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 $\\frac{-b}{a}$) or (using calculus) as the value of its derivative set equal to $0$.\n\n\n"
] | 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:
$\textbf{(A)}\ 1.25, 8.75 \qquad \textbf{(B)}\ 2.75,7.25 \qquad \textbf{(C)}\ 2,8 \qquad \textbf{(D)}\ 4,6 \qquad \textbf{(E)}\ \text{none of these}$
|
[
"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 x = \\frac{50}{8} = \\frac{25}{4}.\\]\nIt follows that $a = \\sqrt{x^2 - 25} = 5\\sqrt{\\frac{25}{16}-1} = \\frac{15}{4}$.\n\n\n$BN = 5 + \\frac{15}{14} = 8.75$, so the answer is $A$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1952_AHSME_Problems/29.json
|
AHSME
|
1952_AHSME_Problems
| 3
| 0
|
Algebra
|
Multiple Choice
|
The expression $a^3-a^{-3}$ equals:
$\textbf{(A) \ }\left(a-\frac{1}{a}\right)\left(a^2+1+\frac{1}{a^2}\right) \qquad \textbf{(B) \ }\left(\frac{1}{a}-a\right)\left(a^2-1+\frac{1}{a^2}\right) \qquad \textbf{(C) \ }\left(a-\frac{1}{a}\right)\left(a^2-2+\frac{1}{a^2}\right) \qquad$
$\textbf{(D) \ }\left(\frac{1}{a}-a\right)\left(\frac{1}{a^2}+1+a^2\right) \qquad \textbf{(E) \ }\text{none of these}$
|
[
"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:
$\textbf{(A)}\ \frac{1-p^2}{200}\qquad \textbf{(B)}\ \frac{\sqrt{1-p^2}}{100}\qquad \textbf{(C)}\ \text{one dollar}\qquad\\ \textbf{(D)}\ 1-\frac{p^2}{10000-p^2}\qquad \textbf{(E)}\ \frac{10000}{10000-p^2}$
|
[
"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 $\\dfrac{10000-p^{2}}{10000}$, we get the value of x. $\\boxed{\\textbf{(E)}\\dfrac{10000}{10000-p^{2}}}$\n\n\n"
] | 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 circles are non-congruent and internally tangent.\n\n\n[asy] pair A=(1,0), B=(-1,0), C=(-5,0), D=(-6,0), E=(5,0), F=(8,0); draw(circle(A,1)); draw(circle(B,1)); draw(circle(C,1)); draw(circle(D,1)); draw(circle(E,1)); draw(circle(F,1)); draw((0,1)--(0,-1),red); draw((-2,1)--(2,1),red); draw((-2,-1)--(2,-1),red); draw((-7,1)--(-4,1),red); draw((-7,-1)--(-4,-1),red); draw((4,1)--(9,1),red); draw((4,-1)--(9,-1),red); draw((5+0.5sqrt(2),0.5sqrt(2))--(8-0.5sqrt(2),-0.5sqrt(2)),red); draw((5+0.5sqrt(2),-0.5sqrt(2))--(8-0.5sqrt(2),0.5sqrt(2)),red); label(\"$3$\",(-1.5,0),E); label(\"$1$\",(0,0.75),N); label(\"$2$\",(0,-0.75),S); label(\"$1$\",(-5.5,0.75),N); label(\"$2$\",(-5.5,-0.75),S); label(\"$1$\",(6.5,0.75),N); label(\"$2$\",(6.5,-0.75),S); label(\"$3$\",(6.5,-0.25),N); label(\"$4$\",(6.5,0.25),S); [/asy]\n\n\n"
] | 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:
$\textbf{(A) \ }\sqrt{b^2+1} \qquad \textbf{(B) \ }\sqrt{b^2-3} \qquad \textbf{(C) \ }\sqrt{b^2+1}+2 \qquad$
$\textbf{(D) \ }b^2+5 \qquad \textbf{(E) \ }\sqrt{b^2+3}$
|
[
"[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(\"$1$\",B--C,S); label(\"$x$\",D--B,W); label(\"$2$\",D--A,N); label(\"$b$\",A--C,E); [/asy]\nLet $\\overline{BD}=x$. According to the Pythagorean Theorem, $x^2+4=b^2+1$. Hence, $x=\\boxed{\\textbf{(B)}\\ \\sqrt{b^2-3}}$.\n\n\n"
] | 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:
$\textbf{(A)}\ \text{none} \qquad \textbf{(B)}\ \text{one} \qquad \textbf{(C)}\ \text{two} \qquad \textbf{(D)}\ \text{three} \qquad \textbf{(E)}\ \text{more than three}$
|
[
"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 answer is $\\fbox{(D) three}$\n\n\n"
] | 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:
$\textbf{(A) \ }C=10+3P \qquad \textbf{(B) \ }C=10P+3 \qquad \textbf{(C) \ }C=10+3(P-1) \qquad$
$\textbf{(D) \ }C=9+3P \qquad \textbf{(E) \ }C=10P-7$
|
[
"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:
$\textbf{(A) \ }\text{no loss or gain} \qquad \textbf{(B) \ }\text{loss of }\textdollar 1000 \qquad \textbf{(C) \ }\text{gain of }\textdollar 1000 \qquad \textbf{(D) \ }\text{gain of }\textdollar 2000 \qquad \textbf{(E) \ }\text{none of these}$
|
[
"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 the current price of the property, $2\\cdot \\textdollar12,000$. Hence, the transaction resulted in a $\\boxed{\\textbf{(B)}\\ \\text{loss of }\\textdollar1000}$.\n\n\n"
] | 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 diameter of the original circle
$\textbf{(C) } \qquad$ greater than the diameter, but less than the semi-circumference of the original circle
$\textbf{(D) } \qquad$ that is infinite
$\textbf{(E) }$ greater than the semi-circumference
|
[
"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 is $n*\\frac{D*pi}{n*2}=\\frac{D*pi}{2}$, and since that is the exact expression for the semi-circumference of the original circle, the answer is $\\boxed{A}$.\n\n\n"
] | 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:
$\text{(A) } D = .PQQQ\ldots \qquad\\ \text{(B) } 10^rD = P.QQQ\ldots \\ \text{(C) } 10^{r + s}D = PQ.QQQ\ldots \qquad\\ \text{(D) } 10^r(10^s - 1)D = Q(P - 1) \\ \text{(E) } 10^r\cdot10^{2s}D = PQQ.QQQ\ldots$
|
[
"$\\fbox{D}$\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:
$\textbf{(A) \ }\text{the triangle is obtuse}$
$\textbf{(B) \ }\text{the angles are in the ratio }6:8:9$
$\textbf{(C) \ }\text{the triangle is acute}$
