competition_id
string | problem_id
int64 | difficulty
int64 | category
string | problem_type
string | problem
string | solutions
list | solutions_count
int64 | source_file
string | competition
string |
|---|---|---|---|---|---|---|---|---|---|
1958_AHSME_Problems
| 44
| 0
|
Other
|
Multiple Choice
|
Given the true statements: (1) If $a$ is greater than $b$, then $c$ is greater than $d$ (2) If $c$ is less than $d$, then $e$ is greater than $f$. A valid conclusion is:
$\textbf{(A)}\ \text{If }{a}\text{ is less than }{b}\text{, then }{e}\text{ is greater than }{f}\qquad \\ \textbf{(B)}\ \text{If }{e}\text{ is greater than }{f}\text{, then }{a}\text{ is less than }{b}\qquad \\ \textbf{(C)}\ \text{If }{e}\text{ is less than }{f}\text{, then }{a}\text{ is greater than }{b}\qquad \\ \textbf{(D)}\ \text{If }{a}\text{ is greater than }{b}\text{, then }{e}\text{ is less than }{f}\qquad \\ \textbf{(E)}\ \text{none of these}$
|
[
"(A) is not valid because $a<b$ does not imply $c\\le d$ (The inverse of Statement 1 is not necessarily true).\n\n\n(B) is not valid because $e>f$ does not imply $c<d$ (The converse of Statement 2 is not necessarily true either).\n\n\nIn (C), $e<f$ does imply $c\\ge d$ (contrapositive of Statement 2) but this does not tell us anything about $a$ and $b$.\n\n\nIn (D), $a>b$ does imply $c>d$, but the inverse of Statement 2 ($e\\le f$) is not necessarily true.\n\n\nSo the answer is $\\boxed{\\textbf{(E) } \\text{None of these}}$.\n(Incidentally, the wording of this problem appears to be wrongly assuming that $x\\not<y$ implies $x>y$ (and vice versa) when in reality $x$ and $y$ could be equal. However, this does not change the fact that none of the four choices follow logically from the given statements.)\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/44.json
|
AHSME
|
1958_AHSME_Problems
| 13
| 0
|
Algebra
|
Multiple Choice
|
The sum of two numbers is $10$; their product is $20$. The sum of their reciprocals is:
$\textbf{(A)}\ \frac{1}{10}\qquad \textbf{(B)}\ \frac{1}{2}\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 4$
|
[
"$x+y=10$\n\n\n$xy=20$\n\n\n$\\frac1x+\\frac1y=\\frac{y}{xy}+\\frac{x}{xy}=\\frac{x+y}{xy}=\\frac{10}{20}=\\boxed{\\frac12\\textbf{ (B)}}$\n\n\n(Which is a lot easier than finding the roots of $x^2-10x+20$.)\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/13.json
|
AHSME
|
1958_AHSME_Problems
| 29
| 0
|
Geometry
|
Multiple Choice
|
In a general triangle $ADE$ (as shown) lines $\overline{EB}$ and $\overline{EC}$ are drawn. Which of the following angle relations is true?
[asy] draw((-8,0)--(-2,0)--(4,0)--(10,0)--(0,10)--cycle,dot); draw((-2,0)--(0,10),dot);draw((4,0)--(0,10),dot); MP("A",(-8,0),S);MP("B",(-2,0),S);MP("C",(4,0),S);MP("D",(10,0),S);MP("E",(0,10),N); MP("x",(-7.9,.4),E);MP("z",(-2,.4),W);MP("m",(-2,.4),E);MP("n",(4,.4),W);MP("c",(4,.4),E);MP("a",(9.9,.4),W); MP("y",(-.2,8.8),SW);MP("w",(.1,8.8),S);MP("b",(.7,9),SE); [/asy]
$\textbf{(A)}\ x \plus{} z \equal{} a \plus{} b\qquad \textbf{(B)}\ y \plus{} z \equal{} a \plus{} b\qquad \textbf{(C)}\ m \plus{} x \equal{} w \plus{} n\qquad \\
\textbf{(D)}\ x \plus{} z \plus{} n \equal{} w \plus{} c \plus{} m\qquad \textbf{(E)}\ x \plus{} y \plus{} n \equal{} a \plus{} b \plus{} m$ (Error compiling LaTeX. Unknown error_msg)
|
[
"$\\fbox{}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/29.json
|
AHSME
|
1958_AHSME_Problems
| 3
| 0
|
Algebra
|
Multiple Choice
|
Of the following expressions the one equal to $\frac{a^{-1}b^{-1}}{a^{-3} - b^{-3}}$ is:
$\textbf{(A)}\ \frac{a^2b^2}{b^2 - a^2}\qquad \textbf{(B)}\ \frac{a^2b^2}{b^3 - a^3}\qquad \textbf{(C)}\ \frac{ab}{b^3 - a^3}\qquad \textbf{(D)}\ \frac{a^3 - b^3}{ab}\qquad \textbf{(E)}\ \frac{a^2b^2}{a - b}$
|
[
"$\\frac{a^{-1}b^{-1}}{a^{-3} - b^{-3}} = \\frac{\\frac{1}{ab}}{\\frac{1}{a^{3}}-\\frac{1}{b^{3}}} = \\frac{\\frac{1}{ab}}{\\frac{1}{a^{3}}-\\frac{1}{b^{3}}}\\cdot\\frac{a^{3}b^{3}}{a^{3}b^{3}} = \\frac{a^{2}b^{2}}{b^{3}-a^{3}}$, $\\boxed{\\text{B}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/3.json
|
AHSME
|
1958_AHSME_Problems
| 34
| 0
|
Algebra
|
Multiple Choice
|
The numerator of a fraction is $6x + 1$, then denominator is $7 - 4x$, and $x$ can have any value between $-2$ and $2$, both included. The values of $x$ for which the numerator is greater than the denominator are:
$\textbf{(A)}\ \frac{3}{5} < x \le 2\qquad \textbf{(B)}\ \frac{3}{5} \le x \le 2\qquad \textbf{(C)}\ 0 < x \le 2\qquad \\ \textbf{(D)}\ 0 \le x \le 2\qquad \textbf{(E)}\ -2 \le x \le 2$
|
[
"$\\fbox{}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/34.json
|
AHSME
|
1958_AHSME_Problems
| 8
| 0
|
Number Theory
|
Multiple Choice
|
Which of these five numbers $\sqrt{\pi^2},\,\sqrt[3]{.8},\,\sqrt[4]{.00016},\,\sqrt[3]{-1}\cdot \sqrt{(.09)^{-1}}$, is (are) rational:
$\textbf{(A)}\ \text{none}\qquad \textbf{(B)}\ \text{all}\qquad \textbf{(C)}\ \text{the first and fourth}\qquad \textbf{(D)}\ \text{only the fourth}\qquad \textbf{(E)}\ \text{only the first}$
|
[
"$\\sqrt{\\pi^2}=\\pi$ is not rational, so eliminate choices $(B)$, $(C)$, and $(E)$. Note that $(-1)^3 = (-1)$, so $\\sqrt[3]{-1}=-1$ is rational. We have found one rational number, so we can eliminate choice $(A)$. The answer is $\\boxed{ \\text{(D)}}$.\n\n\n\n\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/8.json
|
AHSME
|
1958_AHSME_Problems
| 22
| 0
|
Algebra
|
Multiple Choice
|
A particle is placed on the parabola $y = x^2- x -6$ at a point $P$ whose $y$-coordinate is $6$. It is allowed to roll along the parabola until it reaches the nearest point $Q$ whose $y$-coordinate is $-6$. The horizontal distance traveled by the particle (the numerical value of the difference in the $x$-coordinates of $P$ and $Q$) is:
$\textbf{(A)}\ 5\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 1$
|
[
"$\\fbox{}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/22.json
|
AHSME
|
1958_AHSME_Problems
| 18
| 0
|
Algebra
|
Multiple Choice
|
The area of a circle is doubled when its radius $r$ is increased by $n$. Then $r$ equals:
$\textbf{(A)}\ n(\sqrt{2} + 1)\qquad \textbf{(B)}\ n(\sqrt{2} - 1)\qquad \textbf{(C)}\ n\qquad \textbf{(D)}\ n(2 - \sqrt{2})\qquad \textbf{(E)}\ \frac{n\pi}{\sqrt{2} + 1}$
|
[
"Since the new circle has twice the area of the original circle, its radius is $\\sqrt{2}$ times the old radius. \nThus, \n\\[r + n = r\\sqrt{2}\\]\n\\[n = r\\sqrt{2} - r\\]\n\\[n = r(\\sqrt{2} - 1)\\]\n\\[r = \\frac{n}{\\sqrt{2} - 1}\\]\nRationalizing the denominator yields\n\\[r = \\frac{n}{\\sqrt{2} - 1} * \\frac{\\sqrt{2} + 1}{\\sqrt{2} + 1} = n(\\sqrt{2} + 1)\\]\n\n\nTherefore, the answer is $\\fbox{(A)}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/18.json
|
AHSME
|
1958_AHSME_Problems
| 38
| 0
|
Geometry
|
Multiple Choice
|
Let $r$ be the distance from the origin to a point $P$ with coordinates $x$ and $y$. Designate the ratio $\frac{y}{r}$ by $s$ and the ratio $\frac{x}{r}$ by $c$. Then the values of $s^2 - c^2$ are limited to the numbers:
$\textbf{(A)}\ \text{less than }{-1}\text{ are greater than }{+1}\text{, both excluded}\qquad\\ \textbf{(B)}\ \text{less than }{-1}\text{ are greater than }{+1}\text{, both included}\qquad \\ \textbf{(C)}\ \text{between }{-1}\text{ and }{+1}\text{, both excluded}\qquad \\ \textbf{(D)}\ \text{between }{-1}\text{ and }{+1}\text{, both included}\qquad \\ \textbf{(E)}\ {-1}\text{ and }{+1}\text{ only}$
|
[
"$\\fbox{}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/38.json
|
AHSME
|
1958_AHSME_Problems
| 4
| 0
|
Algebra
|
Multiple Choice
|
In the expression $\frac{x + 1}{x - 1}$ each $x$ is replaced by $\frac{x + 1}{x - 1}$. The resulting expression, evaluated for $x = \frac{1}{2}$, equals:
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ -3\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ -1\qquad \textbf{(E)}\ \text{none of these}$
|
[
"When $x=\\frac{1}{2}$, $\\frac{x+1}{x-1}=-3$, substituting $-3$ for $x$ in the original equation we get:\n\n\n$\\frac{-3+1}{-3-1}=\\frac{-2}{-4}=\\frac{1}{2}\\implies \\boxed{\\mathbf{(E)}\\text{ None of these}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/4.json
|
AHSME
|
1958_AHSME_Problems
| 14
| 0
|
Algebra
|
Multiple Choice
|
At a dance party a group of boys and girls exchange dances as follows: The first boy dances with $5$ girls, a second boy dances with $6$ girls, and so on, the last boy dancing with all the girls. If $b$ represents the number of boys and $g$ the number of girls, then:
$\textbf{(A)}\ b = g\qquad \textbf{(B)}\ b = \frac{g}{5}\qquad \textbf{(C)}\ b = g - 4\qquad \textbf{(D)}\ b = g - 5\qquad \\ \textbf{(E)}\ \text{It is impossible to determine a relation between }{b}\text{ and }{g}\text{ without knowing }{b + g.}$
|
[
"After inspection, we notice a general pattern: the $nth$ boy dances with $n + 4$ girls. \nSince the last boy dances with all the girls, there must be four more girls than guys.\n\n\nTherefore, the equation that relates them is $\\fbox{(C) b = g - 4}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/14.json
|
AHSME
|
1958_AHSME_Problems
| 43
| 0
|
Geometry
|
Multiple Choice
|
$\overline{AB}$ is the hypotenuse of a right triangle $ABC$. Median $\overline{AD}$ has length $7$ and median $\overline{BE}$ has length $4$. The length of $\overline{AB}$ is:
$\textbf{(A)}\ 10\qquad \textbf{(B)}\ 5\sqrt{3}\qquad \textbf{(C)}\ 5\sqrt{2}\qquad \textbf{(D)}\ 2\sqrt{13}\qquad \textbf{(E)}\ 2\sqrt{15}$
|
[
"[asy] import geometry; unitsize(50); pair C = (0,0), B = (3,0), A = (0, 4); pair AC = midpoint(A--C), BC = midpoint(B--C); draw(A--B--C--A); draw(A--C, StickIntervalMarker(2, 1)); draw(B--C, StickIntervalMarker(2, 2)); draw(B--AC); draw(A--BC); dot(AC); dot(BC); MP(\"$A$\", A, W); MP(\"$B$\", B, E); MP(\"$C$\", C, W); MP(\"$E$\", AC, W); MP(\"$D$\", BC, S); label(\"$y$\", A--AC, W); label(\"$y$\", AC--C, W); label(\"$x$\", B--BC, S); label(\"$x$\", BC--C, S); draw(rightanglemark(A, C, B)); [/asy]\n\n\nBy the Pythagorean Theorem, $(2x)^2+y^2=BE^2$, and $x^2+(2y)^2=AD^2$.\n\n\nPlugging in, $4x^2+y^2=16$, and $x^2+4y^2=49$.\n\n\nAdding the equations, $5x^2+5y^2=65$, and dividing by $5$, $x^2+y^2=13$.\n\n\n\n\n\n\nNote that the length $AB^2$ is equal to $(2x)^2+(2y)^2=4x^2+4y^2=4(x^2+y^2)$.\n\n\nTherefore, the answer is $\\sqrt{4(13)}=2\\sqrt{13}$ $\\fbox{D}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/43.json
|
AHSME
|
1958_AHSME_Problems
| 42
| 0
|
Geometry
|
Multiple Choice
|
In a circle with center $O$, chord $\overline{AB}$ equals chord $\overline{AC}$. Chord $\overline{AD}$ cuts $\overline{BC}$ in $E$. If $AC = 12$ and $AE = 8$, then $AD$ equals:
$\textbf{(A)}\ 27\qquad \textbf{(B)}\ 24\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 20\qquad \textbf{(E)}\ 18$
