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 |
|---|---|---|---|---|---|---|---|---|---|
1972_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | [asy] real t=pi/8;real u=7*pi/12;real v=13*pi/12; real ct=cos(t);real st=sin(t);real cu=cos(u);real su=sin(u); draw(unitcircle); draw((ct,st)--(-ct,st)--(cos(v),sin(v))); draw((cu,su)--(cu,st)); label("A",(-ct,st),W);label("B",(ct,st),E); label("M",(cu,su),N);label("P",(cu,st),S); label("C",(cos(v),sin(v)),W); //Credit... | [
"$\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/26.json | AHSME |
1972_AHSME_Problems | 30 | 0 | Geometry | Multiple Choice | [asy] real h = 7; real t = asin(6/h)/2; real x = 6-h*tan(t); real y = x*tan(2*t); draw((0,0)--(0,h)--(6,h)--(x,0)--cycle); draw((x,0)--(0,y)--(6,h)); draw((6,h)--(6,0)--(x,0),dotted); label("L",(3.75,h/2),W); label("$\theta$",(6,h-1.5),W);draw(arc((6,h),2,270,270-degrees(t)),Arrow(2mm)); label("6''",(3,0),S); draw((2.5... | [
"[asy] real h = 7; real t = asin(6/h)/2; real x = 6-h*tan(t); real y = x*tan(2*t); draw((0,0)--(0,h)--(6,h)--(x,0)--cycle); draw((x,0)--(0,y)--(6,h)); draw((6,h)--(6,0)--(x,0),dotted); label(\"A\",(0,h),NW);label(\"B\",(6,h),NE);label(\"C\",(6,0),SE);label(\"D\",(0,0),SW);label(\"E\",(x,0),N);label(\"F\",(0,y),W); ... | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/30.json | AHSME |
1972_AHSME_Problems | 31 | 0 | Number Theory | Multiple Choice | When the number $2^{1000}$ is divided by $13$, the remainder in the division is
$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }7\qquad \textbf{(E) }11$
| [
"By Fermat's Little Theorem, we know that $2^{100} \\equiv 2^{1000 \\pmod{12}}\\pmod{13}$. However, we find that $1000 \\equiv 4 \\pmod{12}$, so $2^{1000} \\equiv 2^4 = 16 \\equiv 3 \\pmod{13}$, so the answer is $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/31.json | AHSME |
1972_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | If the area of $\triangle ABC$ is $64$ square units and the geometric mean (mean proportional) between sides $AB$ and $AC$ is $12$ inches, then $\sin A$ is equal to
$\textbf{(A) }\dfrac{\sqrt{3}}{2}\qquad \textbf{(B) }\frac{3}{5}\qquad \textbf{(C) }\frac{4}{5}\qquad \textbf{(D) }\frac{8}{9}\qquad \textbf{(E) }\frac{1... | [
"Draw Diagram later\n\n\nWe can let $AB=s$ and $AC=r$. We can also say that the area of a triangle is $\\frac{1}{2}rs\\sin A$, which we know, is $64$. We also know that the geometric mean of $r$ and $s$ is 12, so $\\sqrt{rs}=12$.\n\n\n\n\nSquaring both sides gives us that $rs=144$. We can substitute this value into... | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/27.json | AHSME |
1972_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | The value(s) of $y$ for which the following pair of equations $x^2+y^2+16=0\text{ and }x^2-3y+12=0$ may have a real common solution, are
$\textbf{(A) }4\text{ only}\qquad \textbf{(B) }-7,~4\qquad \textbf{(C) }0,~4\qquad \textbf{(D) }\text{no }y\qquad \textbf{(E) }\text{all }y$
| [
"Because x\\textsuperscript{2} + y\\textsuperscript{2} + 16 = 0 has no real solutions, ∀ sets containing x\\textsuperscript{2} + y\\textsuperscript{2} + 16 = 0, no real solutions may exist.\n\n\n∴ the solution is $\\fbox{D}$\n\n\n\n\n– TylerO_1.618\n\n\n"
] | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/11.json | AHSME |
1972_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | If a dealer could get his goods for $8$% less while keeping his selling price fixed,
his profit, based on cost, would be increased to $(x+10)$% from his present profit of $x$%, which is
$\textbf{(A) }12\%\qquad \textbf{(B) }15\%\qquad \textbf{(C) }30\%\qquad \textbf{(D) }50\%\qquad \textbf{(E) }75\%$
| [
"If $c$ is the cost, then $0.92c$ is the reduced cost. Selling price is equal to cost plus profit:\n\n\n\n\n\\[c(1+0.01x) = 0.92c(1 + 0.01(x+10))\\]\n\n\nSolving for $x$ yields $x = 15.$ The answer is $\\textbf{(B)}.$\n\n\n-edited by coolmath34\n\n\n"
] | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/2.json | AHSME |
1972_AHSME_Problems | 28 | 0 | Geometry | Multiple Choice | A circular disc with diameter $D$ is placed on an $8\times 8$ checkerboard with width $D$ so that the centers coincide. The number of checkerboard squares which are completely covered by the disc is
$\textbf{(A) }48\qquad \textbf{(B) }44\qquad \textbf{(C) }40\qquad \textbf{(D) }36\qquad \textbf{(E) }32$
| [
"Consider the upper right half of the grid, which consists of a $4\\times4$ section of the checkerboard and a quarter-circle of radius $4$. We can draw this as a coordinate grid and shade in the complete squares. There are $8$ squares in the upper right corner, so there are $8 \\cdot 4 = \\boxed{32}$ whole squares ... | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/28.json | AHSME |
1972_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | The number of cubic feet in the volume of a cube is the same as the number of square inches in its surface area. The length of the edge expressed as a number of feet is
$\textbf{(A) }6\qquad \textbf{(B) }864\qquad \textbf{(C) }1728\qquad \textbf{(D) }6\times 1728\qquad \textbf{(E) }2304$
| [
"If the side length of a cube is $s$ feet the volume is $s^3$ cubic feet and the surface area is $6(12s^2)$ square inches. The volume and surface are numerically equal, so we can write\n\\[s^3 = 6(12s^2)\\]\nSolving yields $s=864.$\n\n\nThe answer is $\\textbf{(B)}.$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/12.json | AHSME |
1972_AHSME_Problems | 32 | 0 | Geometry | Multiple Choice | [asy] real t=pi/12;real u=8*t; real cu=cos(u);real su=sin(u); draw(unitcircle); draw((cos(-t),sin(-t))--(cos(13*t),sin(13*t))); draw((cu,su)--(cu,-su)); label("A",(cos(13*t),sin(13*t)),W); label("B",(cos(-t),sin(-t)),E); label("C",(cu,su),N); label("D",(cu,-su),S); label("E",(cu,sin(-t)),NE); label("2",((cu-1)/2,sin(-t... | [
"$\\boxed{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/32.json | AHSME |
1972_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | A man walked a certain distance at a constant rate. If he had gone $\textstyle\frac{1}{2}$ mile per hour faster, he would have walked the distance in four-fifths of the time; if he had gone $\textstyle\frac{1}{2}$ mile per hour slower, he would have been $2\textstyle\frac{1}{2}$ hours longer on the road. The distance i... | [
"We can make three equations out of the information, and since the distances are the same, we can equate these equations.\n\n\n\\[\\frac{4t}{5}(x+\\frac{1}{2})=xt=(t+\\frac{5}{2})(x-\\frac{1}{2})\\]\nwhere $x$ is the man's rate and $t$ is the time it takes him. \n\n\nLooking at the first two parts of the equations... | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/24.json | AHSME |
1972_AHSME_Problems | 25 | 0 | Geometry | Multiple Choice | Inscribed in a circle is a quadrilateral having sides of lengths $25,~39,~52$, and $60$ taken consecutively. The diameter of this circle has length
$\textbf{(A) }62\qquad \textbf{(B) }63\qquad \textbf{(C) }65\qquad \textbf{(D) }66\qquad \textbf{(E) }69$
| [
"We note that $25^2+60^2=65^2$ and $39^2+52^2=65^2$ so our answer is $\\boxed{C}$.\n\n\n-Pleaseletmewin\n\n\nAlternate Solution:\n\n\nLet's call $\\overline{AB}=25$, $\\overline{BC}=39$, $\\overline{CD}=52$, $\\overline{DA}=60$. Let's call $\\overline{BD}=x$ and $\\angle{DAB}=y$. By LoC we get the relations\n\n\n\\... | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/25.json | AHSME |
1972_AHSME_Problems | 33 | 0 | Algebra | Multiple Choice | The minimum value of the quotient of a (base ten) number of three different non-zero digits divided by the sum of its digits is
$\textbf{(A) }9.7\qquad \textbf{(B) }10.1\qquad \textbf{(C) }10.5\qquad \textbf{(D) }10.9\qquad \textbf{(E) }20.5$
| [
"The answer we are looking for can be expressed as $\\dfrac{100a+10b+c}{a+b+c}$. This is equivalent to $1 + \\dfrac{99a+9b}{a+b+c}$. Because we are trying to minimize our solution, we set $c$ = $9$, so we have $1 + \\dfrac{99a+9b}{a+b+9}$. This is equal to $1 + \\dfrac{9a+9b+81}{a+b+9} + \\dfrac{90a-81}{a+b+9}$, wh... | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/33.json | AHSME |
1972_AHSME_Problems | 13 | 0 | Geometry | Multiple Choice | [asy] draw(unitsquare);draw((0,0)--(.4,1)^^(0,.6)--(1,.2)); label("D",(0,1),NW);label("E",(.4,1),N);label("C",(1,1),NE); label("P",(0,.6),W);label("M",(.25,.55),E);label("Q",(1,.2),E); label("A",(0,0),SW);label("B",(1,0),SE); //Credit to Zimbalono for the diagram [/asy]
Inside square $ABCD$ (See figure) with sides of... | [
"Let the line passing through $M$ parallel to $AB$ intersect $AD$ and $BC$ and $S$ and $T$ respectively. Since $M$ is the midpoint of $AE$, $SM=\\frac{5}{2}$ and $TM=12-\\frac{5}{2}=\\frac{19}{2}$. Since $\\triangle PSM\\sim \\triangle QTM$, $PM:MQ=SM:MT=5:9$, hence our answer is $\\fbox{C}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/13.json | AHSME |
1972_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | If $f(x)=\log \left(\frac{1+x}{1-x}\right)$ for $-1<x<1$, then $f\left(\frac{3x+x^3}{1+3x^2}\right)$ in terms of $f(x)$ is
$\textbf{(A) }-f(x)\qquad \textbf{(B) }2f(x)\qquad \textbf{(C) }3f(x)\qquad \textbf{(D) }\left[f(x)\right]^2\qquad \\ \textbf{(E) }[f(x)]^3-f(x)$
| [
"$\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/29.json | AHSME |
1972_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | If $x=\dfrac{1-i\sqrt{3}}{2}$ where $i=\sqrt{-1}$, then $\dfrac{1}{x^2-x}$ is equal to
$\textbf{(A) }-2\qquad \textbf{(B) }-1\qquad \textbf{(C) }1+i\sqrt{3}\qquad \textbf{(D) }1\qquad \textbf{(E) }2$
| [
"Using DeMoivre's theorem, we can calculate $x^2=\\frac{1+i\\sqrt{3}}{2}$ The denominator is therefore $-1$ which makes the answer \\[\\boxed{\\textbf{(C) }-1}.\\] ~lopkiloinm\n\n\n"
] | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/3.json | AHSME |
1972_AHSME_Problems | 34 | 0 | Algebra | Multiple Choice | Three times Dick's age plus Tom's age equals twice Harry's age.
