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If $64$ is divided into three parts proportional to $2$, $4$, and $6$, the smallest part is:
$\textbf{(A)}\ 5\frac{1}{3}\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 10\frac{2}{3}\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these answers}$ | If the three numbers are in proportion to $2:4:6$, then they should also be in proportion to $1:2:3$. This implies that the three numbers can be expressed as $x$, $2x$, and $3x$. Add these values together to get:
\[x+2x+3x=6x=64\]
Divide each side by 6 and get that
\[x=\frac{64}{6}=\frac{32}{3}=10 \frac{2}{3}\]
which... | 10\frac{2}{3} | Algebra | MCQ | Yes | Yes | amc_aime | false |
After rationalizing the numerator of $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}$, the denominator in simplest form is:
$\textbf{(A)}\ \sqrt{3}(\sqrt{3}+\sqrt{2})\qquad\textbf{(B)}\ \sqrt{3}(\sqrt{3}-\sqrt{2})\qquad\textbf{(C)}\ 3-\sqrt{3}\sqrt{2}\qquad\\ \textbf{(D)}\ 3+\sqrt6\qquad\textbf{(E)}\ \text{None of these answers}$ | To rationalize the numerator, multiply by the conjugate.
\[\frac{\sqrt3 - \sqrt2}{\sqrt3}\cdot \frac{\sqrt3 + \sqrt2}{\sqrt3 + \sqrt2} = \frac{1}{3+\sqrt6}.\]
The denominator is $\boxed{\mathrm{(D)}\text{ } 3+\sqrt6 }.$ | 3+\sqrt{6} | Algebra | MCQ | Yes | Yes | amc_aime | false |
If in the formula $C =\frac{en}{R+nr}$, where $e$, $n$, $R$ and $r$ are all positive, $n$ is increased while $e$, $R$ and $r$ are kept constant, then $C$:
$\textbf{(A)}\ \text{Increases}\qquad\textbf{(B)}\ \text{Decreases}\qquad\textbf{(C)}\ \text{Remains constant}\qquad\textbf{(D)}\ \text{Increases and then decreases}... | Divide both the numerator and denominator by $n$, to get $C=\frac{e}{\frac{R}{n}+r}$. If $n$ increases then the denominator decreases; so that $C$ $\boxed{\mathrm{(A)}\text{ }\mathrm{ Increases}.}$
but what if $n\leq 0$ | \mathrm{(A)} | Algebra | MCQ | Yes | Yes | amc_aime | false |
As the number of sides of a polygon increases from $3$ to $n$, the sum of the exterior angles formed by extending each side in succession:
$\textbf{(A)}\ \text{Increases}\qquad\textbf{(B)}\ \text{Decreases}\qquad\textbf{(C)}\ \text{Remains constant}\qquad\textbf{(D)}\ \text{Cannot be predicted}\qquad\\ \textbf{(E)}\ \t... | By the Exterior Angles Theorem, the exterior angles of all convex polygons add up to $360^\circ,$ so the sum $\boxed{\mathrm{(C)}\text{ remains constant}.}$ | \mathrm{(C)} | Geometry | MCQ | Yes | Yes | amc_aime | false |
The roots of $(x^{2}-3x+2)(x)(x-4)=0$ are:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 0\text{ and }4\qquad\textbf{(C)}\ 1\text{ and }2\qquad\textbf{(D)}\ 0,1,2\text{ and }4\qquad\textbf{(E)}\ 1,2\text{ and }4$ | Factor $x^2-3x+2$ to get $(x-2)(x-1).$ The roots are $\boxed{\mathrm{(D)}\ 0,1,2\text{ and }4.}$ | 0,1,24 | Algebra | MCQ | Yes | Yes | amc_aime | false |
For the simultaneous equations \[2x-3y=8\]
\[6y-4x=9\]
$\textbf{(A)}\ x=4,y=0\qquad\textbf{(B)}\ x=0,y=\frac{3}{2}\qquad\textbf{(C)}\ x=0,y=0\qquad\\ \textbf{(D)}\ \text{There is no solution}\qquad\textbf{(E)}\ \text{There are an infinite number of solutions}$ | Try to solve this system of equations using the elimination method.
\begin{align*} -2(2x-3y)&=-2(8)\\ 6y-4x&=-16\\ 6y-4x&=9\\ 0&=-7 \end{align*}
Something is clearly contradictory so $\boxed{\mathrm{(D)}\text{ There is no solution}.}$
Alternatively, note that the second equation is a multiple of the first except that t... | \mathrm{(D)} | Algebra | MCQ | Yes | Yes | amc_aime | false |
The real roots of $x^2+4$ are:
$\textbf{(A)}\ (x^{2}+2)(x^{2}+2)\qquad\textbf{(B)}\ (x^{2}+2)(x^{2}-2)\qquad\textbf{(C)}\ x^{2}(x^{2}+4)\qquad\\ \textbf{(D)}\ (x^{2}-2x+2)(x^{2}+2x+2)\qquad\textbf{(E)}\ \text{Non-existent}$ | Solution 1
This looks similar to a difference of squares, so we can write it as $(x+2i)(x-2i).$ Neither of these factors are real.
Also, looking at the answer choices, there is no way multiplying two polynomials of degree $2$ will result in a polynomial of degree $2$ as well. Therefore the real factors are $\boxed{\ma... | \mathrm{(E)} | Algebra | MCQ | Yes | Yes | amc_aime | false |
The number of terms in the expansion of $[(a+3b)^{2}(a-3b)^{2}]^{2}$ when simplified is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | Use properties of exponents to move the squares outside the brackets use difference of squares.
\[[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4\]
Using the binomial theorem, we can see that the number of terms is $\boxed{\mathrm{(B)}\ 5}$. | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The formula which expresses the relationship between $x$ and $y$ as shown in the accompanying table is:
\[\begin{tabular}[t]{|c|c|c|c|c|c|}\hline x&0&1&2&3&4\\\hline y&100&90&70&40&0\\\hline\end{tabular}\]
$\textbf{(A)}\ y=100-10x\qquad\textbf{(B)}\ y=100-5x^{2}\qquad\textbf{(C)}\ y=100-5x-5x^{2}\qquad\\ \textbf{(D)}\ ... | Plug in the points $(0,100)$ and $(4,0)$ into each equation. The only one that works for both points is $\mathrm{(C)}.$ Plug in the rest of the points to confirm the answer is indeed $\boxed{\mathrm{(C)}\ y=100-5x-5x^2}$. | 100-5x-5x^2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Of the following
(1) $a(x-y)=ax-ay$
(2) $a^{x-y}=a^x-a^y$
(3) $\log (x-y)=\log x-\log y$
(4) $\frac{\log x}{\log y}=\log{x}-\log{y}$
(5) $a(xy)=ax \cdot ay$
$\textbf{(A)}\text{Only 1 and 4 are true}\qquad\\\textbf{(B)}\ \text{Only 1 and 5 are true}\qquad\\\textbf{(C)}\ \text{Only 1 and 3 are true}\qquad\\\textbf{(D)}\... | The distributive property doesn't apply to logarithms or in the ways illustrated, and only applies to addition and subtraction. Also, $a^{x-y} = \frac{a^x}{a^y}$, so $\boxed{\textbf{(E)} \text{ Only 1 is true}}$. | Only1istrue | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $m$ men can do a job in $d$ days, then $m+r$ men can do the job in:
$\textbf{(A)}\ d+r \text{ days}\qquad\textbf{(B)}\ d-r\text{ days}\qquad\textbf{(C)}\ \frac{md}{m+r}\text{ days}\qquad\\ \textbf{(D)}\ \frac{d}{m+r}\text{ days}\qquad\textbf{(E)}\ \text{None of these}$ | The number of men is inversely proportional to the number of days the job takes. Thus, if $m$ men can do a job in $d$ days, we have that it will take $md$ days for $1$ man to do the job. Thus, $m + r$ men can do the job in $\boxed{\textbf{(C)}\ \frac{md}{m+r}\text{ days}}$. | \frac{md}{+r} | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $R=gS-4$. When $S=8$, $R=16$. When $S=10$, $R$ is equal to:
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 21\qquad\textbf{(E)}\ \text{None of these}$ | Our first procedure is to find the value of $g$. With the given variables' values, we can see that $8g-4=16$ so $g=\frac{20}{8}=\frac{5}{2}$.
With that, we can replace $g$ with $\frac{5}{2}$. When $S=10$, we can see that $10\times\frac{5}{2}-4=\frac{50}{2}-4=25-4=\boxed{\text{(D) 21}}$. | 21 | Algebra | MCQ | Yes | Yes | amc_aime | false |
When $x^{13}+1$ is divided by $x-1$, the remainder is:
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$ | Solution 1
Using synthetic division, we get that the remainder is $\boxed{\textbf{(D)}\ 2}$.