$\textbf{(D) \ }\text{the angle opposite the largest side is double the angle opposite the smallest side}$
$\textbf{(E) \ }\text{none of these}$
|
[
"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 $c<10$, $\\boxed{\\textbf{(C)}\\ \\text{the triangle is acute}}$.\n\n\n"
] | 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:
$\textbf{(A)}\ \frac {\sqrt {8d^2 - p^2}}{2} \qquad \textbf{(B)}\ \frac {\sqrt {8d^2 + p^2}}{2} \qquad \textbf{(C)}\ \frac{\sqrt{6d^2-p^2}}{2}\qquad\\ \textbf{(D)}\ \frac {\sqrt {6d^2 + p^2}}{2} \qquad \textbf{(E)}\ \frac {8d^2 - p^2}{4}$
|
[
"[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(\"$y$\",H); [/asy]\nLet the sides of the rectangle be x and y. WLOG, assume x>y. Then, $2x+2y=p \\Rightarrow x+y=\\frac{p}{2}$.\n\n\nBy pythagorean theorem, $x^2 + y^2 =d^2$.\n\n\nSince $x+y=\\frac{p}{2}$, $(x+y)^2=\\frac{p^2}{4} \\Rightarrow x^2+2xy+y^2=\\frac{p^2}{4}$.\n\n\nRearranging to solve for $2xy$ gives $2xy = \\frac{p^2}{4}-d^2$.\n\n\nRearranging $(x-y)^2$ in terms of the defined variables becomes $(x-y)^2 = d^2 - (\\frac{p^2}{4}-d^2)$. \n\n\nIn order to get (x-y), we have to take the square root of the expression and simplify. \n\n\n$(x-y)=\\sqrt{2d^2-\\frac{p^2}{4}} \\Rightarrow (x-y)=\\sqrt{\\frac{8d^2-p^2}{4}}$ $\\Rightarrow$ $(x-y)=\\frac{\\sqrt{8d^2+p^2}}{2}$ $\\Rightarrow$\n$\\fbox{A}$.\n\n\n"
] | 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:
$\textbf{(A) \ }\frac{AD}{EC}=\frac{AE}{DC} \qquad \textbf{(B) \ }\frac{AD}{EC}=\frac{AB}{BC} \qquad \textbf{(C) \ }\frac{AD}{EC}=\frac{BD}{BE} \qquad$
$\textbf{(D) \ }\frac{AD}{EC}=\frac{(AB)(BD)}{(BE)(BC)} \qquad \textbf{(E) \ }\frac{AD}{EC}=\frac{(AE)(BD)}{(DC)(BE)}$
|
[
"[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(anglemark(D,B,E)); draw(anglemark(E,B,C)); [/asy]\nUsing the Angle Bisector Theorem on $\\angle ABE$ and $\\angle DBC$, we find that $\\frac{AD}{DE}=\\frac{AB}{BE}$ and $\\frac{DE}{EC}=\\frac{BD}{BC}$. Rewriting this to better fit our answer choices gives $AD=\\frac{(DE)(AB)}{BE}$ and $EC=\\frac{(DE)(BC)}{BD}$. Hence, $\\frac{AD}{EC}=\\frac{(DE)(AB)}{BE}\\cdot \\frac{BD}{(DE)(BC)}\\implies \\boxed{\\textbf{(D)}\\ \\frac{AD}{EC}=\\frac{(AB)(BD)}{(BE)(BC)}}$.\n\n\n"
] | 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:
$\textbf{(A)}\ \frac {3 + \sqrt {6} + \sqrt {15}}{6} \qquad \textbf{(B)}\ \frac {\sqrt {6} - 2 + \sqrt {10}}{6} \qquad \textbf{(C)}\ \frac{2+\sqrt{6}+\sqrt{10}}{10} \qquad\\ \textbf{(D)}\ \frac {2 + \sqrt {6} - \sqrt {10}}{6} \qquad \textbf{(E)}\ \text{none of these}$
|
[
"Let $k=\\sqrt{2}+\\sqrt{3}$\nThen $\\frac{\\sqrt{2}}{k-\\sqrt{5}}\\implies \\frac{\\sqrt{2}(k+\\sqrt{5})}{k^2-\\sqrt{5}^2}\\implies\\frac{\\sqrt{2}k+\\sqrt{10}}{k^2-5}\\implies \\frac{\\sqrt{2}(\\sqrt{2}+\\sqrt{3})+\\sqrt{10}}{(\\sqrt{2}+\\sqrt{3})^2-5}\\implies \\frac{2+\\sqrt{6}+\\sqrt{10}}{2\\sqrt{6}}\\implies\\frac{2\\sqrt{6}+6+\\sqrt{60}}{12}\\implies \\frac{\\sqrt{6}+3+\\sqrt{15}}{6}\\fbox{A}$\n\n\n"
] | 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 $x^{13} - 1$ is divisible by $x-1$, meaning $(x^{13} - 1) + 2$ leaves a remainder of $\\boxed{\\mathrm{(D)}\\ 2.}$\n\n"
] | 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?
$\textbf{(A)}\ 125\% \qquad \textbf{(B)}\ 100\% \qquad \textbf{(C)}\ 120\% \qquad \textbf{(D)}\ 80\% \qquad \textbf{(E)}\ 75\%$
|
[
"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 $a(-\\frac{b}{2a})^2+b(-\\frac{b}{2a})+c=a(\\frac{b^2}{4a^2})-\\frac{b^2}{2a}+c=\\frac{b^2}{4a}-\\frac{2b^2}{4a}+c=-\\frac{b^2}{4a}+c=\\frac{4ac}{4a}-\\frac {b^2}{4a}=\\frac{4ac-b^2}{4a} \\Rightarrow \\mathrm{(D)}$.\n\n\n",
"Using calculus, we find that the derivative of the function is $2ax + b$, which has a critical point at $\\frac{-b}{2a}$. The second derivative of the function is $2a$; since $a > 0$, this critical point is a minimum. As in Solution 1, plug this value into the function to obtain $\\boxed{\\textbf{(D)}\\ \\dfrac{4ac-b^2}{4a}}$.\n\n\n~ cxsmi\n\n\n"
] | 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 0\\\\ (y+3)^2 -6y-18+10y+2&=0\\\\ y^2+6y+9-6y-18+10y+2&=0\\\\ y^2+10y-7&=0 \\end{align*}\n\n\nTherefore the answer is $\\boxed{\\textbf{(C)}\\ y^{2}+10y-7=0}$\n\n\n"
] | 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:
\[\begin{tabular}[t]{|c|c|c|c|c|c|}\hline x&0&1&2&3&4\\\hline y&100&90&70&40&0\\\hline\end{tabular}\]
$\textbf{(A)}\ y=100-10x\qquad\textbf{(B)}\ y=100-5x^{2}\qquad\textbf{(C)}\ y=100-5x-5x^{2}\qquad\\ \textbf{(D)}\ y=20-x-x^{2}\qquad\textbf{(E)}\ \text{None of these}$
|
[
"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 factored as $(x+1)(x-1)$. The $x-1$ terms in the numerator and denominator cancel, so the expression is equal to $x+1$ so long as $x$ does not equal $1$. Looking at the function's behavior near 1, we see that as $x$ approaches one, the expression approaches $\\boxed{\\textbf{(D)}\\ 2}$.\n\n\n\n\nNote: Alternatively, we can ignore the domain restriction and just plug in $x = 1$ into the reduced expression. \n\n\n~ cxsmi\n\n"
] | 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?