Double the cube of Harry's age is equal to three times the cube of Dick's age added to the cube of Tom's age.
Their respective ages are relatively prime to each other. The sum of the squares of their ages is
$\textbf{(A) }42\qquad \textbf{(B) }46\qquad... | [
"\\[t=2h-3d\\]\n\\[3d^3+t^3=2h^3\\]\n\n\nFirst, substitute in t into the second equation and get $3d^3+8h^3-36h^2d+54hd^2-27d^3=2h^3$. That turns into $h^3-6h^2d+9hd^2-4d^3=0$ which is factored into $(h-4d)(h-d)^2 =0.$ WLOG, $d=1$ and consequently $h=4$. Then $t=8-3=5$. Everything appears to be relatively prime alr... | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/34.json | AHSME |
1972_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | If $|x-\log y|=x+\log y$ where $x$ and $\log y$ are real, then
$\textbf{(A) }x=0\qquad \textbf{(B) }y=1\qquad \textbf{(C) }x=0\text{ and }y=1\qquad\\ \textbf{(D) }x(y-1)=0\qquad \textbf{(E) }\text{None of these}$
| [
"We have two cases: $x$ is positive and $x$ is negative.\n\\[\\text{Positive} \\qquad x - \\log y = x + \\log y \\qquad \\rightarrow \\qquad 2\\log y = 0 \\rightarrow y = 1\\]\n\\[\\text{Negative} \\qquad -x + \\log y = x + \\log y \\qquad \\rightarrow \\qquad 2x = 0 \\rightarrow x = 0\\]\n\n\nWe can write $x(y-1)=... | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/8.json | AHSME |
1972_AHSME_Problems | 18 | 0 | Geometry | Multiple Choice | Let $ABCD$ be a trapezoid with the measure of base $AB$ twice that of base $DC$, and let $E$ be the point of intersection of the diagonals. If the measure of diagonal $AC$ is $11$, then that of segment $EC$ is equal to
$\textbf{(A) }3\textstyle\frac{2}{3}\qquad \textbf{(B) }3\frac{3}{4}\qquad \textbf{(C) }4\qquad \te... | [
"We begin with a diagram:\n\n\n[asy] pair A, B, C, D, E; A = (0, 0); B = (8, 0); C = (7, 4); D = (3, 4); E = intersectionpoint(A--C, B--D); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); label(\"$A$\", A, W); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$E$\", E, 3W); label(\"$1... | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/18.json | AHSME |
1972_AHSME_Problems | 4 | 0 | Counting | Multiple Choice | The number of solutions to $\{1,~2\}\subseteq~X~\subseteq~\{1,~2,~3,~4,~5\}$, where $X$ is a subset of $\{1,~2,~3,~4,~5\}$ is
$\textbf{(A) }2\qquad \textbf{(B) }4\qquad \textbf{(C) }6\qquad \textbf{(D) }8\qquad \textbf{(E) }\text{None of these}$
| [
"$X$ has to contain $\\{1,~2\\}$, so only $\\{3,~4,~5\\}$ matters. There are two choices for the elements; the element is either in $X$ or outside of $X$. With this combinatorics in mind, the answer is simply $2^3=\\boxed{\\textbf{(D) }8}.$ ~lopkiloinm\n\n\n"
] | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/4.json | AHSME |
1972_AHSME_Problems | 14 | 0 | Geometry | Multiple Choice | A triangle has angles of $30^\circ$ and $45^\circ$. If the side opposite the $45^\circ$ angle has length $8$, then the side opposite the $30^\circ$ angle has length
$\textbf{(A) }4\qquad \textbf{(B) }4\sqrt{2}\qquad \textbf{(C) }4\sqrt{3}\qquad \textbf{(D) }4\sqrt{6}\qquad \textbf{(E) }6$
| [
"This triangle can be split into smaller 30-60-90 and 45-45-90 triangles. The side opposite the $45^\\circ$ angle has length $8,$ so the 30-60-90 triangle has sides $4, 4\\sqrt3,$ and $8.$\n\n\nOne of the legs of the 45-45-90 triangles is $4,$ so the hypotenuse is $4\\sqrt2.$ This is also the side opposite the $30^... | 2 | ./CreativeMath/AHSME/1972_AHSME_Problems/14.json | AHSME |
1972_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | A contractor estimated that one of his two bricklayers would take $9$ hours to build a certain wall and the other $10$ hours.
However, he knew from experience that when they worked together, their combined output fell by $10$ bricks per hour.
Being in a hurry, he put both men on the job and found that it took exactly... | [
"Every hour, the contractors lay $x/9$ and $x/10$ bricks, assuming a total of $x$ bricks in the wall. Together, $x/5$ bricks are laid per hour:\n\\[x = 5(x/9 + x/10 - 10)\\]\nThere are a total of $x=900$ bricks, and the answer is $\\textbf{(C)} \\quad 900$.\n\n\n-edited by coolmath34\n\n\n"
] | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/15.json | AHSME |
1972_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | From among $2^{1/2}, 3^{1/3}, 8^{1/8}, 9^{1/9}$ those which have the greatest and the next to the greatest values, in that order, are
$\textbf{(A) } 3^{1/3},\ 2^{1/2}\quad \textbf{(B) } 3^{1/3},\ 8^{1/8}\quad \textbf{(C) } 3^{1/3},\ 9^{1/9}\quad \textbf{(D) } 8^{1/8},\ 9^{1/9}\quad \\ \text{(E) None of these}$
... | [
"$8^{1/8}$ and $9^{1/9}$ are obviously too small. We must then compare $3^{1/3}$ with $2^{1/2}$. Raising both to the power of $6$ gives $9$ and $8$ respectively. Our answer is therefore $\\boxed{\\textbf{(A) } 3^{1/3},\\ 2^{1/2}}.$ ~lopkiloinm\n\n\n"
] | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/5.json | AHSME |
1972_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | Ann and Sue bought identical boxes of stationery. Ann used hers to write $1$-sheet letters and Sue used hers to write $3$-sheet letters.
Ann used all the envelopes and had $50$ sheets of paper left, while Sue used all of the sheets of paper and had $50$ envelopes left.
The number of sheets of paper in each box was
... | [
"Let $S$ be the number of sheets of paper and $E$ be the number of envelopes. We can write two equations:\n\\[S - E = 50\\]\n\\[E - S/3 = 50\\]\n\n\nSolving yields $(E, S) = (100, 150).$\n\n\nThe answer is $\\textbf{(A)}.$\n\n\n-edited by coolmath34\n\n\n"
] | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/9.json | AHSME |
1972_AHSME_Problems | 35 | 0 | Geometry | Multiple Choice | [asy] draw(unitsquare);draw((0,0)--(.25,sqrt(3)/4)--(.5,0)); label("Z",(0,1),NW);label("Y",(1,1),NE);label("A",(0,0),SW);label("X",(1,0),SE);label("B",(.5,0),S);label("P",(.25,sqrt(3)/4),N); //Credit to Zimbalono for the diagram [/asy]
Equilateral triangle $ABP$ (see figure) with side $AB$ of length $2$ inches is pla... | [
"$\\fbox{D}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/35.json | AHSME |
1953_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | If $y=x+\frac{1}{x}$, then $x^4+x^3-4x^2+x+1=0$ becomes:
$\textbf{(A)}\ x^2(y^2+y-2)=0 \qquad \textbf{(B)}\ x^2(y^2+y-3)=0\\ \textbf{(C)}\ x^2(y^2+y-4)=0 \qquad \textbf{(D)}\ x^2(y^2+y-6)=0\\ \textbf{(E)}\ \text{none of these}$
| [
"We multiply each of the answers to get: $x^2(y^2)+x^2(y)+nx^2$, where $n$ is either $-2,-3,-4,$ or $-6$. Looking at the first term, we have to square $y$, or $x+\\frac{1}{x}$, doing so, we get the equation $x^2+\\frac{1}{x^2}+2$. Multiplying that by $x^2$, we get $x^4+2x^2+1$. Multiplying $y$ by $x^2$, we get the ... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/20.json | AHSME |
1953_AHSME_Problems | 36 | 0 | Algebra | Multiple Choice | Determine $m$ so that $4x^2-6x+m$ is divisible by $x-3$. The obtained value, $m$, is an exact divisor of:
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 64$
| [
"Since the given expression is a quadratic, the factored form would be $(x-3)(4x+y)$, where $y$ is a value such that $-12x+yx=-6x$ and $-3(y)=m$. The only number that fits the first equation is $y=6$, so $m=-18$. The only choice that is a multiple of 18 is $\\boxed{\\textbf{(C) }36}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/36.json | AHSME |
1953_AHSME_Problems | 41 | 0 | Geometry | Multiple Choice | A girls' camp is located $300$ rods from a straight road. On this road, a boys' camp is located $500$ rods from the girls' camp.
