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 |
|---|---|---|---|---|---|---|---|---|---|
1954_AHSME_Problems | 49 | 0 | Algebra | Multiple Choice | The difference of the squares of two odd numbers is always divisible by $8$. If $a>b$, and $2a+1$ and $2b+1$ are the odd numbers, to prove the given statement we put the difference of the squares in the form:
$\textbf{(A)}\ (2a+1)^2-(2b+1)^2\\ \textbf{(B)}\ 4a^2-4b^2+4a-4b\\ \textbf{(C)}\ 4[a(a+1)-b(b+1)]\\ \textbf{(... | [
"Although all of the forms listed can be used to show that the difference of $(2a+1)^2$ and $(2b+1)^2$ is necessarily divisible by $8$, we should use $\\boxed{\\textbf{(C)}}$ because at least one of the second and third factors are necessarily of different parities, so that their product is necessarily even and we ... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/49.json | AHSME |
1954_AHSME_Problems | 48 | 0 | Algebra | Multiple Choice | A train, an hour after starting, meets with an accident which detains it a half hour, after which it proceeds at $\frac{3}{4}$ of its former rate and arrives $3\tfrac{1}{2}$ hours late. Had the accident happened $90$ miles farther along the line, it would have arrived only $3$ hours late. The length of the trip in mile... | [
"Let the speed of the train be $x$ miles per hour, and let $D$ miles be the total distance of the trip, where $x$ and $D$ are unit-less quantities. Then for the trip that actually occurred, the train travelled 1 hour before the crash, and then travelled $D-x$ miles after the crash. In other words, the train travell... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/48.json | AHSME |
1954_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | The two roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are $1$ and:
$\textbf{(A)}\ \frac{b(c-a)}{a(b-c)}\qquad\textbf{(B)}\ \frac{a(b-c)}{c(a-b)}\qquad\textbf{(C)}\ \frac{a(b-c)}{b(c-a)}\qquad\textbf{(D)}\ \frac{c(a-b)}{a(b-c)}\qquad\textbf{(E)}\ \frac{c(a-b)}{b(c-a)}$
| [
"Let the other root be $k$. Then by Vieta's Formulas, $k\\cdot 1=\\frac{c(a-b)}{a(b-c)}$, $\\fbox{D}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/25.json | AHSME |
1954_AHSME_Problems | 33 | 0 | Arithmetic | Multiple Choice | A bank charges $\textdollar{6}$ for a loan of $\textdollar{120}$. The borrower receives $\textdollar{114}$ and repays the loan in $12$ easy installments of $\textdollar{10}$ a month. The interest rate is approximately:
$\textbf{(A)}\ 5 \% \qquad \textbf{(B)}\ 6 \% \qquad \textbf{(C)}\ 7 \% \qquad \textbf{(D)}\ 9\% \q... | [
"The borrower pays $\\textdollar{120}$ in a single year for a loan of $\\textdollar{114}$. This means that the bank charges an interest of $\\textdollar{6}$ for a loan of $\\textdollar{114}$ over a single year, so that the annual interest rate is $100*\\frac{\\textdollar{6}}{\\textdollar{114}} = \\frac{100}{19} \\a... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/33.json | AHSME |
1954_AHSME_Problems | 44 | 0 | Algebra | Multiple Choice | A man born in the first half of the nineteenth century was $x$ years old in the year $x^2$. He was born in:
$\textbf{(A)}\ 1849 \qquad \textbf{(B)}\ 1825 \qquad \textbf{(C)}\ 1812 \qquad \textbf{(D)}\ 1836 \qquad \textbf{(E)}\ 1806$
| [
"If a man born in the 19th century was $x$ years of in the year $x^2$, it implies that the year the man was born was $x^2-x$. So, if the man was born in the first half of the 19th century, it means that $x^2-x < 1850$. Noticing that $40^2 - 40 = 1560$ and $50^2-50 = 2450$, we see that $40 < x < 50$. We can guess va... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/44.json | AHSME |
1954_AHSME_Problems | 13 | 0 | Geometry | Multiple Choice | A quadrilateral is inscribed in a circle. If angles are inscribed in the four arcs cut off
by the sides of the quadrilateral, their sum will be:
$\textbf{(A)}\ 180^\circ \qquad \textbf{(B)}\ 540^\circ \qquad \textbf{(C)}\ 360^\circ \qquad \textbf{(D)}\ 450^\circ\qquad\textbf{(E)}\ 1080^\circ$
| [
"The arcs created by all sides of the quadrilateral will sum to 360 because they cover the entire circle. The inscribed angles created by each side of the trapezoid will all be half the measure of each arc created by their respective side. Thus, the sum of the all the inscribed angles is half the sum of all the arc... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/13.json | AHSME |
1954_AHSME_Problems | 29 | 0 | Geometry | Multiple Choice | If the ratio of the legs of a right triangle is $1: 2$, then the ratio of the corresponding segments of
the hypotenuse made by a perpendicular upon it from the vertex is:
$\textbf{(A)}\ 1: 4\qquad\textbf{(B)}\ 1:\sqrt{2}\qquad\textbf{(C)}\ 1: 2\qquad\textbf{(D)}\ 1:\sqrt{5}\qquad\textbf{(E)}\ 1: 5$
| [
"Let $\\triangle ABC$ be a right triangle with right angle at $B$, $AB = 2$, and $BC = 1$. Let the altitude from $B$ to $AC$ intersect $AC$ at $D$. From the Pythagorean Theorem, $AC = \\sqrt{1^2 + 2^2} = \\sqrt5$, so $BD = \\frac{1\\cdot 2}{\\sqrt5} = \\frac{2\\sqrt5}{5}$. Since $\\triangle ADB \\sim \\triangle BDC... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/29.json | AHSME |
1954_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | If $x$ varies as the cube of $y$, and $y$ varies as the fifth root of $z$, then $x$ varies as the nth power of $z$, where n is:
$\textbf{(A)}\ \frac{1}{15} \qquad\textbf{(B)}\ \frac{5}{3} \qquad\textbf{(C)}\ \frac{3}{5} \qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 8$
| [
"$x=k\\cdot y^3$, $y=j\\cdot z^{\\frac{1}{5}}\\implies y^3=j^3\\cdot z^{\\frac{3}{5}}$, so $x=k\\cdot j^3\\cdot z^{\\frac{3}{5}}$, $\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/3.json | AHSME |
1954_AHSME_Problems | 34 | 0 | Arithmetic | Multiple Choice | The fraction $\frac{1}{3}$:
$\textbf{(A)}\ \text{equals 0.33333333}\qquad\textbf{(B)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10^8}\\ \textbf{(C)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10^9}\\ \textbf{(D)}\ \text{is greater than 0.33333333 by }\frac{1}{3\cdot 10^8}\\ \textbf{(E)}\ \text{is gr... | [
"$\\frac{333333333}{10^8}-\\frac{1}{3}\\implies \\frac{3\\cdot 10^7+3\\cdot 10^6+\\dots+3\\cdot 10^0}{10^8}-\\frac{1}{3}\\implies\\frac{3(10^{8}-1)}{9\\cdot 10^{8}}-\\frac{1}{3}\\implies\\frac{10^8-1}{3\\cdot 10^8}-\\frac{1}{3}\\implies\\frac{10^8}{3\\cdot 10^8}-\\frac{1}{3\\cdot 10^8}-\\frac{1}{3}\\implies\\frac{-... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/34.json | AHSME |
1954_AHSME_Problems | 8 | 0 | Geometry | Multiple Choice | The base of a triangle is twice as long as a side of a square and their areas are the same.
Then the ratio of the altitude of the triangle to the side of the square is:
$\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 4$
| [
"Let the base and altitude of the triangle be $b, h$, the common area $A$, and the side of the square $s$. Then $\\frac{2sh}{2}=s^2\\implies sh=s^2\\implies s=h$, so $\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/8.json | AHSME |
1954_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | The expression $\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}$ cannot be evaluated for $x=-1$ or $x=2$,
since division by zero is not allowed. For other values of $x$:
$\textbf{(A)}\ \text{The expression takes on many different values.}\\ \textbf{(B)}\ \text{The expression has only the value 2.}\\ \textbf{(C)}\ ... | [
"$\\frac{2x^2-x}{(x+1)(x-2)}-\\frac{4+x}{(x+1)(x-2)} = \\frac{2x^2-2x-4}{(x+1)(x-2)}$\n\n\nThis can be factored as $\\frac{(2)(x^2-x-2)}{(x+1)(x-2)} \\implies \\frac{(2)(x+1)(x-2)}{(x+1)(x-2)}$, which cancels out to $2 \\implies \\boxed{\\textbf{(B)} \\text{The expression has only the value 2}}$.\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/22.json | AHSME |
1954_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | Of the following sets, the one that includes all values of $x$ which will satisfy $2x - 3 > 7 - x$ is:
$\textbf{(A)}\ x > 4 \qquad \textbf{(B)}\ x < \frac {10}{3} \qquad \textbf{(C)}\ x = \frac {10}{3} \qquad \textbf{(D)}\ x >\frac{10}{3}\qquad\textbf{(E)}\ x < 0$
| [
"$2x-3>7-x\\implies 3x>10\\implies \\boxed{x>\\frac{10}{3} (\\textbf{D})}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/18.json | AHSME |
1954_AHSME_Problems | 38 | 0 | Algebra | Multiple Choice | If $\log 2 = .3010$ and $\log 3 = .4771$, the value of $x$ when $3^{x+3} = 135$ is approximately
$\textbf{(A) \ } 5 \qquad \textbf{(B) \ } 1.47 \qquad \textbf{(C) \ } 1.67 \qquad \textbf{(D) \ } 1.78 \qquad \textbf{(E) \ } 1.63$
| [
"Taking the logarithm in base $3$ of both sides, we get $x+3 = \\log_3 135$. Using the property $\\log ab = \\log a + \\log b$, we get $x+3 = \\log_3 5 + \\log_3 3^3$, or $x = \\log_3 5$. Converting into base $10$ gives $x = \\frac{\\log 5}{\\log 3} = \\frac{1 - \\log 2}{\\log 3}$. Now, plugging in the values yeild... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/38.json | AHSME |
1954_AHSME_Problems | 4 | 0 | Number Theory | Multiple Choice | If the Highest Common Divisor of $6432$ and $132$ is diminished by $8$, it will equal:
$\textbf{(A)}\ -6 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ -2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
| [
"$\\gcd(6432, 132)$\n$13\\cdot 12=132$\n$\\frac{6432}{6}=1072\\implies\\frac{1072}{4}=268\\implies\\frac{268}{4}=67$, so $\\gcd(2^5\\cdot 3\\cdot 67, 2^2\\cdot 3\\cdot 13)=2^2\\cdot 3=12\\implies 12-8=4, \\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/4.json | AHSME |
1954_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | When simplified $\sqrt{1+ \left (\frac{x^4-1}{2x^2} \right )^2}$ equals:
$\textbf{(A)}\ \frac{x^4+2x^2-1}{2x^2} \qquad \textbf{(B)}\ \frac{x^4-1}{2x^2} \qquad \textbf{(C)}\ \frac{\sqrt{x^2+1}}{2}\\ \textbf{(D)}\ \frac{x^2}{\sqrt{2}}\qquad\textbf{(E)}\ \frac{x^2}{2}+\frac{1}{2x^2}$
| [
"$\\sqrt{\\frac{4x^4}{4x^4}+\\frac{(x^4-1)^2}{4x^4}}\\implies\\sqrt{\\frac{x^8-2x^4+1+4x^4}{4x^4}}\\implies \\sqrt{\\frac{(x^4+1)^2}{(2x^2)^2}}\\implies \\frac{x^4+1}{2x^2}\\implies\\frac{x^2}{2}+\\frac{1}{2x^2}$, $\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/14.json | AHSME |
1954_AHSME_Problems | 43 | 0 | Geometry | Multiple Choice | The hypotenuse of a right triangle is $10$ inches and the radius of the inscribed circle is $1$ inch. The perimeter of the triangle in inches is:
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 30$
| [
"To begin, let's notice that the inscribed circle of the right triangle is its incircle, and that the radius of the incircle is the right triangle's inradius. In this case, the hypotenuse is 10, and the inradius is 1. The formula for the inradius of a right triangle is $r = (a+b-c)/2$, where $r$ is the length of th... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/43.json | AHSME |
1954_AHSME_Problems | 42 | 0 | Algebra | Multiple Choice | Consider the graphs of
\[(1)\qquad y=x^2-\frac{1}{2}x+2\]
and
\[(2)\qquad y=x^2+\frac{1}{2}x+2\]
on the same set of axis.