$\textbf{(A)}\ \text{If }x=1,y=0 \qquad\\ \textbf{(B)}\ \text{If }x=a,y=1 \qquad\\ \textbf{(C)}\ \text{If }x=-1,y\text{ is imaginary (complex)} \qquad\\ \textbf{(D)}\ \text{If }0<x<1,y\text{ is always less than 0 and decreases without limit as }x\text{ approaches zero} \qquad\\ \textbf{(E)}\ \text{Only some of the above statements are correct}$
|
[
"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$\\textbf{(C)}\\ \\text{If }x=-1,y\\text{ is imaginary (complex)}$. Rewriting gives $a^{\\text{complex number}}=-1$. Because $a>1$, therefore positive, there is no real solution to $y$, but there is imaginary.\n\n\n$\\textbf{(D)}\\ \\text{If }0<x<1,y\\text{ is always less than 0 and decreases without limit as }x\\text{ approaches zero}$. Rewriting: $a^y=x$ such that $x<a$. Well, a power of $a$ can be less than $a$ only if $y<1$. And we observe, $y$ has no lower asymptote, because it is perfectly possible to have $y$ be $-100000000$; in fact, the lower $y$ gets, $x$ approaches $0$. This is also correct.\n\n\n$\\textbf{(E)}\\ \\text{Only some of the above statements are correct}$. This is the last option, so it follows that our answer is $\\boxed{\\textbf{(E)}\\ \\text{Only some of the above statements are correct}}$\n\n\n"
] | 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:
$\textbf{(A)}\ 576\text{ in}^{3}\qquad\textbf{(B)}\ 24\text{ in}^{3}\qquad\textbf{(C)}\ 9\text{ in}^{3}\qquad\textbf{(D)}\ 104\text{ in}^{3}\qquad\textbf{(E)}\ \text{None of these}$
|
[
"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) &= (12)(8)(6)\\\\ a^2b^2c^2 &= 576\\\\ abc &= \\boxed{\\mathrm{(B)}\\ 24} \\end{align*}\n\n\n"
] | 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:
$\textbf{(A)}\ x=\dfrac{1}{2}h \qquad \textbf{(B)}\ x=\dfrac{bh}{b+h} \qquad \textbf{(C)}\ x=\dfrac{bh}{2h+b} \qquad \textbf{(D)}\ x=\sqrt{\dfrac{hb}{2}} \qquad \textbf{(E)}\ x=\dfrac{1}{2}b$
|
[
"Draw the triangle, and note that the small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle, so the proportions of the heights is equal to the proportions of the sides. In particular, we get $\\dfrac{2x}{b} = \\dfrac{h - x}{h} \\implies 2xh = bh - bx \\implies (2h + b)x = bh \\implies x = \\dfrac{bh}{2h + b}$. The answer is $\\boxed{\\textbf{(C)}}$.\n\n\n",
"You can also use area to solve this problem. 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. Therefore the smaller triangle (similar to the big one) has an area of $\\dfrac{2x(h-x)}{2}=x(h-x)$. Now note that because the rectangle is right, and the two other pieces are complementary triangles, we can add them together to create one triangle. Therefore the area of these two triangles is $\\dfrac{x(b-2x)}{2}$. This is due to the fact that we subtract $2x$ from the base of the triangle. Lastly the area of the rectangle is $2x^2$. These areas together sum to the area of the big rectangle, which is $\\dfrac{bh}{2}$. Solving we get that $x = \\dfrac{bh}{2h + b}$. The answer is $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 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",
"adding $\\log_{10} n$ to both sides:\n\\[\\log_{10} m + \\log_{10} n=b\\]\nusing the logarithm property: $\\log_a {b} + \\log_a {c}=\\log_a{bc}$\n\\[\\log_{10} {mn}=b\\]\nrewriting in exponential notation:\n\\[10^b=mn\\]\n\\[m=\\boxed{\\mathrm{(E) }\\dfrac{10^b}{n}}\\]\n~Vndom\n\n\n",
"More simply, we can just simulate the problem, if we have $m = 10$, that means the right side must be 1, so the only way we can achieve that with distinct $n$, is if $b = 3$, and $n = 100$. With this we can look through the different answer choices substituting in $b$, $n$, and $m$, and find that $\\boxed{\\mathrm{(E)}}$ is the only one that satisfies the question.\n\n\n~Shadow-18\n\n\n"
] | 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:
$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 43 \qquad \textbf{(C)}\ 29 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ \text{None of these}$
|
[
"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 $5(2g-30)=g+210$. Solving for $g$, we get $g = \\boxed{\\textbf{(A)}\\ 40}$.\n\n\n"
] | 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 of black socks to the number of pairs of blue socks in the original order was:
$\textbf{(A)}\ 4:1 \qquad \textbf{(B)}\ 2:1 \qquad \textbf{(C)}\ 1:4 \qquad \textbf{(D)}\ 1:2 \qquad \textbf{(E)}\ 1:8$
|
[
"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 $1.5(2(4)+1(b))=1(4)+2(b)$. Simplifying gives $b=16$. This means the ratio of the number of pairs of black socks and the number of pairs of blue socks is $\\boxed{\\textbf{(C)}\\ 1:4}$.\n\n\n"
] | 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:
$\textbf{(A)}\ 33\text{ mph}\qquad\textbf{(B)}\ 34\text{ mph}\qquad\textbf{(C)}\ 35\text{ mph}\qquad\textbf{(D)}\ 36\text{ mph}\qquad\textbf{(E)}\ 37\text{ mph}$
|
[
"The car takes $120 \\text{ miles }\\cdot\\dfrac{1 \\text{ hr }}{30 \\text{ miles }}=4 \\text{ hr}$ to get from $A$ to $B$. Also, it takes $120 \\text{ miles }\\cdot\\dfrac{1 \\text{ hr }}{40 \\text{ miles }}=3 \\text{ hr}$ to get from $B$ to $A$. Therefore, the average speed is $\\dfrac{240\\text{ miles }}{7 \\text{ hr}}=34\\dfrac{2}{7}\\text{ mph}$, which is closest to $\\boxed{\\textbf{(B)}\\ 34\\text{ mph}}$.\n\n\n"
] | 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{2}{3}\\]\nwhich is $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 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 merchantman makes $15$. The privateer will overtake the merchantman at:
