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 |
|---|---|---|---|---|---|---|---|---|---|
1952_AHSME_Problems | 44 | 0 | Algebra | Multiple Choice | If an integer of two digits is $k$ times the sum of its digits, the number formed by interchanging the digits is the sum of the digits multiplied by
$\textbf{(A) \ } 9-k \qquad \textbf{(B) \ } 10-k \qquad \textbf{(C) \ } 11-k \qquad \textbf{(D) \ } k-1 \qquad \textbf{(E) \ } k+1$
| [
"Let $n = 10a+b$. The problem states that $10a+b=k(a+b)$. We want to find $x$, where $10b+a=x(a+b)$. Adding these two equations gives $11(a+b) = (k+x)(a+b)$. Because $a+b \\neq 0$, we have $11 = k + x$, or $x = \\boxed{\\textbf{(C) \\ } 11-k}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/44.json | AHSME |
1952_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | The function $x^2+px+q$ with $p$ and $q$ greater than zero has its minimum value when:
$\textbf{(A) \ }x=-p \qquad \textbf{(B) \ }x=\frac{p}{2} \qquad \textbf{(C) \ }x=-2p \qquad \textbf{(D) \ }x=\frac{p^2}{4q} \qquad$
$\textbf{(E) \ }x=\frac{-p}{2}$
| [
"The minimum value of this parabola is found at its turning point, on the line $\\boxed{\\textbf{(E)}\\ x=\\frac{-p}{2}}$.\nIndeed, the turning point of any function of the form $ax^2+bx+c$ has an x-coordinate of $\\frac{-b}{2a}$. This can be seen at the average of the quadratic's two roots (whose sum is $\\frac{-b... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/13.json | AHSME |
1952_AHSME_Problems | 29 | 0 | Geometry | Multiple Choice | In a circle of radius $5$ units, $CD$ and $AB$ are perpendicular diameters. A chord $CH$ cutting $AB$ at $K$ is $8$ units long. The diameter $AB$ is divided into two segments whose dimensions are:
$\textbf{(A)}\ 1.25, 8.75 \qquad \textbf{(B)}\ 2.75,7.25 \qquad \textbf{(C)}\ 2,8 \qquad \textbf{(D)}\ 4,6 \qquad \textb... | [
"Let the intersection of $CH$ and $AB$ be $N$, $O$ be the center of the circle, $ON = a$ and $CN = x$. By power of a point on $N$, we have\n\\[(5+a)(5-a) = 25-a^2 = x(8-x).\\]\n$\\triangle CON$ is a right triangle, so we also know that $x^2 = a^2 + 25$, thus\n\\[25 - a^2 = 25-(x^2-25) = 50-x^2 = 8x-x^2 \\Rightarrow... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/29.json | AHSME |
1952_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | The expression $a^3-a^{-3}$ equals:
$\textbf{(A) \ }\left(a-\frac{1}{a}\right)\left(a^2+1+\frac{1}{a^2}\right) \qquad \textbf{(B) \ }\left(\frac{1}{a}-a\right)\left(a^2-1+\frac{1}{a^2}\right) \qquad \textbf{(C) \ }\left(a-\frac{1}{a}\right)\left(a^2-2+\frac{1}{a^2}\right) \qquad$
$\textbf{(D) \ }\left(\frac{1}{a}-a\... | [
"Recall that $x^3-y^3=(x-y)(x^2+xy+y^2)$. Letting $a=x$ and $a^{-1}=y$, we find that $a^3-a^{-3}= \\boxed{\\textbf{(A)}\\ \\left(a-\\frac{1}{a}\\right)\\left(a^2+1+\\frac{1}{a^2}\\right)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/3.json | AHSME |
1952_AHSME_Problems | 34 | 0 | Algebra | Multiple Choice | The price of an article was increased $p\%$. Later the new price was decreased $p\%$. If the last price was one dollar, the original price was:
$\textbf{(A)}\ \frac{1-p^2}{200}\qquad \textbf{(B)}\ \frac{\sqrt{1-p^2}}{100}\qquad \textbf{(C)}\ \text{one dollar}\qquad\\ \textbf{(D)}\ 1-\frac{p^2}{10000-p^2}\qquad \text... | [
"Converting the word problem in to an equation, we take x as the initial amount, we get \\[x(1+\\dfrac{p}{100})(1-\\dfrac{p}{100}) = 1\\] Using, $(a+b)(a-b) = a^{2} - b^{2}$. We now simplify it to, \\[x(1-\\dfrac{p^{2}}{10000})=1\\] Simplifying to, \\[x(\\dfrac{10000-p^{2}}{10000}) = 1\\] Dividing both sides by $\\... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/34.json | AHSME |
1952_AHSME_Problems | 8 | 0 | Geometry | Multiple Choice | Two equal circles in the same plane cannot have the following number of common tangents.
$\textbf{(A) \ }1 \qquad \textbf{(B) \ }2 \qquad \textbf{(C) \ }3 \qquad \textbf{(D) \ }4 \qquad \textbf{(E) \ }\text{none of these}$
| [
"Two congruent coplanar circles will either be tangent to one another (resulting in $3$ common tangents), intersect one another (resulting in $2$ common tangents), or be separate from one another (resulting in $4$ common tangents).\nHaving only $\\boxed{\\textbf{(A)}\\ 1}$ common tangent is impossible, unless the c... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/8.json | AHSME |
1952_AHSME_Problems | 22 | 0 | Geometry | Multiple Choice | On hypotenuse $AB$ of a right triangle $ABC$ a second right triangle $ABD$ is constructed with hypotenuse $AB$. If $\overline{BC}=1$, $\overline{AC}=b$, and $\overline{AD}=2$, then $\overline{BD}$ equals:
$\textbf{(A) \ }\sqrt{b^2+1} \qquad \textbf{(B) \ }\sqrt{b^2-3} \qquad \textbf{(C) \ }\sqrt{b^2+1}+2 \qquad$
$... | [
"[asy] size(150); import olympiad; pair A,B,C,D; A=(1,3); B=origin; C=(1,0); D=sqrt(6)*dir(aTan(2/sqrt(6))+aTan(9)); draw(A--C--B--A--D--B); markscalefactor=0.02; draw(rightanglemark(A,C,B)); draw(rightanglemark(A,D,B)); label(\"$A$\",A,N); label(\"$B$\",B,SW); label(\"$C$\",C,SE); label(\"$D$\",D,W); label(\"$1$\"... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/22.json | AHSME |
1952_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | $\log p+\log q=\log(p+q)$ only if:
$\textbf{(A) \ }p=q=\text{zero} \qquad \textbf{(B) \ }p=\frac{q^2}{1-q} \qquad \textbf{(C) \ }p=q=1 \qquad$
$\textbf{(D) \ }p=\frac{q}{q-1} \qquad \textbf{(E) \ }p=\frac{q}{q+1}$
| [
"$\\log p+\\log q=\\log(p+q)\\implies \\log pq=\\log(p+q)\\implies pq=p+q\\implies \\boxed{\\textbf{(D)}\\ p=\\frac{q}{q-1}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/18.json | AHSME |
1952_AHSME_Problems | 38 | 0 | Geometry | Multiple Choice | The area of a trapezoidal field is $1400$ square yards. Its altitude is $50$ yards. Find the two bases, if the number of yards in each base is an integer divisible by $8$. The number of solutions to this problem is:
$\textbf{(A)}\ \text{none} \qquad \textbf{(B)}\ \text{one} \qquad \textbf{(C)}\ \text{two} \qquad \te... | [
"Let us denote $8m$ and $8n$ to be our bases. Without loss of generality, let $m \\le n$.\n\n\nThus,\n\\[50 * \\frac{8m + 8n}{2} = 1400\\]\n\\[4m + 4n = 28\\]\n\\[m + n = 7\\]\n\n\nSince $m$ & $n$ are integers, we see that the only solutions to this equation are $(1,6)$, $(2,5)$, and $(3,4)$. \nTherefore, the answe... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/38.json | AHSME |
1952_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | The cost $C$ of sending a parcel post package weighing $P$ pounds, $P$ an integer, is $10$ cents for the first pound and $3$ cents for each additional pound. The formula for the cost is:
$\textbf{(A) \ }C=10+3P \qquad \textbf{(B) \ }C=10P+3 \qquad \textbf{(C) \ }C=10+3(P-1) \qquad$
$\textbf{(D) \ }C=9+3P \qquad \te... | [
"We know that the first pound has a constant price, $10$ cents, and that the other $P-1$ pounds cost $3$ cents apiece. This leaves us with $\\boxed{\\textbf{(C)}\\ C=10+3(P-1)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/4.json | AHSME |
1952_AHSME_Problems | 14 | 0 | Arithmetic | Multiple Choice | A house and store were sold for $\textdollar 12,000$ each. The house was sold at a loss of $20\%$ of the cost, and the store at a gain of $20\%$ of the cost. The entire transaction resulted in:
$\textbf{(A) \ }\text{no loss or gain} \qquad \textbf{(B) \ }\text{loss of }\textdollar 1000 \qquad \textbf{(C) \ }\text{ga... | [