Solution 2
By the remainder theorem, the remainder is equal to the expression $x^{13}+1$ when $x=1.$ This gives the answer of $\boxed{(\mathrm{D})\ 2.}$
Solution 3
Note that $x^{13} - 1 = (x - 1)(x^{12} + x^{11} \cdots + 1)$,... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The volume of a rectangular solid each of whose side, front, and bottom faces are $12\text{ in}^{2}$, $8\text{ in}^{2}$, and $6\text{ in}^{2}$ respectively is:
$\textbf{(A)}\ 576\text{ in}^{3}\qquad\textbf{(B)}\ 24\text{ in}^{3}\qquad\textbf{(C)}\ 9\text{ in}^{3}\qquad\textbf{(D)}\ 104\text{ in}^{3}\qquad\textbf{(E)}\ ... | If the sidelengths of the cubes are expressed as $a, b,$ and $c,$ then we can write three equations:
\[ab=12, bc=8, ac=6.\]
The volume is $abc.$ Notice symmetry in the equations. We can find $abc$ my multiplying all the equations and taking the positive square root.
\begin{align*} (ab)(bc)(ac) &= (12)(8)(6)\\ a^2b^2c^2... | 24 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Successive discounts of $10\%$ and $20\%$ are equivalent to a single discount of:
$\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 72\%\qquad\textbf{(D)}\ 28\%\qquad\textbf{(E)}\ \text{None of these}$ | Without loss of generality, assume something costs $100$ dollars. Then with each successive discount, it would cost $90$ dollars, then $72$ dollars. This amounts to a total of $28$ dollars off, so the single discount would be $\boxed{\mathrm{(D)}\ 28\%.}$ | 28 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A man buys a house for $10,000 and rents it. He puts $12\frac{1}{2}\%$ of each month's rent aside for repairs and upkeep; pays $325 a year taxes and realizes $5\frac{1}{2}\%$ on his investment. The monthly rent (in dollars) is:
$\textbf{(A)} \ \ 64.82\qquad\textbf{(B)} \ \ 83.33\qquad\textbf{(C)} \ \ 72.08\qquad\textb... | $12\frac{1}{2}\%$ is the same as $\frac{1}{8}$, so the man sets one eighth of each month's rent aside, so he only gains $\frac{7}{8}$ of his rent. He also pays $325 each year, and he realizes $5.5\%$, or $550, on his investment. Therefore he must have collected a total of $325 +$550 = $875 in rent. This was for the who... | 83.33 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The equation $x + \sqrt{x-2} = 4$ has:
$\textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root}$ | Solution 1
$x + \sqrt{x-2} = 4$ Original Equation
$\sqrt{x-2} = 4 - x$ Subtract x from both sides
$x-2 = 16 - 8x + x^2$ Square both sides
$x^2 - 9x + 18 = 0$ Get all terms on one side
$(x-6)(x-3) = 0$ Factor
$x = \{6, 3\}$
If you put down A as your answer, it's wrong. You need to check for extraneous roots.
$6 + \sqrt{... | \textbf{(E)} | Algebra | MCQ | Yes | Yes | amc_aime | false |
The value of $\log_{5}\frac{(125)(625)}{25}$ is equal to:
$\textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these}$ | Solution 1
$\log_{5}\frac{(125)(625)}{25}$ can be simplified to $\log_{5}\ (125)(25)$ since $25^2 = 625$. $125 = 5^3$ and $5^2 = 25$ so $\log_{5}\ 5^5$ would be the simplest form. In $\log_{5}\ 5^5$, $5^x = 5^5$. Therefore, $x = 5$ and the answer is $\boxed{\mathrm{(D)}\ 5}$.
Solution 2
$\log_{5}\frac{(125)(625)}{25}$... | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $\log_{10}{m}= b-\log_{10}{n}$, then $m=$
$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$ | We have $b=\log_{10}{10^b}$. Substituting, we find $\log_{10}{m}= \log_{10}{10^b}-\log_{10}{n}$. Using $\log{a}-\log{b}=\log{\dfrac{a}{b}}$, the left side becomes $\log_{10}{\dfrac{10^b}{n}}$. Because $\log_{10}{m}=\log_{10}{\dfrac{10^b}{n}}$, $m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}$. | \frac{10^b}{n} | Algebra | MCQ | Yes | Yes | amc_aime | false |
A car travels $120$ miles from $A$ to $B$ at $30$ miles per hour but returns the same distance at $40$ miles per hour. The average speed for the round trip is closest to:
$\textbf{(A)}\ 33\text{ mph}\qquad\textbf{(B)}\ 34\text{ mph}\qquad\textbf{(C)}\ 35\text{ mph}\qquad\textbf{(D)}\ 36\text{ mph}\qquad\textbf{(E)}\ 37... | The car takes $120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{30 \text{ miles }}=4 \text{ hr}$ to get from $A$ to $B$. Also, it takes $120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{40 \text{ miles }}=3 \text{ hr}$ to get from $B$ to $A$. Therefore, the average speed is $\dfrac{240\text{ miles }}{7 \text{ hr}}=34\dfrac{2}{... | \textbf{(B)}\34 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Two boys $A$ and $B$ start at the same time to ride from Port Jervis to Poughkeepsie, $60$ miles away. $A$ travels $4$ miles an hour slower than $B$. $B$ reaches Poughkeepsie and at once turns back meeting $A$ $12$ miles from Poughkeepsie. The rate of $A$ was:
$\textbf{(A)}\ 4\text{ mph}\qquad \textbf{(B)}\ 8\text{ mph... | Let the speed of boy $A$ be $a$, and the speed of boy $B$ be $b$. Notice that $A$ travels $4$ miles per hour slower than boy $B$, so we can replace $b$ with $a+4$.
Now let us see the distances that the boys each travel. Boy $A$ travels $60-12=48$ miles, and boy $B$ travels $60+12=72$ miles. Now, we can use $d=rt$ to ma... | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A manufacturer built a machine which will address $500$ envelopes in $8$ minutes. He wishes to build another machine so that when both are operating together they will address $500$ envelopes in $2$ minutes. The equation used to find how many minutes $x$ it would require the second machine to address $500$ envelopes al... | Solution 1
We first represent the first machine's speed in per $2$ minutes: $125 \text{ envelopes in }2\text{ minutes}$. Now, we know that the speed per $2$ minutes of the second machine is \[500-125=375 \text{ envelopes in }2\text{ minutes}\]
Now we can set up a proportion to find out how many minutes it takes for the... | \textbf{(B)}\\frac{1}{8}+\frac{1}{x}=\frac{1}{2} | Algebra | MCQ | Yes | Yes | amc_aime | false |
The sum of the roots of the equation $4x^{2}+5-8x=0$ is equal to:
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ -5\qquad\textbf{(C)}\ -\frac{5}{4}\qquad\textbf{(D)}\ -2\qquad\textbf{(E)}\ \text{None of these}$ | We can divide by 4 to get:
$x^{2}-2x+\dfrac{5}{4}=0.$
The Vieta's formula states that in quadratic equation $ax^2+bx+c$, the sum of the roots of the equation is $-\frac{b}{a}$.
Using Vieta's formula, we find that the roots add to $2$ or $\boxed{\textbf{(E)}\ \text{None of these}}$. | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
From a group of boys and girls, $15$ girls leave. There are then left two boys for each girl. After this $45$ boys leave. There are then $5$ girls for each boy. The number of girls in the beginning was:
$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 43 \qquad \textbf{(C)}\ 29 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ \text{... | Let us represent the number of boys $b$, and the number of girls $g$.
From the first sentence, we get that $2(g-15)=b$.
From the second sentence, we get $5(b-45)=g-15$.
Expanding both equations and simplifying, we get $2g-30 = b$ and $5b = g+210$.
Substituting $b$ for $2g-30$, we get $5(2g-30)=g+210$. Solving for $g$, ... | 40 | Algebra | MCQ | Yes | Yes | amc_aime | false |
John ordered $4$ pairs of black socks and some additional pairs of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by $50\%$. The ratio of the number of pairs o... | Let the number of blue socks be represented as $b$. We are informed that the price of the black sock is twice the price of a blue sock; let us assume that the price of one pair of blue socks is $1$. That means the price of one pair of black socks is $2$.
Now from the third and fourth sentence, we see that $1.5(2(4)+1(b... | 1:4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A $25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:
$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \tex... | By the Pythagorean triple $(7,24,25)$, the point where the ladder meets the wall is $24$ feet above the ground. When the ladder slides, it becomes $20$ feet above the ground. By the $(15,20,25)$ Pythagorean triple, The foot of the ladder is now $15$ feet from the building. Thus, it slides $15-7 = \boxed{\textbf{(D)}\ 8... | 8 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The number of circular pipes with an inside diameter of $1$ inch which will carry the same amount of water as a pipe with an inside diameter of $6$ inches is:
$\textbf{(A)}\ 6\pi \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 36\pi$ | It must be assumed that the pipes have an equal height.
We can represent the amount of water carried per unit time by cross sectional area.
Cross sectional of Pipe with diameter $6 in$,
\[\pi r^2 = \pi \cdot 3^2 = 9\pi\]
Cross sectional area of pipe with diameter $1 in$
\[\pi r^2 = \pi \cdot 0.5 \cdot 0.5 = \frac{\pi}... | 36 | Geometry | MCQ | Yes | Yes | amc_aime | false |
When the circumference of a toy balloon is increased from $20$ inches to $25$ inches, the radius is increased by:
$\textbf{(A)}\ 5\text{ in} \qquad \textbf{(B)}\ 2\dfrac{1}{2}\text{ in} \qquad \textbf{(C)}\ \dfrac{5}{\pi}\text{ in} \qquad \textbf{(D)}\ \dfrac{5}{2\pi}\text{ in} \qquad \textbf{(E)}\ \dfrac{\pi}{5}\text{... | Solution 1
When the circumference of a circle is increased by a percentage, the radius is also increased by the same percentage (or else the ratio of the circumference to the diameter wouldn't be $\pi$ anymore)
We see that the circumference was increased by $25\%$. This means the radius was also increased by $25\%$. Th... | \frac{5}{2\pi} | Geometry | MCQ | Yes | Yes | amc_aime | false |
In triangle $ABC$, $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:
$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$ | The inradius is equal to the area divided by semiperimeter. The area is $\frac{(10)(24)}{2} = 120$ because it's a right triangle, as it's side length satisfies the Pythagorean Theorem. The semiperimeter is $30$. Therefore the inradius is $\boxed{\textbf{(B)}\ 4}$. | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
A merchant buys goods at $25\%$ off the list price. He desires to mark the goods so that he can give a discount of $20\%$ on the marked price and still clear a profit of $25\%$ on the selling price. What percent of the list price must he mark the goods?
$\textbf{(A)}\ 125\% \qquad \textbf{(B)}\ 100\% \qquad \textbf{(C)... | Without loss of generality, we can set the list price equal to $100$. The merchant buys the goods for $100*.75=75$. Let $x$ be the marked price.
We then use the equation $0.8x-75=25$ to solve for $x$ and get a marked price of $\boxed{125}$.
$\text{Answer: }\boxed{\mathbf{(A)}}$ | 125 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $y = \log_{a}{x}$, $a > 1$, which of the following statements is incorrect?