$\textbf{(A)}\ 3\text{:}45\text{ p.m.} \qquad \textbf{(B)}\ 3\text{:}30\text{ p.m.} \qquad \textbf{(C)}\ 5\text{:}00\text{ p.m.} \qquad \textbf{(D)}\ 2\text{:}45\text{ p.m.} \qquad \textbf{(E)}\ 5\text{:}30\text{ p.m.}$
|
[
"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 merchantman is unaffected; he still travels at $8$ miles per hour. Therefore the privateer travels at $17\\cdot \\frac{8}{15}=\\frac{136}{15}=9\\frac{1}{15}$ miles per hour. The remaining time $t$ in hours it takes for the privateer to catch up to the merchantman satisfied the equation\n\n\n\\[26+8t=22+\\frac{136}{15}t\\]\n\n\nSimplification yields the equation $\\frac{16}{15}t=4$, which shows that $t=\\frac{15}{4}=3\\frac{3}{4}$. It therefore takes a total of $5\\frac{3}{4}$ hours for the privateer to catch the merchantman, so this will happen at $\\boxed{\\textbf{(E)}\\ 5\\text{:}30\\text{ p.m.}}$\n\n\n\n\n\n\n"
] | 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 $C$:
$\textbf{(A)}\ \text{Increases}\qquad\textbf{(B)}\ \text{Decreases}\qquad\textbf{(C)}\ \text{Remains constant}\qquad\textbf{(D)}\ \text{Increases and then decreases}\qquad\\ \textbf{(E)}\ \text{Decreases and then increases}$
|
[
"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:
$\textbf{(A)}\ \text{The area is doubled} \qquad\\ \textbf{(B)}\ \text{The altitude is doubled} \qquad\\ \textbf{(C)}\ \text{The area is four times the original area} \qquad\\ \textbf{(D)}\ \text{The median is unchanged} \qquad\\ \textbf{(E)}\ \text{The area of the triangle is 0}$
|
[
"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:
$\textbf{(A)}\ 4\text{ mph}\qquad \textbf{(B)}\ 8\text{ mph} \qquad \textbf{(C)}\ 12\text{ mph} \qquad \textbf{(D)}\ 16\text{ mph} \qquad \textbf{(E)}\ 20\text{ mph}$
|
[
"Let the speed of boy $A$ be $a$, and the speed of boy $B$ be $b$. Notice that $A$ travels $4$ miles per hour slower than boy $B$, so we can replace $b$ with $a+4$.\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 $d=rt$ to make an equation, where we set the time to be equal: \\[\\frac{48}{a}=\\frac{72}{a+4}\\]\nCross-multiplying gives $48a+192=72a$. Isolating the variable $a$, we get the equation $24a=192$, so $a=\\boxed{\\textbf{(B) }8 \\text{ mph}}$.\n\n\n",
"Note that $A$ travels $60-12=48$ miles in the time it takes $B$ to travel $60+12=72$ miles. Thus, $B$ travels $72-48=24$ more miles in the given time, meaning $\\frac{24\\text{miles}}{4\\text{miles}/\\text{hour}} = 6 \\text{hours}$ have passed, as $B$ goes $4$ miles per hour faster. Thus, $A$ travels $48$ miles per $6$ hours, or $8$ miles per hour.\n\n\n"
] | 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:
$\textbf{(A)}\ \text{Increases}\qquad\textbf{(B)}\ \text{Decreases}\qquad\textbf{(C)}\ \text{Remains constant}\qquad\textbf{(D)}\ \text{Cannot be predicted}\qquad\\ \textbf{(E)}\ \text{Becomes }(n-3)\text{ straight angles}$
|
[
"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 choose $100 - 3 = 97$ vertices for each vertex to draw the diagonal, as we cannot connect a vertex to itself or any of its two adjacent vertices. Thus, there are $(100)(97)/2=4850$ diagonals, because we are overcounting by a factor of $2$ (we are counting each diagonal twice - one for each endpoint). So, our answer is $\\fbox{A}$.\n\n\n",
"The formula for the number of diagonals of a polygon with $n$ sides is $n(n-3)/2$. Taking $n=100$, we see that the number of diagonals that may be drawn in this polygon is $100(97)/2$ or $\\boxed{\\textbf{(A)}\\ 4850 }$.\n\n\n~ cxsmi\n\n\n"
] | 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:
$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \textbf{(D)}\ 8\text{ ft} \qquad \textbf{(E)}\ 4\text{ ft}$
|
[
"By the Pythagorean triple $(7,24,25)$, the point where the ladder meets the wall is $24$ feet above the ground. When the ladder slides, it becomes $20$ feet above the ground. By the $(15,20,25)$ Pythagorean triple, The foot of the ladder is now $15$ feet from the building. Thus, it slides $15-7 = \\boxed{\\textbf{(D)}\\ 8\\text{ ft}}$.\n\n\n",
"We can observe that the above setup forms a right angled triangles whose base is 7ft and whose hypotenuse is 25ft taking the height to be x ft.\n\n\n\\[x^2 + 7^2 = 25^2\\]\n\\[x^2 = 625 - 49\\]\n\\[x^2 = 576\\]\n\\[x = 24\\]\n\n\nSince the top of the ladder slipped by 4 ft the new height is $24 - 4 = 20 ft$. The base of the ladder has moved so the new base is say $(7+y)$. The hypotenuse remains the same at 25ft. So,\n\n\n\\[20^2 + (7+y)^2 = 25^2\\]\n\\[400 + 49 + y^2 + 14y = 625\\]\n\\[y^2 + 14y - 176 = 0\\]\n\\[y^2 + 22y - 8y - 176\\]\n\\[x(y+22) - 8(y+22)\\]\n\\[(y-8)(y+22)\\]\n\n\nDisregarding the negative solution to equation the solution to the problem is $\\boxed{\\textbf{(D)}\\ 8\\text{ ft}}$.\n\n\n"
] | 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 extraneous roots.\n\n\n$6 + \\sqrt{6 - 2} = 6 + \\sqrt{4} = 6 + 2 = 8 \\ne 4$\n\n\n$3 + \\sqrt{3-2} = 3 + \\sqrt{1} = 3 + 1 = 4 \\checkmark$\n\n\nThere is $\\boxed{\\textbf{(E)} \\text{1 real root}}$\n\n",
"It's not hard to note that $x=3$ simply works, as $3 + \\sqrt{1} = 4$. But, $x$ is increasing, and $\\sqrt{x-2}$ is increasing, so $3$ is the only root. If $x < 3$, $x + \\sqrt{x-2} < 4$, and similarly if $x > 3$, then $x + \\sqrt{x-2} > 4$. Imaginary roots don't have to be considered because squaring the equation gives a quadratic, and quadratics either have only imaginary or only real roots.\n\n",