"Denote the original price of the house and the store as $h$ and $s$, respectively. It is given that $\\frac{4h}{5}=\\textdollar 12,000$, and that $\\frac{6s}{5}=\\textdollar 12,000$. Thus, $h=\\textdollar 15,000$, $s=\\textdollar10,000$, and $h+s=\\textdollar25,000$. This value is $\\textdollar1000$ higher than th... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/14.json | AHSME |
1952_AHSME_Problems | 43 | 0 | Geometry | Multiple Choice | The diameter of a circle is divided into $n$ equal parts. On each part a semicircle is constructed. As $n$ becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length:
$\textbf{(A) } \qquad$ equal to the semi-circumference of the original circle
$\textbf{(B) } \qquad$ equal to the dia... | [
"Note that the half the circumference of a circle with diameter $d$ is $\\frac{\\pi*d}{2}$. \n\n\nLet's call the diameter of the circle D. Dividing the circle's diameter into n parts means that each semicircle has diameter $\\frac{D}{n}$, and thus each semicircle measures $\\frac{D*pi}{n*2}$. The total sum of those... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/43.json | AHSME |
1952_AHSME_Problems | 42 | 0 | Algebra | Multiple Choice | Let $D$ represent a repeating decimal. If $P$ denotes the $r$ figures of $D$ which do not repeat themselves, and $Q$ denotes the $s$ figures of $D$ which do repeat themselves, then the incorrect expression is:
$\text{(A) } D = .PQQQ\ldots \qquad\\ \text{(B) } 10^rD = P.QQQ\ldots \\ \text{(C) } 10^{r + s}D = PQ.QQQ\l... | [
"$\\fbox{D}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/42.json | AHSME |
1952_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | The sides of a triangle are in the ratio $6:8:9$. Then:
$\textbf{(A) \ }\text{the triangle is obtuse}$
$\textbf{(B) \ }\text{the angles are in the ratio }6:8:9$
$\textbf{(C) \ }\text{the triangle is acute}$
$\textbf{(D) \ }\text{the angle opposite the largest side is double the angle opposite the smallest side}... | [
"In a triangle with sides $a$, $b$, and $c$, where $a\\le b\\le c$, $\\bigtriangleup ABC$ is acute if $a^2+b^2>c^2$, right if $a^2+b^2=c^2$, and obtuse if $a^2+b^2<c^2$. If $c$ were equal to $10$, $\\bigtriangleup ABC$ would be right, as a multiple of the Pythagorean triple $(3,4,5)$. Because $c<10$, $\\boxed{\\tex... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/15.json | AHSME |
1952_AHSME_Problems | 5 | 0 | Geometry | Multiple Choice | The points $(6,12)$ and $(0,-6)$ are connected by a straight line. Another point on this line is:
$\textbf{(A) \ }(3,3) \qquad \textbf{(B) \ }(2,1) \qquad \textbf{(C) \ }(7,16) \qquad \textbf{(D) \ }(-1,-4) \qquad \textbf{(E) \ }(-3,-8)$
| [
"The slope of this line is $\\frac{y_2-y_1}{x_2-x_1}=\\frac{12+6}{6-0}=3$. Hence, its equation is $y=3x-6$. The only given point which satisfies these conditions is $\\boxed{\\textbf{(A)}\\ (3,3)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/5.json | AHSME |
1952_AHSME_Problems | 39 | 0 | Geometry | Multiple Choice | If the perimeter of a rectangle is $p$ and its diagonal is $d$, the difference between the length and width of the rectangle is:
$\textbf{(A)}\ \frac {\sqrt {8d^2 - p^2}}{2} \qquad \textbf{(B)}\ \frac {\sqrt {8d^2 + p^2}}{2} \qquad \textbf{(C)}\ \frac{\sqrt{6d^2-p^2}}{2}\qquad\\ \textbf{(D)}\ \frac {\sqrt {6d^2 + p^... | [
"[asy] pair A,B,C,D,E,F,G,H; A=(0,0); B=(10,0); C=(10,5); D=(0,5); E=(5,5.5); F=(5,-0.5); G=(-0.5,2.75); H=(10.5,2.75); draw(A--B--C--D--cycle); draw (B--D); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,NE); label(\"$D$\",(-0.5,5),N); label(\"$x$\",E); label(\"$x$\",F); label(\"$y$\",G); label(\"$y$\"... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/39.json | AHSME |
1952_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | Angle $B$ of triangle $ABC$ is trisected by $BD$ and $BE$ which meet $AC$ at $D$ and $E$ respectively. Then:
$\textbf{(A) \ }\frac{AD}{EC}=\frac{AE}{DC} \qquad \textbf{(B) \ }\frac{AD}{EC}=\frac{AB}{BC} \qquad \textbf{(C) \ }\frac{AD}{EC}=\frac{BD}{BE} \qquad$
$\textbf{(D) \ }\frac{AD}{EC}=\frac{(AB)(BD)}{(BE)(BC)... | [
"[asy] import olympiad; pair A=(0,0), B=(3,3), C=(6,0), D=(2,0), E=(4,0); dot(A); dot(B); dot(C); dot(D); dot(E); label(\"$A$\",A,W); label(\"$B$\",B,N); label(\"$C$\",(5.5,0),E); label(\"$D$\",D,S); label(\"$E$\",E,S); draw(A--B--C--cycle); draw(B--D); draw(B--E); markscalefactor=0.1; draw(anglemark(A,B,D)); draw(... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/19.json | AHSME |
1952_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | If $\frac{x^2-bx}{ax-c}=\frac{m-1}{m+1}$ has roots which are numerically equal but of opposite signs, the value of $m$ must be:
$\textbf{(A) \ }\frac{a-b}{a+b} \qquad \textbf{(B) \ }\frac{a+b}{a-b} \qquad \textbf{(C) \ }c \qquad \textbf{(D) \ }\frac{1}{c} \qquad \textbf{(E) \ }1$
| [
"Cross-multiplying, we find that $(m+1)x^2-(bm+am+b-a)x+c(m-1)=0$. Because the roots of this quadratic are additive inverses, their sum is $0$. According to Vieta's Formulas, $\\frac{bm+am+b-a}{m+1}=0$, or $m(a+b)=a-b$. Hence, $m=\\boxed{\\textbf{(A)}\\ \\frac{a-b}{a+b}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/23.json | AHSME |
1952_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | If $m=\frac{cab}{a-b}$, then $b$ equals:
$\textbf{(A) \ }\frac{m(a-b)}{ca} \qquad \textbf{(B) \ }\frac{cab-ma}{-m} \qquad \textbf{(C) \ }\frac{1}{1+c} \qquad \textbf{(D) \ }\frac{ma}{m+ca} \qquad \textbf{(E) \ }\frac{m+ca}{ma}$
| [
"$m=\\frac{cab}{a-b}\\implies ma-mb=cab\\implies ma=b(m+ac)\\implies b=\\boxed{\\textbf{(D)}\\ \\frac{ma}{m+ca}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/9.json | AHSME |
1952_AHSME_Problems | 35 | 0 | Algebra | Multiple Choice | With a rational denominator, the expression $\frac {\sqrt {2}}{\sqrt {2} + \sqrt {3} - \sqrt {5}}$ is equivalent to:
$\textbf{(A)}\ \frac {3 + \sqrt {6} + \sqrt {15}}{6} \qquad \textbf{(B)}\ \frac {\sqrt {6} - 2 + \sqrt {10}}{6} \qquad \textbf{(C)}\ \frac{2+\sqrt{6}+\sqrt{10}}{10} \qquad\\ \textbf{(D)}\ \frac {2 + \... | [
"Let $k=\\sqrt{2}+\\sqrt{3}$\nThen $\\frac{\\sqrt{2}}{k-\\sqrt{5}}\\implies \\frac{\\sqrt{2}(k+\\sqrt{5})}{k^2-\\sqrt{5}^2}\\implies\\frac{\\sqrt{2}k+\\sqrt{10}}{k^2-5}\\implies \\frac{\\sqrt{2}(\\sqrt{2}+\\sqrt{3})+\\sqrt{10}}{(\\sqrt{2}+\\sqrt{3})^2-5}\\implies \\frac{2+\\sqrt{6}+\\sqrt{10}}{2\\sqrt{6}}\\implies\... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/35.json | AHSME |
1950_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | When $x^{13}+1$ is divided by $x-1$, the remainder is:
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$
| [
"Using synthetic division, we get that the remainder is $\\boxed{\\textbf{(D)}\\ 2}$.\n\n",
"By the remainder theorem, the remainder is equal to the expression $x^{13}+1$ when $x=1.$ This gives the answer of $\\boxed{(\\mathrm{D})\\ 2.}$\n\n",
"Note that $x^{13} - 1 = (x - 1)(x^{12} + x^{11} \\cdots + 1)$, so $... | 3 | ./CreativeMath/AHSME/1950_AHSME_Problems/20.json | AHSME |
1950_AHSME_Problems | 36 | 0 | Algebra | Multiple Choice | A merchant buys goods at $25\%$ off the list price. He desires to mark the goods so that he can give a discount of $20\%$ on the marked price and still clear a profit of $25\%$ on the selling price. What percent of the list price must he mark the goods?