It is desired to build a canteen on the road which shall be exactly the same distance from each camp.
The distance of the canteen from each of the camps is:
$\textbf{(A)}\ 400\text{ rods... | [
"[asy] draw((0,0)--(6,0)); draw((1,0)--(1,3)--(5,0)); draw((1,3)--(1.875,0)); label(\"$A$\",(1,0),S); label(\"$B$\",(5,0),S); label(\"$C$\",(1.875,0),S); label(\"$G$\",(1,3),NW); label(\"$r$\",(0,0),SW); [/asy]\nLet $r$ be the straight road, $G$ be the girls' camp, $B$ be the boys' camp, and $C$ be the water cantee... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/41.json | AHSME |
1953_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | Adams plans a profit of $10$ % on the selling price of an article and his expenses are $15$ % of sales. The rate of markup on an article that sells for $ $5.00$ is:
$\textbf{(A)}\ 20\% \qquad \textbf{(B)}\ 25\% \qquad \textbf{(C)}\ 30\% \qquad \textbf{(D)}\ 33\frac {1}{3}\% \qquad \textbf{(E)}\ 35\%$
| [
"Let the cost of the product be C, and the selling price be S. Since Adams wants a 10% profit on the sale, and his expenses are 15% of the sale, we add $0.10S$ and $0.15S$ to the cost to get the selling price.\nThen, $S=C+0.15S+0.10S \\Rightarrow 0.75S=C \\Rightarrow S=\\frac{4}{3}C \\Rightarrow \\fbox{D}$.\n\n\n"
... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/16.json | AHSME |
1953_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | Charles has $5q + 1$ quarters and Richard has $q + 5$ quarters. The difference in their money in dimes is:
$\textbf{(A)}\ 10(q - 1) \qquad \textbf{(B)}\ \frac {2}{5}(4q - 4) \qquad \textbf{(C)}\ \frac {2}{5}(q - 1) \\ \textbf{(D)}\ \frac{5}{2}(q-1)\qquad \textbf{(E)}\ \text{none of these}$
| [
"If Charles has $5q +1$ quarters and Richard has $q+5$ quarters, then the difference between them is $4q-4$ quarters. Converting quarters to dimes, the difference between them is $\\frac{5}{2}\\cdot (4)(q-1)$, or $10(q-1)$ dimes, $\\implies \\fbox{A}$\n\n\nNote that $q$ does not stand for the number of quarters th... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/6.json | AHSME |
1953_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | The fraction $\frac{\sqrt{a^2+x^2}-\frac{x^2-a^2}{\sqrt{a^2+x^2}}}{a^2+x^2}$ reduces to:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{2a^2}{a^2+x^2} \qquad \textbf{(C)}\ \frac{2x^2}{(a^2+x^2)^{\frac{3}{2}}}\qquad \textbf{(D)}\ \frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}\qquad \textbf{(E)}\ \frac{2x^2}{a^2+x^2}$
| [
"Multiplying the numerator and denominator by $\\sqrt{a^2+x^2}$ results in \n\\[\\frac{a^2+x^2-x^2+a^2}{(a^2+x^2)(\\sqrt{a^2+x^2)}}=\\frac{2a^2}{(a^2+x^2)(\\sqrt{a^2+x^2)}}.\\] \nSince $\\sqrt{a^2+x^2}=(a^2+x^2)^{\\frac{1}{2}}$, \nthe denominator is $(a^2+x^2)^2\\cdot (a^2+x^2)^{\\frac{1}{2}} = (a^2+x^2)^{\\frac{3}... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/7.json | AHSME |
1953_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | A man has part of $$4500$ invested at $4$% and the rest at $6$%. If his annual return on each investment is the same, the average rate of interest which he realizes of the $4500 is:
$\textbf{(A)}\ 5\% \qquad \textbf{(B)}\ 4.8\% \qquad \textbf{(C)}\ 5.2\% \qquad \textbf{(D)}\ 4.6\% \qquad \textbf{(E)}\ \text{none of ... | [
"You are trying to find $\\frac{2(0.06x)}{4500}$, where $x$ is the principle for one investment. To find $x$, solve $0.04(4500-x) = 0.06x$. $X$ will come out to be $1800$. Then, plug in x into the first equation, $\\frac{2(0.06)(1800)}{4500}$, to get $0.048$. Finally, convert that to a percentage and you get $\\box... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/17.json | AHSME |
1953_AHSME_Problems | 40 | 0 | Other | Multiple Choice | The negation of the statement "all men are honest," is:
$\textbf{(A)}\ \text{no men are honest} \qquad \textbf{(B)}\ \text{all men are dishonest} \\ \textbf{(C)}\ \text{some men are dishonest}\qquad \textbf{(D)}\ \text{no men are dishonest}\\ \textbf{(E)}\ \text{some men are honest}$
| [
"Note that $\\textbf{(A)}$ and $\\textbf{(B)}$ are the same, and $\\textbf{(D)}$ is the same as the quote in the statement \"all men are honest\" (that we have to found its negation). If not all men are honest, then that means that there is at least 1 dishonest (could be just one that spoils it for all the rest, or... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/40.json | AHSME |
1953_AHSME_Problems | 37 | 0 | Geometry | Multiple Choice | The base of an isosceles triangle is $6$ inches and one of the equal sides is $12$ inches.
The radius of the circle through the vertices of the triangle is:
$\textbf{(A)}\ \frac{7\sqrt{15}}{5} \qquad \textbf{(B)}\ 4\sqrt{3} \qquad \textbf{(C)}\ 3\sqrt{5} \qquad \textbf{(D)}\ 6\sqrt{3}\qquad \textbf{(E)}\ \text{none... | [
"[asy] draw((0,0)--(3,3sqrt(15))--(6,0)--cycle); draw((3,3sqrt(15))--(3,0)); label(\"$A$\",(3,3sqrt(15)),N); label(\"$B$\",(0,0),SW); label(\"$C$\",(6,0),SE); label(\"$D$\",(3,0),S); label(\"12\",(1.5,5.8),WNW); label(\"12\",(4.5,5.8),ENE); label(\"3\",(1.5,0),N); [/asy]\nLet $\\triangle ABC$ be an isosceles triang... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/37.json | AHSME |
1953_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | If $\log_{10} (x^2-3x+6)=1$, the value of $x$ is:
$\textbf{(A)}\ 10\text{ or }2 \qquad \textbf{(B)}\ 4\text{ or }-2 \qquad \textbf{(C)}\ 3\text{ or }-1 \qquad \textbf{(D)}\ 4\text{ or }-1\\ \textbf{(E)}\ \text{none of these}$
| [
"We know that $x^2-3x+6=10^1$, after expanding the logarithm. Taking the 10 to the other side, $x^2-3x-4=0$. Factoring, we get $(x-4)(x+1)=0$, so $x=\\boxed{4,-1\\Rightarrow \\text{(D)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/21.json | AHSME |
1953_AHSME_Problems | 47 | 0 | Algebra | Multiple Choice | If $x>0$, then the correct relationship is:
$\textbf{(A)}\ \log (1+x) = \frac{x}{1+x} \qquad \textbf{(B)}\ \log (1+x) < \frac{x}{1+x} \\ \textbf{(C)}\ \log(1+x) > x\qquad \textbf{(D)}\ \log (1+x) < x\qquad \textbf{(E)}\ \text{none of these}$
| [
"Plug in $x=9$. Then, you can see that the answer is $\\fbox{D}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/47.json | AHSME |
1953_AHSME_Problems | 10 | 0 | Geometry | Multiple Choice | The number of revolutions of a wheel, with fixed center and with an outside diameter of $6$ feet, required to cause a point on the rim to go one mile is:
$\textbf{(A)}\ 880 \qquad\textbf{(B)}\ \frac{440}{\pi} \qquad\textbf{(C)}\ \frac{880}{\pi} \qquad\textbf{(D)}\ 440\pi\qquad\textbf{(E)}\ \text{none of these}$
| [
"We know that the radius of the wheel is $3$ feet, so the total circumference of the wheel is $6\\pi$ feet. We also know that one mile is equivalent to $5280$ feet. It takes $\\frac{5280}{6\\pi}$ revolutions for any one point on the wheel to travel a mile. Simplifying, we find that the answer is $\\boxed{\\textbf{(... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/10.json | AHSME |
1953_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | The base of a triangle is $15$ inches. Two lines are drawn parallel to the base, terminating in the other two sides, and dividing the triangle into three equal areas. The length of the parallel closer to the base is:
$\textbf{(A)}\ 5\sqrt{6}\text{ inches} \qquad \textbf{(B)}\ 10\text{ inches} \qquad \textbf{(C)}\ 4\... | [
"Let the triangle be $\\triangle ABC$ where $BC$ is the base. Then let the parallels be $MN$ and $PQ$, where $PQ$ is closer to our base $BC$. \n\n\nIt's obvious that $\\triangle APQ \\sim \\triangle ABC$, where $|\\triangle APQ|:|\\triangle ABC|=2:3$, so $\\frac{PQ}{BC}=\\sqrt{\\frac{2}{3}}$. Since we know $BC=15$,... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/26.json | AHSME |
1953_AHSME_Problems | 30 | 0 | Arithmetic | Multiple Choice | A house worth $ $9000$ is sold by Mr. A to Mr. B at a $10$ % loss. Mr. B sells the house back to Mr. A at a $10$ % gain.
The result of the two transactions is:
$\textbf{(A)}\ \text{Mr. A breaks even} \qquad \textbf{(B)}\ \text{Mr. B gains }$900 \qquad \textbf{(C)}\ \text{Mr. A loses }$900\\ \textbf{(D)}\ \text{Mr. ... | [
"When Mr.A sells the house at a $10$% loss, he sells it for $9000(1 - .1) = 8100$. When Mr.B sells the house back to Mr. A at a $10$ % gain he sells it for $8100(1 + .1) = 8910$. Therefore Mr. A has lost $8100-8910 = 810$ dollars, so the answer is $\\boxed{D}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/30.json | AHSME |
1953_AHSME_Problems | 31 | 0 | Arithmetic | Multiple Choice | The rails on a railroad are $30$ feet long. As the train passes over the point where the rails are joined, there is an audible click.
