 5 and angles 3 and 6 are interior
angles on the same side of the transversal. Since a || b, ∠4 is supplementary to ∠5 and ∠3 is supplementary to ∠6. Thus, ∠3 + ∠6 = 180◦ and ∠4+
∠5 = 180◦ .
If the measure of ∠1 is 140◦ , you can find the measure of all the other
angles in Figure 10.3.
∠1 and ∠3 are a pair of vertical angles so ∠1 = ∠3, then ∠3 = 140◦ .
∠1 and ∠5 are corresponding angles, ∠1 = ∠5, then ∠5 = 140◦ .
∠5 and ∠7 are vertical angles, ∠5 = ∠7, then ∠7 = 140◦ .
∠3 and ∠6 are interior angles on the same side of the transversal t, so ∠3
supplements ∠6, ∠3 + ∠6 = 180◦ , 140◦ + ∠6 = 180◦ , ∠6 = 180◦ − 140◦ = 40◦ .
∠6 and ∠2 are corresponding angles, so ∠6 = ∠2 and ∠2 = 40◦ .
∠6 and ∠4 are alternate interior angles, so ∠6 = ∠4 and ∠4 = 40◦ .
∠4 and ∠8 are corresponding angles, so ∠4 = ∠8 and ∠8 = 40◦ .
When a || b and ∠1 = 140◦ , ∠3 = ∠5 = ∠7 = 140◦ and ∠2 = ∠4 = ∠6 =
∠8 = 40◦ .
Angles 1 and 2 are adjacent angles, and the noncommon sides form a
straight line; thus, ∠1 and ∠2 are a linear pair and are supplementary.
270
CONQUERING GMAT MATH AND INTEGRATED REASONING
Types of Polygons
Number of Sides
3
4
5
6
8
Name
Triangle
Quadrilateral
Pentagon
Hexagon
Octagon
The sum of the interior angles of a polygon with n sides is S = (n−2)(180◦ ).
Example 3
Find the sum of the interior angles of a polygon with the given number of
sides.
A. 3
B. 4
C. 5
D. 10
E. 15
Solution
A.
B.
C.
D.
E.
n = 3, S = (n − 2)(180◦ ) = (3 − 2)(180◦ ) = 1 × 180◦ = 180◦
n = 4, S = (n − 2)(180◦ ) = (4 − 2)(180◦ ) = 2 × 180◦ = 360◦
n = 5, S = (n − 2)(180◦ ) = (5 − 2)(180◦ ) = 3 × 180◦ = 540◦
◦
n = 10, S = (n − 2)(180◦ ) = (10 − 2)(180◦ ) = 8 × 180◦ = 1,440
◦
n = 15, S = (n − 2)(180◦ ) = (15 − 2)(180◦ ) = 13 × 180◦ = 2,340
• Two polygons are congruent if the angles of the first polygon are equal to
the corresponding angles of the second polygon and the sides of the first
polygon are equal to the corresponding sides of the second polygon.
• Two polygons are similar if the angles of the first polygon are equal to
the corresponding angles of the second polygon and the sides of the first
polygon are proportional to the corresponding sides of the second polygon.
• The altitude, or height, of a polygon is a line segment from one vertex of
a polygon that is perpendicular to the opposite side, or to the opposite side
extended.
To show that sides are of the same length, mark them with the same number of tick marks. To show that angles have the same measure, mark them
with the same number of arcs.
C
D
A
B
Figure 10.5
In Figure 10.5, the quadrilateral ABCD has sides that are marked to show
AD = BC and AB = CD. The angles are marked to show ∠ A = ∠C and
∠ B = ∠ D.
CHAPTER 10 / GEOMETRY
271
PRACTICE PROBLEMS
A. Equiangular
triangle
C. Regular
pentagon
1. Find the sum of the interior angles of each
polygon.
A. Quadrilateral B. Hexagon C. Octagon
B. Equiangular
quadrilateral
D. Regular octagon
2. Find the measure of one angle for each
polygon.
SOLUTIONS
1. A. A quadrilateral has 4 sides, so n = 4.
S = (n − 2)(180◦ ) = (4 − 2)(180◦ ) = 2 ×
(180◦ ) = 360◦
B. A hexagon has 6 sides, so n = 6.
S = (n − 2)(180◦ ) = (6 − 2)(180◦ ) = 4 ×
180◦ = 720◦
C. An octagon has 8 sides, so n = 8.
S = (n − 2)(180◦ ) = (8 − 2)(180◦ ) = 6 ×
180◦ = 1080◦
2. A. An equiangular triangle is a polygon with 3
sides, and the angles are equal.
S = (3 − 2)180◦ = 180◦ , each angle is equal,
so each is one-third of the sum. 180◦ ÷ 3 =
60◦ . Each angle is 60◦ .
B. An equiangular quadrilateral is a polygon
with 4 sides and all four angles are equal.
S = (4 − 2)(180◦ ) = 360◦ , 360◦ ÷ 4 = 90◦
Each angle is 90◦ .
C. A regular pentagon is a polygon with 5
sides that are equal and 5 angles that are
equal.
S = (5 − 2)(180◦ ) = 540◦ , 540◦ ÷ 5 = 108◦
Each angle is 108◦ .
D. A regular octagon is a polygon with 8 equal
sides and 8 equal angles.
S = (8−2)(180◦ ) = 1080◦ , 1080◦ ÷8 = 135◦
Each angle is 135◦ .
TRIANGLES
A triangle is a polygon with three sides. Figure 10.6 shows triangle ABC
(written as ABC ).
C
A
B
Figure 10.6
Triangles can be classified by the number of equal sides they have. If no
two sides have the same length, the triangle is a scalene triangle. If at least
two sides have the same length, it is an isosceles triangle. When all three
sides have the same length, the triangle is an equilateral triangle.
In Figure 10.6, if AC = BC, then ABC is isosceles. The equal sides are
called the legs of the isosceles triangle. Side AB, the unequal side, is called
its base. The angles A and B are the base angles, and ∠C is the vertex angle.
In an isosceles triangle, the base angles are equal, ∠ A = ∠ B.
Triangles can also be classified by the size of their largest angle. If a
triangle has an obtuse angle, the triangle is called an obtuse triangle. If
a triangle has a right angle, the triangle is a right triangle. When all three
angles of the triangle are acute, the triangle is an acute triangle.
272
CONQUERING GMAT MATH AND INTEGRATED REASONING
A
B
C
Figure 10.7
In Figure 10.7, ABC is a right triangle since angle C is marked with the
block . This is the symbol used to indicate right angles. The sides AC and BC
are the legs of the right triangle. Side AB is the hypotenuse. In a right triangle,
the sides that form the right angle are the legs, and the side opposite the right
angle is the hypotenuse.
Properties of Triangles
In a triangle, the sum of the angles is 180◦ . In ABC, if AB > BC > AC, then
∠C > ∠ A > ∠ B. That is, if the sides of a triangle are unequal, then the
angles opposite the angles are unequal in the same order. Also, in ABC, if
∠ A > ∠ B > ∠C, then BC > AC > AB. In any ABC, AB + BC > AC, AC + CB
> AB, and CA + AB > BC; that is, in any triangle the sum of the lengths of
any two sides is greater than the third side.
Example 4
Can the three given lengths be the sides of a triangle?
A. 3, 8, 12
B. 3, 8, 11
C. 3, 8, 10
Solution
A. The first check is to add the two smaller numbers to see if they exceed the
length of the third side. Here 3 + 8 = 11 and 11 < 12. A triangle cannot
have sides of lengths 3, 8, and 12.
B. 3 + 8 = 11 and 11 = 11. A triangle cannot have sides of lengths 3, 8, and
11.
C. 3+8 = 11 and 11 > 10. 3+10 = 13 and 13 > 10. 8+10 = 18 and 18 > 10.
A triangle can have sides of lengths 3, 8, and 10.
Example 5
Can these be the angles of a triangle?
A. 20◦ , 50◦ , 130◦
D. 35◦ , 50◦ , 95◦
B. 30◦ , 60◦ , 90◦
E. 40◦ , 90◦ , 110◦
C. 40◦ , 40◦ , 100◦
F. 61◦ , 60◦ , 50◦
Solution
A. 20◦ + 50◦ + 130◦ = 200◦
200◦ = 180◦
B. 30◦ + 60◦ + 90◦ = 180◦
180◦ = 180◦
These are not the angles of a
triangle.
These are the angles of a
triangle.
CHAPTER 10 / GEOMETRY
273
C. 40◦ + 40◦ + 100◦ = 180◦
180◦ = 180◦
D. 35◦ + 50◦ + 95◦ = 180◦
180◦ = 180◦
E. 40◦ + 90◦ + 110◦ = 240◦
240◦ = 180◦
F. 61◦ + 60◦ + 50◦ = 171◦
171◦ = 180◦
These are the angles of a
triangle.