$\textbf{(A)}\ 125\% \qquad \textbf{(B)}\ 100\% \qquad \textbf{(... | [
"Without loss of generality, we can set the list price equal to $100$. The merchant buys the goods for $100*.75=75$. Let $x$ be the marked price.\nWe then use the equation $0.8x-75=25$ to solve for $x$ and get a marked price of $\\boxed{125}$.\n\n\n$\\text{Answer: }\\boxed{\\mathbf{(A)}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/36.json | AHSME |
1950_AHSME_Problems | 41 | 0 | Algebra | Multiple Choice | The least value of the function $ax^2 + bx + c$ with $a>0$ is:
$\textbf{(A)}\ -\dfrac{b}{a} \qquad \textbf{(B)}\ -\dfrac{b}{2a} \qquad \textbf{(C)}\ b^2-4ac \qquad \textbf{(D)}\ \dfrac{4ac-b^2}{4a}\qquad \textbf{(E)}\ \text{None of these}$
| [
"The vertex of a parabola is at $x=-\\frac{b}{2a}$ for $ax^2+bx+c$. Because $a>0$, the vertex is a minimum. Therefore $a(-\\frac{b}{2a})^2+b(-\\frac{b}{2a})+c=a(\\frac{b^2}{4a^2})-\\frac{b^2}{2a}+c=\\frac{b^2}{4a}-\\frac{2b^2}{4a}+c=-\\frac{b^2}{4a}+c=\\frac{4ac}{4a}-\\frac {b^2}{4a}=\\frac{4ac-b^2}{4a} \\Rightarro... | 2 | ./CreativeMath/AHSME/1950_AHSME_Problems/41.json | AHSME |
1950_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | The number of terms in the expansion of $[(a+3b)^{2}(a-3b)^{2}]^{2}$ when simplified is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$
| [
"Use properties of exponents to move the squares outside the brackets use difference of squares.\n\n\n\\[[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4\\]\n\n\nUsing the binomial theorem, we can see that the number of terms is $\\boxed{\\mathrm{(B)}\\ 5}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/16.json | AHSME |
1950_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | The values of $y$ which will satisfy the equations $2x^{2}+6x+5y+1=0, 2x+y+3=0$ may be found by solving:
$\textbf{(A)}\ y^{2}+14y-7=0\qquad\textbf{(B)}\ y^{2}+8y+1=0\qquad\textbf{(C)}\ y^{2}+10y-7=0\qquad\\ \textbf{(D)}\ y^{2}+y-12=0\qquad \textbf{(E)}\ \text{None of these equations}$
| [
"If we solve the second equation for $x$ in terms of $y$, we find $x=-\\dfrac{y+3}{2}$ which we can substitute to find:\n\n\n\\[2(-\\dfrac{y+3}{2})^2+6(-\\dfrac{y+3}{2})+5y+1=0\\]\n\n\nMultiplying by two and simplifying, we find:\n\n\n\\begin{align*} 2\\cdot[2(-\\dfrac{y+3}{2})^2+6(-\\dfrac{y+3}{2})+5y+1]&=2\\cdot ... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/6.json | AHSME |
1950_AHSME_Problems | 7 | 0 | Arithmetic | Multiple Choice | If the digit $1$ is placed after a two digit number whose tens' digit is $t$, and units' digit is $u$, the new number is:
$\textbf{(A)}\ 10t+u+1\qquad\textbf{(B)}\ 100t+10u+1\qquad\textbf{(C)}\ 1000t+10u+1\qquad\textbf{(D)}\ t+u+1\qquad\\ \textbf{(E)}\ \text{None of these answers}$
| [
"By placing the digit $1$ after a two digit number, you are changing the units place to $1$ and moving everything else up a place. Therefore the answer is $\\boxed{\\textbf{(B)}\\ 100t+10u+1}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/7.json | AHSME |
1950_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | The formula which expresses the relationship between $x$ and $y$ as shown in the accompanying table is:
\[\begin{tabular}[t]{|c|c|c|c|c|c|}\hline x&0&1&2&3&4\\\hline y&100&90&70&40&0\\\hline\end{tabular}\]
$\textbf{(A)}\ y=100-10x\qquad\textbf{(B)}\ y=100-5x^{2}\qquad\textbf{(C)}\ y=100-5x-5x^{2}\qquad\\ \textbf{(D... | [
"Plug in the points $(0,100)$ and $(4,0)$ into each equation. The only one that works for both points is $\\mathrm{(C)}.$ Plug in the rest of the points to confirm the answer is indeed $\\boxed{\\mathrm{(C)}\\ y=100-5x-5x^2}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/17.json | AHSME |
1950_AHSME_Problems | 40 | 0 | Algebra | Multiple Choice | The limit of $\frac {x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \text{Indeterminate} \qquad \textbf{(C)}\ x-1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$
| [
"Both $x^2-1$ and $x-1$ approach 0 as $x$ approaches $1$, using the L'Hôpital's rule, we have $\\lim \\limits_{x\\to 1}\\frac{x^2-1}{x-1} = \\lim \\limits_{x\\to 1}\\frac{2x}{1} = 2$.\nThus, the answer is $\\boxed{\\textbf{(D)}\\ 2}$.\n\n\n~ MATH__is__FUN\n\n\n",
"The numerator of $\\frac {x^2-1}{x-1}$ can be fac... | 2 | ./CreativeMath/AHSME/1950_AHSME_Problems/40.json | AHSME |
1950_AHSME_Problems | 37 | 0 | Algebra | Multiple Choice | If $y = \log_{a}{x}$, $a > 1$, which of the following statements is incorrect?
$\textbf{(A)}\ \text{If }x=1,y=0 \qquad\\ \textbf{(B)}\ \text{If }x=a,y=1 \qquad\\ \textbf{(C)}\ \text{If }x=-1,y\text{ is imaginary (complex)} \qquad\\ \textbf{(D)}\ \text{If }0<x<1,y\text{ is always less than 0 and decreases without limi... | [
"Let us first check \n\n\n$\\textbf{(A)}\\ \\text{If }x=1,y=0$. Rewriting into exponential form gives $a^0=1$. This is certainly correct.\n\n\n$\\textbf{(B)}\\ \\text{If }x=a,y=1$. Rewriting gives $a^1=a$. This is also certainly correct.\n\n\n$\\textbf{(C)}\\ \\text{If }x=-1,y\\text{ is imaginary (complex)}$. Rewri... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/37.json | AHSME |
1950_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | The volume of a rectangular solid each of whose side, front, and bottom faces are $12\text{ in}^{2}$, $8\text{ in}^{2}$, and $6\text{ in}^{2}$ respectively is:
$\textbf{(A)}\ 576\text{ in}^{3}\qquad\textbf{(B)}\ 24\text{ in}^{3}\qquad\textbf{(C)}\ 9\text{ in}^{3}\qquad\textbf{(D)}\ 104\text{ in}^{3}\qquad\textbf{(E)}... | [
"If the sidelengths of the cubes are expressed as $a, b,$ and $c,$ then we can write three equations:\n\n\n\\[ab=12, bc=8, ac=6.\\]\n\n\nThe volume is $abc.$ Notice symmetry in the equations. We can find $abc$ my multiplying all the equations and taking the positive square root.\n\n\n\\begin{align*} (ab)(bc)(ac) &=... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/21.json | AHSME |
1950_AHSME_Problems | 47 | 0 | Geometry | Multiple Choice | A rectangle inscribed in a triangle has its base coinciding with the base $b$ of the triangle. If the altitude of the triangle is $h$, and the altitude $x$ of the rectangle is half the base of the rectangle, then:
$\textbf{(A)}\ x=\dfrac{1}{2}h \qquad \textbf{(B)}\ x=\dfrac{bh}{b+h} \qquad \textbf{(C)}\ x=\dfrac{bh}{... | [
"Draw the triangle, and note that the small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle, so the proportions of the heights is equal to the proportions of the sides. In particular, we get $\\dfrac{2x}{b} = \\dfrac{h - x}{h} \\implies 2xh = bh - bx \\imp... | 2 | ./CreativeMath/AHSME/1950_AHSME_Problems/47.json | AHSME |
1950_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | After rationalizing the numerator of $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}$, the denominator in simplest form is:
$\textbf{(A)}\ \sqrt{3}(\sqrt{3}+\sqrt{2})\qquad\textbf{(B)}\ \sqrt{3}(\sqrt{3}-\sqrt{2})\qquad\textbf{(C)}\ 3-\sqrt{3}\sqrt{2}\qquad\\ \textbf{(D)}\ 3+\sqrt6\qquad\textbf{(E)}\ \text{None of these answers}... | [
"To rationalize the numerator, multiply by the conjugate.\n\n\n\\[\\frac{\\sqrt3 - \\sqrt2}{\\sqrt3}\\cdot \\frac{\\sqrt3 + \\sqrt2}{\\sqrt3 + \\sqrt2} = \\frac{1}{3+\\sqrt6}.\\]\n\n\nThe denominator is $\\boxed{\\mathrm{(D)}\\text{ } 3+\\sqrt6 }.$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/10.json | AHSME |
1950_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | If $\log_{10}{m}= b-\log_{10}{n}$, then $m=$
$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$
| [
"We have $b=\\log_{10}{10^b}$. Substituting, we find $\\log_{10}{m}= \\log_{10}{10^b}-\\log_{10}{n}$. Using $\\log{a}-\\log{b}=\\log{\\dfrac{a}{b}}$, the left side becomes $\\log_{10}{\\dfrac{10^b}{n}}$. Because $\\log_{10}{m}=\\log_{10}{\\dfrac{10^b}{n}}$, $m=\\boxed{\\mathrm{(E) }\\dfrac{10^b}{n}}$.\n\n\n",
"ad... | 3 | ./CreativeMath/AHSME/1950_AHSME_Problems/26.json | AHSME |
1950_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | From a group of boys and girls, $15$ girls leave. There are then left two boys for each girl. After this $45$ boys leave. There are then $5$ girls for each boy. The number of girls in the beginning was:
$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 43 \qquad \textbf{(C)}\ 29 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ \tex... | [