These parabolas are exactly the same shape. Then:
$\textbf{(A)}\ \text{the graphs coincide.}\\ \textbf{(B)}\ \text{the graph of (1) is lower than the graph of (2).}\\ \textbf{(C)}\ \text{the... | [
"Let us consider the vertices of the two parabolas. The x-coordinate of parabola 1 is given by $\\frac{\\frac{1}{2}}{2} = \\frac{1}{4}$.\n\n\nSimilarly, the x-coordinate of parabola 2 is given by $\\frac{\\frac{-1}{2}}{2} = -\\frac{1}{4}$.\n\n\nFrom this information, we can deduce that $\\textbf{(D)}\\ \\text{the g... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/42.json | AHSME |
1954_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | $\log 125$ equals:
$\textbf{(A)}\ 100 \log 1.25 \qquad \textbf{(B)}\ 5 \log 3 \qquad \textbf{(C)}\ 3 \log 25 \\ \textbf{(D)}\ 3 - 3\log 2 \qquad \textbf{(E)}\ (\log 25)(\log 5)$
| [
"$\\log(1000)=\\log(125)+\\log(8)\\implies 3=\\log(125)+\\log(2^3)\\implies \\log(125)=3-3\\log(2)$, $\\fbox{D}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/15.json | AHSME |
1954_AHSME_Problems | 5 | 0 | Geometry | Multiple Choice | A regular hexagon is inscribed in a circle of radius $10$ inches. Its area is:
$\textbf{(A)}\ 150\sqrt{3} \text{ sq. in.} \qquad \textbf{(B)}\ \text{150 sq. in.} \qquad \textbf{(C)}\ 25\sqrt{3}\text{ sq. in.}\qquad\textbf{(D)}\ \text{600 sq. in.}\qquad\textbf{(E)}\ 300\sqrt{3}\text{ sq. in.}$
| [
"Using the formula for the area of a hexagon given the circumradius: $\\frac{3\\sqrt{3}}{2\\pi}=\\frac{A}{10^2\\pi}\\implies 100\\cdot3\\sqrt{3}\\pi=2A\\pi\\implies 50\\cdot3\\sqrt{3}=A\\implies 150\\sqrt{3}\\ \\boxed{(\\textbf{A})}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/5.json | AHSME |
1954_AHSME_Problems | 39 | 0 | Geometry | Multiple Choice | The locus of the midpoint of a line segment that is drawn from a given external point $P$ to a given circle with center $O$ and radius $r$, is:
$\textbf{(A)}\ \text{a straight line perpendicular to }\overline{PO}\\ \textbf{(B)}\ \text{a straight line parallel to }\overline{PO}\\ \textbf{(C)}\ \text{a circle with cent... | [
"Note that the midpoint of $P$ to the point $Q$ is the image of $Q$ under a homothety of factor $\\frac{1}{2}$ with center $P$. Since homotheties preserve circles, the image of the midpoint as $Q$ varies over the circle is a circle centered at the midpoint of $P$ and the original center and radius half the original... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/39.json | AHSME |
1954_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | If the three points of contact of a circle inscribed in a triangle are joined, the angles of the resulting triangle:
$\textbf{(A)}\ \text{are always equal to }60^\circ\\ \textbf{(B)}\ \text{are always one obtuse angle and two unequal acute angles}\\ \textbf{(C)}\ \text{are always one obtuse angle and two equal acute... | [
"For the sake of clarity, let the outermost triangle be $ABC$ with incircle tangency points $D$, $E$, and $F$ on $BC$, $AC$ and $AB$ respectively. Let $\\angle A=\\alpha$, and $\\angle B=\\beta$. Because $\\triangle AFE$ and $\\triangle BDF$ are isosceles, $\\angle AFE=\\frac{180-\\alpha}{2}$ and $\\angle BFD=\\fra... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/19.json | AHSME |
1954_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | If the margin made on an article costing $C$ dollars and selling for $S$ dollars is $M=\frac{1}{n}C$, then the margin is given by:
$\textbf{(A)}\ M=\frac{1}{n-1}S\qquad\textbf{(B)}\ M=\frac{1}{n}S\qquad\textbf{(C)}\ M=\frac{n}{n+1}S\\ \textbf{(D)}\ M=\frac{1}{n+1}S\qquad\textbf{(E)}\ M=\frac{n}{n-1}S$
| [
"We are given the margin in terms of the cost of the article. Looking at the answers, it appears we need to find the margin in terms of the selling price. The relationship between cost and selling price is that selling price minus the margin is the cost, $S-M=C$.\n\n\nSince $M=\\frac{1}{n}C$ and $C=S-M$, \n$M=\\fra... | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/23.json | AHSME |
1954_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | A point $P$ is outside a circle and is $13$ inches from the center. A secant from $P$ cuts the circle at $Q$ and $R$
so that the external segment of the secant $PQ$ is $9$ inches and $QR$ is $7$ inches. The radius of the circle is:
$\textbf{(A)}\ 3" \qquad \textbf{(B)}\ 4" \qquad \textbf{(C)}\ 5" \qquad \textbf{(D)... | [
"Using the Secant-Secant Power Theorem, you can get $9(16)=(13-r)(13+r)$, where $r$ is the radius of the given circle. Solving the equation, you get a quadratic: $r^2-25$. A radius cannot be negative so the answer is $\\boxed{\\textbf{(C) }5\"}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1954_AHSME_Problems/9.json | AHSME |
1954_AHSME_Problems | 35 | 0 | Algebra | Multiple Choice | In the right triangle shown the sum of the distances $BM$ and $MA$ is equal to the sum of the distances $BC$ and $CA$.
If $MB = x, CB = h$, and $CA = d$, then $x$ equals:
[asy] defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; draw((0,0)--(8,0)--(0,5)--cycle); label("C",(0,0),SW); label("A",(8,0),SE); label(... | [
"The question states that \\[h+d = x+\\sqrt{(x+h)^2+d^2}\\]\n\n\nWe move $x$ to the left: \\[h+d-x = \\sqrt{(x+h)^2+d^2}\\]\n\n\nWe square both sides: \\[h^2 + d^2 + x^2 - 2xh - 2xd + 2hd = x^2 + 2xh + h^2 + d^2\\]\n\n\nCancelling and moving terms, we get: \\[4xh + 2xd = 2hd\\]\n\n\nFactoring $x$: \\[x(4h+2d) = 2hd... | 2 | ./CreativeMath/AHSME/1954_AHSME_Problems/35.json | AHSME |
1955_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | The expression $\sqrt{25-t^2}+5$ equals zero for:
$\textbf{(A)}\ \text{no real or imaginary values of }t\qquad\textbf{(B)}\ \text{no real values of }t\text{ only}\\ \textbf{(C)}\ \text{no imaginary values of }t\text{ only}\qquad\textbf{(D)}\ t=0\qquad\textbf{(E)}\ t=\pm 5$
| [
"We can make the following equation: $\\sqrt{25-t^2}+5 = 0$, which means that $\\sqrt{25-t^2} = -5$ Square both sides, and we get $25 - t^2 = 25$, which would mean that $t = 0$, but that is an extraneous solution and doesn't work in the original equation.\n\n\nTherefore, the correct answer is $\\textbf{(A)}$\n\n\n... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/20.json | AHSME |
1955_AHSME_Problems | 36 | 0 | Geometry | Multiple Choice | A cylindrical oil tank, lying horizontally, has an interior length of $10$ feet and an interior diameter of $6$ feet.