These are the angles of a
triangle.
These are not the angles of a
triangle.
These are not the angles of a
triangle.
C
A
B
D
Figure 10.8
• An exterior angle of a triangle is created when a side of the triangle is
extended through a vertex. In Figure 10.8, ∠ CBD is an exterior angle for
ABC.
The interior and the exterior angles at the same vertex are supplementary.
In Figure 10.8, ∠ABC supp ∠CBD. The exterior angle at one vertex of a triangle
has the same measure as the sum of the measures of the interior angles at
the other two vertices. In Figure 10.8, ∠CBD = ∠ A + ∠C.
C
A
D
B
Figure 10.9
In Figure 10.9, ABC is isosceles with altitude CD. The base is AB. Line CD
bisects the base AB and the vertex ∠C. Also ∠ ACD = ∠BCD and AD = BD.
Since AC and BC are marked as the equal sides, ∠ A = ∠ B.
An equilateral triangle is also an equiangular triangle. The measure of
each angle of an equilateral triangle is 60◦ . An equilateral triangle is also an
isosceles triangle.
The right triangle in Figure 10.10 has legs a and b and hypotenuse c. In a
right triangle ABC with right angle at C, ∠ A and ∠ B are complementary.
• Pythagorean theorem: In a right triangle, the square of the hypotenuse
is equal to the sum of the squares of the legs. For Figure 10.10, c2 = a2 + b2 .
In a 30◦ –60◦ –90◦ right triangle, the hypotenuse is twice the length of the
side opposite the 30◦ angle, and the side opposite the 60◦ angle is 3 multiplied by the side opposite the 30◦ angle. If ∠ A = 30◦ in Figure 10.10, then
∠ B = 60◦ and c = 2a and b = a 3.
CHAPTER 10 / GEOMETRY
281
PERIMETER AND AREA
Perimeter and area are measurements that are commonly the subject of
GMAT math problems.
• The perimeter of a polygon is the distance around the polygon. Hence,
the perimeter is the sum of the lengths of the sides of the polygon.
If a polygon has n sides, the perimeter P of the polygon is P = s1 + s2 + s3 +
· · · + sn, where each s is the length of a side. The perimeter P of a rectangle
with length l and width w is P = 2l + 2w. The perimeter P of a square with
sides of length s is P = 4s.
• The area of a polygon is the amount of the surface enclosed by the polygon.
The area is expressed as the number of square units of surface inside the
polygon.
Areas of Polygons
Polygon
Triangle
Trapezoid
Parallelogram
Rectangle
Square
Right triangle
Symbols Used in Formula
b = base, h = altitude to base
h = distance between parallel sides a and b
b = base side of parallelogram, altitude to base
l = length, w = width
s = side of square
a and b = legs of right side of triangle
Area
A = 0.5bh
A = 0.5h(a + b)
A = bh
A = lw
A = s2
A = 0.5ab
The area of a general polygon can be found by dividing it into smaller
regions, each of which is a polygon whose area can be found.
B
A
C
D
E
Figure 10.23
In Figure 10.23, polygon ABCDE was divided into three nonoverlapping
regions by drawing all the diagonals from one vertex E. The area of polygon
ABCDE equals the area of triangle I plus the area of triangle II plus the area
of triangle III.
PRACTICE PROBLEMS
1. Find the perimeter of a square with side s.
A. s = 25 cm
D. s = 2 ft 9 in
B. s = 76 mm
C. s = 7.5 ft
2. Find the perimeter of a rectangle with length
l and width w.
A. l = 16 cm, w = 9 cm
B. l = 63 ft, w = 49 ft
C. l = 3.5 m, w = 9.75 m
CHAPTER 10 / GEOMETRY
283
C. A = 0.5h(a + b) = 0.5(18 in)(29 in +
36 in) = 0.5(18 in)(65 in) = 585 in2
D. A = 0.5h(a + b) = 0.5(7 ft)(8 ft + 14 ft) =
0.5(7 ft)(22 ft) = 77 ft2
10. A. A = 0.5ab
330 in2
B. A = 0.5ab
84 cm2
11. Draw FX ⊥ ED, draw CY ⊥ GF, AB + CD =
GX = 6 + 4 = 10
GY = AB = 6, YX = CD = 4, FX = 10 − 7 = 3,
CY = GA − BC = 10 − 4 = 6
Area of ABCDEFG = area of ABYG + area
CDXY + area FXE
Area ABYG = lw = (10)(6) = 60
Area CDXY = lw = (6)(4) = 24
Area FXE = 0.5ab = 0.5(3)(7) = 10.5 ED −
YD = 13 − 6 = 7
Area ABCDEFG = 60+24+10.5 = 94.5 square
units
A = 0.5(11 in)(60 in) =
A = 0.5(7 cm)(24 cm) =
CIRCLES
A circle is the set of all points in a plane that are at a fixed distance, or
radius, from a given point, the center.
•
•
•
•
•
Two circles are concentric if they have the same center.
An arc is a part of a circle.
A semicircle is an arc that is one-half of a circle.
A minor arc is an arc that is less than a semicircle.
A major arc is an arc that is greater than a semicircle.
B
X
D
A
O
E
C
F
Figure 10.25
In circle O (Figure 10.25), AO is a radius, BC is a diameter, DF is a chord,
∠AOB is a central angle, ∠FDE is an inscribed angle, line t is a tangent to
circle O at x, and line ED is a secant. Arc BAC is a semicircle, arc AC is a
minor arc, and FDA is a major arc. In circle O, the region enclosed by arc
AC, radius CO, and AD is called a sector of the circle.
• The circumference of a circle is the distance around the circle. C = πd.
Pi (π) is the ratio of the circumference of a circle to its diameter. π is
1
approximately 3.14 or 3 .
7
• A chord of a circle is a line segment joining two points on the circle.
• A diameter is a chord that goes through the center of the circle.
• A radius is a line segment from the center of a circle to a point on the
circle. A diameter is the same length as two radii.
284
CONQUERING GMAT MATH AND INTEGRATED REASONING
• A secant is a line that intersects a circle in two points.
• A tangent is a line that intersects a circle in exactly one point.
• An inscribed angle is an angle whose sides are chords of a circle, and with
its vertex on the circle.
• A central angle is an angle with its vertex at the center of the circle, and
with sides that are radii.
The measure of a central angle is the same as the degree measure of its arc.
However, the degree measure of an inscribed angle is one-half the measure
of its arc. ∠AOC = arc AC and ∠FDE = 0.5 arc EF. If an angle is inscribed in
a semicircle, it is a right angle.
A diameter separates a circle into two semicircles. A semicircle is one-half
of a circle, so it is one-half of 360◦ , or 180◦ . A minor arc is a part of a circle
that is less than a semicircle, so it has a measure that is less then 180◦ .
The circumference of circle C equals π multiplied by the diameter. C = πd,
or C = 2πr.
The area A of a circle is equal to π multiplied by the square of the radius.
A = πr 2 .
The area A of a sector of a circle is to the area of the circle as the central
angle of the sector is to 360◦ . That is, find the ratio of the central angle of the
sector to 360◦ and multiply that ratio by the area of the circle.
When two chords intersect inside a circle, the product of the segments of
one chord is equal to the product of the segments of the other chord.
For Figure 10.26, AE × EB = CE × ED. Also, ∠ AED = 0.5(arc AD + arc
BC).
C
B
E
O
A
D
Figure 10.26
Example 9
If the diameter of a circle is 30 cm, what is the circumference of the circle?
Solution
C = πd
C = 30π cm
Example 10
If the diameter of a circle is 40 feet, what is the area of the circle?
CHAPTER 10 / GEOMETRY
285
Solution
d = 40 ft, so r = 20 ft and use π = 3.14
A = πr 2
A = (π) (20 ft)2
A = 400π ft2
Example 11
In Figure 10.26, if arc AD = 42◦ , arc AC = 72◦ , and arc BC = 46◦ , what is the
measure of arc BD?
Solution
arc AD + arc AC + arc CD + arc BD = 360◦
42◦ + 72◦ + 46◦ + arc BD = 360◦
160◦ + arc BD = 360◦
arc BD = 200◦
PRACTICE PROBLEMS
1. In Figure 10.27, if AC = 12 and BC = 35, what
is the radius of the circle?
5. In Figure 10.26, if CE = ED, AE = 2, and EB =
6, what is the length of CD?
6. In Figure 10.26, if arc CB = 35◦ and arc AD =
95◦ , what is the measure of ∠AED?
C
7. In Figure 10.26, if arc BC = 125◦ and arc AD =
105◦ , what is the measure of ∠BEC?
A
O
B
8. What is the circumference of a circle whose
radius is 10 cm?
9. What is the area of a circle whose diameter is
10 inches?
Figure 10.27
2. In Figure 10.27, if AC = BC = 10 cm, what is
the length of the diameter of the circle?