"Let us represent the number of boys $b$, and the number of girls $g$.\n\n\nFrom the first sentence, we get that $2(g-15)=b$.\n\n\nFrom the second sentence, we get $5(b-45)=g-15$.\n\n\nExpanding both equations and simplifying, we get $2g-30 = b$ and $5b = g+210$.\n\n\nSubstituting $b$ for $2g-30$, we get $5(2g-30)=... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/30.json | AHSME |
1950_AHSME_Problems | 31 | 0 | Algebra | Multiple Choice | John ordered $4$ pairs of black socks and some additional pairs of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by $50\%$. The ratio of the number of pairs o... | [
"Let the number of blue socks be represented as $b$. We are informed that the price of the black sock is twice the price of a blue sock; let us assume that the price of one pair of blue socks is $1$. That means the price of one pair of black socks is $2$.\n\n\nNow from the third and fourth sentence, we see that $1.... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/31.json | AHSME |
1950_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | A car travels $120$ miles from $A$ to $B$ at $30$ miles per hour but returns the same distance at $40$ miles per hour. The average speed for the round trip is closest to:
$\textbf{(A)}\ 33\text{ mph}\qquad\textbf{(B)}\ 34\text{ mph}\qquad\textbf{(C)}\ 35\text{ mph}\qquad\textbf{(D)}\ 36\text{ mph}\qquad\textbf{(E)}\ ... | [
"The car takes $120 \\text{ miles }\\cdot\\dfrac{1 \\text{ hr }}{30 \\text{ miles }}=4 \\text{ hr}$ to get from $A$ to $B$. Also, it takes $120 \\text{ miles }\\cdot\\dfrac{1 \\text{ hr }}{40 \\text{ miles }}=3 \\text{ hr}$ to get from $B$ to $A$. Therefore, the average speed is $\\dfrac{240\\text{ miles }}{7 \\tex... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/27.json | AHSME |
1950_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | If $64$ is divided into three parts proportional to $2$, $4$, and $6$, the smallest part is:
$\textbf{(A)}\ 5\frac{1}{3}\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 10\frac{2}{3}\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these answers}$
| [
"If the three numbers are in proportion to $2:4:6$, then they should also be in proportion to $1:2:3$. This implies that the three numbers can be expressed as $x$, $2x$, and $3x$. Add these values together to get: \n\\[x+2x+3x=6x=64\\]\nDivide each side by 6 and get that \n\\[x=\\frac{64}{6}=\\frac{32}{3}=10 \\frac... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/1.json | AHSME |
1950_AHSME_Problems | 50 | 0 | Algebra | Multiple Choice | A privateer discovers a merchantman $10$ miles to leeward at 11:45 a.m. and with a good breeze bears down upon her at $11$ mph, while the merchantman can only make $8$ mph in her attempt to escape. After a two hour chase, the top sail of the privateer is carried away; she can now make only $17$ miles while the merchant... | [
"Assume that the two boats are traveling along the positive real number line, with the merchantman starting at the number $10$ and the privateer starting at the number $0$. After two hours the merchantman is at $26$ while the privateer is at $22$. When the top sail of the privateer is carried away, the speed of the... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/50.json | AHSME |
1950_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | If in the formula $C =\frac{en}{R+nr}$, where $e$, $n$, $R$ and $r$ are all positive, $n$ is increased while $e$, $R$ and $r$ are kept constant, then $C$:
$\textbf{(A)}\ \text{Increases}\qquad\textbf{(B)}\ \text{Decreases}\qquad\textbf{(C)}\ \text{Remains constant}\qquad\textbf{(D)}\ \text{Increases and then decrease... | [
"Divide both the numerator and denominator by $n$, to get $C=\\frac{e}{\\frac{R}{n}+r}$. If $n$ increases then the denominator decreases; so that $C$ $\\boxed{\\mathrm{(A)}\\text{ }\\mathrm{ Increases}.}$\n\n\nbut what if $n\\leq 0$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/11.json | AHSME |
1950_AHSME_Problems | 46 | 0 | Geometry | Multiple Choice | In triangle $ABC$, $AB=12$, $AC=7$, and $BC=10$. If sides $AB$ and $AC$ are doubled while $BC$ remains the same, then:
$\textbf{(A)}\ \text{The area is doubled} \qquad\\ \textbf{(B)}\ \text{The altitude is doubled} \qquad\\ \textbf{(C)}\ \text{The area is four times the original area} \qquad\\ \textbf{(D)}\ \text{The... | [
"If you double sides $AB$ and $AC$, they become $24$ and $14$ respectively. If $BC$ remains $10$, then this triangle has area $0$ because ${14} + {10} = {24}$, so two sides overlap the third side. Therefore the answer is $\\boxed{\\textbf{(E)}\\ \\text{The area of the triangle is 0}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/46.json | AHSME |
1950_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | Let $R=gS-4$. When $S=8$, $R=16$. When $S=10$, $R$ is equal to:
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 21\qquad\textbf{(E)}\ \text{None of these}$
| [
"Our first procedure is to find the value of $g$. With the given variables' values, we can see that $8g-4=16$ so $g=\\frac{20}{8}=\\frac{5}{2}$.\n\n\nWith that, we can replace $g$ with $\\frac{5}{2}$. When $S=10$, we can see that $10\\times\\frac{5}{2}-4=\\frac{50}{2}-4=25-4=\\boxed{\\text{(D) 21}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/2.json | AHSME |
1950_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | Two boys $A$ and $B$ start at the same time to ride from Port Jervis to Poughkeepsie, $60$ miles away. $A$ travels $4$ miles an hour slower than $B$. $B$ reaches Poughkeepsie and at once turns back meeting $A$ $12$ miles from Poughkeepsie. The rate of $A$ was:
$\textbf{(A)}\ 4\text{ mph}\qquad \textbf{(B)}\ 8\text{ m... | [
"Let the speed of boy $A$ be $a$, and the speed of boy $B$ be $b$. Notice that $A$ travels $4$ miles per hour slower than boy $B$, so we can replace $b$ with $a+4$.\n\n\nNow let us see the distances that the boys each travel. Boy $A$ travels $60-12=48$ miles, and boy $B$ travels $60+12=72$ miles. Now, we can use $d... | 2 | ./CreativeMath/AHSME/1950_AHSME_Problems/28.json | AHSME |
1950_AHSME_Problems | 12 | 0 | Geometry | Multiple Choice | As the number of sides of a polygon increases from $3$ to $n$, the sum of the exterior angles formed by extending each side in succession:
$\textbf{(A)}\ \text{Increases}\qquad\textbf{(B)}\ \text{Decreases}\qquad\textbf{(C)}\ \text{Remains constant}\qquad\textbf{(D)}\ \text{Cannot be predicted}\qquad\\ \textbf{(E)}\ ... | [
"By the Exterior Angles Theorem, the exterior angles of all convex polygons add up to $360^\\circ,$ so the sum $\\boxed{\\mathrm{(C)}\\text{ remains constant}.}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/12.json | AHSME |
1950_AHSME_Problems | 45 | 0 | Counting | Multiple Choice | The number of diagonals that can be drawn in a polygon of 100 sides is:
$\textbf{(A)}\ 4850 \qquad \textbf{(B)}\ 4950\qquad \textbf{(C)}\ 9900 \qquad \textbf{(D)}\ 98 \qquad \textbf{(E)}\ 8800$
| [
"Each diagonal has its two endpoints as vertices of the 100-gon. Each pair of vertices determines exactly one diagonal. Therefore the answer should be $\\binom{100}{2}=4950$. However this also counts the 100 sides of the polygon, so the actual answer is $4950-100=\\boxed{\\textbf{(A)}\\ 4850 }$.\n\n\n",
"We can c... | 3 | ./CreativeMath/AHSME/1950_AHSME_Problems/45.json | AHSME |
1950_AHSME_Problems | 32 | 0 | Geometry | Multiple Choice | A $25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:
$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \t... | [
"By the Pythagorean triple $(7,24,25)$, the point where the ladder meets the wall is $24$ feet above the ground. When the ladder slides, it becomes $20$ feet above the ground. By the $(15,20,25)$ Pythagorean triple, The foot of the ladder is now $15$ feet from the building. Thus, it slides $15-7 = \\boxed{\\textbf{... | 2 | ./CreativeMath/AHSME/1950_AHSME_Problems/32.json | AHSME |
1950_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | The equation $x + \sqrt{x-2} = 4$ has:
$\textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root}$
| [
"$x + \\sqrt{x-2} = 4$ Original Equation\n\n\n$\\sqrt{x-2} = 4 - x$ Subtract x from both sides\n\n\n$x-2 = 16 - 8x + x^2$ Square both sides\n\n\n$x^2 - 9x + 18 = 0$ Get all terms on one side\n\n\n$(x-6)(x-3) = 0$ Factor\n\n\n$x = \\{6, 3\\}$\n\n\nIf you put down A as your answer, it's wrong. You need to check for e... | 3 | ./CreativeMath/AHSME/1950_AHSME_Problems/24.json | AHSME |