$\textbf{(A)}\ \text{If }x=1,y=0 \qquad\\ \textbf{(B)}\ \text{If }x=a,y=1 \qquad\\ \textbf{(C)}\ \text{If }x=-1,y\text{ is imaginary (complex)} \qquad\\ \textbf{(D)}\ \text{If }0<x<1,y\text{ is always less than 0 and decreases without limit ... | Let us first check
$\textbf{(A)}\ \text{If }x=1,y=0$. Rewriting into exponential form gives $a^0=1$. This is certainly correct.
$\textbf{(B)}\ \text{If }x=a,y=1$. Rewriting gives $a^1=a$. This is also certainly correct.
$\textbf{(C)}\ \text{If }x=-1,y\text{ is imaginary (complex)}$. Rewriting gives $a^{\text{complex n... | \textbf{(E)}\ | Algebra | MCQ | Yes | Yes | amc_aime | false |
If the expression $\begin{pmatrix}a & c\\ d & b\end{pmatrix}$ has the value $ab-cd$ for all values of $a, b, c$ and $d$, then the equation $\begin{pmatrix}2x & 1\\ x & x\end{pmatrix}= 3$:
$\textbf{(A)}\ \text{Is satisfied for only 1 value of }x\qquad\\ \textbf{(B)}\ \text{Is satisified for only 2 values of }x\qquad\\ \... | By $\begin{pmatrix}a & c\\ d & b\end{pmatrix}=ab-cd$, we have $2x^2-x=3$. Subtracting $3$ from both sides, giving $2x^2-x-3=0$. This factors to $(2x-3)(x+1)=0$. Thus, $x=\dfrac{3}{2},-1$, so the equation is $\boxed{\textbf{(B)}\ \text{satisified for only 2 values of }x}$.
Note: Alternatively, one may note that the equa... | \textbf{(B)}\satisifiedforonly2valuesofx | Algebra | MCQ | Yes | Yes | amc_aime | false |
Given the series $2+1+\frac {1}{2}+\frac {1}{4}+\cdots$ and the following five statements:
(1) the sum increases without limit
(2) the sum decreases without limit
(3) the difference between any term of the sequence and zero can be made less than any positive quantity no matter how small
(4) the difference between the ... | This series is a geometric series with common ratio $\frac{1}{2}$. Using the well-known formula for the sum of an infinite geometric series, we obtain that this series has a value of $2\cdot \frac{1}{1-\frac{1}{2}}=4$. It immediately follows that statements 1 and 2 are false while statements 4 and 5 are true. 3 is fals... | \textbf{(E)}\Only45 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Reduced to lowest terms, $\frac{a^{2}-b^{2}}{ab} - \frac{ab-b^{2}}{ab-a^{2}}$ is equal to:
$\textbf{(A)}\ \frac{a}{b}\qquad\textbf{(B)}\ \frac{a^{2}-2b^{2}}{ab}\qquad\textbf{(C)}\ a^{2}\qquad\textbf{(D)}\ a-2b\qquad\textbf{(E)}\ \text{None of these}$ | We start off by factoring the second fraction.
\[-\frac{ab-b^2}{ab-a^2} = -\frac{b(a-b)}{-a(a-b)} = \frac{b}{a}.\]
Now create a common denominator and simplify.
\[\frac{a^2-b^2}{ab}+\frac{b}{a}=\frac{a^2-b^2}{ab}+\frac{b^2}{ab} = \frac{a^2}{ab} = \boxed{\mathrm{(A) }\frac{a}{b}}\]
obs: Assume that $a \not = 0, b\not = ... | \frac{}{b} | Algebra | MCQ | Yes | Yes | amc_aime | false |
The limit of $\frac {x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \text{Indeterminate} \qquad \textbf{(C)}\ x-1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$ | Both $x^2-1$ and $x-1$ approach 0 as $x$ approaches $1$, using the L'Hôpital's rule, we have $\lim \limits_{x\to 1}\frac{x^2-1}{x-1} = \lim \limits_{x\to 1}\frac{2x}{1} = 2$.
Thus, the answer is $\boxed{\textbf{(D)}\ 2}$.
~ MATH__is__FUN | 2 | Calculus | MCQ | Yes | Yes | amc_aime | false |
The least value of the function $ax^2 + bx + c$ with $a>0$ is:
$\textbf{(A)}\ -\dfrac{b}{a} \qquad \textbf{(B)}\ -\dfrac{b}{2a} \qquad \textbf{(C)}\ b^2-4ac \qquad \textbf{(D)}\ \dfrac{4ac-b^2}{4a}\qquad \textbf{(E)}\ \text{None of these}$ | The vertex of a parabola is at $x=-\frac{b}{2a}$ for $ax^2+bx+c$. Because $a>0$, the vertex is a minimum. Therefore $a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c=a(\frac{b^2}{4a^2})-\frac{b^2}{2a}+c=\frac{b^2}{4a}-\frac{2b^2}{4a}+c=-\frac{b^2}{4a}+c=\frac{4ac}{4a}-\frac {b^2}{4a}=\frac{4ac-b^2}{4a} \Rightarrow \mathrm{(D)}$. | \frac{4ac-b^2}{4a} | Algebra | MCQ | Yes | Yes | amc_aime | false |
The equation $x^{x^{x^{.^{.^.}}}}=2$ is satisfied when $x$ is equal to:
$\textbf{(A)}\ \infty \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt[4]{2} \qquad \textbf{(D)}\ \sqrt{2} \qquad \textbf{(E)}\ \text{None of these}$ | empty | Algebra | MCQ | Yes | Problem not solved | amc_aime | false | |
The sum to infinity of $\frac{1}{7}+\frac {2}{7^2}+\frac{1}{7^3}+\frac{2}{7^4}+\cdots$ is:
$\textbf{(A)}\ \frac{1}{5} \qquad \textbf{(B)}\ \dfrac{1}{24} \qquad \textbf{(C)}\ \dfrac{5}{48} \qquad \textbf{(D)}\ \dfrac{1}{16} \qquad \textbf{(E)}\ \text{None of these}$ | Note that this is $\frac{1}{7}(1+\frac{1}{49}+\frac{1}{49^2}+...)+\frac{2}{49}(1+\frac{1}{49}+...)=\frac{9}{49}(1+\frac{1}{49}+...)$. Using the formula for a geometric series, we find that this is $\frac{9}{49}(\frac{1}{1-\frac{1}{49}})=\frac{9}{49}(\frac{1}{\frac{48}{49}})=\frac{9}{49}(\frac{49}{48})=\frac{9}{48}=\fra... | \frac{3}{16} | Algebra | MCQ | Yes | Yes | amc_aime | false |
The graph of $y=\log x$
$\textbf{(A)}\ \text{Cuts the }y\text{-axis} \qquad\\ \textbf{(B)}\ \text{Cuts all lines perpendicular to the }x\text{-axis} \qquad\\ \textbf{(C)}\ \text{Cuts the }x\text{-axis} \qquad\\ \textbf{(D)}\ \text{Cuts neither axis} \qquad\\ \textbf{(E)}\ \text{Cuts all circles whose center is at the o... | The domain of $\log x$ is the set of all $\underline{positive}$ reals, so the graph of $y=\log x$ clearly doesn't cut the $y$-axis. It therefore doesn't cut every line perpendicular to the $x$-axis. It does however cut the $x$-axis at $(1,0)$. In addition, if one examines the graph of $y=\log x$, one can clearly see th... | \textbf{(C)}\ | Algebra | MCQ | Yes | Yes | amc_aime | false |
The number of diagonals that can be drawn in a polygon of 100 sides is:
$\textbf{(A)}\ 4850 \qquad \textbf{(B)}\ 4950\qquad \textbf{(C)}\ 9900 \qquad \textbf{(D)}\ 98 \qquad \textbf{(E)}\ 8800$ | Each diagonal has its two endpoints as vertices of the 100-gon. Each pair of vertices determines exactly one diagonal. Therefore the answer should be $\binom{100}{2}=4950$. However this also counts the 100 sides of the polygon, so the actual answer is $4950-100=\boxed{\textbf{(A)}\ 4850 }$. | 4850 | Geometry | MCQ | Yes | Yes | amc_aime | false |
In triangle $ABC$, $AB=12$, $AC=7$, and $BC=10$. If sides $AB$ and $AC$ are doubled while $BC$ remains the same, then:
$\textbf{(A)}\ \text{The area is doubled} \qquad\\ \textbf{(B)}\ \text{The altitude is doubled} \qquad\\ \textbf{(C)}\ \text{The area is four times the original area} \qquad\\ \textbf{(D)}\ \text{The m... | If you double sides $AB$ and $AC$, they become $24$ and $14$ respectively. If $BC$ remains $10$, then this triangle has area $0$ because ${14} + {10} = {24}$, so two sides overlap the third side. Therefore the answer is $\boxed{\textbf{(E)}\ \text{The area of the triangle is 0}}$. | \textbf{(E)}\ | Geometry | MCQ | Yes | Yes | amc_aime | false |
A rectangle inscribed in a triangle has its base coinciding with the base $b$ of the triangle. If the altitude of the triangle is $h$, and the altitude $x$ of the rectangle is half the base of the rectangle, then:
$\textbf{(A)}\ x=\dfrac{1}{2}h \qquad \textbf{(B)}\ x=\dfrac{bh}{b+h} \qquad \textbf{(C)}\ x=\dfrac{bh}{2h... | Draw the triangle, and note that the small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle, so the proportions of the heights is equal to the proportions of the sides. In particular, we get $\dfrac{2x}{b} = \dfrac{h - x}{h} \implies 2xh = bh - bx \implies (2h ... | \frac{bh}{2h+b} | Geometry | MCQ | Yes | Yes | amc_aime | false |
A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is:
$\textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}\qquad\\ \textbf{(B)}\ \text{Greater than the altitude of the triangle}\qquad\\ \t... | Begin by drawing the triangle, the point, the altitudes from the point to the sides, and the segments connecting the point to the vertices. Let the triangle be $ABC$ with $AB=BC=AC=s$. We will call the aforementioned point $P$. Call altitude from $P$ to $BC$ $PA'$. Similarly, we will name the other two altitudes $P... | PA'+PB'+PC'= | Geometry | MCQ | Yes | Yes | amc_aime | false |
A triangle has a fixed base $AB$ that is $2$ inches long. The median from $A$ to side $BC$ is $1\frac{1}{2}$ inches long and can have any position emanating from $A$. The locus of the vertex $C$ of the triangle is:
$\textbf{(A)}\ \text{A straight line }AB,1\dfrac{1}{2}\text{ inches from }A \qquad\\ \textbf{(B)}\ \text{... | The locus of the median's endpoint on $BC$ is the circle about $A$ and of radius $1\frac{1}{2}$ inches. The locus of the vertex $C$ is then the circle twice as big and twice as far from $B$, i.e. of radius $3$ inches and with center $4$ inches from $B$ along $BA$ which means that our answer is: $\textbf{(D)}$. | D | Geometry | MCQ | Yes | Yes | amc_aime | false |
If five geometric means are inserted between $8$ and $5832$, the fifth term in the geometric series:
$\textbf{(A)}\ 648\qquad\textbf{(B)}\ 832\qquad\textbf{(C)}\ 1168\qquad\textbf{(D)}\ 1944\qquad\textbf{(E)}\ \text{None of these}$ | We can let the common ratio of the geometric sequence be $r$. $5832$ is given to be the seventh term in the geometric sequence as there are five terms between it and $a_1$ if we consider $a_1=8$.