|
[
"Let $X$ be a point on $BC$ so $AX \\perp BC$. Let $AX = h$, $EX = \\sqrt{64 - h^2}$ and $BX = \\sqrt{144 - h^2}$. $CE = CX - EX = \\sqrt{144 - h^2} - \\sqrt{64 - h^2}$. Using Power of a Point on $E$, $(BE)(EC) = (AE)(ED)$ (there isn't much information about the circle so I wanted to use PoP). \n\n\n\\[(\\sqrt{144 - h^2} - \\sqrt{64 - h^2})(\\sqrt{144 - h^2} + \\sqrt{64 - h^2}) = 8(ED)\\]\n\n\n\\[(144 - h^2) - (64 - h^2) = 8(ED)\\]\n\n\n\\[80 = 8(ED)\\]\n\n\n\\[ED = 10\\]\n\n\nAdding up $AD$ and $ED$ we get $\\fbox{E}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/42.json
|
AHSME
|
1958_AHSME_Problems
| 15
| 0
|
Geometry
|
Multiple Choice
|
A quadrilateral is inscribed in a circle. If an angle is inscribed into each of the four segments outside the quadrilateral, the sum of these four angles, expressed in degrees, is:
$\textbf{(A)}\ 1080\qquad \textbf{(B)}\ 900\qquad \textbf{(C)}\ 720\qquad \textbf{(D)}\ 540\qquad \textbf{(E)}\ 360$
|
[
"$\\fbox{}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/15.json
|
AHSME
|
1958_AHSME_Problems
| 5
| 0
|
Algebra
|
Multiple Choice
|
The expression $2 + \sqrt{2} + \frac{1}{2 + \sqrt{2}} + \frac{1}{\sqrt{2} - 2}$ equals:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 2 - \sqrt{2}\qquad \textbf{(C)}\ 2 + \sqrt{2}\qquad \textbf{(D)}\ 2\sqrt{2}\qquad \textbf{(E)}\ \frac{\sqrt{2}}{2}$
|
[
"To make this problem easier to solve, lets get the radicals out of the denominator. For $\\frac{1}{2 + \\sqrt2}$, we will multiply the numerator and denominator by $2- \\sqrt2$ so,\n\n\n$\\frac{1}{2 + \\sqrt2} \\cdot \\frac{2 - \\sqrt2}{2 - \\sqrt2} \\Rightarrow \\frac{2 - \\sqrt2}{2}$.\n\n\nNow, the other fraction we need to get the radical out of the denominator is $\\frac{1}{\\sqrt2 - 2}$. Here, we will multiply by the conjugate again, $\\sqrt2 + 2$. So that simplifies to\n\n\n$\\frac{1}{\\sqrt2 - 2} \\cdot \\frac{\\sqrt2 + 2}{\\sqrt2 + 2} \\Rightarrow \\frac{\\sqrt2 + 2}{-2}$.\n\n\nSo now our simplified equation is\n\n\n$2 + \\sqrt2 + \\frac{2- \\sqrt2}{2} + \\frac{\\sqrt2 + 2}{-2} \\Rightarrow 2+ \\sqrt2 + \\frac{2 - \\sqrt2}{2} - \\frac{\\sqrt2 +2}{2}$\n\n\nBringing everything to the same denominator and combining like terms, we get\n\n\n$\\frac{4 + 2\\sqrt2 + 2 - \\sqrt2 - \\sqrt2 - 2}{2} \\Rightarrow \\frac{4}{2} \\Rightarrow 2 \\Rightarrow \\boxed{A}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/5.json
|
AHSME
|
1958_AHSME_Problems
| 39
| 0
|
Algebra
|
Multiple Choice
|
We may say concerning the solution of $|x|^2 + |x| - 6 =0$ that:
$\textbf{(A)}\ \text{there is only one root}\qquad \textbf{(B)}\ \text{the sum of the roots is }{+1}\qquad \textbf{(C)}\ \text{the sum of the roots is }{0}\qquad \\ \textbf{(D)}\ \text{the product of the roots is }{+4}\qquad \textbf{(E)}\ \text{the product of the roots is }{-6}$
|
[
"Note that for all roots $x$, $-x$ will also be a root. Therefore, the sum of all of the roots will be $0$, making the answer $\\fbox{C}$\n\n\n\n\n\n\nWe can find all roots $x$ by setting $|x| = y$. This gives us the equation $y^2+y-6=0$, which has the solutions $y=-3, 2$. However, $|x|$ cannot equal $-3$, so the roots for $x$ are $2$ and $-2$. The sum of the two roots is $0$, making the answer $\\fbox{C}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/39.json
|
AHSME
|
1958_AHSME_Problems
| 19
| 0
|
Geometry
|
Multiple Choice
|
The sides of a right triangle are $a$ and $b$ and the hypotenuse is $c$. A perpendicular from the vertex divides $c$ into segments $r$ and $s$, adjacent respectively to $a$ and $b$. If $a : b = 1 : 3$, then the ratio of $r$ to $s$ is:
$\textbf{(A)}\ 1 : 3\qquad \textbf{(B)}\ 1 : 9\qquad \textbf{(C)}\ 1 : 10\qquad \textbf{(D)}\ 3 : 10\qquad \textbf{(E)}\ 1 : \sqrt{10}$
|
[
"$\\fbox{}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/19.json
|
AHSME
|
1958_AHSME_Problems
| 23
| 0
|
Algebra
|
Multiple Choice
|
If, in the expression $x^2 - 3$, $x$ increases or decreases by a positive amount of $a$, the expression changes by an amount:
$\textbf{(A)}\ {\pm 2ax + a^2}\qquad \textbf{(B)}\ {2ax \pm a^2}\qquad \textbf{(C)}\ {\pm a^2 - 3} \qquad \textbf{(D)}\ {(x + a)^2 - 3}\qquad\\ \textbf{(E)}\ {(x - a)^2 - 3}$
|
[
"Let us represent the increase or decrease in $x$ by $(x \\pm a)$\n\n\nThus our original expression becomes\n\\[(x \\pm a)^2 - 3\\]\n\\[x^2 \\pm 2ax + a^2 - 3\\]\nThe absolute difference between these two expressions is $\\pm 2ax + a^2$.\n\n\nTherefore, the answer is $\\fbox{(A)}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/23.json
|
AHSME
|
1958_AHSME_Problems
| 9
| 0
|
Algebra
|
Multiple Choice
|
A value of $x$ satisfying the equation $x^2 + b^2 = (a - x)^2$ is:
$\textbf{(A)}\ \frac{b^2 + a^2}{2a}\qquad \textbf{(B)}\ \frac{b^2 - a^2}{2a}\qquad \textbf{(C)}\ \frac{a^2 - b^2}{2a}\qquad \textbf{(D)}\ \frac{a - b}{2}\qquad \textbf{(E)}\ \frac{a^2 - b^2}{2}$
|
[
"Solve for x:\n\n\n\\[x^2+b^2=(a-x)^2\\]\n\\[x^2+b^2=x^2-2ax+a^2\\]\n\\[b^2=-2ax+a^2\\]\n\\[2ax=a^2-b^2\\]\n\\[x=\\frac{a^2-b^2}{2a} \\to \\boxed{\\text{(C)}}\\]\n\n\n\n\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/9.json
|
AHSME
|
1958_AHSME_Problems
| 35
| 0
|
Geometry
|
Multiple Choice
|
A triangle is formed by joining three points whose coordinates are integers. If the $x$-coordinate and the $y$-coordinate each have a value of $1$, then the area of the triangle, in square units:
$\textbf{(A)}\ \text{must be an integer}\qquad \textbf{(B)}\ \text{may be irrational}\qquad \textbf{(C)}\ \text{must be irrational}\qquad \textbf{(D)}\ \text{must be rational}\qquad \\ \textbf{(E)}\ \text{will be an integer only if the triangle is equilateral.}$
|
[
"$\\fbox{}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1958_AHSME_Problems/35.json
|
AHSME
|
1957_AHSME_Problems
| 16
| 0
|
Algebra
|
Multiple Choice
|
Goldfish are sold at $15$ cents each. The rectangular coordinate graph showing the cost of $1$ to $12$ goldfish is:
$\textbf{(A)}\ \text{a straight line segment} \qquad \\ \textbf{(B)}\ \text{a set of horizontal parallel line segments}\qquad\\ \textbf{(C)}\ \text{a set of vertical parallel line segments}\qquad\\ \textbf{(D)}\ \text{a finite set of distinct points} \qquad\textbf{(E)}\ \text{a straight line}$
|
[
"$1$ goldfish costs $1 \\cdot 15 = 15$ cents, $2$ goldfish cost $2 \\cdot 15 = 30$ cents, and so on until $12$ fish. One may think that it is a straight line segment at first, but note that there cannot be a third of a goldfish sold (if so then that would be creepy). Thus our answer is $\\boxed{\\textbf{(B)}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/16.json
|
AHSME
|
1957_AHSME_Problems
| 6
| 0
|
Algebra
|
Multiple Choice
|
An open box is constructed by starting with a rectangular sheet of metal $10$ in. by $14$ in. and cutting a square of side $x$ inches from each corner. The resulting projections are folded up and the seams welded. The volume of the resulting box is:
$\textbf{(A)}\ 140x - 48x^2 + 4x^3 \qquad \textbf{(B)}\ 140x + 48x^2 + 4x^3\qquad \\ \textbf{(C)}\ 140x+24x^2+x^3\qquad \textbf{(D)}\ 140x-24x^2+x^3\qquad \textbf{(E)}\ \text{none of these}$
|
[
"The resulting metal piece looks something like this where the white parts are squares of length $x$:\n\n\n[asy] fill((0,4)--(4,4)--(4,0)--(6,0)--(6,4)--(10,4)--(10,10)--(6,10)--(6,14)--(4,14)--(4,10)--(0,10)--cycle,grey); draw((0,0)--(14-4,0)--(10,14)--(0,14)--cycle); draw((0,0)--(0,4)--(4,4)--(4,0)--cycle); draw((10-4,0)--(10,0)--(10,4)--(6,4)--cycle); draw((0,14)--(4,14)--(4,10)--(0,10)--cycle); draw((6,14)--(6,10)--(10,10)--(10,14)--cycle); [/asy]\n\n\nFrom here, try to visualize the rectangular prism coming together and realize the height is $x$, the length is $14-2x$, and the width is $10-2x$. Therefore, the volume is $x(14-2x)(10-2x)=x(4x^2-48x+40)= \\boxed{\\textbf{(A) } 140x - 48x^2 + 4x^3}$.\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/6.json
|
AHSME
|
1957_AHSME_Problems
| 7
| 0
|
Geometry
|
Multiple Choice
|
The area of a circle inscribed in an equilateral triangle is $48\pi$. The perimeter of this triangle is:
$\textbf{(A)}\ 72\sqrt{3} \qquad \textbf{(B)}\ 48\sqrt{3}\qquad \textbf{(C)}\ 36\qquad \textbf{(D)}\ 24\qquad \textbf{(E)}\ 72$
|
[
"[asy] draw((-3,-sqrt(3))--(3,-sqrt(3))--(0,2sqrt(3))--cycle); draw(circle((0,0),sqrt(3))); dot((0,0)); draw((0,0)--(0,-sqrt(3))); [/asy]\nWe can see that the radius of the circle is $4\\sqrt{3}$. We know that the radius is $\\frac{1}{3}$ of each median line of the triangle; each median line is therefore $12\\sqrt{3}$. Since the median line completes a $30$-$60$-$90$ triangle, we can conclude that one of the sides of the triangle is $24$. Triple the side length and we get our answer, $\\boxed{\\textbf{(E)}72}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/7.json
|
AHSME
|
1957_AHSME_Problems
| 1
| 0
|
Geometry
|
Multiple Choice
|
The number of distinct lines representing the altitudes, medians, and interior angle bisectors of a triangle that is isosceles, but not equilateral, is:
$\textbf{(A)}\ 9\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 5\qquad \textbf{(E)}\ 3$
|
[
"[asy] size(2cm); draw((-3,0)--(0,4)--(3,0)--cycle); draw((0,0)--(0,4), red); draw((-3,0)--(0.84, 2.88), green); draw((-3,0)--(1.5, 2), green); draw((-3,0)--(1.636, 1.818), green); draw((3,0)--(-0.84, 2.88), blue); draw((3,0)--(-1.5, 2), blue); draw((3,0)--(-1.636, 1.818), blue); [/asy]\n\n\nAs shown in the diagram above, all nine altitudes, medians, and interior angle bisectors are distinct, except for the three coinciding lines from the vertex opposite to the base. Thusly, there are $7$ distinct lines, so our answer is $\\boxed{\\textbf{(B)}}$, and we are done.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/1.json
|
AHSME
|
1957_AHSME_Problems
| 11
| 0
|
Geometry
|
Multiple Choice
|
The angle formed by the hands of a clock at $2:15$ is:
$\textbf{(A)}\ 30^\circ \qquad \textbf{(B)}\ 27\frac{1}{2}^\circ\qquad \textbf{(C)}\ 157\frac{1}{2}^\circ\qquad \textbf{(D)}\ 172\frac{1}{2}^\circ\qquad \textbf{(E)}\ \text{none of these}$
|
[
"To find the angle of the clock, we first have to determine where the hands are. The time is $2.25$ hours, so the hour hand would be $62.5$ degrees clockwise. As the clock is a quarter of the way through the hour, the minute hand is $90$ degrees clockwise.\n\n\nThus, we can say that the angle formed by the hands is $90 - 62.5 = \\boxed{\\textbf{(B) } 27\\frac{1}{2}}$ degrees.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/11.json
|
AHSME
|
1957_AHSME_Problems
| 48
| 0
|
Geometry
|
Multiple Choice
|
Let $ABC$ be an equilateral triangle inscribed in circle $O$. $M$ is a point on arc $BC$.