1950_AHSME_Problems | 49 | 0 | Geometry | Multiple Choice | A triangle has a fixed base $AB$ that is $2$ inches long. The median from $A$ to side $BC$ is $1\frac{1}{2}$ inches long and can have any position emanating from $A$. The locus of the vertex $C$ of the triangle is:
$\textbf{(A)}\ \text{A straight line }AB,1\dfrac{1}{2}\text{ inches from }A \qquad\\ \textbf{(B)}\ \tex... | [
"The locus of the median's endpoint on $BC$ is the circle about $A$ and of radius $1\\frac{1}{2}$ inches. The locus of the vertex $C$ is then the circle twice as big and twice as far from $B$, i.e. of radius $3$ inches and with center $4$ inches from $B$ along $BA$ which means that our answer is: $\\textbf{(D)}$.\... | 2 | ./CreativeMath/AHSME/1950_AHSME_Problems/49.json | AHSME |
1950_AHSME_Problems | 48 | 0 | Geometry | Multiple Choice | A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is:
$\textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}\qquad\\ \textbf{(B)}\ \text{Greater than the altitude of the triangle}\qquad\\ ... | [
"Begin by drawing the triangle, the point, the altitudes from the point to the sides, and the segments connecting the point to the vertices. Let the triangle be $ABC$ with $AB=BC=AC=s$. We will call the aforementioned point $P$. Call altitude from $P$ to $BC$ $PA'$. Similarly, we will name the other two altitud... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/48.json | AHSME |
1950_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | The value of $\log_{5}\frac{(125)(625)}{25}$ is equal to:
$\textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these}$
| [
"$\\log_{5}\\frac{(125)(625)}{25}$ can be simplified to $\\log_{5}\\ (125)(25)$ since $25^2 = 625$. $125 = 5^3$ and $5^2 = 25$ so $\\log_{5}\\ 5^5$ would be the simplest form. In $\\log_{5}\\ 5^5$, $5^x = 5^5$. Therefore, $x = 5$ and the answer is $\\boxed{\\mathrm{(D)}\\ 5}$.\n\n",
"$\\log_{5}\\frac{(125)(625)}{... | 2 | ./CreativeMath/AHSME/1950_AHSME_Problems/25.json | AHSME |
1950_AHSME_Problems | 33 | 0 | Geometry | Multiple Choice | The number of circular pipes with an inside diameter of $1$ inch which will carry the same amount of water as a pipe with an inside diameter of $6$ inches is:
$\textbf{(A)}\ 6\pi \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 36\pi$
| [
"It must be assumed that the pipes have an equal height.\n\n\nWe can represent the amount of water carried per unit time by cross sectional area.\nCross sectional of Pipe with diameter $6 in$, \n\\[\\pi r^2 = \\pi \\cdot 3^2 = 9\\pi\\]\n\n\nCross sectional area of pipe with diameter $1 in$\n\n\n\\[\\pi r^2 = \\pi \... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/33.json | AHSME |
1950_AHSME_Problems | 44 | 0 | Algebra | Multiple Choice | The graph of $y=\log x$
$\textbf{(A)}\ \text{Cuts the }y\text{-axis} \qquad\\ \textbf{(B)}\ \text{Cuts all lines perpendicular to the }x\text{-axis} \qquad\\ \textbf{(C)}\ \text{Cuts the }x\text{-axis} \qquad\\ \textbf{(D)}\ \text{Cuts neither axis} \qquad\\ \textbf{(E)}\ \text{Cuts all circles whose center is at the... | [
"The domain of $\\log x$ is the set of all $\\underline{positive}$ reals, so the graph of $y=\\log x$ clearly doesn't cut the $y$-axis. It therefore doesn't cut every line perpendicular to the $x$-axis. It does however cut the $x$-axis at $(1,0)$. In addition, if one examines the graph of $y=\\log x$, one can clear... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/44.json | AHSME |
1950_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | The roots of $(x^{2}-3x+2)(x)(x-4)=0$ are:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 0\text{ and }4\qquad\textbf{(C)}\ 1\text{ and }2\qquad\textbf{(D)}\ 0,1,2\text{ and }4\qquad\textbf{(E)}\ 1,2\text{ and }4$
| [
"Factor $x^2-3x+2$ to get $(x-2)(x-1).$ The roots are $\\boxed{\\mathrm{(D)}\\ 0,1,2\\text{ and }4.}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/13.json | AHSME |
1950_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | A manufacturer built a machine which will address $500$ envelopes in $8$ minutes. He wishes to build another machine so that when both are operating together they will address $500$ envelopes in $2$ minutes. The equation used to find how many minutes $x$ it would require the second machine to address $500$ envelopes al... | [
"We first represent the first machine's speed in per $2$ minutes: $125 \\text{ envelopes in }2\\text{ minutes}$. Now, we know that the speed per $2$ minutes of the second machine is \\[500-125=375 \\text{ envelopes in }2\\text{ minutes}\\]\nNow we can set up a proportion to find out how many minutes it takes for th... | 2 | ./CreativeMath/AHSME/1950_AHSME_Problems/29.json | AHSME |
1950_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | The sum of the roots of the equation $4x^{2}+5-8x=0$ is equal to:
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ -5\qquad\textbf{(C)}\ -\frac{5}{4}\qquad\textbf{(D)}\ -2\qquad\textbf{(E)}\ \text{None of these}$
| [
"We can divide by 4 to get:\n$x^{2}-2x+\\dfrac{5}{4}=0.$\n\n\nThe Vieta's formula states that in quadratic equation $ax^2+bx+c$, the sum of the roots of the equation is $-\\frac{b}{a}$.\nUsing Vieta's formula, we find that the roots add to $2$ or $\\boxed{\\textbf{(E)}\\ \\text{None of these}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/3.json | AHSME |
1950_AHSME_Problems | 34 | 0 | Geometry | Multiple Choice | When the circumference of a toy balloon is increased from $20$ inches to $25$ inches, the radius is increased by:
$\textbf{(A)}\ 5\text{ in} \qquad \textbf{(B)}\ 2\dfrac{1}{2}\text{ in} \qquad \textbf{(C)}\ \dfrac{5}{\pi}\text{ in} \qquad \textbf{(D)}\ \dfrac{5}{2\pi}\text{ in} \qquad \textbf{(E)}\ \dfrac{\pi}{5}\tex... | [
"When the circumference of a circle is increased by a percentage, the radius is also increased by the same percentage (or else the ratio of the circumference to the diameter wouldn't be $\\pi$ anymore)\nWe see that the circumference was increased by $25\\%$. This means the radius was also increased by $25\\%$. The ... | 3 | ./CreativeMath/AHSME/1950_AHSME_Problems/34.json | AHSME |
1950_AHSME_Problems | 8 | 0 | Geometry | Multiple Choice | If the radius of a circle is increased $100\%$, the area is increased:
$\textbf{(A)}\ 100\%\qquad\textbf{(B)}\ 200\%\qquad\textbf{(C)}\ 300\%\qquad\textbf{(D)}\ 400\%\qquad\textbf{(E)}\ \text{By none of these}$
| [
"Increasing by $100\\%$ is the same as doubling the radius. If we let $r$ be the radius of the old circle, then the radius of the new circle is $2r.$\n\n\nSince the area of the circle is given by the formula $\\pi r^2,$ the area of the new circle is $\\pi (2r)^2 = 4\\pi r^2.$ The area is quadrupled, or increased by... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/8.json | AHSME |
1950_AHSME_Problems | 22 | 0 | Arithmetic | Multiple Choice | Successive discounts of $10\%$ and $20\%$ are equivalent to a single discount of:
$\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 72\%\qquad\textbf{(D)}\ 28\%\qquad\textbf{(E)}\ \text{None of these}$
| [
"Without loss of generality, assume something costs $100$ dollars. Then with each successive discount, it would cost $90$ dollars, then $72$ dollars. This amounts to a total of $28$ dollars off, so the single discount would be $\\boxed{\\mathrm{(D)}\\ 28\\%.}$\n\n\n",
"Let the object cost $x$ dollars. After the $... | 2 | ./CreativeMath/AHSME/1950_AHSME_Problems/22.json | AHSME |
1950_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | Of the following
(1) $a(x-y)=ax-ay$
(2) $a^{x-y}=a^x-a^y$
(3) $\log (x-y)=\log x-\log y$
(4) $\frac{\log x}{\log y}=\log{x}-\log{y}$
(5) $a(xy)=ax \cdot ay$
$\textbf{(A)}\text{Only 1 and 4 are true}\qquad\\\textbf{(B)}\ \text{Only 1 and 5 are true}\qquad\\\textbf{(C)}\ \text{Only 1 and 3 are true}\qquad\\\... | [
"The distributive property doesn't apply to logarithms or in the ways illustrated, and only applies to addition and subtraction. Also, $a^{x-y} = \\frac{a^x}{a^y}$, so $\\boxed{\\textbf{(E)} \\text{ Only 1 is true}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/18.json | AHSME |
1950_AHSME_Problems | 38 | 0 | Algebra | Multiple Choice | If the expression $\begin{pmatrix}a & c\\ d & b\end{pmatrix}$ has the value $ab-cd$ for all values of $a, b, c$ and $d$, then the equation $\begin{pmatrix}2x & 1\\ x & x\end{pmatrix}= 3$:
$\textbf{(A)}\ \text{Is satisfied for only 1 value of }x\qquad\\ \textbf{(B)}\ \text{Is satisified for only 2 values of }x\qquad\\... | [