If the rectangular surface of the oil has an area of $40$ square feet, the depth of the oil is:
$\textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ 2\sqrt{5}\qquad\textbf{(C)}\ 3-\sqrt{5}\qquad\textbf{(D)}\ ... | [
"In order to complete the rectangle area of the oil tank, the chord on the circle must have a length of $4$. Since it's not the diameter $6$, there are two possible outcomes. The only choice that reflects this is $\\boxed{\\textbf{(E)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/36.json | AHSME |
1955_AHSME_Problems | 16 | 0 | Arithmetic | Multiple Choice | The value of $\frac{3}{a+b}$ when $a=4$ and $b=-4$ is:
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{3}{8}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \text{any finite number}\qquad\textbf{(E)}\ \text{meaningless}$
| [
"As the denominator is $4 - (-4) = 0$, the value of the fraction $\\frac{3}{4 - (-4)}$ is $\\textbf{(E)}\\ \\text{meaningless}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/16.json | AHSME |
1955_AHSME_Problems | 6 | 0 | Arithmetic | Multiple Choice | A merchant buys a number of oranges at $3$ for $10$ cents and an equal number at $5$ for $20$ cents. To "break even" he must sell all at:
$\textbf{(A)}\ \text{8 for 30 cents}\qquad\textbf{(B)}\ \text{3 for 11 cents}\qquad\textbf{(C)}\ \text{5 for 18 cents}\\ \textbf{(D)}\ \text{11 for 40 cents}\qquad\textbf{(E)}\ \te... | [
"Since we are buying at $3$ for $10$ cents and $5$ for $20$ cents, let's assume that together, we are buying 15 oranges.\nThat means that we are getting a total of $30$ oranges for $(10\\times5) + (20\\times3)$ cents.\nThat comes to a total of $30$ oranges for $110$ cents. $110/30$ = $11/3$. This leads us to $3$ fo... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/6.json | AHSME |
1955_AHSME_Problems | 7 | 0 | Arithmetic | Multiple Choice | If a worker receives a $20$% cut in wages, he may regain his original pay exactly by obtaining a raise of:
$\textbf{(A)}\ \text{20\%}\qquad\textbf{(B)}\ \text{25\%}\qquad\textbf{(C)}\ 22\frac{1}{2}\text{\%}\qquad\textbf{(D)}\ \textdollar{20}\qquad\textbf{(E)}\ \textdollar{25}$
| [
"Since the worker receives a $20$% cut in wages, his present wage would be $\\frac{4}{5}$ of the original wage before the cut. To regain his original pay he would have to obtain a raise of $1$ $\\div$ $\\frac{4}{5}$ = $1$$\\frac{1}{4}$.\n\n\nTherefore, the worker would have to get a $\\fbox{{\\bf(B)} 25\\%}$ raise ... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/7.json | AHSME |
1955_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | If $\log x-5 \log 3=-2$, then $x$ equals:
$\textbf{(A)}\ 1.25\qquad\textbf{(B)}\ 0.81\qquad\textbf{(C)}\ 2.43\qquad\textbf{(D)}\ 0.8\qquad\textbf{(E)}\ \text{either 0.8 or 1.25}$
\subsection{Contents}
\begin{itemize}
| [
"\\subsubsection{Definitions and Properties}\n$\\log n$ is defined as the value of $x$ that satisfies the equation $n = 10^x$. Note that other bases can be applied as well, so $\\log_b n$ would be defined as the answer to $n = b^x$\n\n\n$y \\log n = \\log n^y$\n\n\n$\\log x - \\log y = \\log (x / y)$\n\n\nLearn mor... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/17.json | AHSME |
1955_AHSME_Problems | 40 | 0 | Algebra | Multiple Choice | The fractions $\frac{ax+b}{cx+d}$ and $\frac{b}{d}$ are unequal if:
$\textbf{(A)}\ a=c=1, x\neq 0\qquad\textbf{(B)}\ a=b=0\qquad\textbf{(C)}\ a=c=0\\ \textbf{(D)}\ x=0\qquad\textbf{(E)}\ ad=bc$
| [
"We can implement each of these options on the existing two fractions:\n$\\textbf{(A)}$: $\\frac{x+b}{x+d}$ vs $\\frac{b}{d}$. Since $x \\neq 0$, we can see that any remaining value of $x$ will make the two unequal. The answer is $\\boxed{\\textbf{(A)}}$\n\n\n$\\textbf{(B)}$: $\\frac{0}{d}$ vs $\\frac{0}{d}$ Both f... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/40.json | AHSME |
1955_AHSME_Problems | 37 | 0 | Arithmetic | Multiple Choice | A three-digit number has, from left to right, the digits $h, t$, and $u$, with $h>u$. When the number with the digits reversed is subtracted from the original number, the units' digit in the difference is 4. The next two digits, from right to left, are:
$\textbf{(A)}\ \text{5 and 9}\qquad\textbf{(B)}\ \text{9 and 5}\... | [
"We can set up the subtraction like this:\n\\[\\text{ h t u}\\]\n\\[- \\text{u t h}\\]\n\\[-------\\]\n\\[\\text{? ? 4}\\]\nSince $u < h$, we need to borrow the one from the tens column. Since the result of the tens column is 0, the taken 1 would result in the tens digit being 9:\n\n\n\\[\\text{ h t u}\\]\n\\[- \\t... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/37.json | AHSME |
1955_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | Represent the hypotenuse of a right triangle by $c$ and the area by $A$. The altitude on the hypotenuse is:
$\textbf{(A)}\ \frac{A}{c}\qquad\textbf{(B)}\ \frac{2A}{c}\qquad\textbf{(C)}\ \frac{A}{2c}\qquad\textbf{(D)}\ \frac{A^2}{c}\qquad\textbf{(E)}\ \frac{A}{c^2}$
| [
"[asy] draw((0,0) -- (9/5,12/5) -- (5,0) -- cycle); draw((9/5,12/5) -- (9/5,0)); [/asy]\nGiven that the area of the triangle is $A$, and the formula for the area of a triangle is $\\frac{bh}{2}=A$, we can replace $b$ (the base) and $h$ (the height) with $c$ (the hypotenuse) and $a$ (the altitude), we can rearrrange... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/21.json | AHSME |
1955_AHSME_Problems | 47 | 0 | Algebra | Multiple Choice | The expressions $a+bc$ and $(a+b)(a+c)$ are:
$\textbf{(A)}\ \text{always equal}\qquad\textbf{(B)}\ \text{never equal}\qquad\textbf{(C)}\ \text{equal whenever }a+b+c=1\\ \textbf{(D)}\ \text{equal when }a+b+c=0\qquad\textbf{(E)}\ \text{equal only when }a=b=c=0$
| [
"Using the FOIL method, we see that $(a+b)(a+c) = a^2 + ab + ac + bc.$ We want to solve \n\\[a + bc = a^2 + ab + ac + bc\\]\n\\[a = a^2 + ab + ac\\]\n\\[a((a + b + c) - 1) = 0\\]\nThese expressions are $\\boxed{\\text{(C) equal whenever } a + b + c = 1.}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/47.json | AHSME |
1955_AHSME_Problems | 10 | 0 | Arithmetic | Multiple Choice | How many hours does it take a train traveling at an average rate of 40 mph between stops to travel a miles it makes n stops of m minutes each?
$\textbf{(A)}\ \frac{3a+2mn}{120}\qquad\textbf{(B)}\ 3a+2mn\qquad\textbf{(C)}\ \frac{3a+2mn}{12}\qquad\textbf{(D)}\ \frac{a+mn}{40}\qquad\textbf{(E)}\ \frac{a+40mn}{40}$
| [
"The train will take $\\frac{a}{40}$ hours to travel $a$ miles, and it takes $\\frac{nm}{60}$. The LCM of $40$ and $60$ is $120$, which allows for the addition of the fractions $\\frac{3a}{120}$ and $\\frac{2mn}{120}$. The end result is $\\boxed{\\textbf{(A)}\\frac{3a+2mn}{120}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/10.json | AHSME |
1955_AHSME_Problems | 26 | 0 | Arithmetic | Multiple Choice | Mr. $A$ owns a house worth $\textdollar{10000}$. He sells it to Mr. $B$ at $10$% profit. Mr. $B$ sells the house back to Mr. $A$ at a $10$% loss. Then:
$\textbf{(A)}\ \text{Mr. A comes out even}\qquad\textbf{(B)}\ \text{Mr. A makes } \textdollar{ 100}\qquad\textbf{(C)}\ \text{Mr. A makes } \textdollar{ 1000}\\ \textb... | [
"Assume that Mr. $A$ has $\\textdollar{10000}$ to start with (the house) and Mr. $B$ has $\\textdollar{11000}$\n\n\n$A$ sells the house to $B$ for $\\textdollar{11000}$, leaving Mr. $A$ with $\\textdollar{11000}$ and Mr. $B$ with $\\textdollar{10000}$. \n\n\nThen, Mr. $B$ sells the house back to Mr. $A$ for $\\text... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/26.json | AHSME |
1955_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | Each of the equations $3x^2-2=25, (2x-1)^2=(x-1)^2, \sqrt{x^2-7}=\sqrt{x-1}$ has:
$\textbf{(A)}\ \text{two integral roots}\qquad\textbf{(B)}\ \text{no root greater than 3}\qquad\textbf{(C)}\ \text{no root zero}\\ \textbf{(D)}\ \text{only one root}\qquad\textbf{(E)}\ \text{one negative root and one positive root}$
... | [
"Since the question asks us about the unifying characteristic of all three equations' roots, we have to first determine them.\n\n\n$3x^2-2 = 25$ can be rewritten as $3x^2 - 27 = 0$, which gives the following roots $+3$ and $-3$.\n\n\n$(2x-1)^2 = (x-1)^2$ can be expanded to $4x^2-4x+1=x^2-2x+1$, which in turn leads ... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/30.json | AHSME |
1955_AHSME_Problems | 31 | 0 | Geometry | Multiple Choice | An equilateral triangle whose side is $2$ is divided into a triangle and a trapezoid by a line drawn parallel to one of its sides. If the area of the trapezoid equals one-half of the area of the original triangle, the length of the median of the trapezoid is:
$\textbf{(A)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(B)}\ \sqrt... | [
"The area of the large equilateral triangle is $2^2*\\frac{\\sqrt{3}}{4}=\\sqrt{3}$ square units, so the smaller triangle is $\\frac{\\sqrt{3}}{2}$. The side length of the smaller length $s$ is $\\sqrt{\\frac{4}{\\sqrt{3}}*\\frac{\\sqrt{3}}{2}} = \\sqrt{2}$. The median of the trapezoid can be determined by the aver... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/31.json | AHSME |
1955_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | If $r$ and $s$ are the roots of $x^2-px+q=0$, then $r^2+s^2$ equals:
$\textbf{(A)}\ p^2+2q\qquad\textbf{(B)}\ p^2-2q\qquad\textbf{(C)}\ p^2+q^2\qquad\textbf{(D)}\ p^2-q^2\qquad\textbf{(E)}\ p^2$
| [
"We can write $r^2+s^2$ in terms of the sum of the roots $(r+s)$ and the products of the roots $(rs):$\n\\[r^2 + s^2 = (r+s)^2 - 2rs = p^2 - 2q\\]\nThe answer is $\\boxed{\\textbf{(B)}}.$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/27.json | AHSME |
1955_AHSME_Problems | 1 | 0 | Arithmetic | Multiple Choice | Which one of the following is not equivalent to $0.000000375$?