Lines $\overline{AM}$, $\overline{BM}$, and $\overline{CM}$ are drawn. Then $AM$ is:
[asy] defaultpen(linewidth(.8pt)); unitsize(2cm); pair O = origin; pair B = (1,0); pair C = dir(120); pair A = dir(240); pair M = dir(90 - 18); draw(Circle(O,1)); draw(A--C--M--B--cycle); draw(B--C); draw(A--M); dot(O); label("$A$",A,SW); label("$B$",B,E); label("$M$",M,NE); label("$C$",C,NW); label("$O$",O,SE);[/asy]
$\textbf{(A)}\ \text{equal to }{BM + CM}\qquad \textbf{(B)}\ \text{less than }{BM + CM}\qquad \\ \textbf{(C)}\ \text{greater than }{BM+CM}\qquad \\ \textbf{(D)}\ \text{equal, less than, or greater than }{BM + CM}\text{, depending upon the position of } {M}\qquad \\ \textbf{(E)}\ \text{none of these}$
|
[
"Since quadrilateral $ABMC$ is inscribed in circle $O$, thus it is a cyclic quadrilateral. By Ptolemy's Theorem, \\[AC \\cdot MB + MC \\cdot AB = BC \\cdot AM.\\] Because $\\triangle ABC$ is equilateral, we cancel out $AB$, $AC$, and $BC$ to get that \\[BM + CM = AM \\implies \\boxed{\\textbf{(A)}}.\\]\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/48.json
|
AHSME
|
1957_AHSME_Problems
| 33
| 0
|
Algebra
|
Multiple Choice
|
If $9^{x + 2} = 240 + 9^x$, then the value of $x$ is:
$\textbf{(A)}\ 0.1 \qquad \textbf{(B)}\ 0.2\qquad \textbf{(C)}\ 0.3\qquad \textbf{(D)}\ 0.4\qquad \textbf{(E)}\ 0.5$
|
[
"$9^{x+2}$ can be rewritten as $9^x*9^2=9^x*81$, which means the equation can be rewritten as $81(9^x)=240+9^x$, or $80(9^x)=240$, or $9^x=3$. Therefore, $x=\\boxed{\\textbf{(E) }0.5}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/33.json
|
AHSME
|
1957_AHSME_Problems
| 3
| 0
|
Algebra
|
Multiple Choice
|
The simplest form of $1 - \frac{1}{1 + \frac{a}{1 - a}}$ is:
$\textbf{(A)}\ {a}\text{ if }{a\not= 0} \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ {a}\text{ if }{a\not=-1}\qquad \textbf{(D)}\ {1-a}\text{ with not restriction on }{a}\qquad \textbf{(E)}\ {a}\text{ if }{a\not= 1}$
|
[
"We have $1 - \\frac{1}{1 + \\frac{a}{1 - a}} = 1 - \\frac{1}{\\frac{1}{1-a}} = 1 - \\frac{1-a}{1} = a$ for almost all $a$. However, the first step is invalid when $a=1$, and each step is valid otherwise, so the answer is (E).\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/3.json
|
AHSME
|
1957_AHSME_Problems
| 8
| 0
|
Algebra
|
Multiple Choice
|
The numbers $x,\,y,\,z$ are proportional to $2,\,3,\,5$. The sum of $x, y$, and $z$ is $100$. The number y is given by the equation $y = ax - 10$. Then a is:
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ \frac{3}{2}\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ \frac{5}{2}\qquad \textbf{(E)}\ 4$
|
[
"In order to solve the problem, we first need to find each of the three variables. We can use the proportions the problem gives us to find the value of one part, and, by extension, the values of the variables (as $x$ would have $2$ parts, $y$ would have $3$, and $z$ would have $5$). One part, after some algebra, equals $10$, so $x$, $y$, and $z$ are $20$, $30$, and $50$, respectively.\n\n\nWe can plug $x$ and $y$ into the equation given to us: $30 = 20a-10$, and then solve to get $a = \\boxed{\\textbf{(A)}2}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/8.json
|
AHSME
|
1957_AHSME_Problems
| 18
| 0
|
Geometry
|
Multiple Choice
|
Circle $O$ has diameters $AB$ and $CD$ perpendicular to each other. $AM$ is any chord intersecting $CD$ at $P$.
Then $AP\cdot AM$ is equal to:
[asy] defaultpen(linewidth(.8pt)); unitsize(2cm); pair O = origin; pair A = (-1,0); pair B = (1,0); pair C = (0,1); pair D = (0,-1); pair M = dir(45); pair P = intersectionpoint(O--C,A--M); draw(Circle(O,1)); draw(A--B); draw(C--D); draw(A--M); label("$A$",A,W); label("$B$",B,E); label("$C$",C,N); label("$D$",D,S); label("$M$",M,NE); label("$O$",O,NE); label("$P$",P,NW);[/asy]
$\textbf{(A)}\ AO\cdot OB \qquad \textbf{(B)}\ AO\cdot AB\qquad \\ \textbf{(C)}\ CP\cdot CD \qquad \textbf{(D)}\ CP\cdot PD\qquad \textbf{(E)}\ CO\cdot OP$
|
[
"Draw $MB$. Since $\\angle AMB$ is inscribed on a diameter, $\\angle AMB$ is $90^\\circ$. By AA Similarity, $\\triangle APO ~ \\triangle ABM$. Setting up ratios, we get $\\frac{AP}{AO}=\\frac{AB}{AM}$. Cross-multiplying, we get $AP\\cdot AM = AO \\cdot AB$, so the answer is $\\textbf{(B)}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/18.json
|
AHSME
|
1957_AHSME_Problems
| 38
| 0
|
Number Theory
|
Multiple Choice
|
From a two-digit number $N$ we subtract the number with the digits reversed and find that the result is a positive perfect cube. Then:
$\textbf{(A)}\ {N}\text{ cannot end in 5}\qquad\\ \textbf{(B)}\ {N}\text{ can end in any digit other than 5}\qquad \\ \textbf{(C)}\ {N}\text{ does not exist}\qquad\\ \textbf{(D)}\ \text{there are exactly 7 values for }{N}\qquad\\ \textbf{(E)}\ \text{there are exactly 10 values for }{N}$
|
[
"The number $N$ can be written as $10a+b$ with $a$ and $b$ representing the digits. The number $N$ with its digits reversed is $10b+a$. Since the problem asks for a positive number as the difference of these two numbers, than $a>b$. Writing this out, we get $10a+b-(10b+a)=9a-9b=9(a-b)$. Therefore, the difference must be a multiple of $9$, and the only perfect cube with less than $3$ digits and is multiple of $9$ is $3^3=27$. Also, that means $a-b=3$, and there are $7$ possibilities of that, so our answer is \n\n\n$\\boxed{\\textbf{(D)}}$ There are exactly $7$ values of $N$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/38.json
|
AHSME
|
1957_AHSME_Problems
| 4
| 0
|
Algebra
|
Multiple Choice
|
The first step in finding the product $(3x + 2)(x - 5)$ by use of the distributive property in the form $a(b + c) = ab + ac$ is:
$\textbf{(A)}\ 3x^2 - 13x - 10 \qquad \textbf{(B)}\ 3x(x - 5) + 2(x - 5)\qquad \\ \textbf{(C)}\ (3x+2)x+(3x+2)(-5)\qquad \textbf{(D)}\ 3x^2-17x-10\qquad \textbf{(E)}\ 3x^2+2x-15x-10$
|
[
"In order to find the product by using this form of the distributive property, we should express $(3x + 2)(x - 5)$ as $(3x + 2)(x) + (3x + 2)(-5)$. Thusly, $\\boxed{\\text{(C)}}$ is our answer, and we are done.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/4.json
|
AHSME
|
1957_AHSME_Problems
| 14
| 0
|
Algebra
|
Multiple Choice
|
If $y = \sqrt{x^2 - 2x + 1} + \sqrt{x^2 + 2x + 1}$, then $y$ is:
$\textbf{(A)}\ 2x\qquad \textbf{(B)}\ 2(x+1)\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ |x-1|+|x+1|\qquad\textbf{(E)}\ \text{none of these}$
|
[
"In order to solve the problem, we will use two properties, namely, that $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$.\n\n\nWe can use this to simplify the equation, as $y = \\sqrt{x^2 - 2x + 1} + \\sqrt{x^2 + 2x + 1}$ turns into $y = x + 1 + x - 1$. However, the square root function only allows for nonnegative inputs and only generates nonnegative outputs, so $y$ is $\\boxed{\\textbf{(D) }|x+1|+|x-1|}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/14.json
|
AHSME
|
1957_AHSME_Problems
| 42
| 0
|
Number Theory
|
Multiple Choice
|
If $S = i^n + i^{-n}$, where $i = \sqrt{-1}$ and $n$ is an integer, then the total number of possible distinct values for $S$ is:
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ \text{more than 4}$
|
[
"We first use the fact that $i^{-n}=\\frac1{i^n}=\\left(\\frac1i\\right)^n=(-i)^n$. Note that $i^4=1$ and $(-i)^4=1$, so $i^n$ and $(-i)^n$ have are periodic with periods at most 4. Therefore, it suffices to check for $n=0,1,2,3$.\n\n\n\n\nFor $n=0$, we have $i^0+(-i)^0=1+1=2$.\n\n\nFor $n=1$, we have $i^1+(-i)^1=i-i=0$.\n\n\nFor $n=2$, we have $i^2+(-i)^2=-1-1=-2$.\n\n\nFor $n=3$, we have $i^3+(-i)^3=-i+i=0$.\n\n\nHence, the answer is $\\boxed{\\textbf{(C)}\\ 3}$.\n\n\n",
"Notice that the powers of $i$ cycle in cycles of 4. So let's see if $S$ is periodic.\n\n\nFor $n=0$: we have $2$. \n\n\nFor $n=1$: we have $0$.\n\n\nFor $n=2$: we have $-2$.\n\n\nFor $n=3$: we have $0$.\n\n\nFor $n=4$: we have $2$ again. Well, it can be seen that $S$ cycles in periods of 4. Select $\\boxed{C}$.\n\n\n~hastapasta\n\n\n"
] | 2
|
./CreativeMath/AHSME/1957_AHSME_Problems/42.json
|
AHSME
|
1957_AHSME_Problems
| 15
| 0
|
Algebra
|
Multiple Choice
|
The table below shows the distance $s$ in feet a ball rolls down an inclined plane in $t$ seconds.