3. In Figure 10.27, if AC = AO = 5, what is the
area of triangle ABC?
4. In Figure 10.26, if AE = 12, BE = 7, and CE =
6, what is the length of CD?
10. What is the circumference of a circle whose
area is 49π cm2 ?
11. What is the area of a sector of a circle whose
central angle is 60◦ and the radius is 12 ft?
12. What is the area of a circle whose radius is
12.2 cm?
13. Find the circumference of a circle when the
radius is 18 cm.
SOLUTIONS
1. Triangle ACB is a right triangle. Line segment
AB is the hypotenuse of the right triangle
and the diameter of the circle. 122 + 352 =
144 + 1, 225 = 1, 369 = 372 so AB = 37.
The radius is one-half of the diameter. So r =
0.5(37) = 18.5.
2. AC = BC = 10 cm in right triangle ABC.
AC2 + BC2 = AB2 , 102 + 102 = AB 2 , 100 +
100 = AB 2 , 200 = AB 2 , 200 = AB , AB =
100 2 = 10 2
CHAPTER 10 / GEOMETRY
287
Figure 10.29
The volume V = e3 and the surface area S = 6e2 .
A right circular cylinder has bases that are circles, and the curved surface is perpendicular to the bases. See Figure 10.30. The volume V = πr 2 h and
the surface area S = 2πrh + 2πr 2 , where h is the height of the cylinder and r
is the radius of the base.
Figure 10.30
A pyramid is a solid with one base. See Figure 10.31. The vertices of the
1
base are joined to the apex or vertex of the pyramid. The volume V = bh,
3
where b is the area of the base of the pyramid and h is the height from the
base to the apex.
Figure 10.31
A right circular cone is a solid that has one circular base. See Figure 10.32.
The vertex of the cone is a point on a line perpendicular to the circle at its
1
center. The volume V of the cone = πr 2h, where r is the radius of the base
3
and h is the height of the cone. The start height of a right circular cone is a
line segment from the vertex to a point on the circumference of the base.
CHAPTER 10 / GEOMETRY
4 3
πr = 288π cm3
3
4πr 3 = 864π cm3
r 3 = 216 cm3
r = 6 cm
The radius of the sphere is 6 centimeters.
18. V =
291
20. S = 6e2 = 6(14 mm)2 = 6(196 mm2 ) =
1,176 mm2
The surface area of the cube is 1,176 square
millimeters.
19. S = 2lw + 2lh + 2wh = 2(17 cm)(19 cm) +
2(17 cm)(20 cm) + 2(19 cm)(20 cm) =
646 cm2 + 680 cm2 + 760 cm2 = 2,086 cm2
The surface area of the rectangular solid is
2,086 square centimeters.
COORDINATE GEOMETRY
Coordinate geometry is the branch of geometry that deals with planes on a
coordinate system. Each point in the plane has a unique coordinate (x, y).
The horizontal number line is the x axis, and the vertical number line is the
y axis. The zeros on each number line match up, and the point is called the
origin for the coordinate system. The plane is divided into four parts, which
are called quadrants and are numbered I, II, III, and IV in counterclockwise
order from the upper right. See Figure 10.34.
O
Figure 10.34
Points in quadrant I have both coordinates positive, (positive, positive).
In quadrant II, the x coordinate is negative, but the y coordinate is positive
(negative, positive). Points in quadrant III have both coordinates negative
(negative, negative). Points in quadrant IV have a positive x coordinate and
a negative y coordinate (positive, negative).
To locate a point, such as (5, −3), in the plane, start at the origin and move
5 units in the positive x direction (right). Then move from that location 3
units in the negative y direction (down). The point (−3, 5) is 3 units left from
the origin on the x axis and then 5 units up on the y axis. See Figure 10.35.
292
CONQUERING GMAT MATH AND INTEGRATED REASONING
O
Figure 10.35
Let P = (x1 , y1 ) and Q = (x2 , y2 ). The distance from P to Q, denoted by
d, can be found by the distance formula:
(x2 − x1 )2 + (y2 − y1 )2
d=
The midpoint M of the line segment PQ is:
M=
x1 + x2 y1 + y2
,
2
2
The slope of the line through the points P and Q is
M=
y2 − y1
,
x2 − x1
when x1 = x2
Let l1 with slope m1 and l2 with slope m2 be any two nonvertical lines.
If m1 = m2 , then l1 || l2 .
If m1 × m2 = −1, then l1 ⊥ l2 .
A vertical line does not have slope and has the form x = k, where k is a
real number.
A horizontal line has a slope of zero and has the form y = h, where h is a
real number.
Any two horizontal lines are parallel to each other.
Any two vertical lines are parallel to each other.
Any vertical line is perpendicular to any horizontal line.
If a line crosses the x axis, the point at which it crosses is called the x
intercept. If a line crosses the y axis, the point at which it crosses is called
the y intercept. The x intercept is the value of x when y = 0 for the line. The
y intercept is the value of y when x = 0.
306
CONQUERING GMAT MATH AND INTEGRATED REASONING
11. B Rectangle
The diagonals of a rectangle are always equal.
12. C Octagon
An octagon is an eight-sided polygon.
13. B 900◦
n = 7, 5 = (n − 2)180◦ = (7 − 2)180◦ = (5)180◦ = 900◦
14. C 135◦
n = 8, 5 = (n − 2)180◦ = 6(180◦ ) = 1080◦ , 1080◦ ÷ 8 = 135◦
15. D 3
A triangle has the fewest sides of any polygon, 3.
16. A Trapezoid
The diagonals of a parallelogram always bisect each other. The trapezoid
is the only nonparallelogram listed.
17. E The diagonals bisect each other.
The diagonals of a rhombus bisect the angles but not for all
parallelograms.
18. B Ray
A ray is a part of a line with one endpoint.
19. C Sum of angles is 180◦ .
Adjacent angles are two angles in the same plane that have a common
vertex and a common side, and the common side separates the angles.
20. D Reflex
A reflex angle is an angle greater than 180◦ and less than 360◦ . No
triangle can have a reflex angle in it.
21. A 11◦ , 79◦ , 90◦
Every right triangle has a right angle, 90◦ , so the other two angles must
have a sum of 90◦ , or 11◦ + 79◦ = 90◦ .
22. C 126◦
23◦ + 31◦ = 54◦ , 180◦ − 54◦ = 126◦ , the third angle is 126◦ .
23. D 5, 7, 9
In a triangle, any two sides must exceed the third side. Since 5 + 7 =
12 > 9, and 9 is the longest side, these can be the sides of a triangle.
24. E 80 cm
Since ∠ A > ∠ B > ∠C, BC > AC > AB. So BC must be greater than AC =
50. BC < AB + AC, so BC < 31 + 50 = 81. Thus, 81 > BC > 50 and only
80 is a choice.
25. C 2
An acute triangle has 3 acute angles. A right triangle has 1 right angle
and 2 acute angles. An obtuse triangle has 1 obtuse angle and 2 acute
angles. So every triangle has at least 2 acute angles.
310
CONQUERING GMAT MATH AND INTEGRATED REASONING
7. N is the number of sides of a polygon. What is N?
1. The sum of the interior angles is 1800◦ .
2. The sum of the exterior angles is 360 K is the area of a rectangle, then what is K?
1. The length is 3 times the width.
2. The perimeter is 80y
E
H
A
F
B
D
C
G
Figure 10.37
9. Which pair of points listed in Figure 10.37 are 5 units apart?
A.
B.
C.
D.
E.
A and F
B and G
B and H
F and G
A and B
10. Which pair of points in Figure 10.37 lies on a line with a slope of
−1/2?
A.
B.
C.
D.
E.
C and D
E and F
C and E
B and F
G and H
CHAPTER 10 / GEOMETRY
311
SOLUTIONS
1. C Supplementary angles are two angles that have a sum of 180◦ . This is
only true for 97◦ and 83◦ , so the answer is C.
2. B Since the prefix oct means 8, an octagon is a polygon with eight sides.
3. E An equilateral triangle has all sides and angles equal. Since an obtuse
angle has a measure greater than 90◦ , having three angles this large
would mean that the sum of the angles was greater than 180◦ , which
is the sum of all three angles in a triangle.
4. C A trapezoid is a quadrilateral with exactly one pair of parallel sides,
so it is not a parallelogram. A regular quadrilateral is one with all sides
equal and all angles equal, so it is a square, which is a parallelogram.
The correct answer is C.
5. A The formula for the circumference of a circle is C = π d, so C =
3.14(20 cm) = 62.8 cm. The answer is A.
6. B The bases of prisms, pyramids, and cubes must be polygons. A cone
has only one base. A cylinder can have two bases that can be circles. The
answer is B.