$\textbf{(A)}\ 3.75\times 10^{-7}\qquad\textbf{(B)}\ 3\frac{3}{4}\times 10^{-7}\qquad\textbf{(C)}\ 375\times 10^{-9}\qquad \textbf{(D)}\ \frac{3}{8}\times 10^{-7}\qquad\textbf{(E)}\ \frac{3}{80000000}$
| [
"First of all, $0.000000375 = 3.75 \\times 10^{-7}$ in scientific notation. This eliminates $\\textbf{(A)}\\ 3.75\\times 10^{-7}$, $\\textbf{(B)}\\ 3\\frac{3}{4}\\times 10^{-7}$, and $\\textbf{(C)}\\ 375\\times 10^{-9}$ immediately. $\\textbf{(E)}\\ \\frac{3}{80000000}$ is a bit harder, but it can be rewritten as $... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/1.json | AHSME |
1955_AHSME_Problems | 50 | 0 | Algebra | Multiple Choice | In order to pass $B$ going $40$ mph on a two-lane highway, $A$, going $50$ mph, must gain $30$ feet.
Meantime, $C, 210$ feet from $A$, is headed toward him at $50$ mph. If $B$ and $C$ maintain their speeds,
then, in order to pass safely, $A$ must increase his speed by:
$\textbf{(A)}\ \text{30 mph}\qquad\textbf{(B)... | [
"Let $V_A, V_B, V_C$ be $A, B, C$'s velocity, respectively. We want to pass $B$ before we collide with $C$. Since $A$ and $B$ are going in the same direction and $V_A>V_B$, $A$ will pass $B$ in $\\frac{30\\mathrm{ft}}{V_A-V_B}$ time. Since $A$ and $C$ are going in opposite directions, their relative velocity is $V_... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/50.json | AHSME |
1955_AHSME_Problems | 11 | 0 | Other | Multiple Choice | The negation of the statement "No slow learners attend this school" is:
$\textbf{(A)}\ \text{All slow learners attend this school}\\ \textbf{(B)}\ \text{All slow learners do not attend this school}\\ \textbf{(C)}\ \text{Some slow learners attend this school}\\ \textbf{(D)}\ \text{Some slow learners do not attend thi... | [
"The negation of the statement is basically the opposite of the statement, which is $\\boxed{\\textbf{(C) some slow learners attend this school.}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/11.json | AHSME |
1955_AHSME_Problems | 46 | 0 | Geometry | Multiple Choice | The graphs of $2x+3y-6=0, 4x-3y-6=0, x=2$, and $y=\frac{2}{3}$ intersect in:
$\textbf{(A)}\ \text{6 points}\qquad\textbf{(B)}\ \text{1 point}\qquad\textbf{(C)}\ \text{2 points}\qquad\textbf{(D)}\ \text{no points}\\ \textbf{(E)}\ \text{an unlimited number of points}$
| [
"We first convert each of the lines into slope-intercept form ($y = mx + b$):\n\n\n$2x+3y-6=0 ==> 3y = -2x + 6 ==> y = -\\frac{2}{3}x + 2$\n\n\n$4x - 3y - 6 = 0 ==> 4x - 6 = 3y ==> y = \\frac{4}{3}x - 2$\n\n\n$x = 2$ stays as is.\n\n\n$y = \\frac{2}{3}$ stays as is\n\n\nWe can graph the four lines here:[1]\n\n\nWhe... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/46.json | AHSME |
1955_AHSME_Problems | 2 | 0 | Geometry | Multiple Choice | The smaller angle between the hands of a clock at $12:25$ p.m. is:
$\textbf{(A)}\ 132^\circ 30'\qquad\textbf{(B)}\ 137^\circ 30'\qquad\textbf{(C)}\ 150^\circ\qquad\textbf{(D)}\ 137^\circ 32'\qquad\textbf{(E)}\ 137^\circ$
| [
"At $12:25$, the minute hand is at $\\frac{25}{60}\\cdot 360^\\circ$, or $150^\\circ$.\nThe hour hand moves $5$ 'minutes' every hour, or $\\frac{5}{60}\\cdot 360 = 30^\\circ$ every hour. At $30^\\circ$ every hour, the hour hand moves $\\frac{1}{2}$ minutes on the clock every minute. At $12:25$, the hour hand is at ... | 2 | ./CreativeMath/AHSME/1955_AHSME_Problems/2.json | AHSME |
1955_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | On the same set of axes are drawn the graph of $y=ax^2+bx+c$ and the graph of the equation obtained by replacing $x$ by $-x$ in the given equation.
If $b \neq 0$ and $c \neq 0$ these two graphs intersect:
$\textbf{(A)}\ \text{in two points, one on the x-axis and one on the y-axis}\\ \textbf{(B)}\ \text{in one point... | [
"Replacing $x$ with $-x,$ we have the graph of $y = a(-x)^2 + b(-x) + c = ax^2 - bx+c.$\n\n\nWe can plug in simple values of $a, b,$ and $c$ for convenient drawing. For example, we can graph $x^2+2x+1$ and $x^2-2x+1.$ We see that the parabolas intersect at $(0, 1),$ which is on the y-axis.\n\n\nThe answer is $\\box... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/28.json | AHSME |
1955_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | The solution of $\sqrt{5x-1}+\sqrt{x-1}=2$ is:
$\textbf{(A)}\ x=2,x=1\qquad\textbf{(B)}\ x=\frac{2}{3}\qquad\textbf{(C)}\ x=2\qquad\textbf{(D)}\ x=1\qquad\textbf{(E)}\ x=0$
| [
"First, square both sides. This gives us\n\n\n\\[\\sqrt{5x-1}^2+2\\cdot\\sqrt{5x-1}\\cdot\\sqrt{x-1}+\\sqrt{x-1}^2=4 \\Longrightarrow 5x-1+2\\cdot\\sqrt{(5x-1)\\cdot(x-1)}+x-1=4 \\Longrightarrow 2\\cdot\\sqrt{5x^2-6x+1}+6x-2=4\\]\nThen, adding $-6x$ to both sides gives us\n\n\n\\[2\\cdot\\sqrt{5x^2-6x+1}+6x-2-6x=4... | 2 | ./CreativeMath/AHSME/1955_AHSME_Problems/12.json | AHSME |
1955_AHSME_Problems | 45 | 0 | Algebra | Multiple Choice | Given a geometric sequence with the first term $\neq 0$ and $r \neq 0$ and an arithmetic sequence with the first term $=0$.
A third sequence $1,1,2\ldots$ is formed by adding corresponding terms of the two given sequences.
The sum of the first ten terms of the third sequence is:
$\textbf{(A)}\ 978\qquad\textbf{(B)... | [
"Let our geometric sequence be $a,ar,ar^2$ and let our arithmetic sequence be $0,d,2d$. We know that \\[\\begin{cases} a+0=1 \\\\ ar+d=1\\\\ ar^2+2d=2\\end{cases}\\]\nThis implies that $a=1$, hence $r+d=1$ and $r^2+2d=2$. We can rewrite $r+d=1$ as $d = 1-r$, plugging this into $r^2+2d=2$, we get $r^2+2-2r = 2$, we ... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/45.json | AHSME |
1955_AHSME_Problems | 32 | 0 | Algebra | Multiple Choice | If the discriminant of $ax^2+2bx+c=0$ is zero, then another true statement about $a, b$, and $c$ is that:
$\textbf{(A)}\ \text{they form an arithmetic progression}\\ \textbf{(B)}\ \text{they form a geometric progression}\\ \textbf{(C)}\ \text{they are unequal}\\ \textbf{(D)}\ \text{they are all negative numbers}\\ \... | [
"The discriminant of a quadratic is\n\n\n\\[\\Delta = b^2 - 4ac = (2b)^2 - 4ac = 4b^2 - 4ac = 0.\\]\nWe know that $b^2 = ac,$ or $\\frac{b}{a} = \\frac{c}{b}$ so $a, b, c$ form a geometric progression.\n\n\n\n\n$\\boxed{\\text{(B)}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/32.json | AHSME |
1955_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | The function $4x^2-12x-1$:
$\textbf{(A)}\ \text{always increases as }x\text{ increases}\\ \textbf{(B)}\ \text{always decreases as }x\text{ decreases to 1}\\ \textbf{(C)}\ \text{cannot equal 0}\\ \textbf{(D)}\ \text{has a maximum value when }x\text{ is negative}\\ \textbf{(E)}\ \text{has a minimum value of-10}$
| [
"We can use the process of elimination to narrow down the field substantially:\n\n\n$\\textbf{(A)}\\ \\text{always increases as } x\\text{ increases}$ is wrong due to the quadratic nature of the function.\n\n\n$\\textbf{(B)}\\ \\text{always decreases as } x\\text{ decreases to 1}$ is wrong due to the vertex being o... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/24.json | AHSME |
1955_AHSME_Problems | 49 | 0 | Algebra | Multiple Choice | The graphs of $y=\frac{x^2-4}{x-2}$ and $y=2x$ intersect in:
$\textbf{(A)}\ \text{1 point whose abscissa is 2}\qquad\textbf{(B)}\ \text{1 point whose abscissa is 0}\\ \textbf{(C)}\ \text{no points}\qquad\textbf{(D)}\ \text{two distinct points}\qquad\textbf{(E)}\ \text{two identical points}$
| [