\[\begin{tabular}{|c|c|c|c|c|c|c|}\hline t & 0 & 1 & 2 & 3 & 4 & 5\\ \hline s & 0 & 10 & 40 & 90 & 160 & 250\\ \hline\end{tabular}\]
The distance $s$ for $t = 2.5$ is:
$\textbf{(A)}\ 45\qquad \textbf{(B)}\ 62.5\qquad \textbf{(C)}\ 70\qquad \textbf{(D)}\ 75\qquad \textbf{(E)}\ 82.5$
|
[
"Looking at the pattern, we can determine that $t=10s^2$. Applying the relationship, we can see that $s = \\boxed{\\textbf{(B) }62.5}$ when $t=2.5$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/15.json
|
AHSME
|
1957_AHSME_Problems
| 5
| 0
|
Algebra
|
Multiple Choice
|
Through the use of theorems on logarithms
\[\log{\frac{a}{b}} + \log{\frac{b}{c}} + \log{\frac{c}{d}} - \log{\frac{ay}{dx}}\]
can be reduced to:
$\textbf{(A)}\ \log{\frac{y}{x}}\qquad \textbf{(B)}\ \log{\frac{x}{y}}\qquad \textbf{(C)}\ 1\qquad \\ \textbf{(D)}\ 140x-24x^2+x^3\qquad \textbf{(E)}\ \text{none of these}$
|
[
"Using the properties $\\log(x)+\\log(y)=\\log(xy)$ and $\\log(x)-\\log(y)=\\log(x/y)$, we have\n\\begin{align*} \\log\\frac ab+\\log\\frac bc+\\log\\frac cd-\\log\\frac{ay}{dx}&=\\log\\left(\\frac ab\\cdot\\frac bc\\cdot\\frac cd\\right)-\\log \\frac {ay}{dx} \\\\ &=\\log \\frac ad-\\log\\frac{ay}{dx} \\\\ &=\\log\\left(\\frac{\\frac ad}{\\frac{ay}{dx}}\\right) \\\\ &=\\log \\frac xy, \\end{align*}\nso the answer is $\\boxed{\\textbf{(B)} \\log\\frac xy}.$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/5.json
|
AHSME
|
1957_AHSME_Problems
| 19
| 0
|
Number Theory
|
Multiple Choice
|
The base of the decimal number system is ten, meaning, for example, that $123 = 1\cdot 10^2 + 2\cdot 10 + 3$. In the binary system, which has base two, the first five positive integers are $1,\,10,\,11,\,100,\,101$. The numeral $10011$ in the binary system would then be written in the decimal system as:
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 40\qquad \textbf{(C)}\ 10011\qquad \textbf{(D)}\ 11\qquad \textbf{(E)}\ 7$
|
[
"Numbers in binary work similar to their decimal counterparts, where the multiplier associated with each place is multiplied by two every single place to the left. For example, $1111_2$ ($1111$ in base $2$) would equate to $1 * 2^3 + 1 * 2^2 + 1 * 2^1 + 1 * 2^0 = 8+4+2+1 = 15$.\n\n\nUsing this same logic, $10011_2$ would be $1*2^4 + 1*2^1 + 1 * 2^0 = \\boxed{\\textbf{(A) }19}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/19.json
|
AHSME
|
1957_AHSME_Problems
| 23
| 0
|
Algebra
|
Multiple Choice
|
The graph of $x^2 + y = 10$ and the graph of $x + y = 10$ meet in two points. The distance between these two points is:
$\textbf{(A)}\ \text{less than 1} \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ \sqrt{2}\qquad \textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{more than 2}$
|
[
"We can merge the two equations to create $x^2+y=x+y$. Using either the quadratic equation or factoring, we get two solutions with $x$-coordinates $0$ and $1$.\n\n\nPlugging this into either of the original equations, we get $(0,10)$ and $(1,9)$. The distance between those two points is $\\boxed{\\textbf{(C) }\\sqrt{2}}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/23.json
|
AHSME
|
1957_AHSME_Problems
| 9
| 0
|
Algebra
|
Multiple Choice
|
The value of $x - y^{x - y}$ when $x = 2$ and $y = -2$ is:
$\textbf{(A)}\ -18 \qquad \textbf{(B)}\ -14\qquad \textbf{(C)}\ 14\qquad \textbf{(D)}\ 18\qquad \textbf{(E)}\ 256$
|
[
"Just plug in the numbers and follow the order of operations:\n\\[2-(-2)^{2-(-2)}\\]\n\\[2-(-2)^4\\]\n\\[2-16\\]\n\\[\\boxed{\\textbf{(B) }-14}\\]\n\n\n"
] | 1
|
./CreativeMath/AHSME/1957_AHSME_Problems/9.json
|
AHSME
|
1956_AHSME_Problems
| 20
| 0
|
Other
|
Multiple Choice
|
If $(0.2)^x = 2$ and $\log 2 = 0.3010$, then the value of $x$ to the nearest tenth is:
$\textbf{(A)}\ - 10.0 \qquad\textbf{(B)}\ - 0.5 \qquad\textbf{(C)}\ - 0.4 \qquad\textbf{(D)}\ - 0.2 \qquad\textbf{(E)}\ 10.0$
|
[
"To be added.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/20.json
|
AHSME
|
1956_AHSME_Problems
| 36
| 0
|
Number Theory
|
Multiple Choice
|
If the sum $1 + 2 + 3 + \cdots + K$ is a perfect square $N^2$ and if $N$ is less than $100$, then the possible values for $K$ are:
$\textbf{(A)}\ \text{only }1\qquad \textbf{(B)}\ 1\text{ and }8\qquad \textbf{(C)}\ \text{only }8\qquad \textbf{(D)}\ 8\text{ and }49\qquad \textbf{(E)}\ 1,8,\text{ and }49$
|
[
"We can plug in 1, 8, and 49 to see which works.\n\n\n$1 = 1 = 1^2$\n\n\n$1 + 2 + 3 + \\cdots + 8 = \\frac{8 \\cdot 9}{2} = 36 = 6^2$\n\n\n$1 + 2 + 3 + \\cdots + 49 = \\frac{49 \\cdot 50}{2} = 1225 = 45^2$\n\n\nAll of these values produce a perfect square for $1 + 2 + \\cdots + K,$ so the answer is $\\boxed{\\textbf{(E)}}.$\n\n\n-coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/36.json
|
AHSME
|
1956_AHSME_Problems
| 41
| 0
|
Algebra
|
Multiple Choice
|
The equation $3y^2 + y + 4 = 2(6x^2 + y + 2)$ where $y = 2x$ is satisfied by:
$\textbf{(A)}\ \text{no value of }x \qquad \textbf{(B)}\ \text{all values of }x \qquad \textbf{(C)}\ x = 0\text{ only} \\ \textbf{(D)}\ \text{all integral values of }x\text{ only} \qquad \textbf{(E)}\ \text{all rational values of }x\text{ only}$
|
[
"Start by plugging in $y=2x:$\n\\[12x^2 + 2x + 4 = 12x^2 + 4x + 4\\]\nThere is no value of $x$ that can satisfy the equation, so the answer is $\\boxed{\\textbf{(A)}}.$\n\n\n-coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/41.json
|
AHSME
|
1956_AHSME_Problems
| 16
| 0
|
Algebra
|
Multiple Choice
|
The sum of three numbers is $98$. The ratio of the first to the second is $\frac {2}{3}$,
and the ratio of the second to the third is $\frac {5}{8}$. The second number is:
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 32 \qquad\textbf{(E)}\ 33$
|
[
"Let the $3$ numbers be $a,$ $b,$ and $c$. We see that \n\\[a+b+c = 98\\]\nand \n\\[\\frac{a}{b} = \\frac{2}{3} \\Rrightarrow 3a = 2b\\]\n\\[\\frac{b}{c} = \\frac{5}{8} \\Rrightarrow 8b = 5c\\]\nWriting $a$ and $c$ in terms of $b$ we have\n$a = \\frac{2}{3} b$ and $c = \\frac{8}{5} b$. \nSubstituting in the sum, we have\n\\[\\frac{2}{3} b + b + \\frac{8}{5} b = 98\\]\n\\[\\frac{49}{15} b = 98\\]\n\\[b = 98 \\cdot \\frac{15}{49} \\Rrightarrow b = 30\\]\n$\\boxed{C}$\n\n\n~JustinLee2017\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/16.json
|
AHSME
|
1956_AHSME_Problems
| 6
| 0
|
Algebra
|
Multiple Choice
|
In a group of cows and chickens, the number of legs was 14 more than twice the number of heads. The number of cows was:
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 14$
|
[
"Let there be $x$ cows and $y$ chickens. Then, there are $4x+2y$ legs and $x+y$ heads. Writing the equation:\n\\[4x+2y=14+2(x+y)\\] \\[4x+2y=14+2x+2y\\] \\[2x=14\\] \\[x=\\boxed{\\textbf{(B) }7}\\]\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/6.json
|
AHSME
|
1956_AHSME_Problems
| 7
| 0
|
Algebra
|
Multiple Choice
|
The roots of the equation $ax^2 + bx + c = 0$ will be reciprocal if:
$\textbf{(A)}\ a = b \qquad\textbf{(B)}\ a = bc \qquad\textbf{(C)}\ c = a \qquad\textbf{(D)}\ c = b \qquad\textbf{(E)}\ c = ab$
|
[
"Dividing both sides of the equation by $a\\quad(a\\neq0)$ gives $x^2+\\frac{b}{a}x+\\frac{c}{a}$. \n\n\nLetting $r$ and $s$ be the respective roots to this quadratic, $r=\\frac{1}{s} \\Rightarrow rs=1$.\n\n\nFrom Vieta's, $rs=\\frac{c}{a}$, so $\\frac{c}{a}=1\\Rightarrow\\boxed{\\text{(C) }c=a}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/7.json
|
AHSME
|
1956_AHSME_Problems
| 17
| 0
|
Algebra
|
Multiple Choice
|
The fraction $\frac {5x - 11}{2x^2 + x - 6}$ was obtained by adding the two fractions $\frac {A}{x + 2}$ and
$\frac {B}{2x - 3}$. The values of $A$ and $B$ must be, respectively:
$\textbf{(A)}\ 5x,-11\qquad\textbf{(B)}\ -11,5x\qquad\textbf{(C)}\ -1,3\qquad\textbf{(D)}\ 3,-1\qquad\textbf{(E)}\ 5,-11$
|
[
"This is essentially asking for the partial fraction decomposition of \\[\\frac{5x-11}{2x^2 + x - 6}\\]\nLooking at $A$ and $B$, we can write\n\\[A(2x-3) + B(x+2) = 5x-11\\].\nSubstituting $x= -2$, we get\n\\[A(-4-3) + B(0) = -21 \\Rrightarrow A = 3\\]\nSubstituting $x = \\frac{3}{2}$, we get\n\\[A(0) + B(\\frac{7}{2}) = \\frac{15}{2} - \\frac{22}{2} \\Rrightarrow B = -1\\]\nThus, our answer is $\\boxed{D}$\n\n\n~JustinLee2017\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/17.json
|
AHSME
|
1956_AHSME_Problems
| 40
| 0
|
Algebra
|
Multiple Choice
|
If $V = gt + V_0$ and $S = \frac {1}{2}gt^2 + V_0t$, then $t$ equals:
$\textbf{(A)}\ \frac{2S}{V+V_0}\qquad \textbf{(B)}\ \frac{2S}{V-V_0}\qquad \textbf{(C)}\ \frac{2S}{V_0-V}\qquad \textbf{(D)}\ \frac{2S}{V}\qquad \textbf{(E)}\ 2S-V$
|
[
"Solve for $t$ in the first equation:\n\\[V - V_0 = gt\\]\n\\[t = \\frac{V - V_0}{g}\\]\nThen use the second equation to solve for $g:$\n\\[2(S - V_0 t) = gt^2\\]\n\\[g = \\frac{2(S - V_0 t)}{t^2}\\]\n\n\nPlug in $g$ in $t$ to get $\\boxed{\\textbf{(A)}}.$\n-coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/40.json
|
AHSME
|
1956_AHSME_Problems
| 37
| 0
|
Geometry
|
Multiple Choice
|
On a map whose scale is $400$ miles to an inch and a half, a certain estate is represented by a rhombus having a $60^{\circ}$ angle. The diagonal opposite $60^{\circ}$ is $\frac {3}{16}$ in. The area of the estate in square miles is:
$\textbf{(A)}\ \frac{2500}{\sqrt{3}}\qquad \textbf{(B)}\ \frac{1250}{\sqrt{3}}\qquad \textbf{(C)}\ 1250\qquad \textbf{(D)}\ \frac{5625\sqrt{3}}{2}\qquad \textbf{(E)}\ 1250\sqrt{3}$
|
[
"When we construct the diagram, we can see that the rhombus is made up of two equilateral triangles. The short diagonal in the rhombus is one side of both equilateral triangles, so we can simply scale the length up and find the area that way.\n\n\nWe know that the scale is $400$ miles to $\\frac{3}{2}$ inches, so we simply divide by $8$ to find that the short diagonal of the estate is $50$ miles.\n\n\nWhen plugged into the formula, we get $2*50^2\\frac{\\sqrt{3}}{4}$ (we have two congruent triangles to work with), which converts to $\\boxed{\\textbf{(E) }1250\\sqrt{3}}$ square miles.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/37.json
|
AHSME
|
1956_AHSME_Problems
| 21
| 0
|
Geometry
|
Multiple Choice
|
If each of two intersecting lines intersects a hyperbola and neither line is tangent to the hyperbola,
then the possible number of points of intersection with the hyperbola is:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 2\text{ or }3\qquad \textbf{(C)}\ 2\text{ or }4\qquad \textbf{(D)}\ 3\text{ or }4\qquad \textbf{(E)}\ 2,3,\text{ or }4$
|
[
"Consider the hyperbola $x^2-y^2=1$.\nIt is possible for the two intersecting lines to intersect the hyperbola at 2 points if one of them has a slope of 1 and only intersects one part of the hyperbola and the other line doesn't intersect the hyperbola at all (Ex. $y=x+3,\\, x=0.$). If the second line is instead $x=4$, it intersects the hyperbola twice, so the lines can intersect the hyperbola 3 times. Finally, if both lines intersect the hyperbola twice, such as $y=2x-4$ and $y=3x-6$, the lines can intersect the hyperbola 4 times. So the answer is $\\textbf{(E)}\\ 2,3,\\text{ or } 4$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/21.json
|
AHSME
|
1956_AHSME_Problems
| 10
| 0
|
Geometry
|
Multiple Choice
|
A circle of radius $10$ inches has its center at the vertex $C$ of an equilateral triangle $ABC$ and
passes through the other two vertices. The side $AC$ extended through $C$ intersects the circle
at $D$. The number of degrees of angle $ADB$ is:
$\text{(A)} 15 \quad \text{(B)}30 \quad \text{(C)}60 \quad \text{(D)}90 \quad \text{(E)}120$
|
[
"[asy] import olympiad; draw(circle((0,0),10)); dot((0,0)); label(\"C\", (1,-1)); dot((5,5sqrt(3))); dot((-5,-5sqrt(3))); draw((-5,-5sqrt(3))--(5,5sqrt(3))); label(\"A\",(6,5sqrt(3)+1)); label(\"D\",(-6,-5sqrt(3)-1)); draw((0,0)--(10,0)); label(\"10\",(1.5,2.5sqrt(3)+1)); dot((10,0)); label(\"B\",(11,-1)); draw((5,5sqrt(3))--(10,0)); draw(anglemark((10,0),(0,0),(5,5sqrt(3)),60)); label( \"60°\", (2.5,1.25)); draw((-5,-5sqrt(3))--(10,0)); draw(anglemark((10,0),(-5,-5sqrt(3)),(5,5sqrt(3)),60)); label(\"?\",(-2.5,-5sqrt(3)+2.5)); [/asy]\n\n\n$ABC$ is an equilateral triangle, so ∠$C$ must be $60$°. Since $D$ is on the circle and ∠$ADB$ contains arc $AB$, we know that ∠$D$ is $30$° $\\implies \\fbox{B}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/10.json
|
AHSME
|
1956_AHSME_Problems
| 26
| 0
|
Geometry
|
Multiple Choice
|
Which one of the following combinations of given parts does not determine the indicated triangle?