By the formula for each term in a geometric sequence, we find that $a_n=a_1r^{n-1}$ or $(5382)=(8)r^6$
We divide by eight to... | 648 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A privateer discovers a merchantman $10$ miles to leeward at 11:45 a.m. and with a good breeze bears down upon her at $11$ mph, while the merchantman can only make $8$ mph in her attempt to escape. After a two hour chase, the top sail of the privateer is carried away; she can now make only $17$ miles while the merchant... | Assume that the two boats are traveling along the positive real number line, with the merchantman starting at the number $10$ and the privateer starting at the number $0$. After two hours the merchantman is at $26$ while the privateer is at $22$. When the top sail of the privateer is carried away, the speed of the merc... | \textbf{(E)}\5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The values of $y$ which will satisfy the equations $2x^{2}+6x+5y+1=0, 2x+y+3=0$ may be found by solving:
$\textbf{(A)}\ y^{2}+14y-7=0\qquad\textbf{(B)}\ y^{2}+8y+1=0\qquad\textbf{(C)}\ y^{2}+10y-7=0\qquad\\ \textbf{(D)}\ y^{2}+y-12=0\qquad \textbf{(E)}\ \text{None of these equations}$ | If we solve the second equation for $x$ in terms of $y$, we find $x=-\dfrac{y+3}{2}$ which we can substitute to find:
\[2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1=0\]
Multiplying by two and simplifying, we find:
\begin{align*} 2\cdot[2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1]&=2\cdot 0\\ (y+3)^2 -6y-18+10y+2&=0\\ ... | y^{2}+10y-7=0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If the digit $1$ is placed after a two digit number whose tens' digit is $t$, and units' digit is $u$, the new number is:
$\textbf{(A)}\ 10t+u+1\qquad\textbf{(B)}\ 100t+10u+1\qquad\textbf{(C)}\ 1000t+10u+1\qquad\textbf{(D)}\ t+u+1\qquad\\ \textbf{(E)}\ \text{None of these answers}$ | By placing the digit $1$ after a two digit number, you are changing the units place to $1$ and moving everything else up a place. Therefore the answer is $\boxed{\textbf{(B)}\ 100t+10u+1}$. | 100t+10u+1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If the radius of a circle is increased $100\%$, the area is increased:
$\textbf{(A)}\ 100\%\qquad\textbf{(B)}\ 200\%\qquad\textbf{(C)}\ 300\%\qquad\textbf{(D)}\ 400\%\qquad\textbf{(E)}\ \text{By none of these}$ | Increasing by $100\%$ is the same as doubling the radius. If we let $r$ be the radius of the old circle, then the radius of the new circle is $2r.$
Since the area of the circle is given by the formula $\pi r^2,$ the area of the new circle is $\pi (2r)^2 = 4\pi r^2.$ The area is quadrupled, or increased by $\boxed{\math... | 300 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The area of the largest triangle that can be inscribed in a semi-circle whose radius is $r$ is:
$\textbf{(A)}\ r^{2}\qquad\textbf{(B)}\ r^{3}\qquad\textbf{(C)}\ 2r^{2}\qquad\textbf{(D)}\ 2r^{3}\qquad\textbf{(E)}\ \frac{1}{2}r^{2}$ | The area of a triangle is $\frac12 bh.$ To maximize the base, let it be equal to the diameter of the semi circle, which is equal to $2r.$ To maximize the height, or altitude, choose the point directly in the middle of the arc connecting the endpoints of the diameter. It is equal to $r.$ Therefore the area is $\frac12 \... | r^2 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The percent that $M$ is greater than $N$ is:
$(\mathrm{A})\ \frac{100(M-N)}{M} \qquad (\mathrm{B})\ \frac{100(M-N)}{N} \qquad (\mathrm{C})\ \frac{M-N}{N} \qquad (\mathrm{D})\ \frac{M-N}{N} \qquad (\mathrm{E})\ \frac{100(M+N)}{N}$ | $M-N$ is the amount by which $M$ is greater than $N$. We divide this by $N$ to get the percent by which $N$ is increased in the form of a decimal, and then multiply by $100$ to make it a percentage. Therefore, the answer is $\boxed{\mathrm{(B)}\ \dfrac{100(M-N)}{N}}$. | \frac{100(M-N)}{N} | Algebra | MCQ | Yes | Yes | amc_aime | false |
Of the following statements, the one that is incorrect is:
$\textbf{(A)}\ \text{Doubling the base of a given rectangle doubles the area.}$
$\textbf{(B)}\ \text{Doubling the altitude of a triangle doubles the area.}$
$\textbf{(C)}\ \text{Doubling the radius of a given circle doubles the area.}$
$\textbf{(D)}\ \text{Doub... | The well-known area formula for a circle is $A = \pi r^2$, so doubling the radius will result it quadrupling the area (since $A' = \pi (2r)^2 = 4 \pi r^2 = 4A$). Statement $\boxed{\textbf{(C)}}$ is therefore incorrect, and is the correct answer choice.
Statements $\textbf{(A)}$ and $\textbf{(B)}$ are evidently correct... | \textbf{(C)} | Algebra | MCQ | Yes | Yes | amc_aime | false |
The limit of the sum of an infinite number of terms in a geometric progression is $\frac {a}{1 - r}$ where $a$ denotes the first term and $- 1 < r < 1$ denotes the common ratio. The limit of the sum of their squares is:
$\textbf{(A)}\ \frac {a^2}{(1 - r)^2} \qquad\textbf{(B)}\ \frac {a^2}{1 + r^2} \qquad\textbf{(C)}\ \... | Let the original geometric series be $a,ar,ar^2,ar^3,ar^4\cdots$. Therefore, their squares are $a^2,a^2r^2,a^2r^4,a^2r^6,\cdots$, which is a [geometric sequence](https://artofproblemsolving.com/wiki/index.php/Geometric_sequence) with first term $a^2$ and common ratio $r^2$. Thus, the sum is $\boxed{\textbf{(C)}\ \frac ... | \frac{^2}{1-r^2} | Algebra | MCQ | Yes | Yes | amc_aime | false |
At $2: 15$ o'clock, the hour and minute hands of a clock form an angle of:
$\textbf{(A)}\ 30^{\circ} \qquad\textbf{(B)}\ 5^{\circ} \qquad\textbf{(C)}\ 22\frac {1}{2}^{\circ} \qquad\textbf{(D)}\ 7\frac {1}{2} ^{\circ} \qquad\textbf{(E)}\ 28^{\circ}$ | Using the formula $\frac{|60h-11m|}{2}$, where $h$ is the hour time and $m$ is the minute time, we get that the angle is $\frac{|60(2)-11(15)|}{2}=\frac{|120-165|}{2}=\frac{45}{2}=\boxed{\textbf{(C)}\ 22\frac {1}{2}^{\circ} }$ | 22\frac{1}{2} | Geometry | MCQ | Yes | Yes | amc_aime | false |
$A$ can do a piece of work in $9$ days. $B$ is $50\%$ more efficient than $A$. The number of days it takes $B$ to do the same piece of work is:
$\textbf{(A)}\ 13\frac {1}{2} \qquad\textbf{(B)}\ 4\frac {1}{2} \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ \text{none of these answers}$ | Because $B$ is $50\%$ more efficient, he can do $1.5$ pieces of work in $9$ days. This is equal to 1 piece of work in $\textbf{(C)}\ 6$ days | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In connection with proof in geometry, indicate which one of the following statements is incorrect:
$\textbf{(A)}\ \text{Some statements are accepted without being proved.}$
$\textbf{(B)}\ \text{In some instances there is more than one correct order in proving certain propositions.}$
$\textbf{(C)}\ \text{Every term used... | After reading the options, it is very apparent that $\textbf{(E)}\ \text{Indirect proof can be used whenever there are two or more contrary propositions.}$ is the correct answer; rigorous proof is needed no matter what. | E | Geometry | MCQ | Yes | Problem not solved | amc_aime | false |
The largest number by which the expression $n^3 - n$ is divisible for all possible integral values of $n$, is:
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6$ | Factoring the polynomial gives $(n+1)(n)(n-1)$ According to the factorization, one of those factors must be a multiple of two because there are more than 2 consecutive integers. In addition, because there are three consecutive integers, one of the integers must be a multiple of 3. Therefore $6$ must divide the given e... | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
If in applying the [quadratic formula](https://artofproblemsolving.com/wiki/index.php/Quadratic_formula) to a [quadratic equation](https://artofproblemsolving.com/wiki/index.php/Quadratic_equation)
\[f(x) \equiv ax^2 + bx + c = 0,\]
it happens that $c = \frac{b^2}{4a}$, then the graph of $y = f(x)$ will certainly:
$\ma... | The [discriminant](https://artofproblemsolving.com/wiki/index.php/Discriminant) of the quadratic equation is $b^2 - 4ac = b^2 - 4a\left(\frac{b^2}{4a}\right) = 0$. This indicates that the equation has only one root (applying the quadratic formula, we get $x = \frac{-b + \sqrt{0}}{2a} = -b/2a$). Thus it follows that $f(... | C | Algebra | MCQ | Yes | Yes | amc_aime | false |
Indicate in which one of the following equations $y$ is neither directly nor inversely proportional to $x$:
$\textbf{(A)}\ x + y = 0 \qquad\textbf{(B)}\ 3xy = 10 \qquad\textbf{(C)}\ x = 5y \qquad\textbf{(D)}\ 3x + y = 10$
$\textbf{(E)}\ \frac {x}{y} = \sqrt {3}$ | Notice that for any directly or inversely proportional values, it can be expressed as $\frac{x}{y}=k$ or $xy=k$. Now we try to convert each into its standard form counterpart.