"By $\\begin{pmatrix}a & c\\\\ d & b\\end{pmatrix}=ab-cd$, we have $2x^2-x=3$. Subtracting $3$ from both sides, giving $2x^2-x-3=0$. This factors to $(2x-3)(x+1)=0$. Thus, $x=\\dfrac{3}{2},-1$, so the equation is $\\boxed{\\textbf{(B)}\\ \\text{satisified for only 2 values of }x}$.\n\n\nNote: Alternatively, one may... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/38.json | AHSME |
1950_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | Reduced to lowest terms, $\frac{a^{2}-b^{2}}{ab} - \frac{ab-b^{2}}{ab-a^{2}}$ is equal to:
$\textbf{(A)}\ \frac{a}{b}\qquad\textbf{(B)}\ \frac{a^{2}-2b^{2}}{ab}\qquad\textbf{(C)}\ a^{2}\qquad\textbf{(D)}\ a-2b\qquad\textbf{(E)}\ \text{None of these}$
| [
"We start off by factoring the second fraction.\n\n\n\\[-\\frac{ab-b^2}{ab-a^2} = -\\frac{b(a-b)}{-a(a-b)} = \\frac{b}{a}.\\]\n\n\nNow create a common denominator and simplify.\n\n\n\\[\\frac{a^2-b^2}{ab}+\\frac{b}{a}=\\frac{a^2-b^2}{ab}+\\frac{b^2}{ab} = \\frac{a^2}{ab} = \\boxed{\\mathrm{(A) }\\frac{a}{b}}\\]\n\n... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/4.json | AHSME |
1950_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | For the simultaneous equations \[2x-3y=8\]
\[6y-4x=9\]
$\textbf{(A)}\ x=4,y=0\qquad\textbf{(B)}\ x=0,y=\frac{3}{2}\qquad\textbf{(C)}\ x=0,y=0\qquad\\ \textbf{(D)}\ \text{There is no solution}\qquad\textbf{(E)}\ \text{There are an infinite number of solutions}$
| [
"Try to solve this system of equations using the elimination method.\n\n\n\\begin{align*} -2(2x-3y)&=-2(8)\\\\ 6y-4x&=-16\\\\ 6y-4x&=9\\\\ 0&=-7 \\end{align*}\n\n\nSomething is clearly contradictory so $\\boxed{\\mathrm{(D)}\\text{ There is no solution}.}$\n\n\nAlternatively, note that the second equation is a mult... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/14.json | AHSME |
1950_AHSME_Problems | 43 | 0 | Algebra | Multiple Choice | The sum to infinity of $\frac{1}{7}+\frac {2}{7^2}+\frac{1}{7^3}+\frac{2}{7^4}+\cdots$ is:
$\textbf{(A)}\ \frac{1}{5} \qquad \textbf{(B)}\ \dfrac{1}{24} \qquad \textbf{(C)}\ \dfrac{5}{48} \qquad \textbf{(D)}\ \dfrac{1}{16} \qquad \textbf{(E)}\ \text{None of these}$
| [
"Note that this is $\\frac{1}{7}(1+\\frac{1}{49}+\\frac{1}{49^2}+...)+\\frac{2}{49}(1+\\frac{1}{49}+...)=\\frac{9}{49}(1+\\frac{1}{49}+...)$. Using the formula for a geometric series, we find that this is $\\frac{9}{49}(\\frac{1}{1-\\frac{1}{49}})=\\frac{9}{49}(\\frac{1}{\\frac{48}{49}})=\\frac{9}{49}(\\frac{49}{48... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/43.json | AHSME |
1950_AHSME_Problems | 42 | 0 | Algebra | Multiple Choice | The equation $x^{x^{x^{.^{.^.}}}}=2$ is satisfied when $x$ is equal to:
$\textbf{(A)}\ \infty \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt[4]{2} \qquad \textbf{(D)}\ \sqrt{2} \qquad \textbf{(E)}\ \text{None of these}$
| [
"Taking the log, we get $\\log_x 2 = x^{x^{x^{.^{.^.}}}}=2$, and $\\log_x 2 = 2$. Solving for x, we get $2=x^2$, and $\\sqrt{2}=x \\Rightarrow \\mathrm{(D)}$\n\n\n\n\n\n\n",
"$x^{x^{x^{.^{.^.}}}}=2$ is the original equation. If we let $y=x^{x^{x^{.^{.^.}}}}$, then the equation can be written as $y=2$. This also m... | 2 | ./CreativeMath/AHSME/1950_AHSME_Problems/42.json | AHSME |
1950_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | The real roots of $x^2+4$ are:
$\textbf{(A)}\ (x^{2}+2)(x^{2}+2)\qquad\textbf{(B)}\ (x^{2}+2)(x^{2}-2)\qquad\textbf{(C)}\ x^{2}(x^{2}+4)\qquad\\ \textbf{(D)}\ (x^{2}-2x+2)(x^{2}+2x+2)\qquad\textbf{(E)}\ \text{Non-existent}$
| [
"This looks similar to a difference of squares, so we can write it as $(x+2i)(x-2i).$ Neither of these factors are real.\n\n\nAlso, looking at the answer choices, there is no way multiplying two polynomials of degree $2$ will result in a polynomial of degree $2$ as well. Therefore the real factors are $\\boxed{\\m... | 3 | ./CreativeMath/AHSME/1950_AHSME_Problems/15.json | AHSME |
1950_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | If five geometric means are inserted between $8$ and $5832$, the fifth term in the geometric series:
$\textbf{(A)}\ 648\qquad\textbf{(B)}\ 832\qquad\textbf{(C)}\ 1168\qquad\textbf{(D)}\ 1944\qquad\textbf{(E)}\ \text{None of these}$
| [
"We can let the common ratio of the geometric sequence be $r$. $5832$ is given to be the seventh term in the geometric sequence as there are five terms between it and $a_1$ if we consider $a_1=8$.\nBy the formula for each term in a geometric sequence, we find that $a_n=a_1r^{n-1}$ or $(5382)=(8)r^6$\nWe divide by e... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/5.json | AHSME |
1950_AHSME_Problems | 39 | 0 | Algebra | Multiple Choice | Given the series $2+1+\frac {1}{2}+\frac {1}{4}+\cdots$ and the following five statements:
\begin{itemize}
\item (1) the sum increases without limit
\item (2) the sum decreases without limit
\item (3) the difference between any term of the sequence and zero can be made less than any positive quantity no matter how ... | [
"This series is a geometric series with common ratio $\\frac{1}{2}$. Using the well-known formula for the sum of an infinite geometric series, we obtain that this series has a value of $2\\cdot \\frac{1}{1-\\frac{1}{2}}=4$. It immediately follows that statements 1 and 2 are false while statements 4 and 5 are true. ... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/39.json | AHSME |
1950_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | If $m$ men can do a job in $d$ days, then $m+r$ men can do the job in:
$\textbf{(A)}\ d+r \text{ days}\qquad\textbf{(B)}\ d-r\text{ days}\qquad\textbf{(C)}\ \frac{md}{m+r}\text{ days}\qquad\\ \textbf{(D)}\ \frac{d}{m+r}\text{ days}\qquad\textbf{(E)}\ \text{None of these}$
| [
"The number of men is inversely proportional to the number of days the job takes. Thus, if $m$ men can do a job in $d$ days, we have that it will take $md$ days for $1$ man to do the job. Thus, $m + r$ men can do the job in $\\boxed{\\textbf{(C)}\\ \\frac{md}{m+r}\\text{ days}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/19.json | AHSME |
1950_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | A man buys a house for $10,000 and rents it. He puts $12\frac{1}{2}\%$ of each month's rent aside for repairs and upkeep; pays $325 a year taxes and realizes $5\frac{1}{2}\%$ on his investment. The monthly rent (in dollars) is:
$\textbf{(A)} \ \ 64.82\qquad\textbf{(B)} \ \ 83.33\qquad\textbf{(C)} \ \ 72.08\qquad\tex... | [
"$12\\frac{1}{2}\\%$ is the same as $\\frac{1}{8}$, so the man sets one eighth of each month's rent aside, so he only gains $\\frac{7}{8}$ of his rent. He also pays $325 each year, and he realizes $5.5\\%$, or $550, on his investment. Therefore he must have collected a total of $325 +$550 = $875 in rent. This was f... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/23.json | AHSME |
1950_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | The area of the largest triangle that can be inscribed in a semi-circle whose radius is $r$ is:
$\textbf{(A)}\ r^{2}\qquad\textbf{(B)}\ r^{3}\qquad\textbf{(C)}\ 2r^{2}\qquad\textbf{(D)}\ 2r^{3}\qquad\textbf{(E)}\ \frac{1}{2}r^{2}$
| [
"The area of a triangle is $\\frac12 bh.$ To maximize the base, let it be equal to the diameter of the semi circle, which is equal to $2r.$ To maximize the height, or altitude, choose the point directly in the middle of the arc connecting the endpoints of the diameter. It is equal to $r.$ Therefore the area is $\\f... | 1 | ./CreativeMath/AHSME/1950_AHSME_Problems/9.json | AHSME |
1950_AHSME_Problems | 35 | 0 | Geometry | Multiple Choice | In triangle $ABC$, $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:
$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$
| [
"The inradius is equal to the area divided by semiperimeter. The area is $\\frac{(10)(24)}{2} = 120$ because it's a right triangle, as it's side length satisfies the Pythagorean Theorem. The semiperimeter is $30$. Therefore the inradius is $\\boxed{\\textbf{(B)}\\ 4}$.\n\n\n",
"Since this is a right triangle, we ... | 3 | ./CreativeMath/AHSME/1950_AHSME_Problems/35.json | AHSME |
1958_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | If $4^x - 4^{x - 1} = 24$, then $(2x)^x$ equals:
$\textbf{(A)}\ 5\sqrt{5}\qquad \textbf{(B)}\ \sqrt{5}\qquad \textbf{(C)}\ 25\sqrt{5}\qquad \textbf{(D)}\ 125\qquad \textbf{(E)}\ 25$
| [
"$\\fbox{}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/20.json | AHSME |