$\textbf{(A)}\ \text{base angle and vertex angle; isosceles triangle} \\ \textbf{(B)}\ \text{vertex angle and the base; isosceles triangle} \\ \textbf{(C)}\ \text{the radius of the circumscribed circle; equilateral triangle} \\ \textbf{(D)}\ \text{one arm and the radius of the inscribed circle; right triangle} \\ \textbf{(E)}\ \text{two angles and a side opposite one of them; scalene triangle}$
|
[
"Let's look at each answer choice one by one. For choice A, you can swap the base and vertex angles to obtain a different isosceles triangle, so these two criteria alone are not enough. $\\boxed{A}$ is our answer. \n\n\n-coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/26.json
|
AHSME
|
1956_AHSME_Problems
| 30
| 0
|
Geometry
|
Multiple Choice
|
If the altitude of an equilateral triangle is $\sqrt {6}$, then the area is:
$\textbf{(A)}\ 2\sqrt{2}\qquad\textbf{(B)}\ 2\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{3}\qquad\textbf{(D)}\ 6\sqrt{2}\qquad\textbf{(E)}\ 12$
|
[
"Drawing the altitude, we see that is opposite the $60^{\\circ}$ angle in a $30-60-90$ triangle. Thus, we find that the shortest leg of that triangle is $\\sqrt{\\frac{6}{3}} = \\sqrt{2}$, and the side of the equilateral triangle is then $2\\sqrt{2}$. Thus, the area is\n\\[\\frac{(2\\sqrt 2)^2 \\cdot \\sqrt{3}}{4}\\] \\[= \\frac{8 \\sqrt{3}}{4}\\] \\[= 2 \\sqrt {3}\\]\n$\\boxed{B}$\n\n\n~JustinLee2017\n\n\n\n\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/30.json
|
AHSME
|
1956_AHSME_Problems
| 31
| 0
|
Number Theory
|
Multiple Choice
|
In our number system the base is ten. If the base were changed to four you would count as follows:
$1,2,3,10,11,12,13,20,21,22,23,30,\ldots$ The twentieth number would be:
$\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 38 \qquad\textbf{(C)}\ 44 \qquad\textbf{(D)}\ 104 \qquad\textbf{(E)}\ 110$
|
[
"The $20^{\\text{th}}$ number will be the value of $20_{10}$ in base $4$. Thus, we see\n\\[20_{10} = (1) \\cdot 4^2 + (1) \\cdot 4^1 + 0 \\cdot 4^0\\]\n\\[= 110_{4}\\]\n\n\n$\\boxed{E}$\n\n\n~JustinLee2017\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/31.json
|
AHSME
|
1956_AHSME_Problems
| 27
| 0
|
Geometry
|
Multiple Choice
|
If an angle of a triangle remains unchanged but each of its two including sides is doubled, then the area is multiplied by:
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ \text{more than }6$
|
[
"Let the angle be $\\theta$ and the sides around it be $a$ and $b$. \nThe area of the triangle can be written as \\[A =\\frac{a \\cdot b \\cdot \\sin(\\theta)}{2}\\]\nThe doubled sides have length $2a$ and $2b$, while the angle is still $\\theta$. Thus, the area is\n\\[\\frac{2a \\cdot 2b \\cdot \\sin(\\theta)}{2}\\]\n\\[\\Rrightarrow \\frac{4ab \\sin \\theta}{2} = 4A\\]\n\\[\\boxed {C}\\]\n\n\n~JustinLee2017 \n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/27.json
|
AHSME
|
1956_AHSME_Problems
| 1
| 0
|
Algebra
|
Multiple Choice
|
The value of $x + x(x^x)$ when $x = 2$ is:
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 16 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 36 \qquad\textbf{(E)}\ 64$
|
[
"Simple substitution yields\n\\[2 + 2(2^2) = 2 + 2(4) = 10\\]\n\n\nTherefore, the answer is $\\fbox{(A) 10}$\n\n\n\n\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/1.json
|
AHSME
|
1956_AHSME_Problems
| 50
| 0
|
Geometry
|
Multiple Choice
|
In $\triangle ABC, \overline{CA} = \overline{CB}$. On $CB$ square $BCDE$ is constructed away from the triangle. If $x$ is the number of degrees in $\angle DAB$, then
$\textbf{(A)}\ x\text{ depends upon }\triangle ABC \qquad \textbf{(B)}\ x\text{ is independent of the triangle} \\ \textbf{(C)}\ x\text{ may equal }\angle CAD \qquad \\ \textbf{(D)}\ x\text{ can never equal }\angle CAB \\ \textbf{(E)}\ x\text{ is greater than }45^{\circ}\text{ but less than }90^{\circ}$
|
[
"To be added :D\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/50.json
|
AHSME
|
1956_AHSME_Problems
| 11
| 0
|
Algebra
|
Multiple Choice
|
The expression $1 - \frac {1}{1 + \sqrt {3}} + \frac {1}{1 - \sqrt {3}}$ equals:
$\textbf{(A)}\ 1-\sqrt{3}\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ -\sqrt{3}\qquad \textbf{(D)}\ \sqrt{3}\qquad \textbf{(E)}\ 1+\sqrt{3}$
|
[
"We can combine the second and third terms using the rules of arithmetic: We have $-\\frac{1}{1 + \\sqrt {3}} + \\frac{1}{1 - \\sqrt {3}} = \\frac{-(1 - \\sqrt{3}) + (1 + \\sqrt{3})}{(1 - \\sqrt{3})(1 + \\sqrt{3})} = \\frac{2\\sqrt{3}}{1 - 3} = -\\sqrt{3}$.\n\n\nTherefore the entire expression is equal to $1 - \\sqrt{3}$, so our answer is $\\boxed{\\textbf{A}}$ and we are done.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/11.json
|
AHSME
|
1956_AHSME_Problems
| 2
| 0
|
Arithmetic
|
Multiple Choice
|
Mr. Jones sold two pipes at $\textdollar{ 1.20}$ each. Based on the cost, his profit on one was
$20$% and his loss on the other was $20$%.
On the sale of the pipes, he:
$\textbf{(A)}\ \text{broke even}\qquad \textbf{(B)}\ \text{lost }4\text{ cents} \qquad\textbf{(C)}\ \text{gained }4\text{ cents}\qquad \\ \textbf{(D)}\ \text{lost }10\text{ cents}\qquad \textbf{(E)}\ \text{gained }10\text{ cents}$
|
[
"For the first pipe, we are told that his profit was 20%. In other words, he sold the pipe for 120% or $\\frac{6}{5}$ of its original value. This tells us that the original price was $\\frac{5}{6}\\cdot1.20 = $1.00$.\n\n\nFor the second pipe, we are told that his loss was 20%. In other words, he sold the pipe for 80% or $\\frac{4}{5}$ of its original value. This tells us that the original price was $\\frac{5}{4}\\cdot1.20 = $1.50$.\n\n\n\n\nThus, his total cost was $$2.50$ and his total revenue was $$2.40$.\n\n\n\n\nTherefore, he $\\boxed{\\textbf{(D)}\\ \\text{lost }10\\text{ cents}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/2.json
|
AHSME
|
1956_AHSME_Problems
| 28
| 0
|
Algebra
|
Multiple Choice
|
Mr. J left his entire estate to his wife, his daughter, his son, and the cook.
His daughter and son got half the estate, sharing in the ratio of $4$ to $3$.
His wife got twice as much as the son. If the cook received a bequest of $\textdollar{500}$, then the entire estate was:
$\textbf{(A)}\ \textdollar{3500}\qquad \textbf{(B)}\ \textdollar{5500}\qquad \textbf{(C)}\ \textdollar{6500}\qquad \textbf{(D)}\ \textdollar{7000}\qquad \textbf{(E)}\ \textdollar{7500}$
|
[
"The wife, daughter, son, and cook received estates in the ratio $6:4:3:1.$ The estate is worth $6+4+3+1 = 14$ units of $$500.,$ which is $\\boxed{\\textbf{(D)}\\ \\textdollar{7000}.}$\n\n\n~coolmath34 \n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/28.json
|
AHSME
|
1956_AHSME_Problems
| 12
| 0
|
Algebra
|
Multiple Choice
|
If $x^{ - 1} - 1$ is divided by $x - 1$ the quotient is:
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ \frac{1}{x-1}\qquad \textbf{(C)}\ \frac{-1}{x-1}\qquad \textbf{(D)}\ \frac{1}{x}\qquad \textbf{(E)}\ -\frac{1}{x}$
|
[
"We write $x^{-1} - 1 = \\frac{1}{x} - 1 = \\frac{1-x}{1} = -\\frac{1}{x}(x - 1)$. This means that our answer is $\\boxed{\\textbf{(E)}}$, and we are done.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/12.json
|
AHSME
|
1956_AHSME_Problems
| 45
| 0
|
Other
|
Multiple Choice
|
A wheel with a rubber tire has an outside diameter of $25$ in. When the radius has been decreased a quarter of an inch, the number of revolutions in one mile will:
$\text{(A)}\ \text{be increased about }2\% \qquad \\ \text{(B)}\ \text{be increased about }1\% \\ \text{(C)}\ \text{be increased about }20\%\qquad \\ \text{(D)}\ \text{be increased about }\frac{1}{2}\%\qquad \\ \text{(E)}\ \text{remain the same}$
|
[
"To be added.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/45.json
|
AHSME
|
1956_AHSME_Problems
| 32
| 0
|
Algebra
|
Multiple Choice
|
George and Henry started a race from opposite ends of the pool. After a minute and a half,
they passed each other in the center of the pool. If they lost no time in turning and
maintained their respective speeds, how many minutes after starting did they pass each other the second time?