$\textbf{(A)}\ x + y = 0\implies \frac{x}{y}=-1$
$\textbf{(B)}\ 3xy = 10\implies xy=\frac{10}{3}$
$\textbf{(C)}\ x = 5y\implies \frac{x}{y}=5$
$... | \textbf{(D)}\3x+y=10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The expression $21x^2 +ax +21$ is to be factored into two linear prime binomial factors with integer coefficients. This can be done if $a$ is:
$\textbf{(A)}\ \text{any odd number} \qquad\textbf{(B)}\ \text{some odd number} \qquad\textbf{(C)}\ \text{any even number}$
$\textbf{(D)}\ \text{some even number} \qquad\textbf{... | We can factor $21x^2 + ax + 21$ as $(7x+3)(3x+7)$, which expands to $21x^2+42x+21$. So the answer is $\textbf{(D)}\ \text{some even number}$ | D | Algebra | MCQ | Yes | Yes | amc_aime | false |
A six place number is formed by repeating a three place number; for example, $256256$ or $678678$, etc. Any number of this form is always exactly divisible by:
$\textbf{(A)}\ 7 \text{ only} \qquad\textbf{(B)}\ 11 \text{ only} \qquad\textbf{(C)}\ 13 \text{ only} \qquad\textbf{(D)}\ 101 \qquad\textbf{(E)}\ 1001$ | We can express any of these types of numbers in the form $\overline{abc}\times 1001$, where $\overline{abc}$ is a 3-digit number. Therefore, the answer is $\textbf{(E)}\ 1001$. | 1001 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
A rectangular field is half as wide as it is long and is completely enclosed by $x$ yards of fencing. The area in terms of $x$ is:
$(\mathrm{A})\ \frac{x^2}2 \qquad (\mathrm{B})\ 2x^2 \qquad (\mathrm{C})\ \frac{2x^2}9 \qquad (\mathrm{D})\ \frac{x^2}{18} \qquad (\mathrm{E})\ \frac{x^2}{72}$ | Let $w$ be the width. Then $l = 2w$, and the perimeter is $x = 2(2w)+2w = 6w \implies w = \frac{x}6$. The area is $wl = w(2w) = 2w^2 = 2\left(\frac{x^2}{36}\right) = \frac{x^2}{18}$, so the answer is $\mathrm{D}$. | \frac{x^2}{18} | Geometry | MCQ | Yes | Yes | amc_aime | false |
When simplified and expressed with negative exponents, the expression $(x + y)^{ - 1}(x^{ - 1} + y^{ - 1})$ is equal to:
$\textbf{(A)}\ x^{ - 2} + 2x^{ - 1}y^{ - 1} + y^{ - 2} \qquad\textbf{(B)}\ x^{ - 2} + 2^{ - 1}x^{ - 1}y^{ - 1} + y^{ - 2} \qquad\textbf{(C)}\ x^{ - 1}y^{ - 1}$
$\textbf{(D)}\ x^{ - 2} + y^{ - 2} \qqu... | Note that $(x + y)^{-1}(x^{-1} + y^{-1}) = \dfrac{1}{x + y}\cdot\left(\dfrac{1}{x} + \dfrac{1}{y}\right) = \dfrac{1}{x + y}\cdot\dfrac{x + y}{xy} = \dfrac{1}{xy} = x^{-1}y^{-1}$. The answer is $\textbf{(C)}$. | x^{-1}y^{-1} | Algebra | MCQ | Yes | Yes | amc_aime | false |
Given: $x > 0, y > 0, x > y$ and $z\ne 0$. The inequality which is not always correct is:
$\textbf{(A)}\ x + z > y + z \qquad\textbf{(B)}\ x - z > y - z \qquad\textbf{(C)}\ xz > yz$
$\textbf{(D)}\ \frac {x}{z^2} > \frac {y}{z^2} \qquad\textbf{(E)}\ xz^2 > yz^2$ | $\textbf{(A)}\ x + z > y + z\implies x>y$, just subtract $z$ from both sides
$\textbf{(B)}\ x - z > y - z\implies x>y$, just add $z$ to both sides
$\textbf{(C)}\ xz > yz\implies x>y\text{ if }x>0$, so that means that our desired answer is $\boxed{\textbf{(C)}\ xz > yz}$.
As a check:
$\textbf{(D)}\ \frac {x}{z^2} > \fr... | \textbf{(C)}\xz>yz | Inequalities | MCQ | Yes | Incomplete | amc_aime | false |
The values of $a$ in the equation: $\log_{10}(a^2 - 15a) = 2$ are:
$\textbf{(A)}\ \frac {15\pm\sqrt {233}}{2} \qquad\textbf{(B)}\ 20, - 5 \qquad\textbf{(C)}\ \frac {15 \pm \sqrt {305}}{2}$
$\textbf{(D)}\ \pm20 \qquad\textbf{(E)}\ \text{none of these}$ | Putting into exponential form, we get that $10^2=a^2-15a\Rightarrow a^2-15a-100=0$
Now we use the quadratic formula to solve for $a$, and we get $a=\frac{15\pm\sqrt{625}}{2}\implies a=\boxed{\textbf{(B)}\ 20, - 5}$ | =\frac{15\\sqrt{625}}{2}\implies= | Algebra | MCQ | Yes | Incomplete | amc_aime | false |
The radius of a cylindrical box is $8$ inches and the height is $3$ inches. The number of inches that may be added to either the radius or the height to give the same nonzero increase in volume is:
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 5\frac {1}{3} \qquad\textbf{(C)}\ \text{any number} \qquad\textbf{(D)}\ \text{non-exi... | Let $x$ be the number of inches increased. We can set up an equation for $x$: \[8^2 \times (3+x)=(8+x)^2\times 3\]
Expanding gives $3x^2+48x+192=64x+192$.
Combining like terms gives the quadratic $3x^2-16x=0$
Factoring out an $x$ gives $x(3x-16)=0$.
So either $x=0$, or $3x-16=0$.
The first equation is not possible, bec... | 5\frac{1}{3} | Algebra | MCQ | Yes | Yes | amc_aime | false |
$\frac{2^{n+4}-2(2^{n})}{2(2^{n+3})}$ when simplified is:
$\textbf{(A)}\ 2^{n+1}-\frac{1}{8}\qquad\textbf{(B)}\ -2^{n+1}\qquad\textbf{(C)}\ 1-2^{n}\qquad\textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{7}{4}$ | We have $2(2^n)=2^{n+1}$, and $2(2^{n+3})=2^{n+4}$. Thus, $\frac{2^{n+4}-2(2^{n})}{2(2^{n+3})}=\dfrac{2^{n+4}-2^{n+1}}{2^{n+4}}$. Factoring out a $2^{n+1}$ in the numerator, we get $\dfrac{2^{n+1}(2^3-1)}{2^{n+4}}=\dfrac{8-1}{2^3}=\boxed{\textbf{(D)}\ \frac{7}{8}}$. | \frac{7}{8} | Algebra | MCQ | Yes | Yes | amc_aime | false |
The apothem of a square having its area numerically equal to its perimeter is compared with the apothem of an equilateral triangle having its area numerically equal to its perimeter. The first apothem will be:
$\textbf{(A)}\ \text{equal to the second}\qquad\textbf{(B)}\ \frac{4}{3}\text{ times the second}\qquad\textbf... | First we try to find the size of the square. Let $s$ be the side length of the square. It states that $s^2=4s$, therefore $s=0,4$. We cross out the trivial case $s=0$, so the side length of the square is $4$. The apothem of the square is simply half its side length, or $2$.
Let the side length of the equilateral triang... | equaltothe | Geometry | MCQ | Yes | Yes | amc_aime | false |
In the equation $\frac {x(x - 1) - (m + 1)}{(x - 1)(m - 1)} = \frac {x}{m}$ the roots are equal when
$\textbf{(A)}\ m = 1\qquad\textbf{(B)}\ m =\frac{1}{2}\qquad\textbf{(C)}\ m = 0\qquad\textbf{(D)}\ m =-1\qquad\textbf{(E)}\ m =-\frac{1}{2}$ | Multiplying both sides by $(x-1)(m-1)m$ gives us
\[xm(x-1)-m(m+1)=x(x-1)(m-1)\]
\[x^2m-xm-m^2-m=x^2m-xm-x^2+x\]
\[-m^2-m=-x^2+x\]
\[x^2-x-(m^2+m)=0\]
The roots of this quadratic are equal if and only if its discriminant
($b^2-4ac$) evaluates to 0.