1958_AHSME_Problems | 36 | 0 | Geometry | Multiple Choice | The sides of a triangle are $30$, $70$, and $80$ units. If an altitude is dropped upon the side of length $80$, the larger segment cut off on this side is:
$\textbf{(A)}\ 62\qquad \textbf{(B)}\ 63\qquad \textbf{(C)}\ 64\qquad \textbf{(D)}\ 65\qquad \textbf{(E)}\ 66$
| [
"Let the shorter segment be $x$ and the altitude be $y$. The larger segment is then $80-x$. By the Pythagorean Theorem\n, \\[30^2-y^2=x^2 \\qquad(1)\\]\nand \\[(80-x)^2=70^2-y^2 \\qquad(2)\\]\nAdding $(1)$ and $(2)$ and simplifying gives $x=15$. Therefore, the answer is $80-15=\\boxed{\\textbf{(D)}~65}$\n\n\n~mega... | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/36.json | AHSME |
1958_AHSME_Problems | 41 | 0 | Algebra | Multiple Choice | The roots of $Ax^2 + Bx + C = 0$ are $r$ and $s$. For the roots of
$x^2+px +q =0$
to be $r^2$ and $s^2$, $p$ must equal:
$\textbf{(A)}\ \frac{B^2 - 4AC}{A^2}\qquad \textbf{(B)}\ \frac{B^2 - 2AC}{A^2}\qquad \textbf{(C)}\ \frac{2AC - B^2}{A^2}\qquad \\ \textbf{(D)}\ B^2 - 2C\qquad \textbf{(E)}\ 2C - B^2$
| [
"By Vieta's, $r + s = -\\frac{B}{A}$, $rs = \\frac{C}{A}$, and $r^2 + s^2 = -p$. Note that $(r+s)^2 = r^2 + s^2 + 2rs$.\n\n\nTherefore, $(r + s)^2 - 2rs = r^2 + s^2$, or $-\\left(\\frac{B}{A}\\right)^2 - \\frac{2C}{A} = -p$.\n\n\nSimplifying, $\\frac{B^2 - 2CA}{A^2} = -p$. \n\n\nFinally, multiply both sides by $-1$... | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/41.json | AHSME |
1958_AHSME_Problems | 16 | 0 | Geometry | Multiple Choice | The area of a circle inscribed in a regular hexagon is $100\pi$. The area of hexagon is:
$\textbf{(A)}\ 600\qquad \textbf{(B)}\ 300\qquad \textbf{(C)}\ 200\sqrt{2}\qquad \textbf{(D)}\ 200\sqrt{3}\qquad \textbf{(E)}\ 120\sqrt{5}$
| [
"We can split the hexagon into 6 equilateral triangles. If the area of the circle is $100pi$, then the radius is $10$. The radius is equal to the height of one of the equilateral triangles. Using $sin60 = \\frac{\\sqrt3}{2}$, we get the hypotenuse is $\\frac{20\\sqrt{3}}{3}$, which is also equal to the side length.... | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/16.json | AHSME |
1958_AHSME_Problems | 6 | 0 | Arithmetic | Multiple Choice | The arithmetic mean between $\frac {x + a}{x}$ and $\frac {x - a}{x}$, when $x \not = 0$, is:
$\textbf{(A)}\ {2}\text{, if }{a \not = 0}\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ {1}\text{, only if }{a = 0}\qquad \textbf{(D)}\ \frac {a}{x}\qquad \textbf{(E)}\ x$
| [
"We have $\\frac{1}{2}\\cdot \\left(\\frac{x + a}{x} + \\frac{x - a}{x}\\right) = \\frac{2}{2} = \\boxed{\\text{(B) }{1}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/6.json | AHSME |
1958_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | A straight line joins the points $(-1,1)$ and $(3,9)$. Its $x$-intercept is:
$\textbf{(A)}\ -\frac{3}{2}\qquad \textbf{(B)}\ -\frac{2}{3}\qquad \textbf{(C)}\ \frac{2}{5}\qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 3$
| [
"The slope of the line is $\\frac{ \\Delta y }{ \\Delta x} = \\frac{9-1}{3-(-1)} = 2$. Using the formula for the point-slope form of a line, we have $y-y_1 = m(x-x_1)$, so $y-1=2(x-(-1)) \\to y-1=2(x+1)$.\n\n\nThe x-intercept is the x-value when $y=0$, so we substitute 0 for y:\n\n\n\\[0-1=2x+2\\]\n\n\n\\[-1=2x+2\\... | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/7.json | AHSME |
1958_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | If $x$ is positive and $\log{x} \ge \log{2} + \frac{1}{2}\log{x}$, then:
$\textbf{(A)}\ {x}\text{ has no minimum or maximum value}\qquad \\ \textbf{(B)}\ \text{the maximum value of }{x}\text{ is }{1}\qquad \\ \textbf{(C)}\ \text{the minimum value of }{x}\text{ is }{1}\qquad \\ \textbf{(D)}\ \text{the maximum value of... | [
"$\\fbox{}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/17.json | AHSME |
1958_AHSME_Problems | 40 | 0 | Algebra | Multiple Choice | Given $a_0 = 1$, $a_1 = 3$, and the general relation $a_n^2 - a_{n - 1}a_{n + 1} = (-1)^n$ for $n \ge 1$. Then $a_3$ equals:
$\textbf{(A)}\ \frac{13}{27}\qquad \textbf{(B)}\ 33\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ -17$
| [
"Using the recursive definition, we find that $a_3=33$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/40.json | AHSME |
1958_AHSME_Problems | 37 | 0 | Algebra | Multiple Choice | The first term of an arithmetic series of consecutive integers is $k^2 + 1$. The sum of $2k + 1$ terms of this series may be expressed as:
$\textbf{(A)}\ k^3 + (k + 1)^3\qquad \textbf{(B)}\ (k - 1)^3 + k^3\qquad \textbf{(C)}\ (k + 1)^3\qquad \\ \textbf{(D)}\ (k + 1)^2\qquad \textbf{(E)}\ (2k + 1)(k + 1)^2$
| [
"$Option A$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/37.json | AHSME |
1958_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | In the accompanying figure $\overline{CE}$ and $\overline{DE}$ are equal chords of a circle with center $O$. Arc $AB$ is a quarter-circle. Then the ratio of the area of triangle $CED$ to the area of triangle $AOB$ is:
[asy] draw(circle((0,0),10),black+linewidth(.75)); draw((-10,0)--(0,0)--(10,0)--(0,10)--cycle,dot); ... | [
"Draw $OE$. Since triangles $AOB$, $COE$, and $DOE$ are congruent, you can fit triangle $AOE$ twice in $CED$. Thus, our answer is $(E)2:1$.\n$\\fbox{}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/21.json | AHSME |
1958_AHSME_Problems | 47 | 0 | Geometry | Multiple Choice | $ABCD$ is a rectangle (see the accompanying diagram) with $P$ any point on $\overline{AB}$. $\overline{PS} \perp \overline{BD}$ and $\overline{PR} \perp \overline{AC}$. $\overline{AF} \perp \overline{BD}$ and $\overline{PQ} \perp \overline{AF}$. Then $PR + PS$ is equal to:
[asy] draw((-2,-1)--(-2,1)--(2,1)--(2,-1)--c... | [
"Since $\\overline{\\rm PQ}$ and $\\overline{\\rm BD}$ are both perpendicular to $\\overline{\\rm AF}$, $\\overline{\\rm PQ} || \\overline{\\rm BD}$. Thus, $\\angle APQ = \\angle ABD$.\n\n\n\n\nAlso, $\\angle ABD$ = $\\angle CAB$ because $ABCD$ is a rectangle. Thus, $\\angle APQ = \\angle ABD = \\angle CAB$.\n\n\n\... | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/47.json | AHSME |
1958_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | For what real values of $k$, other than $k = 0$, does the equation $x^2 + kx + k^2 = 0$ have real roots?
$\textbf{(A)}\ {k < 0}\qquad \textbf{(B)}\ {k > 0} \qquad \textbf{(C)}\ {k \ge 1} \qquad \textbf{(D)}\ \text{all values of }{k}\qquad \textbf{(E)}\ \text{no values of }{k}$
| [
"An expression of the form $ax^2+bx+c$ has at least one real root when $b^2-4ac \\geq 0$.\n\n\nSubstituting $k$ for $b$ and $k^2$ for $c$, we have \n\n\n\\[k^2-4k^2 \\geq 0\\]\n\n\n\\[-3k^2 \\geq 0\\]\n\n\nbut the range of $-3k^2$ is $(-\\infty,0]$, so the answer is $\\boxed{\\text{(E)}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/10.json | AHSME |
1958_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | A set of $n$ numbers has the sum $s$. Each number of the set is increased by $20$, then multiplied by $5$, and then decreased by $20$. The sum of the numbers in the new set thus obtained is:
$\textbf{(A)}\ s \plus{} 20n\qquad
\textbf{(B)}\ 5s \plus{} 80n\qquad
\textbf{(C)}\ s\qquad
\textbf{(D)}\ 5s\qquad
\textbf{... | [
"$\\fbox{}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/26.json | AHSME |
1958_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | If $xy \equal{} b$ (Error compiling LaTeX. Unknown error_msg) and $\frac{1}{x^2} \plus{} \frac{1}{y^2} \equal{} a$ (Error compiling LaTeX. Unknown error_msg), then $(x \plus{} y)^2$ (Error compiling LaTeX. Unknown error_msg) equals:
$\textbf{(A)}\ (a \plus{} 2b)^2\qquad
\textbf{(B)}\ a^2 \plus{} b^2\qquad
\textbf{(... | [
"$\\fbox{}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/30.json | AHSME |
1958_AHSME_Problems | 31 | 0 | Geometry | Multiple Choice | The altitude drawn to the base of an isosceles triangle is $8$, and the perimeter $32$. The area of the triangle is:
$\textbf{(A)}\ 56\qquad \textbf{(B)}\ 48\qquad \textbf{(C)}\ 40\qquad \textbf{(D)}\ 32\qquad \textbf{(E)}\ 24$
| [
"Consider the half of the triangle that is left of the altitude. Using the information given in the problem, we can determine that this is a right triangle with one leg of length 8 whose other two sides sum to 16. Through either setting the other leg as $x$ and the hypotenuse as $16-x$ and using the Pythagorean The... | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/31.json | AHSME |