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\frac{1}{2}\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\frac{1}{2}\qquad \textbf{(E)}\ 9$
|
[
"Because it took George and Henry $1 \\frac{1}{2}$ minutes to meet each other, it took each of them $1 \\frac{1}{2}$ minutes to travel half the length of the pool. In order for them to meet again, each of them needs to travel $1 \\frac{1}{2}$ lengths of the pool, which takes $\\boxed{\\textbf{(B)} \\quad 4\\frac{1}{2}}$ minutes.\n\n\n-coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/32.json
|
AHSME
|
1956_AHSME_Problems
| 49
| 0
|
Geometry
|
Multiple Choice
|
Triangle $PAB$ is formed by three tangents to circle $O$ and $\angle APB = 40^{\circ}$; then $\angle AOB$ equals:
$\textbf{(A)}\ 45^{\circ}\qquad \textbf{(B)}\ 50^{\circ}\qquad \textbf{(C)}\ 55^{\circ}\qquad \textbf{(D)}\ 60^{\circ}\qquad \textbf{(E)}\ 70^{\circ}$
|
[
"First, from triangle $ABO$, $\\angle AOB = 180^\\circ - \\angle BAO - \\angle ABO$. Note that $AO$ bisects $\\angle BAT$ (to see this, draw radii from $O$ to $AB$ and $AT,$ creating two congruent right triangles), so $\\angle BAO = \\angle BAT/2$. Similarly, $\\angle ABO = \\angle ABR/2$.\n\n\nAlso, $\\angle BAT = 180^\\circ - \\angle BAP$, and $\\angle ABR = 180^\\circ - \\angle ABP$. Hence, \n\n\n$\\angle AOB = 180^\\circ - \\angle BAO - \\angle ABO = 180^\\circ - \\frac{\\angle BAT}{2} - \\frac{\\angle ABR}{2} = 180^\\circ - \\frac{180^\\circ - \\angle BAP}{2} - \\frac{180^\\circ - \\angle ABP}{2}= \\frac{\\angle BAP + \\angle ABP}{2}.$\n\n\nFinally, from triangle $ABP$, $\\angle BAP + \\angle ABP = 180^\\circ - \\angle APB = 180^\\circ - 40^\\circ = 140^\\circ$, so \\[\\angle AOB = \\frac{\\angle BAP + \\angle ABP}{2} = \\frac{140^\\circ}{2} = \\boxed{70^\\circ}.\\]\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/49.json
|
AHSME
|
1956_AHSME_Problems
| 48
| 0
|
Algebra
|
Multiple Choice
|
If $p$ is a positive integer, then $\frac {3p + 25}{2p - 5}$ can be a positive integer, if and only if $p$ is:
$\textbf{(A)}\ \text{at least }3\qquad \textbf{(B)}\ \text{at least }3\text{ and no more than }35\qquad \\ \textbf{(C)}\ \text{no more than }35 \qquad \textbf{(D)}\ \text{equal to }35 \qquad \textbf{(E)}\ \text{equal to }3\text{ or }35$
|
[
"Lets begin by noticing that:\n\\[\\frac{3p+25}{2p-5} = 1 + \\frac{p+30}{2p-5}\\]\n\n\nTherefore, in order for $\\frac{3p+25}{2p-5}$ to be a positive integer, $\\frac{p+30}{2p-5}$ must be a non-negative integer. Since the bottom the the fraction is an odd number, we can multiply the top of $\\frac{p+30}{2p-5}$ by 2 without changing whether it is an integer or not. Therefore, in order for $\\frac{3p+25}{2p-5}$ to be an integer, $\\frac{2p+60}{2p-5} = 1 + \\frac{65}{2p-5}$ must also be an integer. As a result, $2p-5$ must be a factor of 65, or $p = 3, 5, 9, 35$. Therefore $p$ must be at least 3, and less than or equal to 35. So the answer which best fits these constraints is $\\boxed{\\textbf{(B)}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/48.json
|
AHSME
|
1956_AHSME_Problems
| 25
| 0
|
Algebra
|
Multiple Choice
|
The sum of all numbers of the form $2k + 1$, where $k$ takes on integral values from $1$ to $n$ is:
$\textbf{(A)}\ n^2\qquad\textbf{(B)}\ n(n+1)\qquad\textbf{(C)}\ n(n+2)\qquad\textbf{(D)}\ (n+1)^2\qquad\textbf{(E)}\ (n+1)(n+2)$
|
[
"The sum of the odd integers $2k-1$ from $1$ to $n$ is $n^2$. However, in this problem, the sum is instead $2k+1$, starting at $3$ rather than $1$. To rewrite this, we note that $2k-1$ is $2$ less than $2k+1$ for every $k$ we add, so for $n$ $k$'s, we subtract $2n$, giving us $n^2+2n$,which factors as $n(n+2) \\implies \\boxed{\\text{(C)} n(n+2)}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/25.json
|
AHSME
|
1956_AHSME_Problems
| 33
| 0
|
Number Theory
|
Multiple Choice
|
The number $\sqrt {2}$ is equal to:
$\textbf{(A)}\ \text{a rational fraction} \qquad \textbf{(B)}\ \text{a finite decimal} \qquad \textbf{(C)}\ 1.41421 \\ \textbf{(D)}\ \text{an infinite repeating decimal} \qquad \textbf{(E)}\ \text{an infinite non - repeating decimal}$
|
[
"$2$ is not a perfect square, so $\\sqrt{2}$ is irrational. The answer is $\\boxed{\\textbf{(E)}\\ \\text{an infinite non - repeating decimal}}.$\n\n\n~coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/33.json
|
AHSME
|
1956_AHSME_Problems
| 44
| 0
|
Algebra
|
Multiple Choice
|
If $x < a < 0$ means that $x$ and $a$ are numbers such that $x$ is less than $a$ and $a$ is less than zero, then:
$\textbf{(A)}\ x^2 < ax < 0 \qquad \textbf{(B)}\ x^2 > ax > a^2 \qquad \textbf{(C)}\ x^2 < a^2 < 0 \\ \textbf{(D)}\ x^2 > ax\text{ but }ax < 0 \qquad \textbf{(E)}\ x^2 > a^2\text{ but }a^2 < 0$
|
[
"Note that since $x$ and $a$ are negative, $x^2, a^2,$ and $ax$ are all positive. Hence the only correct answer is $\\fbox{B}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/44.json
|
AHSME
|
1956_AHSME_Problems
| 13
| 0
|
Algebra
|
Multiple Choice
|
Given two positive integers $x$ and $y$ with $x < y$. The percent that $x$ is less than $y$ is:
$\textbf{(A)}\ \frac{100(y-x)}{x}\qquad \textbf{(B)}\ \frac{100(x-y)}{x}\qquad \textbf{(C)}\ \frac{100(y-x)}{y}\qquad \\ \textbf{(D)}\ 100(y-x)\qquad \textbf{(E)}\ 100(x - y)$
|
[
"Suppose that $x$ is $p$ percent less than $y$. Then $x = \\frac{100 - p}{100}y$, so that $y - x = \\frac{p}{100}y$. Solving for $p$, we get $p = \\frac{100(y-x)}{y}$, or $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/13.json
|
AHSME
|
1956_AHSME_Problems
| 29
| 0
|
Geometry
|
Multiple Choice
|
The points of intersection of $xy = 12$ and $x^2 + y^2 = 25$ are joined in succession. The resulting figure is:
$\textbf{(A)}\ \text{a straight line}\qquad \textbf{(B)}\ \text{an equilateral triangle}\qquad \textbf{(C)}\ \text{a parallelogram} \\ \textbf{(D)}\ \text{a rectangle} \qquad \textbf{(E)}\ \text{a square}$
|
[
"Graphing both equations, we see that the graphs intersect at $(-4, -3),$ $(-3, -4),$ $(3, 4),$ and $(4, 3).$ These points form a rectangle, so the answer is $\\boxed{\\textbf{(D)}}.$\n\n\n~coolmath34 \n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/29.json
|
AHSME
|
1956_AHSME_Problems
| 3
| 0
|
Arithmetic
|
Multiple Choice
|
The distance light travels in one year is approximately $5,870,000,000,000$ miles. The distance light travels in $100$ years is:
$\textbf{(A)}\ 587\cdot10^8\text{ miles}\qquad \textbf{(B)}\ 587\cdot10^{10}\text{ miles}\qquad \textbf{(C)}\ 587\cdot10^{-10}\text{ miles} \\ \textbf{(D)}\ 587\cdot10^{12} \text{ miles} \qquad \textbf{(E)}\ 587\cdot10^{ - 12} \text{ miles}$
|
[
"The distance light travels in one year can also be written as $587\\cdot10^{10}$. In 100 years, light will travel $(587\\cdot10^{10})\\cdot100=587\\cdot10^{12}$.\n\n\nTherefore, our answer is $\\boxed{\\textbf{(D) }587\\cdot10^{12}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/3.json
|
AHSME
|
1956_AHSME_Problems
| 34
| 0
|
Number Theory
|
Multiple Choice
|
If $n$ is any whole number, $n^2(n^2 - 1)$ is always divisible by
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 24\qquad \textbf{(C)}\ \text{any multiple of }12\qquad \textbf{(D)}\ 12-n\qquad \textbf{(E)}\ 12\text{ and }24$
|
[
"Suppose $n$ is even. So, we have $n^2(n+1)(n-1).$ Out of these three numbers, at least one of them is going to be a multiple of 3. $n^2$ is also a multiple of 4. Therefore, this expression is always divisible by $\\boxed{\\textbf{(A)} \\quad 12}.$\n\n\n-coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/34.json
|
AHSME
|
1956_AHSME_Problems
| 8
| 0
|
Algebra
|
Multiple Choice
|
If $8\cdot2^x = 5^{y + 8}$, then when $y = - 8,x =$
$\textbf{(A)}\ - 4 \qquad\textbf{(B)}\ - 3 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 8$
|
[
"Simple substitution yields\n\\[8 \\cdot 2^{x} = 5^{0}\\]\nReducing the equation gives\n\\[8 \\cdot 2^{x} = 1\\]\nDividing by 8 gives\n\\[2^{x}=\\frac{1}{8}\\]\nThis simply gives that $x=-3$.\nTherefore, the answer is $\\fbox{(B) -3}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/8.json
|
AHSME
|
1956_AHSME_Problems
| 22
| 0
|
Algebra
|
Multiple Choice
|
Jones covered a distance of $50$ miles on his first trip. On a later trip he traveled $300$ miles while going three times as fast.
His new time compared with the old time was:
$\textbf{(A)}\ \text{three times as much} \qquad \textbf{(B)}\ \text{twice as much} \qquad \textbf{(C)}\ \text{the same} \\ \textbf{(D)}\ \text{half as much} \qquad \textbf{(E)}\ \text{a third as much}$
|
[
"WLOG Jones travels at $50$ miles per hour. Then, it took him $1$ hour to travel $50$ miles at $50$ miles per hour.\n\n\nOn his second trip, he travels $300$ miles at $150$ miles per hour, which takes $\\boxed{\\textbf{(B)}\\ \\text{twice as much}}$ time as before.\n\n\n~coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/22.json
|
AHSME
|
1956_AHSME_Problems
| 18
| 0
|
Algebra
|
Multiple Choice
|
If $10^{2y} = 25$, then $10^{ - y}$ equals:
$\textbf{(A)}\ -\frac{1}{5}\qquad \textbf{(B)}\ \frac{1}{625}\qquad \textbf{(C)}\ \frac{1}{50}\qquad \textbf{(D)}\ \frac{1}{25}\qquad \textbf{(E)}\ \frac{1}{5}$
|
[
"If $10^{2y}=(10^{y})^2 = 25$, then $10^y = 5$ and $10^{-y} = \\boxed{\\textbf{(E)} \\quad \\frac{1}{5}}.$\n\n\n\n\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/18.json
|
AHSME
|
1956_AHSME_Problems
| 38
| 0
|
Geometry
|
Multiple Choice
|
In a right triangle with sides $a$ and $b$, and hypotenuse $c$, the altitude drawn on the hypotenuse is $x$. Then:
$\textbf{(A)}\ ab = x^2 \qquad \textbf{(B)}\ \frac {1}{a} + \frac {1}{b} = \frac {1}{x} \qquad \textbf{(C)}\ a^2 + b^2 = 2x^2 \\ \textbf{(D)}\ \frac {1}{x^2} = \frac {1}{a^2} + \frac {1}{b^2} \qquad \textbf{(E)}\ \frac {1}{x} = \frac {b}{a}$
|
[
"The area of this triangle can be expressed in two ways; the first being $\\frac{ab}{2}$ (as this is a right triangle), and the second being $\\frac{cx}{2}$. But by the Pythagorean Theorem, $c=\\sqrt{a^2+b^2}$. Thus a second way of finding the area of the triangle is $\\frac{x\\sqrt{a^2+b^2}}{2}$. By setting them equal to each other we get $x=\\frac{ab}{\\sqrt{a^2+b^2}}$, and we can observe that the correct answer is $\\boxed{\\textbf{(D) \\ }}$.\n\n\n~anduran\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/38.json
|
AHSME
|
1956_AHSME_Problems
| 4
| 0
|
Arithmetic
|
Multiple Choice
|
A man has $\textdollar{10,000 }$ to invest. He invests $\textdollar{4000}$ at 5% and $\textdollar{3500}$ at 4%.