"We can create symmetry in the equation:\n\\[x+\\sqrt{x-2} = 4\\]\n\\[x-2+\\sqrt{x-2} = 2.\\]\nLet $y = \\sqrt{x-2}$, then we have\n\\[y^2+y-2 = 0\\]\n\\[(y+2)(y-1) = 0\\]\nThe two roots are $\\sqrt{x-2} = -2, 1$.\n\n\nNotice, that the first root is extraneous as the range for the square root function is always the non-negative numbers (remember, negative numbers in square roots give imaginary numbers - imaginary numbers in square roots don't give negative numbers); thus, the only real root for $x$ occurs for the second root; squaring both sides and solving for $x$ gives $x=3 \\Rightarrow \\boxed{\\textbf{(E)} \\text{1 real root}}$.\n\n\n~Vndom\n\n\n"
] | 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:
$\textbf{(A)}\ \text{A straight line }AB,1\dfrac{1}{2}\text{ inches from }A \qquad\\ \textbf{(B)}\ \text{A circle with }A\text{ as center and radius }2\text{ inches} \qquad\\ \textbf{(C)}\ \text{A circle with }A\text{ as center and radius }3\text{ inches} \qquad\\ \textbf{(D)}\ \text{A circle with radius }3\text{ inches and center }4\text{ inches from }B\text{ along } BA \qquad\\ \textbf{(E)}\ \text{An ellipse with }A\text{ as focus}$
|
[
"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: $\\textbf{(D)}$.\n\n\n\n\n\n\n",
"Let $A(a,y_A)$, $B(b,y_B)$ and $C(x,y)$.\n\n\nHence, $D\\left(\\frac{x+b}{2},\\frac{y+y_B}{2}\\right)$ is a midpoint of $BC$.\n\n\nThus, the equation of needed locus is\n\\[\\left(\\frac{x+b}{2}-a\\right)^2+\\left(\\frac{y+y_B}{2}-y_A\\right)^2=\\left(\\frac{3}{2}\\right)^2,\\]\nwhich is equation of the circle:\n\\[(x-(2a-b))^2+(y-(2y_A-y_B))^2=3^2.\\]\n\n\nThus, D) is valid because \\[\\sqrt{(2a-b-b)^2+(2y_A-y_B-y_B)^2}=2\\sqrt{(a-b)^2+(y_A-y_b)^2}=2AB=4.\\]\n\n\n\n\n\n\n"
] | 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:
$\textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}\qquad\\ \textbf{(B)}\ \text{Greater than the altitude of the triangle}\qquad\\ \textbf{(C)}\ \text{Equal to the altitude of the triangle}\qquad\\ \textbf{(D)}\ \text{One-half the sum of the sides of the triangle}\qquad\\ \textbf{(E)}\ \text{Greatest when the point is the center of gravity}$
|
[
"Begin by drawing the triangle, the point, the altitudes from the point to the sides, and the segments connecting the point to the vertices. Let the triangle be $ABC$ with $AB=BC=AC=s$. We will call the aforementioned point $P$. Call altitude from $P$ to $BC$ $PA'$. Similarly, we will name the other two altitudes $PB'$ and $PC'$. We can see that \n\\[\\frac{1}{2}sPA'+\\frac{1}{2}sPB'+\\frac{1}{2}sPC'=\\frac{1}{2}sh\\]\nWhere h is the altitude. Multiplying both sides by $2$ and dividing both sides by $s$ gives us\n\\[PA'+PB'+PC'=h\\]\nThe answer is $\\textbf{(C)}$\n\n\n"
] | 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",
"$\\log_{5}\\frac{(125)(625)}{25}$ can be also represented as $\\log_{5}\\frac{(5^3)(5^4)}{5^2}= \\log_{5}\\frac{(5^7)}{5^2}= \\log_{5} 5^5$ which can be solved to get $\\boxed{\\mathrm{(D)}\\ 5}$.\n\n\n"
] | 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 \\cdot 0.5 \\cdot 0.5 = \\frac{\\pi}{4}\\]\n\n\nSo number of 1 in pipes required is the number obtained by dividing their cross sectional areas\n\n\n\\[\\frac{9\\pi}{\\frac{\\pi}{4}} = 36\\]\n\n\nSo the answer is $\\boxed{\\textbf{(D)}\\ 36}$.\n\n\n"
] | 1
|
./CreativeMath/AHSME/1950_AHSME_Problems/33.json
|
AHSME
|
1950_AHSME_Problems
| 44
| 0
|
Algebra
|
Multiple Choice
|
The graph of $y=\log x$
$\textbf{(A)}\ \text{Cuts the }y\text{-axis} \qquad\\ \textbf{(B)}\ \text{Cuts all lines perpendicular to the }x\text{-axis} \qquad\\ \textbf{(C)}\ \text{Cuts the }x\text{-axis} \qquad\\ \textbf{(D)}\ \text{Cuts neither axis} \qquad\\ \textbf{(E)}\ \text{Cuts all circles whose center is at the origin}$
|
[
"The domain of $\\log x$ is the set of all $\\underline{positive}$ reals, so the graph of $y=\\log x$ clearly doesn't cut the $y$-axis. It therefore doesn't cut every line perpendicular to the $x$-axis. It does however cut the $x$-axis at $(1,0)$. In addition, if one examines the graph of $y=\\log x$, one can clearly see that there are many circles centered at the origin that do not intersect the graph of $y=\\log x$. Therefore the answer is $\\boxed{\\textbf{(C)}\\ \\text{Cuts the }x\\text{-axis}}$.\n\n\n"
] | 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 alone is:
$\textbf{(A)}\ 8-x=2 \qquad \textbf{(B)}\ \dfrac{1}{8}+\dfrac{1}{x}=\dfrac{1}{2} \qquad \textbf{(C)}\ \dfrac{500}{8}+\dfrac{500}{x}=500 \qquad \textbf{(D)}\ \dfrac{x}{2}+\dfrac{x}{8}=1 \qquad\\ \textbf{(E)}\ \text{None of these answers}$
|
[
"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 the second machine to address $500$ papers:\n\n\n$\\frac{375}{2}=\\frac{500}{x}$. Solving for $x$, we get $x=\\frac{8}{3}$ minutes.\nNow that we know the speed of the second machine, we can just plug it in each option to see if it equates. We see that $\\boxed{\\textbf{(B)}\\ \\dfrac{1}{8}+\\dfrac{1}{x}=\\dfrac{1}{2}}$ works.\n\n",
"First, notice that the number of envelopes addressed does not matter, because it stays constant throughout the problem. \nNext, notice that we are talking about combining two speeds, so we use the formula $\\frac{1}{\\frac{1}{a}+\\frac{1}{b}}=c$, where $a,b$ are the respective times independently, and $c$ is the combined time. Plugging in for $a$ and $c$, we get $\\frac{1}{\\frac{1}{8}+\\frac{1}{b}}=2$.\n\n\nTaking a reciprocal, we finally get $\\boxed{\\textbf{(B)}\\ \\dfrac{1}{8}+\\dfrac{1}{x}=\\dfrac{1}{2}}$.\n\n\n"