The speed of the train in miles per hour is approximately the number of clicks heard in:
$\textbf{(A)}\ 20\text{ seconds} \qquad \textbf{(B)}\ 2\text{ minutes} \qquad \textbf{(C)}\ 1... | [
"We assume that the clicks are heard at the head of the train. Then if the train's speed in miles per hour is $x$, we can convert it to clicks per minute:\n\\[\\frac{x\\text{ mile}}{\\text{hr}}\\cdot\\left(\\frac{1\\text{ hr}}{60\\text{ min}}\\right)\\cdot\\left(\\frac{5280\\text{ ft}}{1\\text{ mile}}\\right)\\cdot... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/31.json | AHSME |
1953_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | The radius of the first circle is $1$ inch, that of the second $\frac{1}{2}$ inch, that of the third $\frac{1}{4}$ inch and so on indefinitely. The sum of the areas of the circles is:
$\textbf{(A)}\ \frac{3\pi}{4} \qquad \textbf{(B)}\ 1.3\pi \qquad \textbf{(C)}\ 2\pi \qquad \textbf{(D)}\ \frac{4\pi}{3}\qquad \textbf{... | [
"Note the areas of these circles is $1\\pi$, $\\frac{\\pi}{4}$, $\\frac{\\pi}{16}, \\dots$. The sum of these areas will thus be $\\pi\\left(1+\\frac{1}{4}+\\frac{1}{16}+\\dots\\right)$. We use the formula for an infinite geometric series to get the sum of the areas will be $\\pi\\left(\\frac{1}{1-\\frac{1}{4}}\\rig... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/27.json | AHSME |
1953_AHSME_Problems | 1 | 0 | Arithmetic | Multiple Choice | A boy buys oranges at $3$ for $10$ cents. He will sell them at $5$ for $20$ cents. In order to make a profit of $$1.00$, he must sell:
$\textbf{(A)}\ 67 \text{ oranges} \qquad \textbf{(B)}\ 150 \text{ oranges} \qquad \textbf{(C)}\ 200\text{ oranges}\\ \textbf{(D)}\ \text{an infinite number of oranges}\qquad \textbf{... | [
"The boy buys $3$ oranges for $10$ cents or $1$ orange for $\\frac{10}{3}$ cents. He sells them at $\\frac{20}{5}=4$ cents each. \nThat means for every orange he sells, he makes a profit of $4-\\frac{10}{3}=\\frac{2}{3}$ cents. \n\n\nTo make a profit of $100$ cents, he needs to sell $\\frac{100}{\\frac{2}{3}}=\\... | 2 | ./CreativeMath/AHSME/1953_AHSME_Problems/1.json | AHSME |
1953_AHSME_Problems | 50 | 0 | Geometry | Multiple Choice | One of the sides of a triangle is divided into segments of $6$ and $8$ units by the point of tangency of the inscribed circle. If the radius of the circle is $4$, then the length of the shortest side is
$\textbf{(A) \ } 12 \mathrm{\ units} \qquad \textbf{(B) \ } 13 \mathrm{\ units} \qquad \textbf{(C) \ } 14 \mathrm{\... | [
"Let the triangle have side lengths $14, 6+x,$ and $8+x$. The area of this triangle can be computed two ways. We have $A = rs$, and $A = \\sqrt{s(s-a)(s-b)(s-c)}$, where $s = 14+x$ is the semiperimeter. Therefore, $4(14+x)=\\sqrt{(14+x)(x)(8)(6)}$. Solving gives $x = 7$ as the only valid solution. This triangle has... | 2 | ./CreativeMath/AHSME/1953_AHSME_Problems/50.json | AHSME |
1953_AHSME_Problems | 11 | 0 | Geometry | Multiple Choice | A running track is the ring formed by two concentric circles. It is $10$ feet wide. The circumference of the two circles differ by about:
$\textbf{(A)}\ 10\text{ feet} \qquad \textbf{(B)}\ 30\text{ feet} \qquad \textbf{(C)}\ 60\text{ feet} \qquad \textbf{(D)}\ 100\text{ feet}\\ \textbf{(E)}\ \text{none of these}$
| [
"Since the track is 10 feet wide, the diameter of the outer circle will be 20 feet more than the inner circle. Since the circumference of a circle is directly proportional to its diameter, the difference in the circles' diameters is simply $20\\pi$ feet. Using $\\pi \\approx 3$, the answer is $\\fbox{C}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/11.json | AHSME |
1953_AHSME_Problems | 46 | 0 | Geometry | Multiple Choice | Instead of walking along two adjacent sides of a rectangular field, a boy took a shortcut along the diagonal of the field and
saved a distance equal to $\frac{1}{2}$ the longer side. The ratio of the shorter side of the rectangle to the longer side was:
$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{2}{3} \q... | [
"Let $x<y$ be the sides of the rectangle. The length of the diagonal is $\\sqrt{x^2+y^2}$, and the length of the two adjacent sides is $x+y$. Then the distance the boy saves is $x+y-\\sqrt{x^2+y^2}$. Setting this equal to $\\frac12y$, we have\n\\[x+y-\\sqrt{x^2+y^2}=\\frac12y\\]\n\\[x+\\frac12y=\\sqrt{x^2+y^2}\\]\n... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/46.json | AHSME |
1953_AHSME_Problems | 2 | 0 | Arithmetic | Multiple Choice | A refrigerator is offered at sale at $250.00 less successive discounts of 20% and 15%. The sale price of the refrigerator is:
$\textbf{(A) } \text{35\% less than 250.00} \qquad \textbf{(B) } \text{65\% of 250.00} \qquad \textbf{(C) } \text{77\% of 250.00} \qquad \textbf{(D) } \text{68\% of 250.00} \qquad \textbf{(E)... | [
"The first discount takes off $20\\%$ of the price, so the cost of the refrigerator is $0.8\\cdot250$. The next discount takes off $15\\%$, so the cost of the refrigerator is now $0.8\\cdot0.85\\cdot250=0.68\\cdot250$. Thus, the sale price of the refrigerator is $\\boxed{\\textbf{(D)}\\ \\text{68\\% of 250.00}}$.\n... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/2.json | AHSME |
1953_AHSME_Problems | 28 | 0 | Geometry | Multiple Choice | In $\triangle ABC$, sides $a,b$ and $c$ are opposite $\angle{A},\angle{B}$ and $\angle{C}$ respectively. $AD$ bisects $\angle{A}$ and meets $BC$ at $D$.
Then if $x = \overline{CD}$ and $y = \overline{BD}$ the correct proportion is:
$\textbf{(A)}\ \frac {x}{a} = \frac {a}{b + c} \qquad \textbf{(B)}\ \frac {x}{b} = \... | [
"[asy]pair A,B,C,D; A = (5,6); B = (9,0); C = (0,0); D = (B+C)/2; draw(D--A--B--C--A); label(\"$A$\",A,N); label(\"$B$\",B,SW); label(\"$C$\",C,SE); label(\"$D$\",D,S); label(\"$b$\",(A+C)/2,NW); label(\"$c$\",(B+A)/2,NE); label(\"$x$\",(C+D)/2,S); label(\"$y$\",(D+B)/2,S);[/asy]\nBy the Angle Bisector Theorem, $\... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/28.json | AHSME |
1953_AHSME_Problems | 12 | 0 | Geometry | Multiple Choice | The diameters of two circles are $8$ inches and $12$ inches respectively. The ratio of the area of the smaller to the area of the larger circle is:
$\textbf{(A)}\ \frac{2}{3} \qquad \textbf{(B)}\ \frac{4}{9} \qquad \textbf{(C)}\ \frac{9}{4} \qquad \textbf{(D)}\ \frac{1}{2}\qquad \textbf{(E)}\ \text{none of these}$
... | [
"The area of a circle can be calculated as $\\pi{r^2}$ where $r$ is the radius. We know that the radii of the circles are $4$ and $6$ inches (half the diameter) so the ratio of the area of the smaller to the area of the larger circle is $\\frac{16\\pi}{36\\pi}=\\boxed{\\textbf{(C) }\\frac{4}{9}}$.\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/12.json | AHSME |
1953_AHSME_Problems | 45 | 0 | Algebra | Multiple Choice | The lengths of two line segments are $a$ units and $b$ units respectively. Then the correct relation between them is:
$\textbf{(A)}\ \frac{a+b}{2} > \sqrt{ab} \qquad \textbf{(B)}\ \frac{a+b}{2} < \sqrt{ab} \qquad \textbf{(C)}\ \frac{a+b}{2}=\sqrt{ab}\\ \textbf{(D)}\ \frac{a+b}{2}\leq\sqrt{ab}\qquad \textbf{(E)}\ \fr... | [
"Since both lengths are positive, the AM-GM Inequality is satisfied. The correct relationship between $a$ and $b$ is $\\boxed{\\textbf{(E)}\\ \\frac{a+b}{2}\\geq\\sqrt{ab}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/45.json | AHSME |
1953_AHSME_Problems | 32 | 0 | Geometry | Multiple Choice | Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form:
$\textbf{(A)}\ \text{a square} \qquad \textbf{(B)}\ \text{a rectangle} \qquad \textbf{(C)}\ \text{a parallelogram with unequal sides}\\ \textbf{(D)}\ \text{a rhombus}\qquad \textbf{(E)}\ \text{... | [
"Draw a diagram:\n\n\n[asy] unitsize(1cm); pair A, B, C, D; A = (0,0); B = (0,3); C = (4.5,3); D = (4.5,0); draw(A--B--C--D--cycle); pair Aa, Ab, Ba, Bb; Aa = A + (D-A)*dir(30); Ab = A + (D-A)*dir(60); Ba = B + (C-B)*dir(-30); Bb = B + (C-B)*dir(-60); draw(Aa--A--Ab, red); draw(Ba--B--Bb, red); dot(intersectionpo... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/32.json | AHSME |
1953_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | If $a,b,c$ are positive integers less than $10$, then $(10a + b)(10a + c) = 100a(a + 1) + bc$ if:
$\textbf{(A) }b+c=10$
$\qquad\textbf{(B) }b=c$
$\qquad\textbf{(C) }a+b=10$
$\qquad\textbf {(D) }a=b$
$\qquad\textbf{(E) }a+b+c=10$
| [
"Multiply out the LHS to get $100a^2+10ac+10ab+bc=100a(a+1)+bc$. Subtract $bc$ and factor to get $10a(10a+b+c)=10a(10a+10)$. Divide both sides by $10a$ and then subtract $10a$ to get $b+c=10$, giving an answer of $\\boxed{A}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/24.json | AHSME |
1953_AHSME_Problems | 49 | 0 | Geometry | Multiple Choice | The coordinates of $A,B$ and $C$ are $(5,5),(2,1)$ and $(0,k)$ respectively.