7. A The sum of the interior angles of a polygon is found by using the formula S = (n − 2)(180◦ ), so if S = 1800◦ , then n − 2 = 10 and n is 12. So
statement 1 alone is sufficient.
The sum of the exterior angles of every polygon is 360◦ , so statement
2 is not sufficient. Thus, the answer is A.
8. C Since in statement 1 the length is 3 times the width, the area of the
rectangle is (3w)(w) = 3w2 , but you can't find a specific value. Statement
1 alone is not sufficient.
In statement 2, the perimeter is 80, so l + w = 40 and l = 40 − w. The
area is (40 − w)(w) = 40w − w2 , but you still can't find a specific value.
Statement 2 alone is not sufficient.
If you take both statements together, you have l = 3w and 2l +2w = 80.
Combine to get 2w + 2(3w) = 80. 2w + 6w = 80. 8w = 80. w = 10.
l = 30.
Thus, the area is 300. The answer is C.
9. C Point B is (2,0), and point H is (−2, 3), so BH =
= (−4)2 + 32 = 16 + 9 = 25 = 5.
(−2 − 2)2 + (3 − 0)2
10. D Point B is (2,0) and point F is (−2,2), so the slope of BF =
−1
2
=
.
−4
2
2−0
=
−2 − 2
312
CONQUERING GMAT MATH AND INTEGRATED REASONING
GMAT PRACTICE PROBLEMS
For each question, select the best answer.
1. If an angle is 5 times its complement, what is the measure of the
angle?
A.
B.
C.
D.
E.
15◦
18◦
72◦
75◦
150◦
2. What is the name of a triangle with exactly two sides equal?
A.
B.
C.
D.
E.
Isosceles
Equilateral
Acute
Obtuse
Scalene
3. If one leg of a right triangle is 3 and the hypotenuse is 4, how long
is the other leg?
A.
B.
C.
D.
E.
1
7
3.5
5
7
4. Which set of lengths could be the sides of a triangle?
A.
B.
C.
D.
E.
2, 2, 7
4, 7, 8
−6, −6, −6
6, 7, 0
−2, −8, −9
5. If one angle of a right triangle is 57◦ , what is the measure of the
other angle?
A.
B.
C.
D.
E.
33◦
43◦
57◦
123◦
147◦
6. Which solid has an apex and a polygon for a base?
A.
B.
C.
D.
E.
Cone
Cylinder
Cube
Prism
Pyramid
CHAPTER 10 / GEOMETRY
313
7. If N is a polygon, what is the shape of the polygon?
1. N has 15 diagonals.
2. When N is regular, each exterior angle is 60 X is the length of the radius of a circle, what is X?
1. The radius is one-half the length of the diameter.
2. The ratio of the circumference to the diameter is π9. In a square with an area of 64, what is the length of the diagonal?
A.
B.
C.
D.
E.
4
4 2
8
8 2
16
10. In a 30◦ –60◦ –90◦ right triangle, the hypotenuse is 12. What is the
length of the side opposite the 30◦ angle?
A.
B.
C.
D.
E.
6
4 3
6 2
6 3
24
SECTION IV
GMAT MATH
PRACTICE
The questions that appear in the following pages are designed to be
just like the questions on the Quantitative and Integrated Reasoning
sections of the GMAT. These questions were written specifically for this
book and are not endorsed by the GMAT.
GMAT MATH PRACTICE TESTS
Each of the two GMAT Math Practice Tests is designed to simulate an
actual GMAT Quantitative section. Each one has a 75-minute time limit
and 37 questions spread across the content areas of arithmetic, algebra,
and geometry. The questions are divided into the following categories:
•
•
15 data-sufficiency questions
22 general problem-solving questions
For best results, treat each practice test like the actual examination.
Here is how to take each test under conditions similar to those on the
actual exam:
•
•
•
Find a place where you can work comfortably.
Time yourself and observe the given time limits. Note how many
questions remain, if any, when time runs out.
Become familiar with the directions to the test and the reference
information provided to test takers. You will save time on the actual
test day if you are already familiar with this information.
Once you have completed a practice test, check your answers
against the answer key provided. Then review the solutions to each
problem, paying particular attention to the problems you missed. For
those problems, you may want to go back and reread the corresponding
topic review section in this book.
GMAT INTEGRATED REASONING PRACTICE SET
The GMAT Integrated Reasoning Practice Set includes 12 questions
based closely on the samples so far released by the test makers. According to the GMAC, the real Integrated Reasoning section contains 12
questions and has a time limit of 30 minutes.
315
This page intentionally left blank
GMAT MATH PRACTICE
TEST 1 certain ( A club wants to mix 20 pounds of candy
worth $8.00 per pound with candy worth
$5.00 per pound to reduce the cost of the
mixture to $6.00 per pound. How many
pounds of the $5.00 per pound candy should
be used?
A.
B.
C.
D.
E.
20
30
40
50
60
2. What is the solution of n + 2 > 8?
(1) n is a positive integer less than 10.
(2) n is a one-digit prime number.
317
GMAT MATH PRACTICE TEST 1
16. Carlos worked 40 hours last week, including
4 hours on Sunday. He earns $10.50 an hour
1
regularly and 1 times that on Sunday. How
2
much did he earn last week?
A.
B.
C.
D.
E.
$357
$399
$420
$441
$483
17. If 5w = x + y, then what is the average (arithmetic mean) of w, x, and y in terms of w?
A. 2w
B. 3w
C. w + 2
D.
1
w
3
E.
1
w
2
18. What is the ratio of P to Q?
(1) P is 4 less twice Q.
(2) 4P equals 7Q.
19. If
y
2+y
2
= 5 and = 6, what is the value of
?
x
3
x+3
A. 11
B.
100
17
C.
40
11
D.
20
17
E.
8
11
319
20. Which has the least value?
A.
8
(23 )(52 )
B.
10
(22 )(53 )
C.
28
(23 )(53 )
D.
16
(22 )(52 )
E.
140
(24 )(53 )
21. What is N when N is an integer and
x < N < y?
(1) y − x = 7
(2) x and y are integers.
22. A woman is 4 times as old as her daughter.
In 3 years she will be 3 times as old as
her daughter. How old is the woman now?
A.
B.
C.
D.
E.
3 years
6 years
12 years
24 years
36 years
23. If a triangle has one side of 3 cm and another
side of 4 cm, how long is the third side?
(1) The triangle is a right triangle.
(2) The third side is the longest side.
24. In PQR, PQ is an integer and PQ > 6. What
is PQ?
(1) PR + QR = 8
(2) PQR is equiangular and PR = 10.
25. What is the measure of the supplement
of an angle with a measure of 42◦ ?
A.
B.
C.
D.
E.
318◦
228◦
138◦
48◦
21◦
320
MATH PRACTICE TEST
Questions 26 and 27 are based on the following graph.
26. Based on the graph, the Nile is longer than
the sum of what two rivers?
A.
B.
C.
D.
E.
Congo and Euphrates
Euphrates and Amazon
Congo and Yangtze
Volga and Congo
Volga and Euphrates
27. Based on the graph, the Nile is how much
longer than the Congo?
A.
B.
C.
D.
E.
100 miles
200 miles
600 miles
800 miles
1,200 miles
28. A hat contains 12 cards marked with a star
and 18 unmarked cards. What is the
probability that one card selected at random
will be marked with a star?
A.
B.
C.
D.
E.
2
5
3
5
2
3
3
2
5
2
29. What is the average (arithmetic mean) of
the values 39, 40, 39, 45, 42, 35, 47?
A.
B.
C.
D.
E.
39
41
42
45
47
30. A rectangular room that is 8 meters by 5
meters is to be carpeted using carpet costing
$12.50 per square meter. How much will the
carpet cost?
A.
B.
C.
D.
E.
$40
$100
$162.50
$480
$500
31. A rectangular box that is 7 inches long by
4 inches wide by 3 inches deep is carefully
packed to hold the maximum number of
blocks that are 1 inch by 1 inch by 2 inches.
How many blocks can be packed into
the box?
A.
B.
C.
D.
E.
84 blocks
42 blocks
28 blocks
14 blocks
2 blocks
GMAT MATH PRACTICE TEST 1
321
32. Is n < 0?
34. Based on the graph, what is the ratio of the
proposed spending on national security
compared to health?
1
<0
n
(2) n2 > 0
(1)
33. The original funding to build a housing
development was $1.75 billion. The funding
was increased to $2.5 billion. By what
percentage was the original funding
increased?
A.
B.
C.
D.
E.
38%
43%
52%
65%
71%
A.
B.
C.
D.
E.
11 to 3
3 to 1
33 to 5
1 to 3
3 to 11
35. Based on the graph, approximately what
part of the proposed budget is for health,
education, and interest combined?
A.
B.
1
10
C.
1
5
D.
1
4
E.
Questions 34 and 35 are based on the following
graph.