"We can simplify $y=\\frac{x^2-4}{x-2}$ in order to help solve this problem: $y=\\frac{x^2-4}{x-2}=\\frac{(x+2)(x-2)}{x-2}=x+2$.\n\n\nIn order to find solutions, we have to have $x+2=2x$. $x=2$, but this turns out to be an extraneous solution, as, when put into the two equations, one of them turns out an undefined... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/49.json | AHSME |
1955_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | One of the factors of $x^4+2x^2+9$ is:
$\textbf{(A)}\ x^2+3\qquad\textbf{(B)}\ x+1\qquad\textbf{(C)}\ x^2-3\qquad\textbf{(D)}\ x^2-2x-3\qquad\textbf{(E)}\ \text{none of these}$
| [
"We can test each of the answer choices by using polynomial division. \n\n\n$x^2 + 3$ leaves behind a remainder, and so does $x^2 - 3$.\n\n\nIn addition, $x + 1$ also fails the test, and that takes down $x^2 - 2x - 3$, which can be expressed as $(x + 1)(x - 3)$. That leaves $\\boxed{(\\textbf{E})}$\n\n\n",
"Notic... | 2 | ./CreativeMath/AHSME/1955_AHSME_Problems/25.json | AHSME |
1955_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | The fraction $\frac{a^{-4}-b^{-4}}{a^{-2}-b^{-2}}$ is equal to:
$\textbf{(A)}\ a^{-6}-b^{-6}\qquad\textbf{(B)}\ a^{-2}-b^{-2}\qquad\textbf{(C)}\ a^{-2}+b^{-2}\\ \textbf{(D)}\ a^2+b^2\qquad\textbf{(E)}\ a^2-b^2$
| [
"By the difference of squares property, $a^{-4} - b^{-4}$ is equivalent to $(a^{-2} + b^{-2})(a^{-2} - b^{-2})$. This means the fraction is now equal to $\\frac{(a^{-2} + b^{-2})(a^{-2} - b^{-2})}{a^{-2} - b^{-2}}$, which simplifies to $\\textbf{(C)}\\ a^{-2}+b^{-2}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/13.json | AHSME |
1955_AHSME_Problems | 29 | 0 | Geometry | Multiple Choice | In the figure, $PA$ is tangent to semicircle $SAR$; $PB$ is tangent to semicircle $RBT$; $SRT$ is a straight line;
the arcs are indicated in the figure. $\angle APB$ is measured by:
[asy] unitsize(1.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=3; pair O1=(0,0), O2=(3,0), Sp=(-2,0), R=(2,0), T=(4,0); p... | [
"[asy] unitsize(1.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=3; pair O1=(0,0), O2=(3,0), Sp=(-2,0), R=(2,0), T=(4,0); pair A=O1+2*dir(60), B=O2+dir(85); pair Pa=rotate(90,A)*O1, Pb=rotate(-90,B)*O2; pair P=extension(A,Pa,B,Pb); pair[] dots={Sp,R,T,A,B,P}; draw(P--P+5*(A-P)); draw(P--P+5*(B-P)); clip... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/29.json | AHSME |
1955_AHSME_Problems | 3 | 0 | Arithmetic | Multiple Choice | If each number in a set of ten numbers is increased by $20$, the arithmetic mean (average) of the ten numbers:
$\textbf{(A)}\ \text{remains the same}\qquad\textbf{(B)}\ \text{is increased by 20}\qquad\textbf{(C)}\ \text{is increased by 200}\\ \textbf{(D)}\ \text{is increased by 10}\qquad\textbf{(E)}\ \text{is increa... | [
"Let the sum of the 10 numbers be x. The mean is then $\\frac{x}{10}$. Then, since you're adding 20 to each number, the new sum of the numbers is x+200, since there are 10 numbers. Then, the new mean is $\\frac{x+200}{10}$, which simplifies to $\\frac{x}{10}+20$, which is $\\fbox{B}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/3.json | AHSME |
1955_AHSME_Problems | 34 | 0 | Geometry | Multiple Choice | A $6$-inch and $18$-inch diameter poles are placed together and bound together with wire.
The length of the shortest wire that will go around them is:
$\textbf{(A)}\ 12\sqrt{3}+16\pi\qquad\textbf{(B)}\ 12\sqrt{3}+7\pi\qquad\textbf{(C)}\ 12\sqrt{3}+14\pi\\ \textbf{(D)}\ 12+15\pi\qquad\textbf{(E)}\ 24\pi$
| [
"[asy] draw(circle((0,0),3)); draw(circle((12,0),9)); draw((-1,2sqrt(2)) -- (9,6sqrt(2))); draw((-1,-2sqrt(2)) -- (9,-6sqrt(2))); draw((0,0) -- (-1,2sqrt(2))); draw((12,0) -- (9,6sqrt(2))); draw((0,0) -- (12,0)); draw((0,0) -- (10,3sqrt(3))); [/asy]\n\\[\\text{A diagram of the problem...aside from a few blips.}\\]\... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/34.json | AHSME |
1955_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | The graph of $x^2-4y^2=0$:
$\textbf{(A)}\ \text{is a hyperbola intersecting only the }x\text{-axis}\\ \textbf{(B)}\ \text{is a hyperbola intersecting only the }y\text{-axis}\\ \textbf{(C)}\ \text{is a hyperbola intersecting neither axis}\\ \textbf{(D)}\ \text{is a pair of straight lines}\\ \textbf{(E)}\ \text{does no... | [
"By difference of squares, we can rewrite the equation as $(x-2y)(x+2y) = 0$, which is just the union of the two lines $x - 2y = 0$ and $x + 2y = 0$. Therefore, our answer is $\\boxed{\\textbf{(D)}}$, and we are done.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/8.json | AHSME |
1955_AHSME_Problems | 22 | 0 | Arithmetic | Multiple Choice | On a $\textdollar{10000}$ order a merchant has a choice between three successive discounts of $20$%, $20$%, and $10$% and
three successive discounts of $40$%, $5$%, and $5$%. By choosing the better offer, he can save:
$\textbf{(A)}\ \text{nothing at all}\qquad\textbf{(B)}\ $440\qquad\textbf{(C)}\ $330\qquad\textbf{... | [
"In order to solve this problem, we can simply try both paths and finding the positive difference between the two.\n\n\nThe first path goes $10,000 * 0.8 = 8,000 * 0.8 = 6400 * 0.9 = \\textbf{5760}$\n\n\nThe second path goes $10,000 * 0.6 = 6,000 * 0.95 \\text{ or } (1 - 0.05) = 5700 * 0.95 = \\textbf{5415}$\n\n\nT... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/22.json | AHSME |
1955_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | The discriminant of the equation $x^2+2x\sqrt{3}+3=0$ is zero. Hence, its roots are:
$\textbf{(A)}\ \text{real and equal}\qquad\textbf{(B)}\ \text{rational and equal}\qquad\textbf{(C)}\ \text{rational and unequal}\\ \textbf{(D)}\ \text{irrational and unequal}\qquad\textbf{(E)}\ \text{imaginary}$
| [
"Since the discriminant is zero, there is one distinct root, or, relevant to this question, two equal roots. The fact that only one solution exists means that the equation can be simplified into $(x + \\sqrt{3})^2=0$. The distinct solution^ to the equation, as we can clearly see, is $-\\sqrt{3}$, which is $\\textbf... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/18.json | AHSME |
1955_AHSME_Problems | 38 | 0 | Algebra | Multiple Choice | Four positive integers are given. Select any three of these integers, find their arithmetic average,
and add this result to the fourth integer. Thus the numbers $29, 23, 21$, and $17$ are obtained. One of the original integers is:
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}... | [
"Define numbers $a, b, c,$ and $d$ to be the four numbers. In order to satisfy the following conditions, the system of equation should be constructed. (It doesn't matter which variable is which.)\n\\[\\frac{a+b+c}{3}+d=29\\]\n\\[\\frac{a+b+d}{3}+c=23\\]\n\\[\\frac{a+c+d}{3}+b=21\\]\n\\[\\frac{b+c+d}{3}+a=17\\]\nAdd... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/38.json | AHSME |
1955_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | The equality $\frac{1}{x-1}=\frac{2}{x-2}$ is satisfied by:
$\textbf{(A)}\ \text{no real values of }x\qquad\textbf{(B)}\ \text{either }x=1\text{ or }x=2\qquad\textbf{(C)}\ \text{only }x=1\\ \textbf{(D)}\ \text{only }x=2\qquad\textbf{(E)}\ \text{only }x=0$
| [
"From the equality, $\\frac{1}{x-1}=\\frac{2}{x-2}$, we get ${(x-1)}\\times2={(x-2)}\\times1$.\n\n\nSolving this, we get, ${2x-2}={x-2}$.\n\n\nThus, the answer is $\\fbox{{\\bf(E)} \\text{only} x = 0}$.\n\n\nSolution by awesomechoco\n\n\n"
] | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/4.json | AHSME |
1955_AHSME_Problems | 14 | 0 | Geometry | Multiple Choice | The length of rectangle $R$ is $10$% more than the side of square $S$. The width of the rectangle is $10$% less than the side of the square.