In order to have a yearly income of $\textdollar{500}$, he must invest the remainder at:
$\textbf{(A)}\ 6\%\qquad\textbf{(B)}\ 6.1\%\qquad\textbf{(C)}\ 6.2\%\qquad\textbf{(D)}\ 6.3\%\qquad\textbf{(E)}\ 6.4\%$
|
[
"The man currently earns $4000 \\cdot \\frac{5}{1000} + 3500 \\cdot \\frac{4}{1000} = 340$ dollars. So, we need to find the value of $x$ such that\n\\[2500 \\cdot \\frac{x}{1000} = 160.\\]\nSolving, we get $x = \\boxed{\\textbf{(E) }6.4\\%.}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/4.json
|
AHSME
|
1956_AHSME_Problems
| 14
| 0
|
Geometry
|
Multiple Choice
|
The points $A,B,C$ are on a circle $O$. The tangent line at $A$ and the secant $BC$ intersect at $P, B$ lying between $C$ and $P$.
If $\overline{BC} = 20$ and $\overline{PA} = 10\sqrt {3}$, then $\overline{PB}$ equals:
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 10\sqrt {3} \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 30$
|
[
"Because $PA$ is a tangent line, angle $\\angle OAP$ is a right angle. Drop a perpendicular from $O$ to $BC$ at $E.$ We find that $BE = EC = 10.$\n\n\nLet $AR = r$ and $PE = a$. We now have a system of equations.\n\n\n\\[(10\\sqrt{3})^2+r^2=PR^2\\]\n\\[(\\sqrt{r^2-10^2})^2+PE^2=r^2-10^2+PE^2=PR^2\\]\n\n\nSet them equal to each other and solve.\n\n\n\\[(10\\sqrt{3})^2+r^2=r^2-10^2+PE^2\\]\n\\[(10\\sqrt{3})^2=-10^2+PE^2\\]\n\\[400=PE^2\\]\n\\[20=PE\\]\n\n\nWe know $BE = 10$, so $PE = 20 - 10 = 10$, which is $\\boxed{B}$.\n\n\n~Revised by MC413551\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/14.json
|
AHSME
|
1956_AHSME_Problems
| 43
| 0
|
Geometry
|
Multiple Choice
|
The number of scalene triangles having all sides of integral lengths, and perimeter less than $13$ is:
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 18$
|
[
"We can write all possible triangles adding up to 12 or less\n\\[(2, 4, 5)=11\\]\n\\[(3, 4, 5)=12\\]\n\\[(2, 3, 4)=9\\]\n\n\nThis leaves $\\boxed{\\textbf{(C)} \\quad 3}$ scalene triangles.\n\n\n-coolmath34\n\n\n-rubslul\n\n\n(If you see any cases I missed out, edit them in.)\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/43.json
|
AHSME
|
1956_AHSME_Problems
| 42
| 0
|
Algebra
|
Multiple Choice
|
The equation $\sqrt {x + 4} - \sqrt {x - 3} + 1 = 0$ has:
$\textbf{(A)}\ \text{no root} \qquad \textbf{(B)}\ \text{one real root} \\ \textbf{(C)}\ \text{one real root and one imaginary root} \\ \textbf{(D)}\ \text{two imaginary roots} \qquad \qquad\textbf{(E)}\ \text{two real roots}$
|
[
"Simplify.\n\\[\\sqrt {x + 4} = \\sqrt{x-3} -1\\]\n\\[x + 4 = x - 3 -2\\sqrt{x-3} + 1\\]\n\\[6 = -2\\sqrt{x-3}\\]\nWe don't need to solve any further. The square root term is negative, therefore there will be no real roots. The answer is $\\boxed{A}.$\n\n\n-coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/42.json
|
AHSME
|
1956_AHSME_Problems
| 15
| 0
|
Algebra
|
Multiple Choice
|
The root(s) of $\frac {15}{x^2 - 4} - \frac {2}{x - 2} = 1$ is (are):
$\textbf{(A)}\ -5\text{ and }3\qquad \textbf{(B)}\ \pm 2\qquad \textbf{(C)}\ 2\text{ only}\qquad \textbf{(D)}\ -3\text{ and }5\qquad \textbf{(E)}\ 3\text{ only}$
|
[
"Use a common denominator: $\\frac{15 - 2(x+2)}{x^2-4} = 1$. After moving the $1$ to the left side, notice that you can factor the numerator:\n\\[\\frac{(x-5)(x+3)}{x^2-4} = 0\\]\nThe roots of this equation are $\\boxed{\\textbf{(D)} \\quad -3, 5}.$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/15.json
|
AHSME
|
1956_AHSME_Problems
| 5
| 0
|
Geometry
|
Multiple Choice
|
A nickel is placed on a table. The number of nickels which can be placed around it, each tangent to it and to two others is:
$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$
|
[
"Arranging the nickels in a hexagonal fashion, we see that only $\\boxed{\\textbf{(C) }6}$ nickels can be placed around the central nickel.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/5.json
|
AHSME
|
1956_AHSME_Problems
| 39
| 0
|
Geometry
|
Multiple Choice
|
The hypotenuse $c$ and one arm $a$ of a right triangle are consecutive integers. The square of the second arm is:
$\textbf{(A)}\ ca\qquad \textbf{(B)}\ \frac{c}{a}\qquad \textbf{(C)}\ c+a\qquad \textbf{(D)}\ c-a\qquad \textbf{(E)}\ \text{none of these}$
|
[
"The sides of the triangle are $a,$ $\\sqrt{c^2-a^2},$ and $c.$ We know that the hypotenuse and one leg are consecutive integers, so we can rewrite the side lengths as $a,$ $\\sqrt{2a+1},$ and $a+1.$\n\n\nSquaring the middle length gets $2a + 1 = a + c,$ so the answer is $\\boxed{\\textbf{(C)}}.$\n\n\n-coolmath34\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/39.json
|
AHSME
|
1956_AHSME_Problems
| 19
| 0
|
Algebra
|
Multiple Choice
|
Two candles of the same height are lighted at the same time. The first is consumed in $4$ hours and the second in $3$ hours.
Assuming that each candle burns at a constant rate, in how many hours after being lighted was the first candle twice the height of the second?
$\textbf{(A)}\ \frac{3}{4}\qquad\textbf{(B)}\ 1\frac{1}{2}\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 2\frac{2}{5}\qquad\textbf{(E)}\ 2\frac{1}{2}$
|
[
"In $x$ hours, the height of the first candle is $1 - \\frac{x}{4}$ and the height of the second candle is $1 - \\frac{x}{3}$ Essentially, we are solving the equation\n\\[1 - \\frac{x}{4} = 2(1 - \\frac{x}{3})\\].\n\n\n$- \\frac{x}{4} = 1 - \\frac{2x}{3}$ | $-1$\n\n\n$- 3x = 12 - 8x$ | $\\cdot 12$\n\n\n$5x = 12$ | $+8x$\n\n\n$x = \\frac{12}{5}$ | $/5$\n\n\n\n\n$x = \\boxed{\\textbf{(D)} \\quad 2 \\frac{2}{5}}$ hours.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/19.json
|
AHSME
|
1956_AHSME_Problems
| 23
| 0
|
Algebra
|
Multiple Choice
|
About the equation $ax^2 - 2x\sqrt {2} + c = 0$, with $a$ and $c$ real constants,
we are told that the discriminant is zero. The roots are necessarily:
$\textbf{(A)}\ \text{equal and integral}\qquad \textbf{(B)}\ \text{equal and rational}\qquad \textbf{(C)}\ \text{equal and real} \\ \textbf{(D)}\ \text{equal and irrational} \qquad \textbf{(E)}\ \text{equal and imaginary}$
|
[
"Plugging into the quadratic formula, we get\n\\[x = \\frac{2\\sqrt{2} \\pm \\sqrt{8-4ac}}{2a}.\\]\nThe discriminant is equal to 0, so this simplifies to $x = \\frac{2\\sqrt{2}}{2a}=\\frac{\\sqrt{2}}{a}.$ Because we are given that $a$ is real, $x$ is always real, and the answer is $\\boxed{\\textbf{(C)}}.$\n\n\n~ cxsmi (significant edits)\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/23.json
|
AHSME
|
1956_AHSME_Problems
| 9
| 0
|
Algebra
|
Multiple Choice
|
When you simplify $\left[ \sqrt [3]{\sqrt [6]{a^9}} \right]^4\left[ \sqrt [6]{\sqrt [3]{a^9}} \right]^4$, the result is:
$\textbf{(A)}\ a^{16} \qquad\textbf{(B)}\ a^{12} \qquad\textbf{(C)}\ a^8 \qquad\textbf{(D)}\ a^4 \qquad\textbf{(E)}\ a^2$
|
[
"This simplifies to \n\\[(a^{\\frac{9}{6}/3})^4 \\cdot (a^{\\frac{9}{3}/6})^4 = (a^{\\frac{1}{2}})^4 \\cdot (a^{\\frac{1}{2}})^4 = (a^2)(a^2) = \\boxed{a^4}\\]\nThe answer is $\\boxed{\\textbf{(D)}}.$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1956_AHSME_Problems/9.json
|
AHSME
|
1959_AHSME_Problems
| 20
| 0
|
Algebra
|
Multiple Choice
|
It is given that $x$ varies directly as $y$ and inversely as the square of $z$, and that $x=10$ when $y=4$ and $z=14$. Then, when $y=16$ and $z=7$, $x$ equals:
\begin{verbatim}
$\textbf{(A)}\ 180\qquad \textbf{(B)}\ 160\qquad \textbf{(C)}\ 154\qquad \textbf{(D)}\ 140\qquad \textbf{(E)}\ 120$
\end{verbatim}
|
[
"$x$ varies directly to $\\frac{y}{z^2}$ (The inverse variation of y and the square of z)\n\n\nWe can write the expression \n\n\n$x = \\frac{ky}{z^2}$\n\n\nNow we plug in the values of $x=10$ when $y=4$ and $z=14$. \n\n\nThis gives us $k=490$\n\n\nWe can use this to find the value of $x$ when $y=4$ and $z=14$\n\n\n$x=\\frac{490\\cdot4}{14^2}$\n\n\nSimplifying this we get,\n\n\n$\\fbox{B) 160}$\n\n\n~lli, awanglnc\n\n\n"
] | 1
|
./CreativeMath/AHSME/1959_AHSME_Problems/20.json
|
AHSME
|
1959_AHSME_Problems
| 16
| 0
|
Algebra
|
Multiple Choice
|
The expression$\frac{x^2-3x+2}{x^2-5x+6}\div \frac{x^2-5x+4}{x^2-7x+12},$ when simplified is:
$\textbf{(A)}\ \frac{(x-1)(x-6)}{(x-3)(x-4)} \qquad\textbf{(B)}\ \frac{x+3}{x-3}\qquad\textbf{(C)}\ \frac{x+1}{x-1}\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2$
|
[
"Factoring each of the binomials in the expression $\\frac{x^2-3x+2}{x^2-5x+6}\\div \\frac{x^2-5x+4}{x^2-7x+12},$ will yield the result of \\[\\frac{(x-2)(x-1)}{(x-3)(x-2)}\\div \\frac{(x-4)(x-1)}{(x-3)(x-4)},\\]\nWe can eliminate like terms to get $\\frac {x-1}{x-3}\\div \\frac{x-1}{x-3}$, which, according to identity property, is equivalent to the answer (D) 1.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1959_AHSME_Problems/16.json
|
AHSME
|
1959_AHSME_Problems
| 6
| 0
|
Other
|
Multiple Choice
|
Given the true statement: If a quadrilateral is a square, then it is a rectangle.