1
20
1
3
36. What is the value of x3 − y3 ?
(1) x − y = x + 3
(2) x − y = 4 − y
37. If the sale price of a television is $216 after a
10% reduction was made, what was the
regular price of the television?
A.
B.
C.
D.
E.
$21.60
$24.00
$194.40
$237.60
$240.00
STOP
DO NOT GO BACK AND CHECK YOUR WORK.
GMAT MATH PRACTICE
TEST 2 certain( In a graduating class of 240 students, 80%
apply to college. Of the students who apply
to college, 75% actually attend college. How
many students from the graduating class
attend college?
A.
B.
C.
D.
E.
48
144
180
192
372
2. If x + 3y − 3 = 2y − 3x, what is the value of
x?
(1) y2 = 25
(2) y = 5
327
328
MATH PRACTICE TEST
3. The tens digit of a two-digit number is twice
as large as the units digit. If the digits are
reversed, the new number is 36 less than the
original number. What is the original
number?
A.
B.
C.
D.
E.
4
8
42
48
84
4. If n is a member of the set {10, 15, 20, 25, 30,
35, 40}, what is the value of n?
(1) n is a multiple of 3.
(2) n is a multiple of 2.
5. If 82x+4 = 44x−3 , what is the value of x?
A.
B.
C.
D.
E.
1
3
6
9
10
6. If n is a positive integer, is n + 1 a prime?
(1) n is a prime number.
(2) n is even.
7. If x2 = 1, which is equal to
x
2
x
?
−
+ 2
x+1 x−1 x −1
9. If P and Q are positive integers, which
CANNOT be the greatest common divisor of
P and Q?
A.
B.
C.
D.
E.
10. In a square, the length of a side is s and the
diagonal is d. What is the perimeter of the
square?
(1) The area is 25 cm2 .
(2) The diagonal is 10 2 in.
11. Is x > 0?
(1) x3 > 0
(2) −3x < x
12. A clothes dryer has a sale price of $203.15
after a 15% reduction. What is the regular
price of the dryer?
A.
B.
C.
D.
E.
−x + 3
x+1
B.
2
x+1
C.
−2
x+1
A.
D.
2
x−1
B.
E.
−2
x−1
C.
(1) x is a multiple of 4.
(2) x is a multiple of 15.
14. Which fraction is equal to the decimal
0.0125?
D.
(1) Twice x equals 4 times y.
(2) xy is the square of a positive integer.
$274.85
$239
$233.62
$172.68
$35.85
13. x is a multiple of 12 that is less than 100.
What is x?
A.
8. If x and y are positive integers, what is the
value of x?
P+Q
PQ
P
Q
1
E.
1
4
1
8
1
16
1
32
1
80
GMAT MATH PRACTICE TEST 2
15. If x, y, and z are positive integers with the
ratio 2 : 4 : 6, what is the value of x?
(1) x + y + z = 60
(2) y + z = 50
16. What value is equal to
A.
B.
C.
D.
E.
3/8
?
3/4 + 2/3
3
4
15
28
9
34
3
16
21
20
17. If x = 1, which expression is equal to
x(x − 1) − 2(x + 1) + 3(x + 5)
?
x−1
A.
B.
C.
D.
E.
x2 + 13
x−1
x2 + 17
x−1
x2 + x + 12
x−1
x2 + x + 5
x−1
13
x−1
18. If t = 0, is r greater than zero?
(1) r − t = 8
(2) −r t = 8
19. What is the sum of the prime factors of 570?
A.
B.
C.
D.
E.
29
30
65
66
67
20. Which number is divisible by 2, 3, 4, and 6
but is not divisible by 5?
A.
B.
C.
D.
E.
138
644
1,020
1,428
4,620
329
21. If l is the length of a rectangle and w is the
width of the rectangle, what is the perimeter
of the rectangle?
(1) l + 2w = 80
(2) l + w = 40
22. If x is an integer and y = 7x + 5, which of
the following CANNOT be a divisor of y?
A.
B.
C.
D.
E.
10
11
12
13
14
23. At a discount wholesale store, a microwave
is priced at a x% discount off the original
warehouse price. Later, during a sale, the
storeowner offers to sell the microwave at
y% off the regular discount price. What was
the original warehouse price?
(1) x = 15
(2) x + y = 20
24. One pump drains one-half of a pond in 3
hours, and then a second pump starts
draining the pond. The two pumps working
together finish emptying the pond in
one-half hour. How long would it take the
second pump to drain the pond if it had to
do the job alone?
A.
B.
C.
D.
E.
1 hour
1.2 hours
3 hours
5 hours
6 hours
25. Is p2 an even integer?
(1) p is an even integer.
√
(2) p is an even number.
26. What is the value of
A.
B.
C.
D.
E.
(−1.6)(1.5) − (1.2)(3.5)
?
30
0.24
0.22
0.06
−0.06
−0.22
27. In a rectangle, the length is l, the width is w,
and the diagonal is d. What is the area?
(1) d = 13 and w = 5
(2) d = 2l
330
MATH PRACTICE TEST
28. Which is equal to 3 −
A.
B.
C.
D.
E.
2 3+
2 ?
11 − 6 2
7−6 2
5
7
11
29. If w = 3x − 4y2 , what is w?
(1) x = 324
(2) y2 > 4
30. The perimeter of a rectangular garden is 80
feet, and the area of the garden is 391
square feet. What is the length of the shorter
side of the garden?
A.
B.
C.
D.
E.
17 feet
23 feet
34 feet
40 feet
46 feet
31. The average of x, y, and z is 40. What is x?
(1) y = 2z
(2) y + z = 75
Questions 32 and 33 are based on the following
graph.
32. Based on the graph, what would be spent on
clothing out of a budget of $2,500?
A.
B.
C.
D.
E.
$250
$300
$375
$500
$625
33. Based on the graph, how much more is
being spent on food than on clothing?
A.
B.
C.
D.
E.
5%
8%
10%
13%
15%
34. Which is equal to
A. 2
B.
C.
D.
E.
5
5 2
10
25
50
25 + 25?
GMAT MATH PRACTICE TEST 2
331
Questions 35 and 36 are based on the following graph.
37. What is the product of the roots of
x2 − 15x + 36?
35. Based on the graph, how many students
were late during August, September, and
October combined?
A.
B.
C.
D.
E.
A.
B.
C.
D.
E.
14
12
10
8
6
36
15
9
−15
−36
36. Based on the graph, between which two
consecutive months was the greatest change
in lateness?
A.
B.
C.
D.
E.
March to April
October to November
May to June
December to January
January to February
STOP
DO NOT GO BACK AND CHECK YOUR WORK.
GMAT INTEGRATED
REASONING PRACTICE SET
DIRECTIONS: Select the best answer or answers for the questions
below. You may use a calculator for this section of the test only.
(Note: On the actual test, you will be provided with an online
calculator. You will not be permitted to bring your own calculator
to the test.)
1. Clark University currently enrolls 8,800 students per year. Talbot
College currently enrolls 15,100 students per year. The numbers of
students enrolled by both schools are increasing each year at a constant rate. If each of these schools continues to enroll an increased
number of students annually at its constant rate, in seven years
both schools will enroll the same number of students for the first
time. Each year after seven years, Clark University will enroll more
students per year than Talbot College.
In the table below, identify the rates of increase, in annual students
enrolled, for each school that together meet the enrollment forecasts
described above. Select only one option in each column.
Clark University
Talbot College
Rate of increase
(enrollments per year)
100
250
400
900
1,200
1,800
2.
The graph shown below is a scatter plot with 48 points, each representing the number of radio ads per hour run by a company each day
over 48 months, and the corresponding monthly sales revenue that
the company earned. The sales revenues, measured in thousands of
dollars, were tallied on the last day of each month that the ads were
run. The solid line is the regression line, and the dashed line is the
line through the points (1, $20,000) and (6, $50,000). Select the best
answer to fill in the blanks in each of the statements below based on
the data shown in the graph.
337
338
GMAT INTEGRATED REASONING PRACTICE SET
The number of months in which the company generated more than
$50,000 of revenue is closest to ________ percent of 48.
A.
B.
C.
D.
E.
0
10
15
17
25
The slope of the regression line is __________ the slope of the dashed
line.
A. less than
B. greater than
C. equal to
The relationship between the radio ads run per hour and sales revenue
is __________.
A. negative
B. zero
C. positive
3.
The table below gives information on the total inventory in 2010 and
the total items sold by an international furniture manufacturer over
a three-year period, from 2008 to 2010. The 19 furniture items were
included in the table because they fall among the top 25 items produced by the company in terms of both total inventory and total
items sold. In addition to listing the total inventory and total number
sold for each furniture type, the table also gives the percent increase
or decrease over the 2009 inventory and 2005–2007 sales numbers
and the rank of each furniture type for total inventory and total
items sold.