The ratio of the areas, $R:S$, is:
$\textbf{(A)}\ 99: 100\qquad\textbf{(B)}\ 101: 100\qquad\textbf{(C)}\ 1: 1\qquad\textbf{(D)}\ 199: 200\qquad\textbf{(E)}\ 201: 200$
| [
"Let each of the square's sides be $x$. The dimensions of the rectangle can be expressed as $1.1x$ and $0.9x$. Therefore, the area of the rectangle is $0.99x^2$, while the square has an area of $x^2$. The ratio of $R : S$ can be defined as $0.99x^2 : x^2$, which ultimately leads to $\\textbf{(A) } 99 : 100$\n\n\n"
... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/14.json | AHSME |
1955_AHSME_Problems | 43 | 0 | Geometry | Multiple Choice | The pairs of values of $x$ and $y$ that are the common solutions of the equations $y=(x+1)^2$ and $xy+y=1$ are:
$\textbf{(A)}\ \text{3 real pairs}\qquad\textbf{(B)}\ \text{4 real pairs}\qquad\textbf{(C)}\ \text{4 imaginary pairs}\\ \textbf{(D)}\ \text{2 real and 2 imaginary pairs}\qquad\textbf{(E)}\ \text{1 real and... | [
"Graph the two equations. $y=(x+1)^2$ produces a parabola with vertex $(-1, 0),$ while $xy+y=1$ produces a hyperbola. They intersect at the point $(0, 1).$\n\n\nThis makes one real solution. Only one answer choice has 1 real solution listed, so the answer is $\\boxed{\\textbf{(E)}}.$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/43.json | AHSME |
1955_AHSME_Problems | 42 | 0 | Algebra | Multiple Choice | If $a, b$, and $c$ are positive integers, the radicals $\sqrt{a+\frac{b}{c}}$ and $a\sqrt{\frac{b}{c}}$ are equal when and only when:
$\textbf{(A)}\ a=b=c=1\qquad\textbf{(B)}\ a=b\text{ and }c=a=1\qquad\textbf{(C)}\ c=\frac{b(a^2-1)}{a}\\ \textbf{(D)}\ a=b\text{ and }c\text{ is any value}\qquad\textbf{(E)}\ a=b\text... | [
"Here, we are told that the two quantities are equal. \n\n\nSquaring both sides, we get: $a+\\frac{b}{c}=a^2*\\frac{b}{c}$.\n\n\nMultiply both sides by $c$: $ac+b=ba^2$.\n\n\nLooking at all answer choices, we can see that $ac=ba^2-b=b(a^2-1)$. \n\n\nThis means that $c=\\frac{b(a^2-1)}{a}$, and this is option C. The... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/42.json | AHSME |
1955_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | The ratio of the areas of two concentric circles is $1: 3$. If the radius of the smaller is $r$, then the difference between the
radii is best approximated by:
$\textbf{(A)}\ 0.41r \qquad \textbf{(B)}\ 0.73 \qquad \textbf{(C)}\ 0.75 \qquad \textbf{(D)}\ 0.73r \qquad \textbf{(E)}\ 0.75r$
| [
"Observe that there are 3 feet in 1 yard, 9 square feet in 1 square yard, and 27 cubic feet in 1 cubic yard.\n\n\nThis means that if the ratio is $x$ in 1D, it is $x^2$ in 2D, and it is $x^3$ in 3D.\n\n\nWe can apply this thinking into our current question, which would have $1:3$ in the 2D-plane. Since the radius i... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/15.json | AHSME |
1955_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | $5y$ varies inversely as the square of $x$. When $y=16, x=1$. When $x=8, y$ equals:
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 128 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ \frac{1}{4} \qquad \textbf{(E)}\ 1024$
| [
"An inverse variation can be expressed in the form $xy = n$, where $n$ is any number (except perhaps zero). Since $5y$ varies inversely with the square of $x$, this particular one will be $5yx^2 = n$. \n\n\nWe can plug in $16$ for $y$ and $1$ for $x$, which makes $n=80$. The equation is now $5yx^2 = 80$.\n\n\nWhen ... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/5.json | AHSME |
1955_AHSME_Problems | 39 | 0 | Algebra | Multiple Choice | If $y=x^2+px+q$, then if the least possible value of $y$ is zero $q$ is equal to:
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac{p^2}{4}\qquad\textbf{(C)}\ \frac{p}{2}\qquad\textbf{(D)}\ -\frac{p}{2}\qquad\textbf{(E)}\ \frac{p^2}{4}-q$
| [
"The least possible value of $y$ is given at the $y$ coordinate of the vertex. The $x$- coordinate is given by \n\\[\\frac{-p}{(2)(1)} = \\frac{-p}{2}\\] Plugging this into the quadratic, we get\n\\[y = \\frac{p^2}{4} - \\frac{p^2}{2} + q\\]\n\\[0 = \\frac{p^2}{4} - \\frac{2p^2}{4} + q\\]\n\\[0 = \\frac{-p^2}{4} + ... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/39.json | AHSME |
1955_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | Two numbers whose sum is $6$ and the absolute value of whose difference is $8$ are roots of the equation:
$\textbf{(A)}\ x^2-6x+7=0\qquad\textbf{(B)}\ x^2-6x-7=0\qquad\textbf{(C)}\ x^2+6x-8=0\\ \textbf{(D)}\ x^2-6x+8=0\qquad\textbf{(E)}\ x^2+6x-7=0$
| [
"The first two hints can be expressed as the following system of equations:\n\\[\\begin{cases} (1) & a + b = 6 \\\\ (2) & a - b = 8 \\end{cases}\\]\nFrom this, we can clearly see that $a = 7$, and that $b = -1$.\n\n\nSince quadratic equations can generally be expressed in the form of $(x - a)(x - b) = 0$, where a a... | 2 | ./CreativeMath/AHSME/1955_AHSME_Problems/19.json | AHSME |
1955_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | In checking the petty cash a clerk counts $q$ quarters, $d$ dimes, $n$ nickels, and $c$ cents. Later he discovers that $x$ of the nickels were counted as quarters and $x$ of the dimes were counted as cents. To correct the total obtained the clerk must:
$\textbf{(A)}\ \text{make no correction}\qquad\textbf{(B)}\ \text... | [
"If the clerk mistook $x$ nickels as quarters, then every mistake inflates the total by $20$ cents. In order to correct this, we have to subtract $20$ cents $x$ times, for a total of $20x$ cents. We can do the same for the $x$ dimes that were turned into pennies (or cents). This exchange would increase the total va... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/23.json | AHSME |
1955_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | A circle is inscribed in a triangle with sides $8, 15$, and $17$. The radius of the circle is:
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 7$
| [
"We know that $A = sr$, where $A$ is the triangle's area, $s$ its semiperimeter, and $r$ its inradius. Since this particular triangle is a right triangle (which we can verify by the Pythagorean theorem), the area is half of $8*15 = 120$, and the semiperimeter is half of $8 + 15 + 17 = 40$. Therefore, the inradius i... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/9.json | AHSME |
1955_AHSME_Problems | 35 | 0 | Algebra | Multiple Choice | Three boys agree to divide a bag of marbles in the following manner. The first boy takes one more than half the marbles. The second takes a third of the number remaining. The third boy finds that he is left with twice as many marbles as the second boy. The original number of marbles:
$\textbf{(A)}\ \text{is none of t... | [
"[asy] draw((0,0)--(13,0)); dot((0,0)); dot((7,0)); dot((9,0)); dot((13,0)); [/asy]\nThe line that represents the marbles is split into three sections: $3x+1, x, \\text{and } 2x$. Since the amount of marbles each boy relies on the variable $x$, which cannot be solved for, the original amount of marbles $\\boxed{\\... | 1 | ./CreativeMath/AHSME/1955_AHSME_Problems/35.json | AHSME |
1952_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | If $\frac{x}{y}=\frac{3}{4}$, then the incorrect expression in the following is:
$\textbf{(A) \ }\frac{x+y}{y}=\frac{7}{4} \qquad \textbf{(B) \ }\frac{y}{y-x}=\frac{4}{1} \qquad \textbf{(C) \ }\frac{x+2y}{x}=\frac{11}{3} \qquad$
$\textbf{(D) \ }\frac{x}{2y}=\frac{3}{8} \qquad \textbf{(E) \ }\frac{x-y}{y}=\frac{1}{... | [
"Let's consider each answer choice.\n\n\n$\\boxed{A.} \\quad \\frac{x+y}{y}=\\frac{x}{y}+\\frac{y}{y}=\\frac{3}{4}+1=\\boxed{\\frac{7}{4}}$\n\n\n$\\boxed{B.} \\quad \\frac{y-x}{y}=\\frac{y}{y}-\\frac{x}{y}=1-\\frac{3}{4}=\\frac{1}{4}\\implies \\left(\\frac{y-x}{y}\\right)^{-1}=\\frac{y}{y-x}=\\boxed{\\frac{4}{1}}$\... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/20.json | AHSME |
1952_AHSME_Problems | 36 | 0 | Algebra | Multiple Choice | To be continuous at $x = - 1$, the value of $\frac {x^3 + 1}{x^2 - 1}$ is taken to be:
$\textbf{(A)}\ - 2 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ \frac {3}{2} \qquad \textbf{(D)}\ \infty \qquad \textbf{(E)}\ -\frac{3}{2}$
| [
"Factoring the numerator using the sum of cubes identity and denominator using the difference of squares identity gives \\[\\dfrac{(x+1)(x^{2}-x+1)}{(x+1)(x-1)}\\]\nCancelling out a factor of $x+1$ from the numerator and denominator gives \\[\\dfrac{(x^{2}-x+1)}{(x-1)}\\]\nPlugging in $x= -1$ gives $\\dfrac{3}{-2}$... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/36.json | AHSME |
1952_AHSME_Problems | 41 | 0 | Algebra | Multiple Choice | Increasing the radius of a cylinder by $6$ units increased the volume by $y$ cubic units. Increasing the height of the cylinder by $6$ units also increases the volume by $y$ cubic units. If the original height is $2$, then the original radius is:
$\text{(A) } 2 \qquad \text{(B) } 4 \qquad \text{(C) } 6 \qquad \text{... | [
"We know that the volume of a cylinder is equal to $\\pi r^2h$, where $r$ and $h$ are the radius and height, respectively. So we know that $2\\pi (r+6)^2-2\\pi r^2=y=\\pi r^2(2+6)-2\\pi r^2$. Expanding and rearranging, we get that $2\\pi (12r+36)=6\\pi r^2$. Divide both sides by $6\\pi$ to get that $4r+12=r^2$, and... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/41.json | AHSME |
1952_AHSME_Problems | 16 | 0 | Geometry | Multiple Choice | If the base of a rectangle is increased by $10\%$ and the area is unchanged, then the altitude is decreased by:
$\textbf{(A) \ }9\% \qquad \textbf{(B) \ }10\% \qquad \textbf{(C) \ }11\% \qquad \textbf{(D) \ }11\frac{1}{9}\% \qquad \textbf{(E) \ }9\frac{1}{11}\%$
| [
"$b\\cdot a=\\frac{11}{10}b\\cdot xa\\implies x=\\frac{10}{11}$. Hence, the altitude is decreased by $1-x=\\frac{1}{11}=\\boxed{\\textbf{(E)}\\ 9\\frac{1}{11}\\%}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/16.json | AHSME |
1952_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | The difference of the roots of $x^2-7x-9=0$ is:
$\textbf{(A) \ }+7 \qquad \textbf{(B) \ }+\frac{7}{2} \qquad \textbf{(C) \ }+9 \qquad \textbf{(D) \ }2\sqrt{85} \qquad \textbf{(E) \ }\sqrt{85}$
| [
"Denote the $2$ roots of this quadratic as $r_1$ and $r_2$. Note that $(r_1-r_2)^2=(r_1+r_2)^2-4r_1r_2$. By Vieta's Formula's, $r_1+r_2=7$, and $r_1r_2=-9$. Thus, $r_1-r_2=\\sqrt{49+4\\cdot 9}=\\boxed{\\textbf{(E)}\\ \\sqrt{85}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/6.json | AHSME |
1952_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | When simplified, $(x^{-1}+y^{-1})^{-1}$ is equal to:
$\textbf{(A) \ }x+y \qquad \textbf{(B) \ }\frac{xy}{x+y} \qquad \textbf{(C) \ }xy \qquad \textbf{(D) \ }\frac{1}{xy} \qquad \textbf{(E) \ }\frac{x+y}{xy}$
| [
"$(x^{-1}+y^{-1})^{-1}=\\left(\\frac{1}{x}+\\frac{1}{y}\\right)^{-1}=\\left(\\frac{x+y}{xy}\\right)^{-1}=\\boxed{\\textbf{(B)}\\ \\frac{xy}{x+y}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/7.json | AHSME |
1952_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | A merchant bought some goods at a discount of $20\%$ of the list price. He wants to mark them at such a price that he can give a discount of $20\%$ of the marked price and still make a profit of $20\%$ of the selling price. The per cent of the list price at which he should mark them is:
$\textbf{(A) \ }20 \qquad \te... | [
"Let $C$ represent the cost of the goods, and let $L$, $S$, and $M$ represent the list, selling, and marked prices of the goods, respectively. Hence, we have three equations, which we need to manipulate in order to relate $M$ and $L$:\n\n\n$C=\\frac{4}{5}L$\n\n\n$S=C+\\frac{1}{5}S$\n\n\n$S=\\frac{4}{5}M$\n\n\nWe fi... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/17.json | AHSME |
1952_AHSME_Problems | 40 | 0 | Algebra | Multiple Choice | In order to draw a graph of $ax^2+bx+c$, a table of values was constructed. These values of the function for a set of equally spaced increasing values of $x$ were $3844, 3969, 4096, 4227, 4356, 4489, 4624$, and $4761$. The one which is incorrect is:
$\text{(A) } 4096 \qquad \text{(B) } 4356 \qquad \text{(C) } 4489 \... | [
"Since the polynomial is quadratic, its second differences must be constant. Taking the first differences, we have\n\\begin{align*} 3969-3844&=125 \\\\ 4096-3969&=127 \\\\ 4227-4096&=131 \\\\ 4356-4227&=129 \\\\ 4489-4356&=133 \\\\ 4624-4489&=135 \\\\ 4761-4624&=137 \\end{align*}\nThis leads to a common second diff... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/40.json | AHSME |
1952_AHSME_Problems | 37 | 0 | Geometry | Multiple Choice | Two equal parallel chords are drawn $8$ inches apart in a circle of radius $8$ inches. The area of that part of the circle that lies between the chords is:
$\textbf{(A)}\ 21\frac{1}{3}\pi-32\sqrt{3}\qquad \textbf{(B)}\ 32\sqrt{3}+21\frac{1}{3}\pi\qquad \textbf{(C)}\ 32\sqrt{3}+42\frac{2}{3}\pi \qquad\\ \textbf{(D)}\ ... | [
"[asy] pair A,B,C,D,E,F,G; A=(0,0); B=(-5,4sqrt(3)+2); C=(5,4sqrt(3)+2); D=(-5,-4sqrt(3)-0.5); E=(5,-4sqrt(3)-0.5); F=(-4,0); G=(4,-sqrt(2)/2); label(\"$A$\",(0,-0.5),S); label(\"$B$\",B,SE); label(\"$C$\",C,SW); label(\"$D$\",D,SE); label(\"$E$\",E,SW); label(\"$F$\",F,W); label(\"$G$\",G,NE); draw(circle(A,8)); ... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/37.json | AHSME |
1952_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | The sides of a regular polygon of $n$ sides, $n>4$, are extended to form a star. The number of degrees at each point of the star is:
$\textbf{(A) \ }\frac{360}{n} \qquad \textbf{(B) \ }\frac{(n-4)180}{n} \qquad \textbf{(C) \ }\frac{(n-2)180}{n} \qquad$
$\textbf{(D) \ }180-\frac{90}{n} \qquad \textbf{(E) \ }\frac{1... | [
"[asy] import olympiad; pair A,B,C,D,E; A=(-1.5,1.5sqrt(3)); B=(-4.5,1.5sqrt(3)); C=origin; D=(1.5,1.5sqrt(3)); E=3*dir(-120); draw(A--D,dashed); draw(D--E,dashed); draw(B--A--C--E); markscalefactor=0.075; draw(anglemark(C,A,D)); draw(anglemark(D,C,A)); [/asy]\n\n\nThe measure of each angle, in degrees, of an $n$-s... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/21.json | AHSME |
1952_AHSME_Problems | 47 | 0 | Algebra | Multiple Choice | In the set of equations $z^x = y^{2x},\quad 2^z = 2\cdot4^x, \quad x + y + z = 16$, the integral roots in the order $x,y,z$ are:
$\textbf{(A) } 3,4,9 \qquad \textbf{(B) } 9,-5,-12 \qquad \textbf{(C) } 12,-5,9 \qquad \textbf{(D) } 4,3,9 \qquad \textbf{(E) } 4,9,3$
| [
"The easiest method, which in this case is not very time consuming, is to guess and check and use process of elimination. \nWe can immediately rule out (B) since it does not satisfy the third equation. \nThe first equation allows us to eliminate (A) and (C) because the bases are different and aren't powers of each ... | 2 | ./CreativeMath/AHSME/1952_AHSME_Problems/47.json | AHSME |
1952_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | An automobile went up a hill at a speed of $10$ miles an hour and down the same distance at a speed of $20$ miles an hour. The average speed for the round trip was:
$\textbf{(A) \ }12\frac{1}{2}\text{mph} \qquad \textbf{(B) \ }13\frac{1}{3}\text{mph} \qquad \textbf{(C) \ }14\frac{1}{2}\text{mph} \qquad \textbf{(D) \... | [
"Let $r$, $t$, and $d$ represent the rate, time, and distance (respectively) of each trip. We know that $rt=d$, or $r_1t_1=r_2t_2$. Because it is given that $2r_1=r_2$, $t_1=2t_2$. Furthermore, $d=10t_1=20t_2$. The average speed can be expressed as $\\frac{2d}{t_1+t_2}=\\frac{2(20t_2)}{2t_2+t_2}=\\frac{40}{3}=\\box... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/10.json | AHSME |
1952_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | If $\left(r+\frac1r\right)^2=3$, then $r^3+\frac1{r^3}$ equals
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 6$
| [
"We know $r+\\frac1r=\\sqrt3$. Cubing this gives $r^3+3r+\\frac3r+\\frac1{r^3}=3\\sqrt3$. But $3r+\\frac3r=3\\left(r+\\frac1r\\right)=3\\sqrt3$, so subtracting this from the first equation gives\n$r^3+\\frac1{r^3}=\\boxed{0\\textbf{ (C)}}$. \n(Actually, $r+\\frac1r$ could have been equal to $-\\sqrt3$ instead of $\... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/26.json | AHSME |
1952_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | When the sum of the first ten terms of an arithmetic progression is four times the sum of the first five terms, the ratio of the first term to the common difference is:
$\textbf{(A)}\ 1: 2 \qquad \textbf{(B)}\ 2: 1 \qquad \textbf{(C)}\ 1: 4 \qquad \textbf{(D)}\ 4: 1 \qquad \textbf{(E)}\ 1: 1$
| [
"Let our first term be $a$ and our common difference be $d$. Thus, the first few terms of the sequence are $a$, $a + d$, $a + 2d$, ...\n\n\nThe sum of the first 5 terms is\n\\[a + (a + d) + (a + 2d) + ... + (a + 4d) = 5a + 10d\\]\nThe sum of the first 10 terms is \n\\[a + (a + d) + (a + 2d) + ... + (a + 9d) = 10a ... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/30.json | AHSME |
1952_AHSME_Problems | 31 | 0 | Counting | Multiple Choice | Given $12$ points in a plane no three of which are collinear, the number of lines they determine is:
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 54 \qquad \textbf{(C)}\ 120 \qquad \textbf{(D)}\ 66 \qquad \textbf{(E)}\ \text{none of these}$
| [
"Since no three points are collinear, every two points must determine a distinct line. Thus, there are $\\dbinom{12}{2} = \\frac{12\\cdot11}{2} = 66$ lines. \n\n\nTherefore, the answer is $\\fbox{(D) 66}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/31.json | AHSME |
1952_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is:
$\textbf{(A)}\ 1:2\qquad\textbf{(B)}\ 1:3\qquad\textbf{(C)}\ 1:\sqrt3\qquad\textbf{(D)}\ \sqrt3:2 \qquad\textbf{(E)}\ 2:3$
| [
"If the radius of the circle is $r$, then the perimeter of the first triangle is $3\\left(\\frac{2r}{\\sqrt3}\\right)=2r\\sqrt3$, and the perimeter of the second is $3r\\sqrt3$. So the ratio is $\\boxed{\\frac23{\\textbf{ (E)}}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/27.json | AHSME |
1952_AHSME_Problems | 1 | 0 | Geometry | Multiple Choice | If the radius of a circle is a rational number, its area is given by a number which is:
$\textbf{(A)\ } \text{rational} \qquad \textbf{(B)\ } \text{irrational} \qquad \textbf{(C)\ } \text{integral} \qquad \textbf{(D)\ } \text{a perfect square }\qquad \textbf{(E)\ } \text{none of these}$
| [
"Let the radius of the circle be the common fraction $\\frac{a}{b}.$ Then the area of the circle is $\\pi \\cdot \\frac{a^2}{b^2}.$\nBecause $\\pi$ is irrational and $\\frac{a^2}{b^2}$ is rational, their product must be irrational. The answer is $\\boxed{B}.$\n\n\n",
"The phrasing of the problem makes it clear th... | 2 | ./CreativeMath/AHSME/1952_AHSME_Problems/1.json | AHSME |
1952_AHSME_Problems | 50 | 0 | Algebra | Multiple Choice | A line initially 1 inch long grows according to the following law, where the first term is the initial length.