It follows that, of the converse and the inverse of this true statement is:
$\textbf{(A)}\ \text{only the converse is true} \qquad \\ \textbf{(B)}\ \text{only the inverse is true }\qquad \\ \textbf{(C)}\ \text{both are true} \qquad \\ \textbf{(D)}\ \text{neither is true} \qquad \\ \textbf{(E)}\ \text{the inverse is true, but the converse is sometimes true}$
|
[
"First, let us list the statement \"If a quadrilateral is a square, then it is a rectangle\" as a statement of the form \"If $p$, then $q$\". In this case, $p$ is \"a quadrilateral is a square\", and $q$ is \"it is a rectangle\".\n\n\nThe converse is then: \"If $q$, then $p$\". Plugging in, we get \"If a quadrilateral is a rectangle, then it is a square\". However, this is obviously false, as a rectangle does not have to have four sides of the same measure.\n\n\nThe inverse is then: \"If not $p$, then not $q$\". Plugging in, we get \"If a quadrilateral is not a square, then it is not a rectangle\". This is also false, since a quadrilateral can be a rectangle but not be a square. \n\n\nTherefore, the answer is $\\boxed{\\textbf{D}}$.\n\n\n\n\n\n\n"
] | 1
|
./CreativeMath/AHSME/1959_AHSME_Problems/6.json
|
AHSME
|
1959_AHSME_Problems
| 17
| 0
|
Algebra
|
Multiple Choice
|
If $y=a+\frac{b}{x}$, where $a$ and $b$ are constants, and if $y=1$ when $x=-1$, and $y=5$ when $x=-5$, then $a+b$ equals:
$\textbf{(A)}\ -1 \qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$
|
[
"Plugging in the x and y values, we obtain the following system of equations:\n\\[\\begin {cases} 1 = a - b \\\\ 5 = a - \\frac{b}{5} \\end {cases}\\]\nWe can then subtract the equations to obtain the equation $4 = 0.8b$, which works out to $b = 5$.\n\n\nPlugging that in to the original system of equations:\n\\[\\begin {cases} 1 = a - 5 \\\\ 5 = a - \\frac{5}{5} \\end {cases}\\]\nIt is easy to see that $a = 6$, and that $a+b$ is (E) 11.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1959_AHSME_Problems/17.json
|
AHSME
|
1959_AHSME_Problems
| 37
| 0
|
Algebra
|
Multiple Choice
|
When simplified the product $\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)\cdots\left(1-\frac1n\right)$ becomes:
$\textbf{(A)}\ \frac1n \qquad\textbf{(B)}\ \frac2n\qquad\textbf{(C)}\ \frac{2(n-1)}{n}\qquad\textbf{(D)}\ \frac{2}{n(n+1)}\qquad\textbf{(E)}\ \frac{3}{n(n+1)}$
|
[
"The product $\\left(1-\\frac13\\right)\\left(1-\\frac14\\right)\\left(1-\\frac15\\right)\\cdots\\left(1-\\frac1n\\right)$ can be simplified to $\\frac{2*3*4\\cdots*(n-2)*(n-1)}{3*4*5\\cdots*(n-1)*n}$, which ultimately works out to $\\textbf{(B)} \\frac{2}{n}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1959_AHSME_Problems/37.json
|
AHSME
|
1959_AHSME_Problems
| 21
| 0
|
Geometry
|
Multiple Choice
|
If$p$ is the perimeter of an equilateral $\triangle$ inscribed in a circle, the area of the circle is:
$\textbf{(A)}\ \frac{\pi p^2}{3} \qquad\textbf{(B)}\ \frac{\pi p^2}{9}\qquad\textbf{(C)}\ \frac{\pi p^2}{27}\qquad\textbf{(D)}\ \frac{\pi p^2}{81}\qquad\textbf{(E)}\ \frac{\pi p^2\sqrt3}{27}$
|
[
"A side length of the triangle is $\\frac{p}3$. An altitude of the triangle, by 30-60-90 triangles, is $\\frac{p\\sqrt{3}}{6}$. Because all classical triangle centers coincide on an equilateral triangle, by centroid properties a circumradius is $\\frac{p\\sqrt{3}}{9}$. Finally, the area of the circumcircle is $\\frac{\\pi p^2}{27}\\rightarrow\\boxed{\\textbf{C}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1959_AHSME_Problems/21.json
|
AHSME
|
1959_AHSME_Problems
| 10
| 0
|
Geometry
|
Multiple Choice
|
In $\triangle ABC$ with $\overline{AB}=\overline{AC}=3.6$, a point $D$ is taken on $AB$ at a distance $1.2$ from $A$. Point $D$ is joined to $E$ in the prolongation of $AC$ so that $\triangle AED$ is equal in area to $ABC$. Then $\overline{AE}$ is: $\textbf{(A)}\ 4.8 \qquad\textbf{(B)}\ 5.4\qquad\textbf{(C)}\ 7.2\qquad\textbf{(D)}\ 10.8\qquad\textbf{(E)}\ 12.6$
|
[
"Note that $\\frac{1}{2}AB * AC *\\sin\\angle BAC = \\frac{1}{2}AD * AE *\\sin\\angle DAE$. Since $\\angle BAC = \\angle DAE$, we have $AB*AC = AD*AE$, so that $3.6*3.6 = 1.2*AE$. Therefore, $AE = \\frac{3.6^2}{1.2} = 10.8$. Thusly, our answer is $\\boxed{\\text{(D)}}$, and we are done.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1959_AHSME_Problems/10.json
|
AHSME
|
1959_AHSME_Problems
| 26
| 0
|
Geometry
|
Multiple Choice
|
The base of an isosceles triangle is $\sqrt 2$. The medians to the leg intersect each other at right angles. The area of the triangle is:
$\textbf{(A)}\ 1.5 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2.5\qquad\textbf{(D)}\ 3.5\qquad\textbf{(E)}\ 4$
|
[
"Drop the altitude $AD$ from the vertex $A$ to base $BC$, and note that it intersects the centroid $M$. Then $BCM$ is an isosceles right triangle, so $MB=1$, so $MD=\\frac{\\sqrt{2}}{2}$. By centroid properties, $AD=3MD=\\frac{3\\sqrt2}{2}$. Then the area of $ABC$ is $\\frac12\\sqrt{2}\\frac{3\\sqrt2}{2}=1.5\\rightarrow\\boxed{\\textbf{A}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1959_AHSME_Problems/26.json
|
AHSME
|
1959_AHSME_Problems
| 1
| 0
|
Geometry
|
Multiple Choice
|
Each edge of a cube is increased by $50$%. The percent of increase of the surface area of the cube is:
$\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 125\qquad\textbf{(C)}\ 150\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 750$
|
[
"Note that increasing the length of each edge by $50$% with result in a cube that is similar to the original cube with scale factor $1.5$. Therefore, the surface area will increase by a factor of $1.5^2$, or $2.25$. Converting this back into a percent, the percent increase will be $125$%. Therefore, the answer is $\\boxed{\\textbf{B}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1959_AHSME_Problems/1.json
|
AHSME
|
1959_AHSME_Problems
| 50
| 0
|
Counting
|
Multiple Choice
|
A club with $x$ members is organized into four committees in accordance with these two rules:
\[\text{(1)}\ \textup{Each member belongs to two and only two committees}\qquad \\ \text{(2)}\ \textup{Each pair of committees has one and only one member in common}\]
Then $x$:
$\textbf{(A)} \ \textup{cannot be determined} \qquad \\ \textbf{(B)} \ \textup{has a single value between 8 and 16} \qquad \\ \textbf{(C)} \ \textup{has two values between 8 and 16} \qquad \\ \textbf{(D)} \ \textup{has a single value between 4 and 8} \qquad \\ \textbf{(E)} \ \textup{has two values between 4 and 8}$
|
[
"We can label each of the $x$ members with the pair of committees that they belong to, which is clearly valid due to rule (1). Then, by rule (2), for each pair of committees, there is exactly one member labeled with that pair. But since we have four committees, there must be $x = \\binom{4}{2} = 6$ members in total. Thusly our choice is $\\boxed{\\textbf{(D)}}$, and we are done.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1959_AHSME_Problems/50.json
|
AHSME
|
1959_AHSME_Problems
| 2
| 0
|
Geometry
|
Multiple Choice
|
Through a point $P$ inside the $\triangle ABC$ a line is drawn parallel to the base $AB$, dividing the triangle into two equal areas.
If the altitude to $AB$ has a length of $1$, then the distance from $P$ to $AB$ is:
$\textbf{(A)}\ \frac12 \qquad\textbf{(B)}\ \frac14\qquad\textbf{(C)}\ 2-\sqrt2\qquad\textbf{(D)}\ \frac{2-\sqrt2}{2}\qquad\textbf{(E)}\ \frac{2+\sqrt2}{8}$
|
[
"$\\boxed{\\textbf{D}}$\n\n\n\n\n\n\n"
] | 1
|
./CreativeMath/AHSME/1959_AHSME_Problems/2.json
|
AHSME
|
1959_AHSME_Problems
| 28
| 0
|
Geometry
|
Multiple Choice
|
In triangle $ABC$, $AL$ bisects angle $A$, and $CM$ bisects angle $C$. Points $L$ and $M$ are on $BC$ and $AB$, respectively. The sides of $\triangle ABC$ are $a$, $b$, and $c$. Then $\frac{AM}{MB}=k\frac{CL}{LB}$ where $k$ is:
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ \frac{bc}{a^2}\qquad\textbf{(C)}\ \frac{a^2}{bc}\qquad\textbf{(D)}\ \frac{c}{b}\qquad\textbf{(E)}\ \frac{c}{a}$
|
[
"By the angle bisector theorem, $AM/AB=AC/BC$ and $CL/LB=AC/AB$, so by rearranging the given equation and noting $AB=c$ and $BC=a$, $k=c/a\\rightarrow\\boxed{\\textbf{E}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1959_AHSME_Problems/28.json
|
AHSME
|
1959_AHSME_Problems
| 12
| 0
|
Algebra
|
Multiple Choice
|
By adding the same constant to $20,50,100$ a geometric progression results. The common ratio is: $\textbf{(A)}\ \frac53 \qquad\textbf{(B)}\ \frac43\qquad\textbf{(C)}\ \frac32\qquad\textbf{(D)}\ \frac12\qquad\textbf{(E)}\ \frac{1}3$
|
[
"Suppose that the constant is $x$. Then $20+x, 50+x, 100+x$ is a geometric progression, so $(20+x)(100+x) = (50+x)^2$. Expanding, we get $2000 + 120x + x^2 = 2500 + 100x + x^2$; therefore, $20x = 500$, so $x=25$.\n\n\nNow we can calculate our geometric progression to be $20+25, 50+25, 100+25 = 45, 75, 125$. Therefore, the common ratio is $\\frac{125}{75} = \\frac{75}{45} = \\boxed{\\frac{5}{3}}$, and our answer is $\\boxed{\\textbf{A}}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1959_AHSME_Problems/12.json
|
AHSME
|
1959_AHSME_Problems
| 24
| 0
|
Algebra
|
Multiple Choice
|
A chemist has m ounces of salt that is $m$% salt. How many ounces of salt must he add to make a solution that is $2m$% salt?
$\textbf{(A)}\ \frac{m}{100+m} \qquad\textbf{(B)}\ \frac{2m}{100-2m}\qquad\textbf{(C)}\ \frac{m^2}{100-2m}\qquad\textbf{(D)}\ \frac{m^2}{100+2m}\qquad\textbf{(E)}\ \frac{2m}{100+2m}$
|
[
"There are $m$ oz of salt, and there is $m\\times\\frac{m}{100}=\\frac{m^2}{100}$ oz of pure salt. We wish to add $k$ oz of pure salt so that $\\frac{m^2/100+k}{m+k}=\\frac{2m}{100}$. Solving this results in $k=\\frac{m^2}{100-2m}\\rightarrow\\boxed{\\textbf{C}}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1959_AHSME_Problems/24.json
|
AHSME
|
1959_AHSME_Problems
| 49
| 0
|
Algebra
|
Multiple Choice
|
For the infinite series $1-\frac12-\frac14+\frac18-\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}-\cdots$ let $S$ be the (limiting) sum. Then $S$ equals:
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac27\qquad\textbf{(C)}\ \frac67\qquad\textbf{(D)}\ \frac{9}{32}\qquad\textbf{(E)}\ \frac{27}{32}$
|
[
"Notice that we can arrange the sequence like so: \\begin{align*}1-\\frac12-\\frac14+\\frac18-\\frac{1}{16}-\\frac{1}{32}+\\frac{1}{64}-\\frac{1}{128}-\\cdots &= 1-\\frac{5}{8}-\\frac{5}{64}-\\frac{5}{512}-\\dots \\\\ &= 1-\\left(\\sum_{i=1}^{\\infty}\\frac{5}{8^n}\\right) \\\\ &= 1-\\frac{5}{7} \\text{ (by the convergence of a geometric series)} \\\\ &=\\frac{2}{7}\\end{align*}\nhence our answer is $\\fbox{B}$\n\n\n"
] | 1
|
./CreativeMath/AHSME/1959_AHSME_Problems/49.json
|
AHSME
|
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