[Note: On the actual exam, you will have the ability to sort the table by
any of its columns. Columns can be sorted in ascending order only. The
table is shown below sorted in different ways.]
342
GMAT INTEGRATED REASONING PRACTICE SET
August 10, 10:04 a.m.
To date we have received 40 computers. We need 100 computers donated
to meet our goal for the new training lab. We have requested help from
all of the students' families, so we should invite local businesses as well.
In all of our past drives, including this one so far, we have received donations from about 20 percent of those who received requests. (Of course,
we might always receive more or less than that average, so we should
consider the possibilities of not meeting the goal or overspending the
budget for the thank-you event.) Each individual or organization donating a computer will receive two invitations to our thank-you event to
celebrate the opening of the lab. Refreshments and supplies for the
event are expected to run $20 per person. What is the total budget for
the allow
us to accommodate 2 attendees for each of the 100 computers donated.
The budget is firm, so we should take care to ensure that the event
costs stay within this amount. Although we do not have resources to
extend the budget, if necessary we could determine ways to reduce
the cost per person if we receive more donations than the original goal
amount.
Suppose that the donations coordinator requests computer donations
from 400 local businesses. If all the information in the three e-mails is
accurate, the number of attendees that will be invited to participate in
the thank-you event is closest to:
135
160
200
240
300
5.
Company X currently owes $200,000 on a business loan. Company Y currently owes $410,000 on a business loan. Both companies repay their loans at a fixed dollar amount per year that
includes both interest and principle. If each company repays its
loan at its fixed dollar amount per year, in three years the companies will owe the same amount. After three years, Company Y
will owe less than Company X until the loans are paid off.
In the table below, identify the fixed dollar annual repayment
amounts for each company that together meet the repayment projections given above. Select only one option in each column.
GMAT INTEGRATED REASONING PRACTICE SET
Company X
343
Company Y
Repayment amount
(dollars per year)
5,000
10,000
25,000
40,000
67,500
80,000
6.
The graph above is a bar graph with seven bars, each representing the
number of complaints received by a telephone company from its new
customers. The customers received follow-up calls anywhere from one to
seven weeks after placing their orders. The customers were grouped by
follow-up call timing, and the number of complaints was recorded for each
group over a one-year period. Select the best answer to fill in the blanks in
each of the statements below based on the data shown in the graph.
The relationship between the number of complaints and the number of
weeks before the call is __________.
A. zero
B. negative
C. positive
The number of complaints made by customers who received a followup call one week after placing their orders is closest to ________ percent
of the number of complaints made by customers who received a followup call seven weeks after placing their orders.
A.
B.
C.
D.
E.
0
15
30
50
75
GMAT INTEGRATED REASONING PRACTICE SET
347
E-mail 1—E-mail from marketing director to research associate
November 12, 1:15 p.m.
What was our return on investment last year from ads placed in various
media? I am developing our marketing budget for next year and would like
to determine whether Internet advertising should be continued as extensively as we have in past years. Also, are there data to show how returns
from various advertising campaigns differ from quarter to quarter?
E-mail 2—E-mail from research associate in response to marketing director's November 12, 1:15 p.m. message
November 12, 1:35 p.m.
Attached is a graph that shows the return on investment from last year's
advertising campaigns. Typically we do not repeat campaigns in media
that return less than 20 percent in any quarter. The return on investment
for Internet ads was strong throughout the year, which supports continuing Internet advertising as we have in the past.
Graph 1—Attached to the research associate's November 12, 1:35 p.m.
message
Consider each of the following statements. Does the information in the
three sources support the inference as stated?
Yes No
The research associate does not believe that the 2010
advertising campaign was successful as a whole.
The marketing director plans to recommend against continuing to use Internet advertising as extensively as the
company has in past years.
To increase total advertising returns, the company's Internet campaign might be expanded during 3rd quarter 2011.
Based on the company's typical practices, radio advertising
should be discontinued in 2011.
348
GMAT INTEGRATED REASONING PRACTICE SET
9. Suppose that the 2011 returns from campaigns in all advertising
media remain the same as those received in 2010. If all the information in the three sources is accurate and the company spends
$100,000 on television advertising during the first quarter of 2011,
the returns received from television advertising during this quarter
will be closest to:
$10,000
$15,500
$25,000
$30,000
$30,750
10. Johnston Booksellers currently generates $50,000 in annual
sales revenue. Its competitor, Trevor Books, currently generates
$490,000 in annual sales revenue. The sales revenue generated by
Johnston Booksellers is increasing each year at a constant rate,
while the sales revenue generated by Trevor Books is decreasing
each year at a constant rate. If Johnston continues to generate
an increased amount of revenue annually at its constant rate
and Trevor continues to generate a decreased amount of revenue
annually at its constant rate, in four years the bookstores will
earn the same amount of annual sales revenue. After the four-year
mark, Johnston Booksellers will receive more sales revenue per
year than Trevor Books.
In the table below, identify the rates of increase or decrease, in
annual revenue earned, for each bookstore that together meet the
revenue forecasts described above. Select only one option in each
column.
Johnston Booksellers
Trevor Books
Rate of increase or decrease
(dollars per year)
10,000
20,000
50,000
60,000
90,000
180,000
11. The table below gives information from a gallery management database regarding the total number of exhibits and the total number
of art pieces maintained for 18 artists from around the world. The
artists in the table are among the top 30 artists internationally
in terms of both total numbers of exhibits and total pieces in the
company's collection. The table ranks the artists according to their
total exhibits and total pieces in the art collection.
[Note: On the real exam, students will have the ability to sort the table by
any of its columns. Columns can be sorted in ascending order only. The
table is shown below sorted in different ways.]
350
GMAT INTEGRATED REASONING PRACTICE SET
Sorted by Rank of Pieces in the Collection (Column 7)
Artist
Exhibits
Pieces in Collection
City
State
Zip Code
Number
% Change
Rank
Number
Des Moines
USA
A743
191
12
103
1
New York
USA
A219
190
13
97
2
Barcelona
Spain
A648
200
4
95
3
London
England
A347
169
20
95
4
London
England
A629
191
11
94
6
Athens
Greece
A724
200
6
92
7
Rome
Italy
B657
188
14
92
9
Paris
France
A935
243
1
90
10
Paris
France
A221
170
18
89
11
Milan
Italy
B309
201
3
87
13
Montreal
Canada
B221
193
10
86
14
New York
USA
B607
186
15
82
15
Chicago
USA
B253
200
7
78
17
Boston
USA
A985
207
2
73
18
Lyons
France
A684
194
9
71
19
Florence
Italy
A223
185
16
67
20
Berlin
Germany
B557
173
17
65
21
Seattle
USA
B681
196
8
59
27
Review each of the statements below. Based on information provided in
the table, indicate whether the statement is true or false.
True False
The top-ranking artists, in terms of both total exhibits
and total number of pieces in the collection, live in the
United States.
The lowest ranking artist, in terms of number of exhibits,
participated in approximately 70 percent of as many
exhibits as did the top-ranking artist in this category.
All of the codes for the top five ranking artists, in terms
of both number of exhibits and pieces in the collection,
begin with the letter A.
Exactly 15 percent of the 18 artists in the table participated in 200 exhibits.
12. The graph shown is a scatter plot with 60 points, each representing the number of quality assurance inspections conducted on
products manufactured at 60 different factories, and the corresponding numbers of product recalls experienced by each factory.
Each factory conducted a consistent number of quality assurance
inspections on all products produced during a one-year period,
and the number of product recalls was measured over that same
period. The solid line is the regression line, and the dashed line is
the line through the points (1, 1) and (7, 5). Select the best answer
GMAT INTEGRATED REASONING PRACTICE SET
351
to fill in the blanks in each of the statements below based on the
data shown in the graph.
The slope of the regression line is __________.
A. positive
B. negative
C. zero
The number of products that received more than seven quality
assurance inspections is closest to ________ percent of 60.
A.
B.
C.
D.
E.
0
10
20
35
50
The relationship between quality assurance inspections and the
number of product recalls is __________.
A. zero
B. negative
C. positive
This page intentionally left blank
SOLUTIONS
1. The correct answer is 1,800 enrollments per year for Clark University
and 900 enrollments per year for Talbot College.
Clark University
Talbot College
Rate of increase
(enrollments per year)
100
250
400
☼
900
1,200
☼
1,800
If Clark University increases its enrollment by 1,800 students per year,
in seven years it will enroll 21,400 students per year. If Talbot College
increases its enrollment by 900 students per year, in seven years it will
enroll 21,400 students per year as well. After the seven-year mark,
Clark University will enroll more students per year than Talbot College.
2. The number of months in which the company generated more than
$50,000 of revenue is closest to 0 percent of 48. According to the
graph, only one month generated $50,000 in revenue; no months
generated more than $50,000. The slope of the regression line is
less than the slope of the dashed line. The regression line slants
downward from left to right, so it has a negative slope. The dotted
line slants upward from left to right, so it has a positive slope. The
relationship between the radio ads run per hour and sales revenue
is negative. As the number of ads per hour increases, the monthly
revenue decreases.