] | 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:
$\textbf{(A)}\ 5\text{ in} \qquad \textbf{(B)}\ 2\dfrac{1}{2}\text{ in} \qquad \textbf{(C)}\ \dfrac{5}{\pi}\text{ in} \qquad \textbf{(D)}\ \dfrac{5}{2\pi}\text{ in} \qquad \textbf{(E)}\ \dfrac{\pi}{5}\text{ in}$
|
[
"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 radius of the original balloon is $\\frac{20}{2\\pi}=\\frac{10}{\\pi}$. With the $25\\%$ increase, it becomes $\\frac{12.5}{\\pi}$. The increase is $\\frac{12.5-10}{\\pi}=\\frac{2.5}{\\pi}=\\boxed{\\textbf{(D)}\\ \\dfrac{5}{2\\pi}\\text{ in}}$.\n\n",
"Circumference of a circle is $2 \\pi r$ so the radius is $\\frac{circumference}{2 \\pi}$\n\n\nSo radius of first circle\n\n\n\\[2 \\pi r = 20\\]\n\\[r = \\frac{10}{\\pi}\\]\n\n\nRadius of second circle\n\\[2 \\pi r = 25\\]\n\\[r = \\frac{25}{2 \\pi}\\]\n\n\nThe difference of these radii is \n\n\n\\[\\frac{25}{2 \\pi} - \\frac{10}{\\pi} = \\frac{5}{2 \\pi}\\]\n\n\nSo the answer is $\\boxed{\\textbf{(D)}\\ \\dfrac{5}{2\\pi}\\text{ in}}$.\n\n",
"Let the radius of the circle with the larger circumference be $r_2$ and the circle with the smaller circumference be $r_1$. Calculating the ratio of the two\n\\[\\frac{r_2}{r_1}=\\frac{25}{20}=\\frac{5}{4}\\]\n\\[4r_2=5r_1\\]\n\\[4(r_2-r_1)=r_1\\]\n\\[r_2-r_1=\\frac{r_1}{4}=\\frac{\\frac{20}{2\\pi}}{4}=\\frac{10}{4\\pi}=\\boxed{\\textbf{(D)}\\ \\dfrac{5}{2\\pi}\\text{ in}}\\]\n\n\n"
] | 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 $\\boxed{\\mathrm{(C) }300\\%.}$\n\n\n"
] | 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 $10\\%$ discount, it's worth $(1-10\\%)x=0.9x$ dollars. After a $20\\%$ discount on that, it's worth $(1-20\\%)(0.9x)=0.72x$ dollars. Say the single discount is of $k$. Then $(1-k)x=0.72x$. So $k=0.28$, or $k=28\\%$. So select $\\boxed{D}$.\n\n\n~hastapasta\n\n\n"
] | 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) $a(xy)=ax \cdot ay$
$\textbf{(A)}\text{Only 1 and 4 are true}\qquad\\\textbf{(B)}\ \text{Only 1 and 5 are true}\qquad\\\textbf{(C)}\ \text{Only 1 and 3 are true}\qquad\\\textbf{(D)}\ \text{Only 1 and 2 are true}\qquad\\\textbf{(E)}\ \text{Only 1 is true}$
|
[
"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 $\begin{pmatrix}2x & 1\\ x & x\end{pmatrix}= 3$:
$\textbf{(A)}\ \text{Is satisfied for only 1 value of }x\qquad\\ \textbf{(B)}\ \text{Is satisified for only 2 values of }x\qquad\\ \textbf{(C)}\ \text{Is satisified for no values of }x\qquad\\ \textbf{(D)}\ \text{Is satisfied for an infinite number of values of }x\qquad\\ \textbf{(E)}\ \text{None of these.}$
|
[
"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 note that the equation is quadratic with a nonzero discriminant, so it will be satisfied for exactly two values of $x$.\n\n\n"
] | 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\nobs: Assume that $a \\not = 0, b\\not = 0, a\\not = b$.\n\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 multiple of the first except that the constants don't match up. So, there is no solution.\n\n\n"
] | 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 $\\frac{9}{49}(\\frac{1}{1-\\frac{1}{49}})=\\frac{9}{49}(\\frac{1}{\\frac{48}{49}})=\\frac{9}{49}(\\frac{49}{48})=\\frac{9}{48}=\\frac{3}{16} \\Rightarrow \\mathrm{(E)}$\n\n\n"
] | 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 means that $x^y=2$, considering that adding one $x$ to the start and then taking that $x$ to the power of $y$ does not have an effect on the equation, since $y$ is infinitely long in terms of $x$ raised to itself forever. It is already known that $y=2$ from what we first started with, so this shows that $x^y=x^2=2$. If $x^2=2$, then that means that $\\sqrt{2}=x \\Rightarrow \\mathrm{(D)}$.\n\n\nThis is just a faster way once you get used to it, instead of taking a log of the function.\n\n\n~mathmagical\n\n\n"
] | 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 $\\boxed{\\mathrm{(E)}\\text{ Non-existent.}}$\n\n",
"Let's try to find all real roots of $x^2+4=0$. If there was a real root (call it $r$), then it would satisfy $r^2+4=0$. Rearranging gives that $r^2=-4$. Therefore a real root $r$ of $x^2+4=0$ would satisfy $r^2<0$. However, the Trivial Inequality states that no real $r$ satisfies $r^2<0$, which must mean that there are no real roots $r$; they are $\\boxed{\\mathrm{(E)}\\text{ Non-existent.}}$\n\n",
"Let us try to factor $x^2+4.$ We can see that the first term has variables only and the second has a constant only. Therefore, we cannot factor out anything and our answer is $\\boxed{\\mathrm{(E)}\\text{ Non-existent.}}$\n\n\n"
] | 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 eight to find:\n$r^6=729$\n$r=\\pm 3$\n\n\nBecause $a_2$ will not be between $8$ and $5832$ if $r=-3$ we can discard it as an extraneous solution. We find $r=3$ and $a_5=a_1r^4=(8)(3)^4=\\boxed{\\textbf{(A)}\\ 648}$\n\n\n"
] | 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 small
\item (4) the difference between the sum and 4 can be made less than any positive quantity no matter how small
\item (5) the sum approaches a limit
\end{itemize}
Of these statments, the correct ones are:
$\textbf{(A)}\ \text{Only }3 \text{ and }4\qquad \textbf{(B)}\ \text{Only }5 \qquad \textbf{(C)}\ \text{Only }2\text{ and }4 \qquad \textbf{(D)}\ \text{Only }2,3\text{ and }4 \qquad \textbf{(E)}\ \text{Only }4\text{ and }5$
|
[
"This series is a geometric series with common ratio $\\frac{1}{2}$. Using the well-known formula for the sum of an infinite geometric series, we obtain that this series has a value of $2\\cdot \\frac{1}{1-\\frac{1}{2}}=4$. It immediately follows that statements 1 and 2 are false while statements 4 and 5 are true. 3 is false because we cannot make \\textbf{any} term of the sequence approach 0, though we can choose some term that is less than any given positive quantity. The correct answer is therefore $\\boxed{\\textbf{(E)}\\ \\text{Only }4\\text{ and }5}$.\n\n\n"
] | 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:
$\textbf{(A)} \ \ 64.82\qquad\textbf{(B)} \ \ 83.33\qquad\textbf{(C)} \ \ 72.08\qquad\textbf{(D)} \ \ 45.83\qquad\textbf{(E)} \ \ 177.08$
|
[
"$12\\frac{1}{2}\\%$ is the same as $\\frac{1}{8}$, so the man sets one eighth of each month's rent aside, so he only gains $\\frac{7}{8}$ of his rent. He also pays $325 each year, and he realizes $5.5\\%$, or $550, on his investment. Therefore he must have collected a total of $325 +$550 = $875 in rent. This was for the whole year, so he collected $\\frac{875}{12}$ dollars each month as rent. This is only $\\frac{7}{8}$ of the monthly rent, so the monthly rent in dollars is $\\frac{875}{12}\\cdot \\frac{8}{7}=\\boxed{\\textbf{(B)}\\ \\ 83.33}$.\n\n\n"
] | 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 $\\frac12 \\cdot 2r \\cdot r = \\boxed{\\mathrm{(A) }r^2}$.\n\n\n"
] | 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 have\n\\[\\frac{a+b-c}{2}=\\boxed{4}\\]\n\n\n- kante314\n\n\n",
"We know that the formula of in radius is area of triangle by its semiperimeter , hence it is a right triangle => area is 1/2×24×10=120 and semiperimeter is 60/2=30 => 120/30=4\nHK🗿\n\n\n"
] | 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~megaboy6679\n\n\n"
] | 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 $-1$ to get $p = \\frac{2CA - B^2}{A^2}$, making the answer $\\fbox{C}$.\n\n\n"
] | 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. Using the hexagon formula (or going back to the equilateral triangle formula $\\cdot 6$), we get \n\n\n$\\frac{6}{4} \\cdot s^2\\sqrt{3} \\Rightarrow$ $\\frac{6}{4} \\cdot (\\frac{20\\sqrt3}{3})^2 \\cdot \\sqrt3 \\Rightarrow$ $\\frac{6}{4} \\cdot \\frac{20\\sqrt3}{3} \\cdot \\frac{20\\sqrt3}{3} \\cdot \\sqrt3 \\Rightarrow$ $\\boxed{200\\sqrt3}$\n\n\n"
] | 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\\]\n\n\n\\[2x = -3\\]\n\n\n\\[x = -\\frac{3}{2} \\to \\boxed{\\text{(A)}}\\]\n\n\n"
] | 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:
$\textbf{(A)}\ {x}\text{ has no minimum or maximum value}\qquad \\ \textbf{(B)}\ \text{the maximum value of }{x}\text{ is }{1}\qquad \\ \textbf{(C)}\ \text{the minimum value of }{x}\text{ is }{1}\qquad \\ \textbf{(D)}\ \text{the maximum value of }{x}\text{ is }{4}\qquad \\ \textbf{(E)}\ \text{the minimum value of }{x}\text{ is }{4}$
|
[
"$\\fbox{}$\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); MP("O",(0,0),N);MP("C",(-10,0),W);MP("D",(10,0),E);;MP("E",(0,10),N); draw((-sqrt(70),-sqrt(30))--(sqrt(30),-sqrt(70))--(0,0)--cycle,dot); MP("A",(-sqrt(70),-sqrt(30)),SW);MP("B",(sqrt(30),-sqrt(70)),SE); [/asy]
$\textbf{(A)}\ \sqrt {2} : 1\qquad \textbf{(B)}\ \sqrt {3} : 1\qquad \textbf{(C)}\ 4 : 1\qquad \textbf{(D)}\ 3 : 1\qquad \textbf{(E)}\ 2 : 1$
|
[
"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)--cycle,dot); draw((-2,-1)--(2,1)--(2,-1)--(-2,1),dot); draw((-2,1)--(-6/5,-3/5),black+linewidth(.75)); draw((6/5,3/5)--(1,1)--(-3/2+1/10,-2/10),black+linewidth(.75)); draw((1,1)--(1-3/5,1-6/5),black+linewidth(.75)); MP("A",(-2,1),NW);MP("B",(2,1),NE);MP("C",(2,-1),SE);MP("D",(-2,-1),SW); MP("Q",(-3/2+1/10,-2/10),W);MP("T",(-2/5,1/5),N);MP("P",(1,1),N); MP("F",(-6/5,-3/5),SE);MP("E",(0,0),S);MP("S",(6/5,3/5),S);MP("R",(1-3/5,1-6/5),S); [/asy]
$\textbf{(A)}\ PQ\qquad \textbf{(B)}\ AE\qquad \textbf{(C)}\ PT + AT\qquad \textbf{(D)}\ AF\qquad \textbf{(E)}\ EF$
\usepackage{amssymb}
|
[
"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\n\nSince $\\angle APQ = \\angle CAB$, $\\triangle APT$ is isosceles with $PT = AT$.\n\n\n\n\n$\\angle ATQ$ and $\\angle PTR$ are vertical angles and are congruent. Thus, by HL congruence, $\\triangle ATQ \\cong \\triangle PTR$, so $AQ = PR$.\n\n\n\n\nWe also know that $PSFQ$ is a rectangle, since $\\overline{\\rm PS} \\perp \\overline{\\rm BD}$, $\\overline{\\rm BF} \\perp \\overline{\\rm AF}$, and $\\overline{\\rm PQ} \\perp \\overline{\\rm AF}$.\n\n\n\n\nSince $PSFQ$ is a rectangle, $QF = PS$. We also found earlier that $AQ = PR$. Thus, $PR + PS = AQ + QF = AF$.\n\n\n\n\nOur answer is $PR + PS = \\fbox{D) AF}$\n\n\n~ lovesummer\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:
$\textbf{(A)}\ s \plus{} 20n\qquad
\textbf{(B)}\ 5s \plus{} 80n\qquad
\textbf{(C)}\ s\qquad
\textbf{(D)}\ 5s\qquad
\textbf{(E)}\ 5s \plus{} 4n$ (Error compiling LaTeX. Unknown error_msg)
|
[
"$\\fbox{}$\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:
$\textbf{(A)}\ (a \plus{} 2b)^2\qquad
\textbf{(B)}\ a^2 \plus{} b^2\qquad
\textbf{(C)}\ b(ab \plus{} 2)\qquad
\textbf{(D)}\ ab(b \plus{} 2)\qquad
\textbf{(E)}\ \frac{1}{a} \plus{} 2b$ (Error compiling LaTeX. Unknown error_msg)