This means
\[(-1)^2-4(-m^2-m)=0\]
\[4m^2+4m+1=0\]
\[(2m+1)^2=0\]
\[m=-\f... | -\frac{1}{2} | Algebra | MCQ | Yes | Yes | amc_aime | false |
Through a point inside a triangle, three lines are drawn from the vertices to the opposite sides forming six triangular sections. Then:
$\textbf{(A)}\ \text{the triangles are similar in opposite pairs}\qquad\textbf{(B)}\ \text{the triangles are congruent in opposite pairs}$
$\textbf{(C)}\ \text{the triangles are equal... | Say we draw 3 different types of triangles as the following
1.An equilateral triangle with side lengths $x$
2.An isosceles triangle with side lengths $x+3,x+3,x-1$
3.A scalene triangle with sides $x+7,x-3,x+4$
1. Gives us 6 congruent $30-60-90$ triangles, which means
$\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)}... | \textbf{(E)} | Geometry | MCQ | Yes | Incomplete | amc_aime | false |
The pressure $(P)$ of wind on a sail varies jointly as the area $(A)$ of the sail and the square of the velocity $(V)$ of the wind. The pressure on a square foot is $1$ pound when the velocity is $16$ miles per hour. The velocity of the wind when the pressure on a square yard is $36$ pounds is:
$\textbf{(A)}\ 10\frac{... | Because $P$ varies jointly as $A$ and $V^2$, that means that there is a number $k$ such that $P=kAV^2$. You are given that $P=1$ when $A=1$ and $V=16$. That means that $1=k(1)(16^2) \rightarrow k=\frac{1}{256}$. Then, substituting into the original equation with $P=36$ and $A=9$ (because a square yard is $9$ times a sq... | C | Algebra | MCQ | Yes | Yes | amc_aime | false |
Of the following sets of data the only one that does not determine the shape of a triangle is:
$\textbf{(A)}\ \text{the ratio of two sides and the inc{}luded angle}\\ \qquad\textbf{(B)}\ \text{the ratios of the three altitudes}\\ \qquad\textbf{(C)}\ \text{the ratios of the three medians}\\ \qquad\textbf{(D)}\ \text{th... | The answer is $\boxed{\textbf{(D)}}$. The ratio of the altitude to the base is insufficient to determine the shape of a triangle; you also need to know the ratio of the two segments into which the altitude divides the base. | \textbf{(D)} | Geometry | MCQ | Yes | Yes | amc_aime | false |
If the length of a [diagonal](https://artofproblemsolving.com/wiki/index.php/Diagonal) of a [square](https://artofproblemsolving.com/wiki/index.php/Square) is $a + b$, then the area of the square is:
$\mathrm{(A) \ (a+b)^2 } \qquad \mathrm{(B) \ \frac{1}{2}(a+b)^2 } \qquad \mathrm{(C) \ a^2+b^2 } \qquad \mathrm{(D) \ \... | Let a side be $s$; then by the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), $s^2 + s^2 = 2s^2 = (a+b)^2$. The area of a square is $s^2 = \frac{(a+b)^2}{2} \Rightarrow \mathrm{(B)}$.
Alternatively, using the area formula for a [kite](https://artofproblemsolving.com/wiki/inde... | \frac{1}{2}(+b)^2 | Geometry | MCQ | Yes | Yes | amc_aime | false |
If two poles $20''$ and $80''$ high are $100''$ apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:
$\textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these}$ | The [two poles formula](https://artofproblemsolving.com/wiki/index.php/Two_poles_formula) says this height is half the harmonic mean of the heights of the two poles. (The distance between the poles is irrelevant.) So the answer is $\frac1{\frac1{20}+\frac1{80}}$, or $\frac1{\frac1{16}}=\boxed{16 \textbf{ (C)}}$. | 16 | Geometry | MCQ | Yes | Yes | amc_aime | false |
A total of $28$ handshakes were exchanged at the conclusion of a party. Assuming that each participant was equally polite toward all the others, the number of people present was:
$\textbf{(A)}\ 14\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 56\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 7$ | The handshake equation is $h=\frac{n(n-1)}{2}$, where $n$ is the number of people and $h$ is the number of handshakes. There were $28$ handshakes, so $28=\frac{n(n-1)}{2}$
$56=n(n-1)$
The factors of $56$ are:
$1, 2, 4, 7, 8, 14, 28, 56$.
As we can see, only $7, 8$ fit the requirements $n(n-1)$ if $n$ was an integer.
Th... | 8 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
If $\triangle ABC$ is inscribed in a semicircle whose diameter is $AB$, then $AC+BC$ must be
$\textbf{(A)}\ \text{equal to }AB\qquad\textbf{(B)}\ \text{equal to }AB\sqrt{2}\qquad\textbf{(C)}\ \geq AB\sqrt{2}\qquad\textbf{(D)}\ \leq AB\sqrt{2}$ $\textbf{(E)}\ AB^{2}$ | Because $AB$ is the diameter of the semi-circle, it follows that $\angle C = 90$. Now we can try to eliminate all the solutions except for one by giving counterexamples.
$\textbf{(A):}$
Set point $C$ anywhere on the perimeter of the semicircle except on $AB$. By triangle inequality, $AC+BC>AB$, so $\textbf{(A)}$ is wr... | \textbf{(D)}\\leqAB\sqrt{2} | Geometry | MCQ | Yes | Yes | amc_aime | false |
The roots of the equation $x^{2}-2x = 0$ can be obtained graphically by finding the abscissas of the points of intersection of each of the following pairs of equations except the pair:
$\textbf{(A)}\ y = x^{2}, y = 2x\qquad\textbf{(B)}\ y = x^{2}-2x, y = 0\qquad\textbf{(C)}\ y = x, y = x-2\qquad\textbf{(D)}\ y = x^{2}... | If you find the intersections of the curves listed in the answers $\textbf{(A)}$, $\textbf{(B)}$, $\textbf{(D)}$, and $\textbf{(E)}$, you will find that their abscissas are $0$ and $2$. Also you can note that the curves in $\textbf{(C)}$ don't actually intersect. Therefore the answer is $\boxed{\textbf{(C)}}$. | \textbf{(C)} | Algebra | MCQ | Yes | Yes | amc_aime | false |
The value of $10^{\log_{10}7}$ is:
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ \log_{10}7\qquad\textbf{(E)}\ \log_{7}10$ | $\log_{10}7=x \Rightarrow 10^x=7$
Substitute $\log_{10}7=x$ in $10^{\log_{10}7} \Rightarrow 10^x=?$
It was already stated that $10^x=7$, so our answer is $\textbf{(A)}\ 7$ | 7 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $a^{x}= c^{q}= b$ and $c^{y}= a^{z}= d$, then
$\textbf{(A)}\ xy = qz\qquad\textbf{(B)}\ \frac{x}{y}=\frac{q}{z}\qquad\textbf{(C)}\ x+y = q+z\qquad\textbf{(D)}\ x-y = q-z$
$\textbf{(E)}\ x^{y}= q^{z}$ | Try solving both equations for $a$. Taking the $x$-th root of both sides in the first equation and the $z$-the root of both sides in the second gives $a=c^{\frac{q}x}$ and $a=c^{\frac{y}z}$.
So $\frac{q}x=\frac{y}z$. Multiplying both sides by $xz$, $qz=xy$. $\boxed{\textbf{(A)}}$. | xy=qz | Algebra | MCQ | Yes | Yes | amc_aime | false |
Which of the following methods of proving a geometric figure a locus is not correct?
$\textbf{(A)}\ \text{Every point of the locus satisfies the conditions and every point not on the locus does}\\ \text{not satisfy the conditions.}$
$\textbf{(B)}\ \text{Every point not satisfying the conditions is not on the locus and... | Statement $\boxed{\textbf{(B)}}$ is wrong because it does not imply that all points that satisfy the conditions are on the locus. | \textbf{(B)} | Geometry | MCQ | Yes | Problem not solved | amc_aime | false |
A number which when divided by $10$ leaves a remainder of $9$, when divided by $9$ leaves a remainder of $8$, by $8$ leaves a remainder of $7$, etc., down to where, when divided by $2$, it leaves a remainder of $1$, is:
$\textbf{(A)}\ 59\qquad\textbf{(B)}\ 419\qquad\textbf{(C)}\ 1259\qquad\textbf{(D)}\ 2519\qquad\text... | If we add $1$ to the number, it becomes divisible by $10, 9, 8, \cdots, 2, 1$. The LCM of $1$ through $10$ is $2520$, therefore the number we want to find is $2520-1=\boxed{\textbf{(D)}\ 2519}$ | 2519 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
A rise of $600$ feet is required to get a railroad line over a mountain. The grade can be kept down by lengthening the track and curving it around the mountain peak. The additional length of track required to reduce the grade from $3\%$ to $2\%$ is approximately:
$\textbf{(A)}\ 10000\text{ ft.}\qquad\textbf{(B)}\ 2000... | A grade is the rise divided by the horizontal length for a given segment of track. This means we can get the horizontal length of the track by dividing the rise by the grade.
At a $3\%$ grade, the horizontal track length is $20000$ feet. At a $2\%$ grade, the horizontal track length is $30000$ feet. The difference is $... | \textbf{(A)} | Geometry | MCQ | Yes | Yes | amc_aime | false |
A stone is dropped into a well and the report of the stone striking the bottom is heard $7.7$ seconds after it is dropped. Assume that the stone falls $16t^2$ feet in t seconds and that the velocity of sound is $1120$ feet per second. The depth of the well is:
$\textbf{(A)}\ 784\text{ ft.}\qquad\textbf{(B)}\ 342\text{ ... | Let $d$ be the depth of the well in feet, let $t_1$ be the number of seconds the rock took to fall to the bottom of the well, and let $t_2$ be the number of seconds the sound took to travel back up the well. We know $t_1+t_2=7.7$. Now we can solve $t_1$ for $d$:
\[d=16t_1^2\]
\[\frac{d}{16}=t_1^2\]
\[t_1=\frac{\sqrt{d}... | 784 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A barn with a roof is rectangular in shape, $10$ yd. wide, $13$ yd. long and $5$ yd. high. It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of sq. yd. to be painted is:
$\mathrm{(A) \ } 360 \qquad \mathrm{(B) \ } 460 \qquad \mathrm{(C) \ } 490 \qquad \mathrm{(D... | The walls are $13*5=65$ and $10*5=50$ in area, and the ceiling has an area of $10*13=130$.