3. The correct answers are shown below.
True False
☼
Exactly 50 percent of the furniture items that experienced a decrease in both total inventory and total
items sold are Red mahogany.
☼
The furniture type experiencing the greatest percentage increase in total inventory from 2009 to 2010
also experienced the greatest percentage decrease in
the total number of items sold.
☼
The furniture type with the highest rank based on
total inventory is the same as the type with the highest rank based on total items sold.
☼
The total inventory of Cherry Red mahogany Tables
in 2009 was approximately 206,400.
353
354
GMAT INTEGRATED REASONING PRACTICE SET
There were 6 furniture items that experienced a decrease in total inventory. Of these, 3 (exactly 50 percent) are Red mahogany. However, all
19 furniture items experienced a decrease in total items sold. Only 6 of
these (32 percent) are Red mahogany, so the statement is false.
The furniture type experiencing the greatest percentage increase in
total inventory from 2009 to 2010 is the Oak Natural Desk, at 9.7
percent. This same item also experienced the greatest percentage
decrease in the total number of items sold (–8.1 percent).
The furniture type with the highest rank based on total inventory
is the Oak Natural Chair. The furniture type with the highest rank
based on total items sold is the Maple Natural Chair.
The total inventory of Cherry Red mahogany Tables in 2010 was
200,407. This furniture type experienced a 3.1 percent decrease in
total inventory, or a drop of about 6,000 items. Its inventory in 2009
was therefore approximately 206,400.
4. The correct answer is 240 invitees.
In E-mail #2, the donations coordinator states that donations are
usually received from about 20 percent of those who receive requests.
The computer drive had already received 40 donations, and each
donor would receive 2 invitations to the thank-you event, for a total
of 80 invitees. If requests were extended to 400 local businesses and
20 percent of those made a donation, the drive would receive 80
computers from businesses. That would add 160 invitations to the
thank-you event, for a total of 240 invitees.
5. The correct answer is $10,000 per year for Company X and $80,000
per year for Company Y.
Company X
Company Y
Repayment amount
(dollars per year)
5,000
☼
10,000
25,000
40,000
67,500
☼
80,000
If Company X repays its loan at a rate of $10,000 per year, in three
years it will owe $170,000. If Company Y repays its loan at a rate of
$80,000 per year, in three years it will also owe $170,000. After the
first three years, Company X will owe more on its loan than Company Y, until the loans are paid off.
6. The relationship between the number of complaints and the number
of weeks before the call is positive. As the number of weeks increases,
the number of complaints also increases. The number of complaints
made by customers who received a follow-up call one week after
placing their orders is closest to 50 percent of the number of complaints made by customers who received a follow-up call seven weeks
GMAT INTEGRATED REASONING PRACTICE SET
355
after placing their orders. Approximately 30 complaints were made
by customers who received a follow-up call one week after placing
their orders, and nearly 60 complaints were made by customers who
received a call after seven weeks.
If the company wishes to limit its complaints to 40 or fewer per
year, it should make follow-up calls no more than two weeks after
customers place their orders. If the company waits three weeks or
longer, it is likely to receive more than 40 complaints per year.
7. The correct answers are shown below.
True False
☼
The tour city with the median rank based on total
tickets sold is the same as the city with the lowest
rank based on total sales revenue.
☼
The total number of tickets sold in Providence, RI, in
2009 was approximately 12,000.
☼
The tour city experiencing the greatest percentage
increase in total sales revenue from 2009 to 2010
made more sales revenue in 2010 than any other city
on the tour.
☼
Approximately 60 percent of the tour cities in the
table experienced a percentage decrease in total tickets sold from 2009 to 2010.
The tour city with the median rank based on total tickets sold is
Denver, CO. Denver has the highest rank based on total sales
revenue, so the statement is false.
The total number of tickets sold in Providence, RI, in 2010 was
10,004. Providence experienced a 2.1 percent increase in ticket sales
from 2009 to 2010. So, the number of tickets sold in 2009 would be
less than 10,000, not greater than 10,000.
Denver, CO, experienced the greatest percentage increase in total
sales revenue from 2009 to 2010, at 9.2 percent. It also made more
sales revenue in 2010 than any other city on the tour, at $688,095.
None of the tour cities in the table experienced a percentage decrease
in total tickets sold from 2009 to 2010. All the cities experienced a
percentage increase in total tickets sold over this period.
8. The correct answers are shown below.
Yes
No
☼
The research associate does not believe that the 2010
advertising campaign was successful as a whole.
☼
The marketing director plans to recommend against
continuing to Internet advertising as extensively as
the company has in past years.
☼
To increase total advertising returns, the company's
Internet campaign might be expanded during 3rd
quarter 2011.
☼
Based on the company's typical practices, radio
advertising should be discontinued in 2011.
356
GMAT INTEGRATED REASONING PRACTICE SET
The first inference is not supported by the information in the three
sources. The research associate does not evaluate the advertising
campaign as a whole in E-mail 2.
The marketing director does not imply an intent to recommend
reducing Internet advertising. In E-mail 1, the director requests
data to determine whether Internet advertising should be continued
as extensively as it has been in past years.
The graph shows that returns from Internet advertising were
strongest during 3rd quarter 2010. To increase total advertising
returns, the company might expand its Internet advertising during
this quarter in order to capitalize on the potential for additional
gains.
In E-mail 2, the research associate explains that the company typically does not repeat campaigns in media that return less than 20
percent in any quarter. Radio advertising returned less than 20 percent in every quarter of 2010, so based on the company's practices,
the radio campaign should be discontinued.
9. The correct answer is $25,000.
The graph shows that television advertising returned 25 percent on
funds invested in first quarter 2010. If the return remains the same
for 2011, the $100,000 invested in television advertising would produce gains of 25 percent, or $25,000.
10. The correct answer is $90,000 per year for Johnston Booksellers
and $20,000 per year for Trevor Books.
Johnston
Booksellers
Trevor
Books
Rate of increase or decrease
(dollars per year)
10,000
☼
20,000
50,000
60,000
☼
90,000
170,000
If Johnston Booksellers increases its sales revenue by $90,000 per year,
in four years it will earn $410,000 in annual revenue. If Trevor Books
decreases its sales revenue by $20,000 per year, in four years it will also
earn $410,000 in annual revenue. After the fourth year, Johnston Booksellers will generate more sales revenue each year than Trevor Books.
11. The correct answers are shown below.
True False
☼
The top-ranking artists, in terms of both total exhibits and total number of pieces in the collection, are
from the United States.
☼
The lowest ranking artist, in terms of number of exhibits, participated in approximately 70 percent of as many
exhibits as did the top-ranking artist in this category.
GMAT INTEGRATED REASONING PRACTICE SET
☼
☼
357
All the codes for the top five ranking artists, in terms
of both pieces in the collection and number of exhibits, begin with the letter A.
Exactly 15 percent of the 18 artists in the table participated in 200 exhibits.
The top-ranking artist for total exhibits is from Paris, France.
The lowest-ranking artist, in terms of number of exhibits, participated in 169 exhibits. The highest-ranking artist in this category
participated in 243 exhibits. The lowest-ranking artist participated
in 69.5 percent as many exhibits as did the top-ranking artist, or
approximately 70 percent.
The third-ranking artist for exhibits is code B309, so the third statement is false.
A total of 3 artists participated in 200 exhibits. There are 18 artists
listed in the table, so 3 artists represents 16.67 percent of the total,
not 15 percent.
12. The slope of the regression line is zero. The solid line is flat, slanting neither upwards nor downwards. This means that the line has
no change; its slope is therefore zero. The number of products that
received more than seven quality assurance inspections is closest
to 10 percent of 60. The graph shows that four products received
more than seven quality assurance inspections. These four products are 0.067 percent of the total, or 6.7 percent. The relationship
between quality assurance inspections and the number of product
recalls is zero. The number of product recalls does not appear to be
affected by the number of quality assurance inspections. As inspections increase, product recalls stay about the same. |
Please Note: Prerequisite Algebra 1
This set includes Geometry Grade 10 PACEs 1109-1120, which cover:
To reason logically and systematically.
Properties and theorems and how to use them for solving problems-converse and construction, parallelograms, tangency, exterior angle; Hinge, Pythagorean, and regular polygon theorems.
Types of shapes, angles, arcs, and chords, and learns to find the circumference and area |
Lacey StatisticsAs with all of my subjects, I prefer the old school approach to learning math -- working problems -- more is better. In addition, I have lots of handouts and notes that synthesize pertinent information germane to a specific topic. Algebra 2 is an important stepping stone to higher levels of mathematics use the book Statistics, a First Course, Seventh Edition. It is an older version for the course, but the information is still accurate and up to date. I would be happy to help you finish the course with a good grade |
Consumer Math
Consumer Math Series. Produced by Pearl Production, published by Steck-Vaughn, 2011. For adolescents and adults.