The value of $k$ that makes $\overline{AC}+\overline{BC}$ as small as possible is:
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4\frac{1}{2} \qquad \textbf{(C)}\ 3\frac{6}{7} \qquad \textbf{(D)}\ 4\frac{5}{6}\qquad \textbf{(E)}\ 2\frac{1}{7}$
| [
"$k$ will be between $1$ and $5$ for $AC+BC$ to be the smallest. If we mirror point $A$ across the y-axis to $A'$, with coordinates $(-5,5),$ the distance $A'C+BC$ will be same as $AC+BC$. The minimum of $A'C+BC$ will occur when $C$ is on the straight line connecting $A'$ and $B$ (i.e., $C$ lies on the line $A'B$).... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/49.json | AHSME |
1953_AHSME_Problems | 48 | 0 | Geometry | Multiple Choice | If the larger base of an isosceles trapezoid equals a diagonal and the smaller base equals the altitude,
then the ratio of the smaller base to the larger base is:
$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{3}{4} \qquad \textbf{(D)}\ \frac{3}{5}\qquad \textbf{(E)}\ \frac{2... | [
"[asy] draw((0,0)--(1,3)--(4,3)--(5,0)--cycle); draw((0,0)--(4,3)); draw((4,3)--(4,0)); draw((3.8,0)--(3.8,0.2)--(4,0.2)); label(\"$A$\",(0,0),W); label(\"$B$\",(1,3),NW); label(\"$C$\",(4,3),NE); label(\"$D$\",(5,0),E); label(\"$E$\",(4,0),S); label(\"1\",(2,1.5),NW); [/asy]\n\n\nLet $a$ be the length of the small... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/48.json | AHSME |
1953_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | In a geometric progression whose terms are positive, any term is equal to the sum of the next two following terms. then the common ratio is:
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ \text{about }\frac{\sqrt{5}}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2}\qquad \textbf{(D)}\ \frac{1-\sqrt{5}}{2}\qquad \textbf{(E)}\ \fra... | [
"Given first term $a$ and common ratio $r$, we have $a=a*r+a*r^2$, and. We divide by $a$ in the first equation to get $1=r+r^2$. Rewriting, we have $r^2+r-1=0$. We use the quadratic formula to get $r = \\frac{-1+-\\sqrt{1^2-4(1)(-1)}}{2(1)}$. Because the terms all have to be positive, we must add the discriminant, ... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/25.json | AHSME |
1953_AHSME_Problems | 33 | 0 | Geometry | Multiple Choice | The perimeter of an isosceles right triangle is $2p$. Its area is:
$\textbf{(A)}\ (2+\sqrt{2})p \qquad \textbf{(B)}\ (2-\sqrt{2})p \qquad \textbf{(C)}\ (3-2\sqrt{2})p^2\\ \textbf{(D)}\ (1-2\sqrt{2})p^2\qquad \textbf{(E)}\ (3+2\sqrt{2})p^2$
| [
"Given leg length $x$, we can write the perimeter of this triangle to be $2x+x\\sqrt{2}=2p$. Thus, $x(2+\\sqrt{2})=2p$. Divide to get $x=\\frac{2p}{2+\\sqrt{2}}$. Multiply by the conjugate and simplify to get $x=p(2-\\sqrt{2})$. Square and divide by two to get the area of the triangle, or $\\frac{p^2*(2-\\sqrt{2}^2... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/33.json | AHSME |
1953_AHSME_Problems | 44 | 0 | Algebra | Multiple Choice | In solving a problem that reduces to a quadratic equation one student makes a mistake only in the constant term of the equation and
obtains $8$ and $2$ for the roots. Another student makes a mistake only in the coefficient of the first degree term and
find $-9$ and $-1$ for the roots. The correct equation was:
$\t... | [
"Let $x^2+bx+c=0$ represent the correct equation. Since the coefficient of the $x^2$ term is $1$, the sum of the roots is $-b$, and the product of the roots is $c$.\n\n\nIf a student only misreads the constant term, he must have the correct sum of roots. Therefore, the sum of the roots is $8+2=10$, so $b=-10$. If a... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/44.json | AHSME |
1953_AHSME_Problems | 13 | 0 | Geometry | Multiple Choice | A triangle and a trapezoid are equal in area. They also have the same altitude. If the base of the triangle is 18 inches, the median of the trapezoid is:
$\textbf{(A)}\ 36\text{ inches} \qquad \textbf{(B)}\ 9\text{ inches} \qquad \textbf{(C)}\ 18\text{ inches}\\ \textbf{(D)}\ \text{not obtainable from these data}\q... | [
"We know that the area of a trapezoid is $mh$ where $m$ is the median and the area of a triangle is $\\frac{bh}{2}$. We know that $b=18$ and the height of the trapezoid is congruent to the height of the triangle. Thus, we get $\\frac{18h}{2}=9h=mh$, so $m=\\textbf{(B) }9\\text{ inches}$.\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/13.json | AHSME |
1953_AHSME_Problems | 29 | 0 | Number Theory | Multiple Choice | The number of significant digits in the measurement of the side of a square whose computed area is $1.1025$ square inches to
the nearest ten-thousandth of a square inch is:
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 1$
| [
"There are 5 significant digits, $1$, $1$, $0$, $2$, and $5$. The answer is $\\boxed{D}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/29.json | AHSME |
1953_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | The factors of the expression $x^2+y^2$ are:
$\textbf{(A)}\ (x+y)(x-y) \qquad \textbf{(B)}\ (x+y)^2 \qquad \textbf{(C)}\ (x^{\frac{2}{3}}+y^{\frac{2}{3}})(x^{\frac{4}{3}}+y^{\frac{4}{3}})\\ \textbf{(D)}\ (x+iy)(x-iy)\qquad \textbf{(E)}\ \text{none of these}$
| [
"Trying each case out, we see \n$(x+iy)(x-iy)=x^2+xyi-xyi+(iy)(-iy)=x^2+(-1)(-y^2)=x^2+y^2$\n\n\nSo $\\boxed{\\text{D}}$ works\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/3.json | AHSME |
1953_AHSME_Problems | 34 | 0 | Geometry | Multiple Choice | If one side of a triangle is $12$ inches and the opposite angle is $30^{\circ}$, then the diameter of the circumscribed circle is:
$\textbf{(A)}\ 18\text{ inches} \qquad \textbf{(B)}\ 30\text{ inches} \qquad \textbf{(C)}\ 24\text{ inches} \qquad \textbf{(D)}\ 20\text{ inches}\\ \textbf{(E)}\ \text{none of these}$
| [
"By the Extended Law of Sines, the diameter, or twice the circumradius $R$, is given by\n\\[2R=\\frac{12\\text{ inches}}{\\sin30^\\circ}=\\boxed{\\textbf{(C)}\\ 24\\text{ inches}}.\\]\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/34.json | AHSME |
1953_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | The value of $x$ at the intersection of $y=\frac{8}{x^2+4}$ and $x+y=2$ is:
$\textbf{(A)}\ -2+\sqrt{5} \qquad \textbf{(B)}\ -2-\sqrt{5} \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \text{none of these}$
| [
"$x+y=2\\implies y=2-x$. Then $2-x=\\frac{8}{x^2+4}\\implies (2-x)(x^2+4)=8$. We now notice that $x=0\\implies (2)(4)=8$, so $\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/8.json | AHSME |
1953_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | The logarithm of $27\sqrt[4]{9}\sqrt[3]{9}$ to the base $3$ is:
$\textbf{(A)}\ 8\frac{1}{2} \qquad \textbf{(B)}\ 4\frac{1}{6} \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ \text{none of these}$
| [
"$27\\sqrt[4]{9}\\sqrt[3]{9}$ can be rewritten as $3^3\\cdot 3^\\frac{1}{2}\\cdot 3^\\frac{2}{3}$. Using exponent rules, this simplifies to $3^\\frac{25}{6}$.\nThe problem wants us to find $\\log_3{27\\sqrt[4]{9}\\sqrt[3]{9}}$. We just found that this is equal to $\\log_3{3^\\frac{25}{6}}$. Using logarithm rules, t... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/22.json | AHSME |
1953_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | One of the factors of $x^4+4$ is:
$\textbf{(A)}\ x^2+2 \qquad \textbf{(B)}\ x+1 \qquad \textbf{(C)}\ x^2-2x+2 \qquad \textbf{(D)}\ x^2-4\\ \textbf{(E)}\ \text{none of these}$
| [
"$(x^2-2x+2)\\cdot(x^2+2x+2)=x^4+4 \\implies \\textbf{(C)}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/18.json | AHSME |
1953_AHSME_Problems | 38 | 0 | Algebra | Multiple Choice | If $f(a)=a-2$ and $F(a,b)=b^2+a$, then $F(3,f(4))$ is:
$\textbf{(A)}\ a^2-4a+7 \qquad \textbf{(B)}\ 28 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 11$
| [
"We find $f(4)=(4)-2=2$, so $F(3,f(4))=F(3,2)=(2)^2+3=\\boxed{\\textbf{(C) }7}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/38.json | AHSME |
1953_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | The roots of $x(x^2+8x+16)(4-x)=0$ are:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 0,4 \qquad \textbf{(C)}\ 0,4,-4 \qquad \textbf{(D)}\ 0,4,-4,-4 \qquad \textbf{(E)}\ \text{none of these}$
| [
"We factor the middle part into $(x+4)^2$.\n\n\nThe equation now becomes $x(x+4)^2(4-x)=0$\n\n\nThe solutions are then $0, -4, 4$. So the answer is $\\boxed{\\text{C}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/4.json | AHSME |
1953_AHSME_Problems | 14 | 0 | Geometry | Multiple Choice | Given the larger of two circles with center $P$ and radius $p$ and the smaller with center $Q$ and radius $q$. Draw $PQ$. Which of the following statements is false?