\[1+\frac{1}{4}\sqrt{2}+\frac{1}{4}+\frac{1}{16}\sqrt{2}+\frac{1}{16}+\frac{1}{64}\sqrt{2}+\frac{1}{64}+\cdots\]
If the growth process continues forever, the limit of the length of the line is:
$\textbf{... | [
"We can rewrite our sum as the sum of two infinite geometric sequences.\n\\[1 + \\frac{1}{4}\\sqrt{2} + \\frac{1}{4} + \\frac{1}{16}\\sqrt{2} + \\frac{1}{16} + ... =\\]\n\\[(1 + \\frac{1}{4} + \\frac{1}{16} + \\frac{1}{64} + ...) + (\\frac{1}{4}\\sqrt{2} + \\frac{1}{16}\\sqrt{2} + \\frac{1}{64}\\sqrt{2} + ...)\\]\n... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/50.json | AHSME |
1952_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | If $y=f(x)=\frac{x+2}{x-1}$, then it is incorrect to say:
$\textbf{(A)\ }x=\frac{y+2}{y-1}\qquad\textbf{(B)\ }f(0)=-2\qquad\textbf{(C)\ }f(1)=0\qquad$
$\textbf{(D)\ }f(-2)=0\qquad\textbf{(E)\ }f(y)=x$
| [
"$f(1)=\\frac{3}{0}$, which is undefined. Hence, the (in)correct answer is $\\boxed{\\textbf{(C)}\\ f(1)=0}$.\nWe can verify that the other statements are valid.\n\n\n$\\textbf{(A).\\ } y=\\frac{x+2}{x-1}\\implies xy-y=x+2\\implies x(y-1)=y+2\\implies x=\\frac{y+2}{y-1}$\n\n\n$\\textbf{(B).\\ } f(0)=\\frac{(0)+2}{(... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/11.json | AHSME |
1952_AHSME_Problems | 46 | 0 | Geometry | Multiple Choice | The base of a new rectangle equals the sum of the diagonal and the greater side of a given rectangle, while the altitude of the new rectangle equals the difference of the diagonal and the greater side of the given rectangle. The area of the new rectangle is:
$\text{(A) greater than the area of the given rectangle} \... | [
"Let the larger side of the rectangle be y and the smaller side of the rectangle be x. Then, by pythagorean theorem, the diagonal of the rectangle has length $\\sqrt{x^{2}+y^{2}}$. \n\n\nBy the definition of the problem, the area of the new rectangle is $(\\sqrt{x^{2}+y^{2}}+y)(\\sqrt{x^{2}+y^{2}}-y)$\n\n\nExpandin... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/46.json | AHSME |
1952_AHSME_Problems | 2 | 0 | Arithmetic | Multiple Choice | Two high school classes took the same test. One class of $20$ students made an average grade of $80\%$; the other class of $30$ students made an average grade of $70\%$. The average grade for all students in both classes is:
$\textbf{(A)}\ 75\%\qquad \textbf{(B)}\ 74\%\qquad \textbf{(C)}\ 72\%\qquad \textbf{(D)}\ 77\... | [
"The desired average can be found by dividing the total number of points earned by the total number of students. There are $20\\cdot 80+30\\cdot 70=3700$ points earned and $20+30=50$ students. Thus, our answer is $\\frac{3700}{50}$, or $\\boxed{\\textbf{(B)}\\ 74\\%}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/2.json | AHSME |
1952_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | In the table shown, the formula relating x and y is:
\[\begin{array}{|c|c|c|c|c|c|}\hline x & 1 & 2 & 3 & 4 & 5\\ \hline y & 3 & 7 & 13 & 21 & 31\\ \hline\end{array}\]
$\text{(A) } y = 4x - 1 \qquad\quad \text{(B) } y = x^3 - x^2 + x + 2 \qquad\\ \text{(C) } y = x^2 + x + 1 \qquad \text{(D) } y = (x^2 + x + 1)(x -... | [
"A simple method of solving the problem is trying each of the answer choices. \nOne can plug in values of x and y for each, because many x-values and their single corresponding y-values are given.\n\n\n\\begin{enumerate}\n\\item Choice A works for (1,3) and (2,7) but fails to work on (3,13) because 3*4=12 and 12-1=... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/28.json | AHSME |
1952_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | The sum to infinity of the terms of an infinite geometric progression is $6$. The sum of the first two terms is $4\frac{1}{2}$. The first term of the progression is:
$\textbf{(A) \ }3 \text{ or } 1\frac{1}{2} \qquad \textbf{(B) \ }1 \qquad \textbf{(C) \ }2\frac{1}{2} \qquad \textbf{(D) \ }6 \qquad \textbf{(E) \ }9\t... | [
"This geometric sequence can be written as $a+ar+ar^2+ar^3+\\cdots$. We are given that $a+ar=4\\frac{1}{2}$. Using the formula for the sum of an infinite geometric series, we know that $\\frac{a}{1-r}=6$. Solving for $r$ in the second equation, we find that $r=\\frac{6-a}{6}$. Plugging this into the first equation ... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/12.json | AHSME |
1952_AHSME_Problems | 45 | 0 | Algebra | Multiple Choice | If $a$ and $b$ are two unequal positive numbers, then:
$\text{(A) } \frac{2ab}{a+b}>\sqrt{ab}>\frac{a+b}{2}\qquad \text{(B) } \sqrt{ab}>\frac{2ab}{a+b}>\frac{a+b}{2} \\ \text{(C) } \frac{2ab}{a+b}>\frac{a+b}{2}>\sqrt{ab}\qquad \text{(D) } \frac{a+b}{2}>\frac{2ab}{a+b}>\sqrt{ab} \\ \text{(E) } \frac {a + b}{2} > \sqr... | [
"Using the RMS-AM-GM-HM inequality, we can see that the answer is $\\fbox{E}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/45.json | AHSME |
1952_AHSME_Problems | 32 | 0 | Algebra | Multiple Choice | $K$ takes $30$ minutes less time than $M$ to travel a distance of $30$ miles. $K$ travels $\frac {1}{3}$ mile per hour faster than $M$. If $x$ is $K$'s rate of speed in miles per hours, then $K$'s time for the distance is:
$\textbf{(A)}\ \dfrac{x + \frac {1}{3}}{30} \qquad \textbf{(B)}\ \dfrac{x - \frac {1}{3}}{30} ... | [
"Using the formula d=rt, and setting d=30 and r=x, we can easily see that the answer is $\\fbox{D}$.\nNote that the first sentence is irrelevant.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/32.json | AHSME |
1952_AHSME_Problems | 24 | 0 | Geometry | Multiple Choice | In the figure, it is given that angle $C = 90^{\circ}$, $\overline{AD} = \overline{DB}$, $DE \perp AB$, $\overline{AB} = 20$, and $\overline{AC} = 12$. The area of quadrilateral $ADEC$ is:
[asy] unitsize(7); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A,B,C,D,E; A=(0,0); B=(20,0); C=(36/5,48/5); D=(10,0); E=(10,7... | [
"$[ADEC]=[BCA]-[BDE]$\n\n\nNote that, as right triangles sharing $\\angle B$, $\\triangle BDE \\sim \\triangle BCA$. \n\n\n$\\overline{BD}=\\frac{20}{2}=10$\n\n\nBecause the sides of $\\triangle BCA$ are in the ratio $3:4:5$, $\\overline{BC}=16$.\n\n\nThe sides of our triangles are in the ratio $\\frac{10}{16}=\\fr... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/24.json | AHSME |
1952_AHSME_Problems | 49 | 0 | Geometry | Multiple Choice | [asy] unitsize(27); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A,B,C,D,E,F,X,Y,Z; A=(3,3); B=(0,0); C=(6,0); D=(4,0); E=(4,2); F=(1,1); draw(A--B--C--cycle); draw(A--D); draw(B--E); draw(C--F); X=intersectionpoint(A--D,C--F); Y=intersectionpoint(B--E,A--D); Z=intersectionpoint(B--E,C--F); label("$A$",A,N); label(... | [
"Let $[ABC]=K.$ Then $[ADC] = \\frac{1}{3}K,$ and hence $[N_1DC] = \\frac{1}{7} [ADC] = \\frac{1}{21}K.$ Similarly, $[N_2EA]=[N_3FB] = \\frac{1}{21}K.$ Then $[N_2N_1CE] = [ADC] - [N_1DC]-[N_2EA] = \\frac{5}{21}K,$ and same for the other quadrilaterals. Then $[N_1N_2N_3]$ is just $[ABC]$ minus all the other regions ... | 6 | ./CreativeMath/AHSME/1952_AHSME_Problems/49.json | AHSME |
1952_AHSME_Problems | 48 | 0 | Algebra | Multiple Choice | Two cyclists, $k$ miles apart, and starting at the same time, would be together in $r$ hours if they traveled in the same direction, but would pass each other in $t$ hours if they traveled in opposite directions. The ratio of the speed of the faster cyclist to that of the slower is:
$\text{(A) } \frac {r + t}{r - t}... | [
"[asy] pair A,B,C; A=(0,0); B=(8,0); C=(4,1); draw((A)--(B)); label(\"$A$\",A,S); label(\"$B$\",B,SE); label(\"$k$\",C,SW); [/asy]\n\n\n",
"We have a simple formula $d = rt$. The times the problem gives us, $r$ and $t$ are kind of annoying, so I will just let $x$ and $y$ be $r$ and $t$, respectively. I will also ... | 2 | ./CreativeMath/AHSME/1952_AHSME_Problems/48.json | AHSME |
1952_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | A powderman set a fuse for a blast to take place in $30$ seconds. He ran away at a rate of $8$ yards per second. Sound travels at the rate of $1080$ feet per second. When the powderman heard the blast, he had run approximately:
$\textbf{(A)}\ \text{200 yd.}\qquad\textbf{(B)}\ \text{352 yd.}\qquad\textbf{(C)}\ \text{300... | [
"Let $p(t)=24t$ be the number of feet the powderman is from the blast at $t$ seconds after the fuse is lit, and let $q(t)=1080t-32400$ be the number of feet the sound has traveled. We want to solve for $p(t)=q(t)$.\n\\[24t=1080t-32400\\]\n\\[1056t=32400\\]\n\\[t=\\frac{32400}{1056}\\]\n\\[t=\\frac{675}{22}=30.6\\ov... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/25.json | AHSME |
1952_AHSME_Problems | 33 | 0 | Geometry | Multiple Choice | A circle and a square have the same perimeter. Then:
$\text{(A) their areas are equal}\qquad\\ \text{(B) the area of the circle is the greater} \qquad\\ \text{(C) the area of the square is the greater} \qquad\\ \text{(D) the area of the circle is } \pi \text{ times the area of the square}\qquad\\ \text{(E) none of t... | [
"Assume that the side length of the square is 4 units. Then the perimeter for both shapes is is 4 * 4 = 16 units. Since the perimeter of a circle is 2πr, then 2πr =16 and r = 8/π, which is about 2.5. The area of a circle is πr^2 so (2.5^2)*π = 6.25π or about 19.6. This is more than the area of the square, which is ... | 1 | ./CreativeMath/AHSME/1952_AHSME_Problems/33.json | AHSME |
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