$((65+50)2)2+130=590 \Rightarrow \boxed{\mathrm{(D)}}$ | 590 | Geometry | MCQ | Yes | Yes | amc_aime | false |
$\left(\frac{(x+1)^{2}(x^{2}-x+1)^{2}}{(x^{3}+1)^{2}}\right)^{2}\cdot\left(\frac{(x-1)^{2}(x^{2}+x+1)^{2}}{(x^{3}-1)^{2}}\right)^{2}$ equals:
$\textbf{(A)}\ (x+1)^{4}\qquad\textbf{(B)}\ (x^{3}+1)^{4}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ [(x^{3}+1)(x^{3}-1)]^{2}$
$\textbf{(E)}\ [(x^{3}-1)^{2}]^{2}$ | First, note that we can pull the exponents out of every factor, since they are all squared. This results in
$\left(\frac{(x+1)(x^{2}-x+1)}{x^{3}+1}\right)^{4}\cdot\left(\frac{(x-1)(x^{2}+x+1)}{x^{3}-1}\right)^{4}$
Now, multiplying the numerators together gives
$\left(\frac{x^3+1}{x^3+1}\right)^{4}\cdot\left(\frac{x^3-1... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The formula expressing the relationship between $x$ and $y$ in the table is:
\[\begin{tabular}{|c|c|c|c|c|c|}\hline x & 2 & 3 & 4 & 5 & 6\\ \hline y & 0 & 2 & 6 & 12 & 20\\ \hline\end{tabular}\]
$\textbf{(A)}\ y = 2x-4\qquad\textbf{(B)}\ y = x^{2}-3x+2\qquad\textbf{(C)}\ y = x^{3}-3x^{2}+2x$
$\textbf{(D)}\ y = x^{2}-4x... | Just plug the $x,y$ pair $(6,20)$ into each of the 5 answer choices:
(A): $2(6)-4=8\ne20$
(B): $6^2-3(6)+2=20$
(C): $6^3-3(6^2)+2(6)=120\ne20$
(D): $6^2-4(6)=12\ne20$
(E): $6^2-4=32\ne20$
The only one that works is $\boxed{\textbf{(B)}}$. | \textbf{(B)} | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $x =\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}$, then:
$\textbf{(A)}\ x = 1\qquad\textbf{(B)}\ 0 2\text{ but finite}$ | We note that $x^2=1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}=1+x$. By the quadratic formula, $x=\dfrac{1\pm \sqrt{5}}{2}$. Because there are only positive square roots in $x$, $x$ must be positive, thus, it is $\dfrac{1+\sqrt{5}}{2}\approx 1.618$. Thus, $\boxed{\textbf{(C)}\ 1 < x < 2}$. | \textbf{(C)}\1<x<2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Of the following statements, the only one that is incorrect is:
$\textbf{(A)}\ \text{An inequality will remain true after each side is increased,}$ $\text{ decreased, multiplied or divided zero excluded by the same positive quantity.}$
$\textbf{(B)}\ \text{The arithmetic mean of two unequal positive quantities is great... | The answer is $\boxed{\textbf{(E)}}$.
Quite the opposite of statement (E) is true--the sum $a+b$ is minimized when $a=b$, but it approaches $\infty$ when one of $a,b$ gets arbitrarily small. | \textbf{(E)} | Inequalities | MCQ | Yes | Yes | amc_aime | false |
If $\frac{xy}{x+y}= a,\frac{xz}{x+z}= b,\frac{yz}{y+z}= c$, where $a, b, c$ are other than zero, then $x$ equals:
$\textbf{(A)}\ \frac{abc}{ab+ac+bc}\qquad\textbf{(B)}\ \frac{2abc}{ab+bc+ac}\qquad\textbf{(C)}\ \frac{2abc}{ab+ac-bc}$
$\textbf{(D)}\ \frac{2abc}{ab+bc-ac}\qquad\textbf{(E)}\ \frac{2abc}{ac+bc-ab}$ | Note that $\frac{1}{a}=\frac{1}{x}+\frac{1}{y}$, $\frac{1}{b}=\frac{1}{x}+\frac{1}{z}$, and $\frac{1}{c}=\frac{1}{y}+\frac{1}{z}$. Therefore
\[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{2}\]
Therefore
\[\frac{1}{x}=\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}-\frac{1}{c}=\frac{1}{2a}+\frac... | \textbf{(E)}\\frac{2abc}{ac+-} | Algebra | MCQ | Yes | Yes | amc_aime | false |
If you are given $\log 8\approx .9031$ and $\log 9\approx .9542$, then the only logarithm that cannot be found without the use of tables is:
$\textbf{(A)}\ \log 17\qquad\textbf{(B)}\ \log\frac{5}{4}\qquad\textbf{(C)}\ \log 15\qquad\textbf{(D)}\ \log 600\qquad\textbf{(E)}\ \log .4$ | While $\log 17 = \log(8 + 9)$, we cannot easily deal with the logarithm of a sum. Furthermore, $17$ is prime, so none of the logarithm rules involving products or differences works. It therefore cannot be found without the use of a table (note: in 1951, calculators were very rare). The correct answer is therefore $\box... | \textbf{(A)}\\log17 | Algebra | MCQ | Yes | Yes | amc_aime | false |
$AB$ is a fixed diameter of a circle whose center is $O$. From $C$, any point on the circle, a chord $CD$ is drawn perpendicular to $AB$. Then, as $C$ moves over a semicircle, the bisector of angle $OCD$ cuts the circle in a point that always:
$\textbf{(A)}\ \text{bisects the arc }AB\qquad\textbf{(B)}\ \text{trisects ... | Draw in diameter $CE$. Note that, as it is inscribed in a semicircle, $\bigtriangleup CDE$ always has a right angle at $D$. Since it is given that $AB \perp CD$, $AB \parallel DE$ by congruent corresponding angles. $CP$ bisects $DE$, as well as the arc subtended by $DE$, $\widehat{DPE}$. Because $AB \parallel DE$, $CP$... | \textbf{(A)}\bisectsthearcAB | Geometry | MCQ | Yes | Yes | amc_aime | false |
If $r$ and $s$ are the roots of the equation $ax^2+bx+c=0$, the value of $\frac{1}{r^{2}}+\frac{1}{s^{2}}$ is:
$\textbf{(A)}\ b^{2}-4ac\qquad\textbf{(B)}\ \frac{b^{2}-4ac}{2a}\qquad\textbf{(C)}\ \frac{b^{2}-4ac}{c^{2}}\qquad\textbf{(D)}\ \frac{b^{2}-2ac}{c^{2}}$
$\textbf{(E)}\ \text{none of these}$ | Note that $\frac{1}{r^2}+\frac{1}{s^2} = \frac{r^2+s^2}{r^2s^2} = \frac{(r+s)^2-2(rs)}{(rs)^2}$.
By [Vieta's](https://artofproblemsolving.com/wiki/index.php/Vieta%27s), this is $\frac{(-\frac{b}{a})^2-2(\frac{c}{a})}{(\frac{c}{a})^2} = \frac{\frac{b^2}{a^2}-\frac{2c}{a}}{\frac{c^2}{a^2}} = \frac{b^2-2ac}{c^2} \implies... | \frac{b^2-2ac}{^2} | Algebra | MCQ | Yes | Yes | amc_aime | false |
The area of a square inscribed in a semicircle is to the area of the square inscribed in the entire circle as:
$\textbf{(A)}\ 1: 2\qquad\textbf{(B)}\ 2: 3\qquad\textbf{(C)}\ 2: 5\qquad\textbf{(D)}\ 3: 4\qquad\textbf{(E)}\ 3: 5$ | Let the radius of the circle be $r$. Let $s_1$ be the side length of the square inscribed in the semicircle and $s_2$ be the side length of the square inscribed in the entire circle. For the square in the semicircle, we have $s_1^2 + (\frac{s_1}{2})^2 = r^2 \Rightarrow s_1^2 = \frac{4}{5}r^2$. For the square in the cir... | 2:5 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The medians of a right triangle which are drawn from the vertices of the acute angles are $5$ and $\sqrt{40}$. The value of the hypotenuse is:
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 2\sqrt{40}\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 2\sqrt{13}\qquad\textbf{(E)}\ \text{none of these}$ | We will proceed by coordinate bashing.
Call the first leg $2a$, and the second leg $2b$ (We are using the double of a variable to avoid any fractions)
Notice that we want to find $\sqrt{(2a)^2+(2b)^2}$
Two equations can be written for the two medians: $a^2 + 4b^2 = 40$ and $4a^2+b^2= 25$.