1958_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | The points $(2,-3)$, $(4,3)$, and $(5, k/2)$ are on the same straight line. The value(s) of $k$ is (are):
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ -12\qquad \textbf{(C)}\ \pm 12\qquad \textbf{(D)}\ {12}\text{ or }{6}\qquad \textbf{(E)}\ {6}\text{ or }{6\frac{2}{3}}$
| [
"First find the slope. Then use the point-slope formula to find the equation of the line. Then substitute 5 for x to find y. \n$\\text{(A)}12$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/27.json | AHSME |
1958_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | The value of $[2 - 3(2 - 3)^{-1}]^{-1}$ is:
$\textbf{(A)}\ 5\qquad \textbf{(B)}\ -5\qquad \textbf{(C)}\ \frac{1}{5}\qquad \textbf{(D)}\ -\frac{1}{5}\qquad \textbf{(E)}\ \frac{5}{3}$
| [
"$\\frac{1}{2-3(-1)}=\\frac{1}{5}$, so the answer is $\\boxed{\\text{C}}$. \n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/1.json | AHSME |
1958_AHSME_Problems | 50 | 0 | Algebra | Multiple Choice | In this diagram a scheme is indicated for associating all the points of segment $\overline{AB}$ with those of segment $\overline{A'B'}$, and reciprocally. To described this association scheme analytically, let $x$ be the distance from a point $P$ on $\overline{AB}$ to $D$ and let $y$ be the distance from the associated... | [
"$\\fbox{}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/50.json | AHSME |
1958_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | The number of roots satisfying the equation $\sqrt{5 - x} = x\sqrt{5 - x}$ is:
$\textbf{(A)}\ \text{unlimited}\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 0$
| [
"Solve the equation for x.\n\n\n\\[\\sqrt{5-x}=x\\sqrt{5-x}\\]\n\n\n\\[x\\sqrt{5-x} - \\sqrt{5-x} = 0\\]\n\n\n\\[(x-1)\\sqrt{5-x}=0\\]\n\n\n\\[x=1,5\\]\n\n\nThere are two solutions $\\to \\boxed{\\textbf{(C)}}$\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/11.json | AHSME |
1958_AHSME_Problems | 46 | 0 | Algebra | Multiple Choice | For values of $x$ less than $1$ but greater than $-4$, the expression
$\frac{x^2 - 2x + 2}{2x - 2}$
has:
$\textbf{(A)}\ \text{no maximum or minimum value}\qquad \\ \textbf{(B)}\ \text{a minimum value of }{+1}\qquad \\ \textbf{(C)}\ \text{a maximum value of }{+1}\qquad \\ \textbf{(D)}\ \text{a minimum value of }{-1}\q... | [
"From $\\frac{x^2 - 2x + 2}{2x - 2}$, we can further factor $\\frac{x^2 - 2x + 2}{2(x - 1)}$ and then $\\frac{(x-1)^{2}+1}{2(x - 1)}$ and finally $\\frac{x-1}{2}+\\frac{1}{2x-2}$. Using $AM-GM$, we can see that $\\frac{x-1}{2}=\\frac{1}{2x-2}$. From there, we can get that $2=2 \\cdot (x-1)^{2}$.\n\n\nFrom there, we... | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/46.json | AHSME |
1958_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | If $\frac {1}{x} - \frac {1}{y} = \frac {1}{z}$, then $z$ equals:
$\textbf{(A)}\ y - x\qquad \textbf{(B)}\ x - y\qquad \textbf{(C)}\ \frac {y - x}{xy}\qquad \textbf{(D)}\ \frac {xy}{y - x}\qquad \textbf{(E)}\ \frac {xy}{x - y}$
| [
"$\\frac{1}{x}-\\frac{1}{y}=\\frac{1}{z}$\n\n\n$\\frac{y}{xy}-\\frac{x}{xy}=\\frac{1}{z}$\n\n\n$\\frac{y-x}{xy}=\\frac{1}{z}$\n\n\n$\\frac{1}{\\frac{y-x}{xy}}=\\frac{1}{\\frac{1}{z}}$\n\n\n$\\frac{xy}{y-x}=z$\n\n\nThe answer is therefore $\\boxed{\\text{D}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/2.json | AHSME |
1958_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | A $16$-quart radiator is filled with water. Four quarts are removed and replaced with pure antifreeze liquid. Then four quarts of the mixture are removed and replaced with pure antifreeze. This is done a third and a fourth time. The fractional part of the final mixture that is water is:
$\textbf{(A)}\ \frac{1}{4}\qqu... | [
"Every time the process is done, $\\frac{3}{4}$ of the mixture is replaced with antifreeze. That means that $\\frac{3}{4}$ of the water is replaced by antifreeze, and the amount of water in the mixture after the fourth time is $\\left(\\frac{3}{4}\\right)^4 = \\boxed{\\textbf{(B) }\\frac{81}{256}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/28.json | AHSME |
1958_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | If $P = \frac{s}{(1 + k)^n}$ then $n$ equals:
$\textbf{(A)}\ \frac{\log{\left(\frac{s}{P}\right)}}{\log{(1 + k)}}\qquad \textbf{(B)}\ \log{\left(\frac{s}{P(1 + k)}\right)}\qquad \textbf{(C)}\ \log{\left(\frac{s - P}{1 + k}\right)}\qquad \\ \textbf{(D)}\ \log{\left(\frac{s}{P}\right)} + \log{(1 + k)}\qquad \textbf{... | [
"\\[P=\\frac{s}{(1+k)^n}\\]\n\n\n\\[(1+k)^n=\\frac{s}{P}\\]\n\n\nTake the $\\log$ of each side.\n\n\n\\[n \\log(1+k) = \\log\\left(\\frac{s}{P}\\right)\\]\n\n\n\\[n = \\frac{\\log\\left(\\frac{s}{P}\\right)}{\\log(1+k)} \\to \\boxed{\\text{(A)}}\\]\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/12.json | AHSME |
1958_AHSME_Problems | 45 | 0 | Algebra | Multiple Choice | A check is written for $x$ dollars and $y$ cents, $x$ and $y$ both two-digit numbers. In error it is cashed for $y$ dollars and $x$ cents, the incorrect amount exceeding the correct amount by $$17.82$. Then:
$\textbf{(A)}\ {x}\text{ cannot exceed }{70}\qquad \\ \textbf{(B)}\ {y}\text{ can equal }{2x}\qquad\\ \te... | [
"$\\fbox{}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/45.json | AHSME |
1958_AHSME_Problems | 32 | 0 | Algebra | Multiple Choice | With $ $1000$ a rancher is to buy steers at $ $25$ each and cows at $ $26$ each. If the number of steers $s$ and the number of cows $c$ are both positive integers, then:
$\textbf{(A)}\ \text{this problem has no solution}\qquad\\ \textbf{(B)}\ \text{there are two solutions with }{s}\text{ exceeding }{c}\qquad \\ \tex... | [
"$\\fbox{}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/32.json | AHSME |
1958_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | A man travels $m$ feet due north at $2$ minutes per mile. He returns due south to his starting point at $2$ miles per minute. The average rate in miles per hour for the entire trip is:
$\textbf{(A)}\ 75\qquad \textbf{(B)}\ 48\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 24\qquad\\ \textbf{(E)}\ \text{impossible to ... | [
"$\\fbox{}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/24.json | AHSME |
1958_AHSME_Problems | 49 | 0 | Counting | Multiple Choice | In the expansion of $(a + b)^n$ there are $n + 1$ dissimilar terms. The number of dissimilar terms in the expansion of $(a + b + c)^{10}$ is:
$\textbf{(A)}\ 11\qquad \textbf{(B)}\ 33\qquad \textbf{(C)}\ 55\qquad \textbf{(D)}\ 66\qquad \textbf{(E)}\ 132$
| [
"Expand the binomial $((a+b)+c)^n$ with the binomial theorem. We have:\n\n\n\\[\\sum\\limits_{k=0}^{10} \\binom{10}{k} (a+b)^k c^{10-k}\\]\n\n\nSo for each iteration of the summation operator, we add k+1 dissimilar terms. Therefore our answer is:\n\n\n\\[\\sum\\limits_{k=0}^{10} k+1 = \\frac{11(1+11)}{2} = 66 \\to ... | 2 | ./CreativeMath/AHSME/1958_AHSME_Problems/49.json | AHSME |
1958_AHSME_Problems | 48 | 0 | Geometry | Multiple Choice | Diameter $\overline{AB}$ of a circle with center $O$ is $10$ units. $C$ is a point $4$ units from $A$, and on $\overline{AB}$. $D$ is a point $4$ units from $B$, and on $\overline{AB}$. $P$ is any point on the circle. Then the broken-line path from $C$ to $P$ to $D$:
$\textbf{(A)}\ \text{has the same length for all p... | [
"If $P$ is on $A$, then the length is 10, eliminating answer choice $(B)$.\n\n\nIf $P$ is equidistant from $C$ and $D$, the length is $2\\sqrt{1^2+5^2}=2\\sqrt{26}>10$, eliminating $(A)$ and $(C)$.\n\n\nIf triangle $CDP$ is right, then angle $CDP$ is right or angle $DCP$ is right. Assume that angle $DCP$ is right. ... | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/48.json | AHSME |
1958_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | If $\log_{k}{x}\cdot \log_{5}{k} = 3$, then $x$ equals:
$\textbf{(A)}\ k^6\qquad \textbf{(B)}\ 5k^3\qquad \textbf{(C)}\ k^3\qquad \textbf{(D)}\ 243\qquad \textbf{(E)}\ 125$
| [
"$\\fbox{}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/25.json | AHSME |
1958_AHSME_Problems | 33 | 0 | Algebra | Multiple Choice | For one root of $ax^2 + bx + c = 0$ to be double the other, the coefficients $a,\,b,\,c$ must be related as follows:
$\textbf{(A)}\ 4b^2 = 9c\qquad \textbf{(B)}\ 2b^2 = 9ac\qquad \textbf{(C)}\ 2b^2 = 9a\qquad \\ \textbf{(D)}\ b^2 - 8ac = 0\qquad \textbf{(E)}\ 9b^2 = 2ac$
| [
"$\\fbox{}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1958_AHSME_Problems/33.json | AHSME |
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