$\textbf{(A)}\ p-q\text{ can be equal to }\overline{PQ}\\ \textbf{(B)}\ p+q\text{ can be equal to }\overline{PQ}\\ \textbf{(C)}\ p+q\text{ can be les... | [
"We will test each option to see if it can be true or not. Links to diagrams are provided.\n\\[\\textbf{(A)}\\ p-q\\text{ can be equal to }\\overline{PQ}\\]\nLet circle $Q$ be inside circle $P$ and tangent to circle $P$, and the point of tangency be $R$. $PR = p$, and $QR = q$, so $PR - QR = PQ = p-q.$\n[asy] pair ... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/14.json | AHSME |
1953_AHSME_Problems | 43 | 0 | Algebra | Multiple Choice | If the price of an article is increased by percent $p$, then the decrease in percent of sales must not exceed $d$ in order to yield the same income. The value of $d$ is:
$\textbf{(A)}\ \frac{1}{1+p} \qquad \textbf{(B)}\ \frac{1}{1-p} \qquad \textbf{(C)}\ \frac{p}{1+p} \qquad \textbf{(D)}\ \frac{p}{p-1}\qquad \textbf{... | [
"It turns out that none of the listed solutions are the true solution for the problem as written, so we'll edit it slightly by replacing \"percent\" with \"proportion\". To solve the modified problem, note that the price of the article is $1+p$ times what it was originally, so that demand must be at least $\\frac{1... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/43.json | AHSME |
1953_AHSME_Problems | 42 | 0 | Geometry | Multiple Choice | The centers of two circles are $41$ inches apart. The smaller circle has a radius of $4$ inches and the larger one has a radius of $5$ inches.
The length of the common internal tangent is:
$\textbf{(A)}\ 41\text{ inches} \qquad \textbf{(B)}\ 39\text{ inches} \qquad \textbf{(C)}\ 39.8\text{ inches} \qquad \textbf{(D... | [
"[asy] size(400); draw((0,0)--(41,0)); draw((0,0)--(45/41,200/41)--(1645/41,-160/41)); draw((0,0)--(1600/41,-360/41)--(41,0)); draw(circle((0,0),5)); draw(circle((41,0),4)); label(\"$A$\",(0,0),W); label(\"$B$\",(41,0),E); label(\"$C$\",(45/41,200/41),N); label(\"$D$\",(1645/41,-160/41),SE); label(\"$E$\",(1600/41,... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/42.json | AHSME |
1953_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | A circular piece of metal of maximum size is cut out of a square piece and then a square piece of maximum size is cut out of the circular piece. The total amount of metal wasted is:
$\textbf{(A)}\ \frac{1}{4} \text{ the area of the original square}\\ \textbf{(B)}\ \frac{1}{2}\text{ the area of the original square}\... | [
"The maximum diameter of the circular piece is the same as the side length of the square piece, so the circle is tangent to the square on all four sides. The maximum size a square piece that you can cut from the circle now has 4 edges that are the same as the square created by connecting the four midpoints of the o... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/15.json | AHSME |
1953_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | If $\log_6 x=2.5$, the value of $x$ is:
$\textbf{(A)}\ 90 \qquad \textbf{(B)}\ 36 \qquad \textbf{(C)}\ 36\sqrt{6} \qquad \textbf{(D)}\ 0.5 \qquad \textbf{(E)}\ \text{none of these}$
| [
"$\\log_6 x=\\frac{5}{2}\\implies 2\\log_6 x=5\\implies \\log_6 x^2=5\\implies x^2=6^5\\implies x=36\\sqrt{6}$, $\\fbox{C}$\n\n\n",
"$\\log_6 x=\\frac{5}{2}\\implies 6^{\\frac{5}{2}}=x \\implies \\sqrt{6^5}=x \\implies 36\\sqrt{6} = x$\n\n\n\n\n\n\n"
] | 2 | ./CreativeMath/AHSME/1953_AHSME_Problems/5.json | AHSME |
1953_AHSME_Problems | 39 | 0 | Algebra | Multiple Choice | The product, $\log_a b \cdot \log_b a$ is equal to:
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ a \qquad \textbf{(C)}\ b \qquad \textbf{(D)}\ ab \qquad \textbf{(E)}\ \text{none of these}$
| [
"\\[a^x=b\\]\n\\[b^y=a\\]\n\\[{a^x}^y=a\\]\n\\[xy=1\\]\n\\[\\log_a b\\log_b a=1\\]\nAs a result, the answer should be $\\boxed{\\textbf{(A) }1}$.\n\n\n",
"Apply the change of base formula to $\\log_a b$ and $\\log_b a$. For simplicity, let us convert to base-10 log.\nBy change of base, the expression becomes $\\... | 2 | ./CreativeMath/AHSME/1953_AHSME_Problems/39.json | AHSME |
1953_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | In the expression $xy^2$, the values of $x$ and $y$ are each decreased $25$ %; the value of the expression is:
$\textbf{(A)}\ \text{decreased } 50\% \qquad \textbf{(B)}\ \text{decreased }75\%\\ \textbf{(C)}\ \text{decreased }\frac{37}{64}\text{ of its value}\qquad \textbf{(D)}\ \text{decreased }\frac{27}{64}\text{ ... | [
"$xy^2$\n\n\n$(\\frac{3}{4}x)(\\frac{3}{4}y)^2$\n\n\n$(\\frac{3}{4})^3xy^2$\n\n\n$\\frac{27}{64}xy^2$\n\n\n$xy^2-\\frac{27}{64}xy^2 = \\frac{37}{64}xy^2 \\implies \\textbf{(C)}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/19.json | AHSME |
1953_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | The equation $\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5$ has:
$\textbf{(A)}\ \text{an extraneous root between } - 5\text{ and } - 1 \\ \textbf{(B)}\ \text{an extraneous root between }-10\text{ and }-6\\ \textbf{(C)}\ \text{a true root between }20\text{ and }25\qquad \textbf{(D)}\ \text{two true roots}\\ \textb... | [
"We can multiply both sides by $\\sqrt{x+10}$ to get $x+4=5\\sqrt{x+10}$. We can now square both sides to get $x^2+8x+16=25x+250$, which yields $x^2-17x-234=0$. We can factor the quadratic as $(x+9)(x-26)=0$, giving us roots of $-9$ and $26$. Plugging in these values, we find that $-9$ is an extraneous root and $26... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/23.json | AHSME |
1953_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | The number of ounces of water needed to reduce $9$ ounces of shaving lotion containing $50$ % alcohol to a lotion containing $30$ % alcohol is:
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$
| [
"Say we add $N$ ounces of water to the shaving lotion. Since half of a $9$ ounce bottle of shaving lotion is alcohol, we know that we have $\\frac{9}{2}$ ounces of alcohol. We want $\\frac{9}{2}=0.3(9+N)$ (because we want the amount of alcohol, $\\frac{9}{2}$, to be $30\\%$, or $0.3$, of the total amount of shaving... | 2 | ./CreativeMath/AHSME/1953_AHSME_Problems/9.json | AHSME |
1953_AHSME_Problems | 35 | 0 | Algebra | Multiple Choice | If $f(x)=\frac{x(x-1)}{2}$, then $f(x+2)$ equals:
$\textbf{(A)}\ f(x)+f(2) \qquad \textbf{(B)}\ (x+2)f(x) \qquad \textbf{(C)}\ x(x+2)f(x) \qquad \textbf{(D)}\ \frac{xf(x)}{x+2}\\ \textbf{(E)}\ \frac{(x+2)f(x+1)}{x}$
| [
"First simplify $f(x+2)$:\n\\[f(x+2)=\\frac{(x+2)(x+1)}{2}.\\]\nThen simplify the answers and see which one matches.\n\n\n$f(x)+f(2)=\\frac{x(x-1)}{2} + 1,$\n\n\n$(x+2)f(x)=\\frac{(x+2)x(x-1)}{2},$\n\n\n$x(x+2)f(x)=\\frac{(x+2)x^2(x-1)}{2},$\n\n\n$\\frac{xf(x)}{x+2}=\\frac{x^2(x-1)}{2(x+2)},$\n\n\n$\\frac{(x+2)f(x+... | 1 | ./CreativeMath/AHSME/1953_AHSME_Problems/35.json | AHSME |
1954_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | The equation $x^3+6x^2+11x+6=0$ has:
$\textbf{(A)}\ \text{no negative real roots}\qquad\textbf{(B)}\ \text{no positive real roots}\qquad\textbf{(C)}\ \text{no real roots}\\ \textbf{(D)}\ \text{1 positive and 2 negative roots}\qquad\textbf{(E)}\ \text{2 positive and 1 negative root}$
| [
"By the rational root theorem, $1, -1, 2, -2, 3, -3, 6, -6$ are possible rational roots. Because $x^3+6x^2+11x+6>0$ for $x>0$, so there are no positive roots. We try $-1, -2, -3, -6$, so $x=-1, x=-3, x=-2$, so there are no positive real roots; $\\fbox{B}$\n\n\n",
"Note that there are no sign changes (all coeffici... | 2 | ./CreativeMath/AHSME/1954_AHSME_Problems/20.json | AHSME |
1954_AHSME_Problems | 36 | 0 | Algebra | Multiple Choice | A boat has a speed of $15$ mph in still water. In a stream that has a current of $5$ mph it travels a certain