Add them together and we get ... | 2\sqrt{13} | Geometry | MCQ | Yes | Yes | amc_aime | false |
Mr. $A$ owns a home worth $$10,000$. He sells it to Mr. $B$ at a $10\%$ profit based on the worth of the house. Mr. $B$ sells the house back to Mr. $A$ at a $10\%$ loss. Then:
$\mathrm{(A) \ A\ comes\ out\ even } \qquad$ $\mathrm{(B) \ A\ makes\ 1100\ on\ the\ deal}$ $\qquad \mathrm{(C) \ A\ makes\ 1000\ on\ the\ de... | Mr. $A$ sells his home for $(1 + 10$%$)$ $\cdot$ $10,000$ dollars $=$ $1.1$ $\cdot$ $10,000$ dollars $=$ $11,000$ dollars to Mr. $B$. Then, Mr. $B$ sells it at a price of $(1-10$%$)$ $\cdot$ $11,000$ dollars $=$ $0.9$ $\cdot$ $11,000$ dollars $=$ $9,900$ dollars, thus $11,000 - 9,900$ $=$ $\boxed{\textrm{(B)}\ \text{A... | 1100 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Tom, Dick and Harry started out on a $100$-mile journey. Tom and Harry went by automobile at the rate of $25$ mph, while Dick walked at the rate of $5$ mph. After a certain distance, Harry got off and walked on at $5$ mph, while Tom went back for Dick and got him to the destination at the same time that Harry arrived. ... | Let $d_1$ be the distance (in miles) that Harry traveled on car, and let $d_2$ be the distance (in miles) that Tom backtracked to get Dick. Let $T$ be the time (in hours) that it took the three to complete the journey. We now examine Harry's journey, Tom's journey, and Dick's journey. These yield, respectively, the equ... | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The bottom, side, and front areas of a rectangular box are known. The product of these areas
is equal to:
$\textbf{(A)}\ \text{the volume of the box} \qquad\textbf{(B)}\ \text{the square root of the volume} \qquad\textbf{(C)}\ \text{twice the volume}$
$\textbf{(D)}\ \text{the square of the volume} \qquad\textbf{(E)}\ \... | Let the length of the edges of this box have lengths $a$, $b$, and $c$. We're given $ab$, $bc$, and $ca$. The product of these values is $a^2b^2c^2$, which is the square of the volume of the box. $\boxed{\textbf{(D)}}$ | \textbf{(D)} | Geometry | MCQ | Yes | Yes | amc_aime | false |
An error of $.02''$ is made in the measurement of a line $10''$ long, while an error of only $.2''$ is made in a measurement of a line $100''$ long. In comparison with the relative error of the first measurement, the relative error of the second measurement is:
$\textbf{(A)}\ \text{greater by }.18 \qquad\textbf{(B)}\ \... | There error percentage of the first measure is $0.2\%$. The error percentage of the second measure is also $0.2\%$. Therefore, the answer is $\textbf{(B)}\ \text{the same}$ | B | Algebra | MCQ | Yes | Yes | amc_aime | false |
The price of an article is cut $10 \%.$ To restore it to its former value, the new price must be increased by:
$\textbf{(A) \ } 10 \% \qquad\textbf{(B) \ } 9 \% \qquad \textbf{(C) \ } 11\frac{1}{9} \% \qquad\textbf{(D) \ } 11 \% \qquad\textbf{(E) \ } \text{none of these answers}$ | [Without loss of generality](https://artofproblemsolving.com/wiki/index.php/Without_loss_of_generality), let the price of the article be 100. Thus, the new price is $.9\cdot 100= 90$. Then, to restore it to the original price, we solve $90x=100$. We find $x=1\dfrac{1}{9}$, thus, the percent increase is $1\dfrac{1}{9}-... | 11\frac{1}{9} | Algebra | MCQ | Yes | Yes | amc_aime | false |
An equilateral triangle is drawn with a side of length $a$. A new equilateral triangle is formed by joining the midpoints of the sides of the first one. Then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. The limit of the sum of the perimeters of all the t... | The perimeter of the first triangle is $3a$. The perimeter of the 2nd triangle is half of that, after drawing a picture. The 3rd triangle's perimeter is half the second's, and so on. Therefore, we are computing $3a+\frac{3a}{2}+\frac{3a}{4}+\cdots$.
The starting term is $3a$, and the common ratio is $1/2$. Therefore, ... | 6a | Geometry | MCQ | Yes | Yes | amc_aime | false |
End of preview. Expand in Data Studio
Competition answer mathematics training pool
Competition mathematics problems that ask for a single final answer, three public datasets gathered
at pinned revisions, shipped twice over. sources/ holds each dataset the way its publisher ships
it, in its own file format with its own fields and nothing renamed, 113045 rows across three
folders. pool/ holds the union of those same datasets in one format, one JSON object per line,
deduplicated by problem text and reduced to 107637 rows, every row labelled with the dataset it
came from, that dataset's revision and its provenance class.
Nothing here is filtered for quality or for difficulty. Problems written by people and problems copied from competition papers sit side by side, labelled, and which of them to use is the reader's decision. Take the union if one format is what you want, take a source folder if you already know that dataset and want its own columns.
Layout
README.md
pool/pool-00000-of-00003.jsonl ... pool-00002-of-00003.jsonl the union, 107637 rows
sources/deepscaler/ 1 file, 38514 rows
sources/omni_math/ 1 file, 4251 rows
sources/numinamath_1_5/ 3 files, 70280 rows
The union is shuffled with a fixed seed, so any prefix of any file is a sample of the whole pool rather than of one source. The pool is about 242 MB on disk, 168 MB of it the union and 74 MB the source folders. One of the three files in the last folder holds no rows at all, because neither of the two subsets kept from that dataset appears in that shard.
The union under pool/
One JSON object per line, with these fields.
| Field | What it holds |
|---|---|
id |
problem_ and a seven digit number, unique across the whole union |
problem |
the problem text, exactly as the source publishes it, LaTeX and all |
answer |
the final answer alone, as a string, or null when the row has none |
solution |
the worked solution, or null when the row has none |
has_answer |
whether answer is not null |
has_solution |
whether solution is not null |
answer_from |
dataset when the source published the answer as its own field, boxed_in_solution when it was read from the last box of the solution because the row has no answer field, null when there is no answer |
source |
the folder under sources/ the row came from |
source_subset |
the value of that source's own subset column, where it has one: the competition for one source, the year for another, the collection for a third |
source_repo |
the Hugging Face repository the source was taken from |
source_revision |
the commit of that repository, so every row is traceable to a pinned revision |
source_file |
the file under sources/ the row came from |
source_row |
the row's position in that file, counting from zero |
provenance |
how the PROBLEM was produced: human, collected or model-generated |
solution_provenance |
how the SOLUTION was produced, on the same three values, null when there is no solution |
Counts over the 107637 rows: 60971 carry an answer and 45625 carry a solution. Of the answers, 60939 are the source's own answer field and 32 were read out of the last box of a solution, and 30239 of them are a plain integer, the rest being closed forms such as fractions, radicals and expressions. By provenance of the problem: 106745 collected, 892 human, none model-generated. Everything is in English.
By source: 69251 rows from numinamath_1_5, 37494 from deepscaler and 892 from omni_math.
The last number is small because the union keeps one row per distinct problem and prefers the
copy that carries the most, an answer first, then a solution, and then the earlier source in the
list below. The deepscaler folder was itself assembled from competition archives and from the
omni_math collection, so most of that collection's problems appear in it as well and are kept
there, with the omni_math folder still holding them in full under sources/.
Roughly a third of the rows have neither an answer nor a solution. Those are the proof and
construction problems of the forum subset, which state a claim to prove rather than a quantity to
compute. If you want only the rows a checker can grade, filter on has_answer, and if you want
only the rows whose answer is an integer, filter on that too.
The sources under sources/
deepscaler, agentica-org/DeepScaleR-Preview-Dataset at
b6ae8c60f5c1f2b594e2140b91c49c9ad0949e29, 38514 rows, one JSON file holding a list of objects
with problem, answer and solution, the answer being the final answer alone. MIT. Provenance
collected, solutions collected. About forty thousand competition problems with a single short
answer, compiled from the papers of the American Invitational Mathematics Examination of 1984 to
2023, the American Mathematics Competitions papers before 2023, the Omni-MATH collection and the
STILL collection. The problems and the worked solutions are the ones the competitions and the
archives published. A model was used only to read the final answer out of each solution, and
answers a symbolic checker could not handle were dropped. Most of its rows carry an answer and no
solution text.
omni_math, KbsdJames/Omni-MATH at 40ba231d8f16e29ecd40e6407e2c8640145a8f62, 4251 rows, one
JSON object per line with problem, solution, answer, domain, difficulty and source, the source
naming the competition. Apache 2.0. Provenance human, solutions human. Olympiad problems collected
from the official competition websites and the Art of Problem Solving wiki, with the solutions
those sites published and a difficulty rating and a subject path added by hand. Problems and
solutions are both written by people. Its source column names more than eighty competitions, the
largest of them the International Mathematical Olympiad shortlists, the Harvard MIT Mathematics
Tournament, the United States of America Mathematical Olympiad and the Putnam competition.
numinamath_1_5, AI-MO/NuminaMath-1.5 at 1b05109f9e5c1ad06c0663519502416c30b300f8, 70280
rows, parquet shards with problem, solution, answer, problem_type, question_type,
problem_is_valid, solution_is_valid, source and synthetic. An answer of proof or notfound means
the row has no short answer. Apache 2.0. Provenance collected, solutions model-generated. Two
subsets of the largest compilation of competition mathematics on the hub, kept and the rest left
out: aops_forum, the problems posted to the Art of Problem Solving forum, 64183 rows of the
union, and amc_aime, the problems of the American Mathematics Competitions and American
Invitational Mathematics Examination archives, 5068 rows of the union. Both are problems people
wrote and posted, gathered by crawling. The solutions were rewritten into a uniform style by a
large language model whatever the problem's origin, so the solution text is model-generated across
the whole file, and the short answers were extracted by the same model. The compilation's school
mathematics, its Chinese curriculum material and its model-written problems are not here: this
pool is competition mathematics with a short answer.
Provenance in plain words
- human: a person wrote the text, with no model in the loop.
- collected: the text is a real artefact of people doing something, a competition paper, an examination, a forum post, gathered rather than written for a dataset.
- model-generated: a language model wrote it. No problem in this pool is in this class. Every solution of the largest folder is, because that compilation rewrote its solutions with a model.
- rule-generated: a program or a template produced it. Nothing in this pool is in this class.
Almost all of the pool is collected: real competition problems, gathered at scale. If you care
that the reasoning was written by a person rather than a model, filter solution_provenance to
human and collected, which leaves the omni_math rows and the deepscaler rows that carry a
solution.
Licences
Each source is redistributed here under the licence its publisher declares for it, and the licence of every row is the licence of its source folder.
| Source | Licence |
|---|---|
| deepscaler | MIT |
| omni_math | Apache 2.0 |
| numinamath_1_5 | Apache 2.0 |
Apache 2.0 is the more demanding of the two, so it governs the pool as a whole. Nothing here is licensed for non-commercial use and nothing here is untagged. The compilations gather material from many places, and the licence above is the one their publisher applies to the compilation.
Several large corpora that would otherwise belong here are deliberately absent. Some competition archives exist on the hub only under a non-commercial share-alike licence and cannot be redistributed at all, so the omni_math folder is what carries the problems of those competitions that it happens to hold. The MATH training corpus is not included: the original upload was disabled on a takedown notice from the rightsholder, which also asked for datasets branched off it to be removed, and of the mirrors that remain one declares no licence at all. The Art of Problem Solving instruction corpus is not included either: its authors disclaim the licence tag the upload carries, and the forum it was crawled from now publishes a machine-readable reservation of rights against training use.
A note on the counts
The counts above are the counts in this pool. Rows were removed from every source while it was assembled, so they do not match the totals the source repositories publish.
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