|
[
"$\\fbox{}$\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 Theorem, or by recognizing this $6-8-10$ triangle, we find that the other leg has length 6. So the triangle's total area is 48, and our answer is $\\fbox{E}$.\n\n\n"
] | 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 point $P'$ of $\overline{A'B'}$ to $D'$. Then for any pair of associated points, if $x = a,\, x + y$ equals:
[asy] draw((0,-3)--(0,3),black+linewidth(.75)); draw((1,-2.5)--(5,-2.5),black+linewidth(.75)); draw((3,2.5)--(4,2.5),black+linewidth(.75)); draw((1,-2.5)--(4,2.5),black+linewidth(.75)); draw((5,-2.5)--(3,2.5),black+linewidth(.75)); draw((2.6,-2.5)--(3.6,2.5),black+linewidth(.75)); dot((0,2.5));dot((1,2.5));dot((2,2.5));dot((3,2.5));dot((4,2.5));dot((5,2.5)); dot((0,-2.5));dot((1,-2.5));dot((2,-2.5));dot((3,-2.5));dot((4,-2.5));dot((5,-2.5)); MP("D",(0,2.5),NW);MP("A",(3,2.5),N);MP("P",(3.5,2.5),N);MP("B",(4,2.5),N); MP("D'",(0,-2.5),NW);MP("B'",(1,-2.5),NW);MP("P'",(2.25,-2.5),N);MP("A'",(5,-2.5),NE); MP("0",(0,2.5),SE);MP("1",(1,2.5),SE);MP("2",(2,2.5),SE);MP("3",(3,2.5),SE);MP("4",(4,2.5),SE);MP("5",(5,2.5),SE); MP("0",(0,-2.5),SE);MP("1",(1,-2.5),SE);MP("2",(2,-2.5),SE);MP("3",(3,-2.5),SE);MP("4",(4,-2.5),SE);MP("5",(5,-2.5),SE); [/asy]
$\textbf{(A)}\ 13a\qquad \textbf{(B)}\ 17a - 51\qquad \textbf{(C)}\ 17 - 3a\qquad \textbf{(D)}\ \frac {17 - 3a}{4}\qquad \textbf{(E)}\ 12a - 34$
|
[
"$\\fbox{}$\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:
$\textbf{(A)}\ \text{no maximum or minimum value}\qquad \\ \textbf{(B)}\ \text{a minimum value of }{+1}\qquad \\ \textbf{(C)}\ \text{a maximum value of }{+1}\qquad \\ \textbf{(D)}\ \text{a minimum value of }{-1}\qquad \\ \textbf{(E)}\ \text{a maximum value of }{-1}$
|
[
"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 get that $x$ is either $2$ or $0$. Substituting both of them in, you get that if $x=2$, then the value is $1$. If you plug in the value of $x=0$, you get the value of $-1$. So the answer is $\\textbf{(E)}$\n\n\nSolution by: the_referee\n\n\n"
] | 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:
$\textbf{(A)}\ \frac{1}{4}\qquad \textbf{(B)}\ \frac{81}{256}\qquad \textbf{(C)}\ \frac{27}{64}\qquad \textbf{(D)}\ \frac{37}{64}\qquad \textbf{(E)}\ \frac{175}{256}$
|
[
"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:
$\textbf{(A)}\ \frac{\log{\left(\frac{s}{P}\right)}}{\log{(1 + k)}}\qquad \textbf{(B)}\ \log{\left(\frac{s}{P(1 + k)}\right)}\qquad \textbf{(C)}\ \log{\left(\frac{s - P}{1 + k}\right)}\qquad \\ \textbf{(D)}\ \log{\left(\frac{s}{P}\right)} + \log{(1 + k)}\qquad \textbf{(E)}\ \frac{\log{(s)}}{\log{(P(1 + k))}}$
|
[
"\\[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:
$\textbf{(A)}\ {x}\text{ cannot exceed }{70}\qquad \\ \textbf{(B)}\ {y}\text{ can equal }{2x}\qquad\\ \textbf{(C)}\ \text{the amount of the check cannot be a multiple of }{5}\qquad \\ \textbf{(D)}\ \text{the incorrect amount can equal twice the correct amount}\qquad \\ \textbf{(E)}\ \text{the sum of the digits of the correct amount is divisible by }{9}$
|
[
"$\\fbox{}$\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:
$\textbf{(A)}\ \text{this problem has no solution}\qquad\\ \textbf{(B)}\ \text{there are two solutions with }{s}\text{ exceeding }{c}\qquad \\ \textbf{(C)}\ \text{there are two solutions with }{c}\text{ exceeding }{s}\qquad \\ \textbf{(D)}\ \text{there is one solution with }{s}\text{ exceeding }{c}\qquad \\ \textbf{(E)}\ \text{there is one solution with }{c}\text{ exceeding }{s}$
|
[
"$\\fbox{}$\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:
$\textbf{(A)}\ 75\qquad \textbf{(B)}\ 48\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 24\qquad\\ \textbf{(E)}\ \text{impossible to determine without knowing the value of }{m}$
|
[
"$\\fbox{}$\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 \\boxed{\\textbf{D}}\\]\n\n\n",
"Each term in the expansion of $(a+b+c)^{10}$ will have the form $a^i \\times b^j \\times c^k$, where $0\\le i, j, k\\le 10$ and $a+b+c=10$. So, we need to find the number of triplets of nonnegative integers $(a, b, c)$ such that $a+b+c=10$. Using Stars and Bars, this value is $\\binom{12}{2}=66$.\n\n\n"
] | 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 $D$:
$\textbf{(A)}\ \text{has the same length for all positions of }{P}\qquad\\ \textbf{(B)}\ \text{exceeds }{10}\text{ units for all positions of }{P}\qquad \\ \textbf{(C)}\ \text{cannot exceed }{10}\text{ units}\qquad \\ \textbf{(D)}\ \text{is shortest when }{\triangle CPD}\text{ is a right triangle}\qquad \\ \textbf{(E)}\ \text{is longest when }{P}\text{ is equidistant from }{C}\text{ and }{D}.$
|
[
"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. Triangle $APB$ is right, so $CP=\\sqrt{4*6}=\\sqrt{24}$. Then, $DP=\\sqrt{28}$, so the length we are looking for is $\\sqrt{24}+\\sqrt{28}>10$, eliminating $(D)$.\n\n\nThus, our answer is $(E)$.\n$\\fbox{}$\n\n\nNote: Say you are not convinced that $\\sqrt{24}+\\sqrt{28}>10$. We can prove this as follows.\n\n\nStart by simplifying the equation: $\\sqrt{6}+\\sqrt{7}>5$.\n\n\nSquare both sides: $6+2\\sqrt{42}+7>25$.\n\n\nSimplify: $\\sqrt{42}>6$\n\n\nSquare both sides again: $42>36$. From here, we can just reverse our steps to get $\\sqrt{24}+\\sqrt{28}>10$.\n\n\n"
] | 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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