Each book in this series covers basic math concepts before exploring more specific topics. You will find clear explanations with ample practice exercises to fix them in the brain. There are also pretests, section reviews, and posttests, practice forms and charts, and 120 worksheet pages, along with answers in the back. Students will also find money tips, problem-solving strategies, group projects, and a glossary to help with vocabulary in each book. Each book is 160 pages.
The comprehensive lessons include a wide variety of exercises and activities to interest and motivate students. They will find problems that they will encounter in real life, and exercises can easily be applied in the real world.
Extension activities offer more challenging problems on the lesson's theme There are also many calculator activities. A "Think About It" section presents real-life problems which will motivate students to participate in class discussions. Everything in the lessons is connected to situations that students realize they will actually be in someday and will need to know how to handle. Here are the individual titles, each covering a very specific area of consumer math.
The Mathematics of Housing and Taxes. Math Concepts covered are whole numbers, fractions, decimals, and percents; mean, median, and mode; basic operations on a calculator; mental computing; and estimating. Taxes topics include reporting income, federal income taxes, using tax form 1040EZ, itemized deductions, and state and city income taxes. Housing topics include renting an apartment, buying a house or condominium, getting a mortgage, real estate taxes, homeowner's insurance, utilities, and decorating and remodeling. BTH-5780. $16.19
The Mathematics of Work: Math Concepts covered are whole numbers, fractions, decimals, and percents; mean, median, and mode; basic operations on a calculator; mental computing; and estimating. Part-time and Summer Jobs topics include finding a job, computing pay, tips, and Social Security. Full-time Work topics are hourly wages and overtime pay; time sheets and time cards; salary; piecework; commission; health insurance; and life insurance. BTH-5783. $16.19
Financial Math Seriesby Steck-Vaughn Staff. Harcourt
Achieve, 2008. This two-book series gives students the financial skills they
need to think critically and make sound financial decisions. These books help
students achieve personal financial literacy, money management skills, and the
ability to solve real-life problems. These books will easily serve as texts for
consumer math.
Life
Skills for Today's World: Money and Consumersby Vivian Bernstein. This was
designed for adult education at reading levels 4-6. But it is also all I have to replace
the out-of-print Consumer Math book that used to be available from Steck-Vaughn. It's a
consumable workbook with the following chapter headings: Using Banking Services; Using a
Checking Account; Credit and Interest (and Completing a Credit Card Application); Paying
Bills; Being a Smart Shopper; Buying Insurance; and Owning a Car. The book is illustrated
with charts and B/W photos. It also contains a glossary, index, and answer key. 96 pages.
BTH-519. $16.38-D Click on image to enlarge it.
We receive a small commission from some links to third party sites with whom we have an affiliate relationship.
We're in the Pacific Time Zone Our normal office hours when we answer the phone are 10: 30 AM - 4 PM Monday -
Friday. We answer email and ship at other times, but normally take Sundays off. You can fax your purchase orders to 805-237-8639, but it's good to follow up with a phone call in a couple of days to make sure we got it, if you did not receive a delivery confirmation email. Please include an email address with your purchase order in case we have questions and so we can send your delivery confirmation number to you. You can also attach a signed purchase order to an email.
We prefer you ask your questions about products
and availability by
e-mail (barbsbooks@sbcglobal.net), since that gives us more time to prepare a thoughtful answer and
keeps us from playing phone tag. Please remember to put our email address in
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unless you also leave a phone number. We cannot usually tell you immediately over the phone if a
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have to call you back or email you anyway. We do check our e-mail frequently
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Hi I have been programming for a while now and I'm about to leave my job to study more. The course I want to study says it requires year 12 / sixth form mathematics or I can take an exam there when I apply. I never did any year 12 so I have no idea what to expect. My maths have been fine for what I need so far but I'm worried that I'll be given all sorts of unfair questions.
Will the math questions likely be programming related for a programming course?
What should I study?
Edit:
Its in New Zealand but I think it would probably be similar to UK sixth formGiven the reference to Y12/Sixth Form, I'm going to assume UK AS level maths, which covers trigonometry, introductory calculus, algebra and geometry.
Your course requirements will almost certainly vary, but in my experience, you'll probably find discrete maths topics to be most relevant to computer science/computing related fields - that is, set theory, prepositional/predicate logic etc. The chances are, if you're at all competent at programming, most of that will come to you fairly easily.
That's not to say that the other stuff is completely useless - particularly coupled with the applied mathematics components (mechanics, statistics, etc.), it's a great way to build your thinking in a different direction.
Programming related maths questions can be broken down pretty much into purely programming related and domain related. On the programming side many algorithms are based on discrete methods, and computer graphics deal with a lot of geometry and trig. Once you start hitting domain specific stuff, you can include most other mathematics, including lots of statistics and numerical methods.
What should I study?
Have a look at the first year cirriculum and talk to the tutors, this should give you a fair idea.
I doubt the math questions will be programming related, since that is a seperate and on the whole much harder area. It is also an area where you really have to know the underlying math before you can tackle.
For example a question like "solve this set of linear equations" is a lot easier to tackle than "write a program that will solve linear equations of this type in a reasonably effective manner". And if you can't do the first you have no real chance of doing the second.
I'm sure you can ask for a copy of some old exams to see what kind of questions they'll be asking. Take a look at those and try to guage if you have the skills necessary.
My CS degree did not have a maths pre-req, but I'm very glad I had A level maths, and I wish my maths syllabus had covered some stuff it hadn't. Those who didn't have A level maths struggled with some of the courses.
Having said that, it's nothing you can't catch up on with some self-study. Working through an A-level revision guide, maybe asking someone to help you with any parts that don't make sense, would quickly fix things.
Maths topics I encountered on my course include
Trigonometry: mostly in graphics work
Calculus: comes up a fair amount, especially in simulation software
Matrices: used a lot in bitmap graphics. Also came up in genetic algorithms and neural nets.
Stats: I needed this for analysing the success of genetic algorithms. It will be useful elsewhere.
(The one I wish I'd been taught in school is matrices. Why WJEC didn't include this in their syllabus at that time is beyond me.)
Further comment, re the exam. I strongly suggest you ask the entrance officer about this.
In my experience, as long as you're motivated and intelligent, universities don't want to turn you away. It's likely that they'd be happy to give you a list of maths subjects to read up on before the exam. |
Mathematics for the Clinical LaboratoryMathematics for the Clinical Laboratory gives you a hands-on, in-depth understanding that goes beyond rote memorization. You'll learn calculation skills that will stay with you throughout your education and into your career. This comprehensive, practical text shows you how to perform the clinical calculations used in each area of the laboratory to ensure accurate results for patients. These features make it all possible: a general introduction to each mathematical calculation, including its theoretical and practical applications; a different color highlighting the progression from calculation to solution to demonstrate how to work through a problem; step-by-step examples of mathematical calculations that can be used as templates to ensure your calculations are correct every time; separate chapters on each area of the laboratory, including immunohematology and microbiology, as well as chapters on statistical concepts; coverage of commonly used clinical laboratory calculations such as molarity, dilutions, and concentration for a better understanding of clinical laboratory mathematics; practice problems at the end of each chapter to further reinforce math concepts; and an answer key that provides detailed rationales for all solutions to calculations. |
Fundamentals of Mathematical Physics by Edgar A. Kraut Indispensable for students of modern physics, this text provides the necessary background in mathematics to study the concepts of electromagnetic theory and quantum mechanics. 1967 edition.
Techniques and Applications of Path Integration by L. S. Schulman Suitable for advanced undergraduates and graduate students, this text develops the techniques of path integration and deals with applications, covering a host of illustrative examples. 26 figures. 1981 edition.
A Bridge to Advanced Mathematics by Dennis Sentilles This helpful "bridge" book offers students the foundations they need to understand advanced mathematics. The two-part treatment provides basic tools and covers sets, relations, functions, mathematical proofs and reasoning, more. 1975 edition.
Elasticity by Robert William Soutas-Little A comprehensive survey of the methods and theories of linear elasticity, this three-part introductory treatment covers general theory, two-dimensional elasticity, and three-dimensional elasticity. Ideal text for a two-course sequence on elasticity. 1984 edition.
Mathematical Tools for Physics by James Nearing Encouraging students' development of intuition, this original work begins with a review of basic mathematics and advances to infinite series, complex algebra, differential equations, Fourier series, and more. 2010 editionA Survey of Industrial Mathematics by Charles R. MacCluer Students learn how to solve problems they'll encounter in their professional lives with this concise single-volume treatment. It employs MATLAB and other strategies to explore typical industrial problems. 2000 |
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