distance downstream and returns. The ratio of the average speed for the round trip to the speed in still water is:
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{1}{1}\qquad\textbf{(C)}\ \frac{8}{9}\qq... | [
"WLOG, let the distance the boat travel be $1$ mile. Then the boat takes $\\frac{1}{15+5}=\\frac{1}{20}=3$ minutes, to travel down a mile, then to travel back up the river, the boat travels $15-5=10$ miles per hour, taking $\\frac{1}{10}=\\frac{1}{10}=6$ minutes to travel up the river. This gives an average speed o... | 2 | ./CreativeMath/AHSME/1954_AHSME_Problems/36.json | AHSME |
1954_AHSME_Problems | 41 | 0 | Algebra | Multiple Choice | The sum of all the roots of $4x^3-8x^2-63x-9=0$ is:
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ -8 \qquad \textbf{(D)}\ -2 \qquad \textbf{(E)}\ 0$
| [
"By Vieta's Formulas, $\\frac{--8}{4}=2$, $\\fbox{B}$\n\n\n",
"$4x^3-8x^2-63x-9=0$\n\n\n$4(x^3-2x^2-\\frac{63}{4}x-\\frac{9}{4})=0$\n\n\nBy Vieta's Formulas:\n$x^3+ax^2+bx+c=(x-r)(x-p)(x-q)$\n\n\n$x^3+ax^2+bx+c=x^3+(-r-p-q)x^2+(rp+pq+rq)x+(-rpq)$\n\n\n$x^3-2x^2-\\frac{63}{4}x-\\frac{9}{4}=x^3+(-r-p-q)x^2+(rp+pq+r... | 2 | ./CreativeMath/AHSME/1954_AHSME_Problems/41.json | AHSME |
1954_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | If $f(x) = 5x^2 - 2x - 1$, then $f(x + h) - f(x)$ equals:
$\textbf{(A)}\ 5h^2 - 2h \qquad \textbf{(B)}\ 10xh - 4x + 2 \qquad \textbf{(C)}\ 10xh - 2x - 2 \\ \textbf{(D)}\ h(10x+5h-2)\qquad\textbf{(E)}\ 3h$
| [
"$5(x+h)^2 - 2(x+h) - 1-(5x^2 - 2x - 1)\\implies 5(x^2+2xh+h^2)-2x-2h-1-5x^2+2x+1\\implies 10xh+5h^2-2h$\n\n\n$\\implies h(10x+5h-2) \\boxed{(\\textbf{D})}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/16.json | AHSME |
1954_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | The value of $\frac{1}{16}a^0+\left (\frac{1}{16a} \right )^0- \left (64^{-\frac{1}{2}} \right )- (-32)^{-\frac{4}{5}}$ is:
$\textbf{(A)}\ 1 \frac{13}{16} \qquad \textbf{(B)}\ 1 \frac{3}{16} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{1}{16}$
| [
"$\\frac{1}{16}a^0+\\left (\\frac{1}{16a} \\right )^0- \\left (64^{-\\frac{1}{2}} \\right )- (-32)^{-\\frac{4}{5}}\\implies \\frac{1}{16}+1-\\frac{1}{8}-((-32)^4)^\\frac{1}{5}\\implies 1-\\frac{1}{16}-\\frac{1}{16}$$\\implies1-\\frac{1}{8}\\implies\\boxed{\\textbf{(D) }\\frac{7}{8}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/6.json | AHSME |
1954_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | A housewife saved $\textdollar{2.50}$ in buying a dress on sale. If she spent $\textdollar{25}$ for the dress, she saved about:
$\textbf{(A)}\ 8 \% \qquad \textbf{(B)}\ 9 \% \qquad \textbf{(C)}\ 10 \% \qquad \textbf{(D)}\ 11 \% \qquad \textbf{(E)}\ 12\%$
| [
"Since she saved $\\textdollar{2.50}$ and spent $\\textdollar{25}$, the original price of the dress was $\\textdollar{27.50}$. The percent saved can be modeled as $27.5x=2.5$, so the answer is $x=\\frac{2.5}{27.5}$ or $\\frac{1}{11}$, approximately $9\\% \\implies \\textbf{(B)}$\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/7.json | AHSME |
1954_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | The graph of the function $f(x) = 2x^3 - 7$ goes:
$\textbf{(A)}\ \text{up to the right and down to the left} \\ \textbf{(B)}\ \text{down to the right and up to the left}\\ \textbf{(C)}\ \text{up to the right and up to the left}\\ \textbf{(D)}\ \text{down to the right and down to the left}\\ \textbf{(E)}\ \text{none ... | [
"What the question is basically asking, is the limit as the function goes to each end of infinity: \n\n\n$\\lim_{x\\to\\infty} f(x)=\\infty$\n\n\n$\\lim_{x\\to-\\infty} f(x)=-\\infty$. \n\n\nThis means it goes up and the right, and down and to the left. $\\boxed{(\\textbf{A})}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/17.json | AHSME |
1954_AHSME_Problems | 40 | 0 | Algebra | Multiple Choice | If $\left (a+\frac{1}{a} \right )^2=3$, then $a^3+\frac{1}{a^3}$ equals:
$\textbf{(A)}\ \frac{10\sqrt{3}}{3}\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 7\sqrt{7}\qquad\textbf{(E)}\ 6\sqrt{3}$
| [
"$a+\\frac{1}{a}=\\sqrt{3}\\implies (a+\\frac{1}{a})^3=3\\sqrt{3}\\implies a^3+3\\frac{a^2}{a}+\\frac{3a}{a^2}+\\frac{1}{a^3}\\implies a^3+3(a+\\frac{1}{a})+\\frac{1}{a^3}=3\\sqrt{3}\\implies a^3+\\frac{1}{a^3}+3\\sqrt{3}=3\\sqrt{3}\\implies a^3+\\frac{1}{a^3}=0$, $\\fbox{C}$\n\n\n",
"$\\left (a+\\frac{1}{a} \\ri... | 3 | ./CreativeMath/AHSME/1954_AHSME_Problems/40.json | AHSME |
1954_AHSME_Problems | 37 | 0 | Geometry | Multiple Choice | Given $\triangle PQR$ with $\overline{RS}$ bisecting $\angle R$, $PQ$ extended to $D$ and $\angle n$ a right angle, then:
[asy] path anglemark2(pair A, pair B, pair C, real t=8, bool flip=false) { pair M,N; path mark; M=t*0.03*unit(A-B)+B; N=t*0.03*unit(C-B)+B; if(flip) mark=Arc(B,t*0.03,degrees(C-B)-360,degre... | [
"Let $\\angle PRS$ be $\\theta$.\n\n\n$p+ q + 2\\theta = 180$\n\n\n$m+\\theta+90=180 \\implies m+\\theta=90 \\implies 2m+2\\theta=180$\n\n\n$p+q+2\\theta=2m+2\\theta \\implies \\frac{p+q}{2}=m \\implies \\boxed{\\textbf{(B) \\ } \\angle m = \\frac{1}{2}(\\angle p + \\angle q)}$\n\n\n",
"[asy] import math; path an... | 2 | ./CreativeMath/AHSME/1954_AHSME_Problems/37.json | AHSME |
1954_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | The roots of the equation $2\sqrt {x} + 2x^{ - \frac {1}{2}} = 5$ can be found by solving:
$\textbf{(A)}\ 16x^2-92x+1 = 0\qquad\textbf{(B)}\ 4x^2-25x+4 = 0\qquad\textbf{(C)}\ 4x^2-17x+4 = 0\\ \textbf{(D)}\ 2x^2-21x+2 = 0\qquad\textbf{(E)}\ 4x^2-25x-4 = 0$
| [
"Make the substitution $t=\\sqrt{x}$. Then $2t+\\frac{2}{t}=5\\implies 2t^2-5t+2=0$. Then $2x-5\\sqrt{x}+2=0\\implies 2x+2=5\\sqrt{x}\\implies (2x+2)^2=25x\\implies 4x^2+8x+4=25x\\implies 4x^2-17x+4=0$, $\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/21.json | AHSME |
1954_AHSME_Problems | 47 | 0 | Algebra | Multiple Choice | At the midpoint of line segment $AB$ which is $p$ units long, a perpendicular $MR$ is erected with length $q$ units. An arc is described from $R$ with a radius equal to $\frac{1}{2}AB$, meeting $AB$ at $T$. Then $AT$ and $TB$ are the roots of:
$\textbf{(A)}\ x^2+px+q^2=0\\ \textbf{(B)}\ x^2-px+q^2=0\\ \textbf{(C)}\ x... | [
"Note that $AT + TB = p$. Note, also, that $AT * TB = (AM - TM) * (BM + TM) = (AM - TM) * (AM + TM) = AM^2 - TM^2 = TR^2 - TM^2 = MR^2 = q^2$. Therefore, by Vieta's formulas, we have $(x - AT)(x - TB) = x^2 - px + q^2$, so our answer is $\\boxed{\\textbf{(B)}}$ and we are done.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/47.json | AHSME |
1954_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | The sum of the numerical coefficients in the expansion of the binomial $(a+b)^6$ is:
$\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 7$
| [
"$P(1,1)=(1+1)^6=2^6=64$, $\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/10.json | AHSME |
1954_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles
The straight line $\overline{AB}$ is divided at $C$ so that $AC=3CB$. Circles are described on $\overline{AC}$ and $\overline{CB}$ as diameters and a common tangent meets $AB$ produced at $D$. Then $BD$ equals:
$\textbf{(A)}\ \text{diameter o... | [
"Let $x=\\overline{BD}$ and let $r$ be the radius of the small circle. Draw the line from the center of each of the circles to the point of contact of the tangent of the circle. By similar triangles, $\\frac{x+r}{r}=\\frac{x+5r}{3r} \\implies x=r$, or $\\boxed{\\textbf{(B)}}$.\n\n\n==Video Solution\n\n\nhttps://you... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/26.json | AHSME |
1954_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | $A$ and $B$ together can do a job in $2$ days; $B$ and $C$ can do it in four days; and $A$ and $C$ in $2\frac{2}{5}$ days.
The number of days required for A to do the job alone is:
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 2.8$
| [
"Let $A$ do $r_A$ of the job per day, $B$ do $r_B$ of the job per day, and $C$ do $r_C$ of the job per day. These three quantities have unit $\\frac{\\text{job}}{\\text{day}}$. Therefore our three conditions give us the three equations:\n\\begin{align*} (2\\text{ days})(r_A+r_B)&=1\\text{ job},\\nonumber\\\\ (4\\te... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/30.json | AHSME |
1954_AHSME_Problems | 31 | 0 | Geometry | Multiple Choice | In $\triangle ABC$, $AB=AC$, $\angle A=40^\circ$. Point $O$ is within the triangle with $\angle OBC \cong \angle OCA$.
The number of degrees in $\angle BOC$ is:
$\textbf{(A)}\ 110^{\circ} \qquad \textbf{(B)}\ 35^{\circ} \qquad \textbf{(C)}\ 140^{\circ} \qquad \textbf{(D)}\ 55^{\circ} \qquad \textbf{(E)}\ 70^{\circ}... | [
"Since $\\triangle ABC$ is an isosceles triangle, $\\angle ABC = \\angle ACB = 70^{\\circ}$. Let $\\angle OBC = \\angle OCA = x$. Since $\\angle ACB = 70$, $\\angle OCB = 70 - x$. The angle of $\\triangle OBC$ add up to $180$, so $\\angle BOC = 180 - (x + 70 - x) = \\boxed{\\textbf{(A) } 110^{\\circ}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/31.json | AHSME |
1954_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | A right circular cone has for its base a circle having the same radius as a given sphere.
The volume of the cone is one-half that of the sphere. The ratio of the altitude of the cone to the radius of its base is:
$\textbf{(A)}\ \frac{1}{1}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D... | [
"Because the circle has the same radius as the sphere, the cylinder and sphere have the same radius. Then from the volume of cylinder and volume of a sphere formulas, we have $\\frac{1}{3} \\pi r^2 h= \\frac{2}{3} \\pi r^3 \\implies h=2r\\implies \\frac{h}{r}=2$ $\\boxed{(\\textbf{D})}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/27.json | AHSME |
1954_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | The square of $5-\sqrt{y^2-25}$ is:
$\textbf{(A)}\ y^2-5\sqrt{y^2-25} \qquad \textbf{(B)}\ -y^2 \qquad \textbf{(C)}\ y^2 \\ \textbf{(D)}\ (5-y)^2\qquad\textbf{(E)}\ y^2-10\sqrt{y^2-25}$
| [
"$(5-\\sqrt{y^2-25})^2 = 5^2-2\\cdot 5\\cdot\\sqrt{y^2-25}+y^2-25 = y^2-10\\sqrt{y^2-25}$ $\\ \\Rightarrow$ $\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/1.json | AHSME |
1954_AHSME_Problems | 50 | 0 | Geometry | Multiple Choice | The times between $7$ and $8$ o'clock, correct to the nearest minute, when the hands of a clock will form an angle of $84^{\circ}$ are:
$\textbf{(A)}\ \text{7: 23 and 7: 53}\qquad \textbf{(B)}\ \text{7: 20 and 7: 50}\qquad \textbf{(C)}\ \text{7: 22 and 7: 53}\\ \textbf{(D)}\ \text{7: 23 and 7: 52}\qquad \textbf{(E)... | [
"At $7$ o'clock, the hour hand is at the position $\\tfrac{7}{12}\\cdot 360^{\\circ}=210^{\\circ}$ clockwise from the $12$ o'clock position, and the minute hand is exactly at the $12$ o'clock position. Thus the minute hand is $360^{\\circ}-210^{\\circ}=150^{\\circ}$ ahead of the hour hand, while it is also $210^{\\... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/50.json | AHSME |
1954_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | A merchant placed on display some dresses, each with a marked price. He then posted a sign “$\frac{1}{3}$ off on these dresses.”
The cost of the dresses was $\frac{3}{4}$ of the price at which he actually sold them. Then the ratio of the cost to the marked price was:
$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ ... | [
"Let $x$ be the marked price. Then $\\frac{2}{3}x$ would be the selling price of the dresses. The cost of the dresses was $\\frac{3}{4}$ of the selling price, or $\\frac{3}{4}\\cdot\\frac{2}{3}x \\implies \\frac{1}{2}x$. So the ratio of the cost to the marked price is $\\frac{\\frac{1}{2}x}{x}$, or $\\frac{1}{2} \\... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/11.json | AHSME |
1954_AHSME_Problems | 46 | 0 | Geometry | Multiple Choice | In the diagram, if points $A, B$ and $C$ are points of tangency, then $x$ equals:
[asy] unitsize(5cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=3; pair A=(-3*sqrt(3)/32,9/32), B=(3*sqrt(3)/32, 9/32), C=(0,9/16); pair O=(0,3/8); draw((-2/3,9/16)--(2/3,9/16)); draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqr... | [
"First we extend the line with $A$ and the line with $B$ so that they both meet the line with $C$, forming an equilateral triangle. Let the vertices of this triangle be $D$, $E$, and $F$. We know it is equilateral because of the angle of $60^\\circ$ shown, and because the tangent lines $\\overline{EF}$ and $\\overl... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/46.json | AHSME |
1954_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | The equation $\frac{2x^2}{x-1}-\frac{2x+7}{3}+\frac{4-6x}{x-1}+1=0$ can be transformed by eliminating fractions to the equation $x^2-5x+4=0$.
The roots of the latter equation are $4$ and $1$. Then the roots of the first equation are:
$\textbf{(A)}\ 4 \text{ and }1 \qquad \textbf{(B)}\ \text{only }1 \qquad \textbf{(... | [
"We plug $x=4, x=1$ into the original equation\n$x=4\\implies\\frac{8}{-3}+\\frac{15}{3}+\\frac{-20}{-3}\\implies\\frac{15+20-8}{4}+1=0$\n$x=1\\implies\\frac{2}{0}-\\frac{9}{3}+\\frac{-2}{0}+1=0$, which is undefined because $\\frac{k}{0}$ is undefined. \n\n\nThus the only solution to the equation is $x=4$, $\\fbox{... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/2.json | AHSME |
1954_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | If $\frac{m}{n}=\frac{4}{3}$ and $\frac{r}{t}=\frac{9}{14}$, the value of $\frac{3mr-nt}{4nt-7mr}$ is:
$\textbf{(A)}\ -5\frac{1}{2}\qquad\textbf{(B)}\ -\frac{11}{14}\qquad\textbf{(C)}\ -1\frac{1}{4}\qquad\textbf{(D)}\ \frac{11}{14}\qquad\textbf{(E)}\ -\frac{2}{3}$
| [
"From $\\frac{m}{n}=\\frac{4}{3}$, we have $3m=4n$. From $\\frac{r}{t}=\\frac{9}{14}$, we have $14r=9t\\implies 7r=4.5t$\n\n\nThis simplifies the fraction to $\\frac{4nr-nt}{4nt-7r\\cdot m}\\implies \\frac{4nr-nt}{4nt-4.5mt}\\implies \\frac{4nr-nt}{4nt-1.5t\\cdot3m}\\implies \\frac{4nr-nt}{4nr-1.5\\cdot t\\cdot 4n}... | 2 | ./CreativeMath/AHSME/1954_AHSME_Problems/28.json | AHSME |
1954_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | The solution of the equations
\begin{align*}2x-3y &=7 \\ 4x-6y &=20\end{align*}
is:
$\textbf{(A)}\ x=18, y=12 \qquad \textbf{(B)}\ x=0, y=0 \qquad \textbf{(C)}\ \text{There is no solution} \\ \textbf{(D)}\ \text{There are an unlimited number of solutions}\qquad \textbf{(E)}\ x=8, y=5$
| [
"$2x-3y=7\\implies 4x-6y=14$\nThen $4x-6y=14=20$, which is false, so there are no solutions; $\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/12.json | AHSME |
1954_AHSME_Problems | 45 | 0 | Geometry | Multiple Choice | In a rhombus, $ABCD$, line segments are drawn within the rhombus, parallel to diagonal $BD$, and terminated in the sides of the rhombus. A graph is drawn showing the length of a segment as a function of its distance from vertex $A$. The graph is:
$\textbf{(A)}\ \text{A straight line passing through the origin.}\\ \te... | [
"Note that the length of the segment increases linearly with distance from $A$, starting at zero, until it passes side $BD$ (this can be shown with similar triangles). From there, it decreases linearly until it reaches zero at point $C$ (this can also be shown with similar triangles). The only shape that matches th... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/45.json | AHSME |
1954_AHSME_Problems | 32 | 0 | Algebra | Multiple Choice | The factors of $x^4+64$ are:
$\textbf{(A)}\ (x^2+8)^2\qquad\textbf{(B)}\ (x^2+8)(x^2-8)\qquad\textbf{(C)}\ (x^2+2x+4)(x^2-8x+16)\\ \textbf{(D)}\ (x^2-4x+8)(x^2-4x-8)\qquad\textbf{(E)}\ (x^2-4x+8)(x^2+4x+8)$
| [
"Notice that: \n\\begin{align*} x^4+64 &= x^4+2^6 \\\\ \t&= x^4+4\\cdot 2^4 \\end{align*}\n\n\nSo, using the Sophie Germain Identity, we can factor the expression as:\n\\begin{align*} x^4+64 & = x^4+4\\cdot 2^4\\\\ \t& = (x^2+2\\cdot2^2-4x)(x^2+2\\cdot2^2+4x)\\\\ & = (x^2-4x+8)(x^2+4x+8) \\end{align*}\nSo... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/32.json | AHSME |
1954_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | The values of $k$ for which the equation $2x^2-kx+x+8=0$ will have real and equal roots are:
$\textbf{(A)}\ 9\text{ and }-7\qquad\textbf{(B)}\ \text{only }-7\qquad\textbf{(C)}\ \text{9 and 7}\\ \textbf{(D)}\ -9\text{ and }-7\qquad\textbf{(E)}\ \text{only 9}$
| [
"$b^2-4ac=0\\implies b^2=4ac\\implies (1-k)^2=4(2)(8)\\implies (1-k)^2=8^2\\implies 1-k=8\\implies k=-7, 1-k=-8\\implies k=9$, $\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/24.json | AHSME |
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