Difference between revisions of "1952 AHSME Problems"
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+ | {{AHSC 50 Problems | ||
+ | |year=1952 | ||
+ | }} | ||
== Problem 1 == | == Problem 1 == | ||
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== Problem 5 == | == Problem 5 == | ||
− | <math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math> | + | |
+ | The points <math> (6,12) </math> and <math> (0,-6) </math> are connected by a straight line. Another point on this line is: | ||
+ | |||
+ | <math> \textbf{(A) \ }(3,3) \qquad \textbf{(B) \ }(2,1) \qquad \textbf{(C) \ }(7,16) \qquad \textbf{(D) \ }(-1,-4) \qquad \textbf{(E) \ }(-3,-8) </math> | ||
[[1952 AHSME Problems/Problem 5|Solution]] | [[1952 AHSME Problems/Problem 5|Solution]] | ||
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== Problem 6 == | == Problem 6 == | ||
− | <math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math> | + | The difference of the roots of <math> x^2-7x-9=0 </math> is: |
+ | |||
+ | <math> \textbf{(A) \ }+7 \qquad \textbf{(B) \ }+\frac{7}{2} \qquad \textbf{(C) \ }+9 \qquad \textbf{(D) \ }2\sqrt{85} \qquad \textbf{(E) \ }\sqrt{85} </math> | ||
[[1952 AHSME Problems/Problem 6|Solution]] | [[1952 AHSME Problems/Problem 6|Solution]] | ||
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== Problem 7 == | == Problem 7 == | ||
− | <math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math> | + | When simplified, <math> (x^{-1}+y^{-1})^{-1} </math> is equal to: |
+ | |||
+ | <math> \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} </math> | ||
[[1952 AHSME Problems/Problem 7|Solution]] | [[1952 AHSME Problems/Problem 7|Solution]] | ||
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== Problem 8 == | == Problem 8 == | ||
− | <math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math> | + | Two equal circles in the same plane cannot have the following number of common tangents. |
+ | |||
+ | <math> \textbf{(A) \ }1 \qquad \textbf{(B) \ }2 \qquad \textbf{(C) \ }3 \qquad \textbf{(D) \ }4 \qquad \textbf{(E) \ }\text{none of these} </math> | ||
[[1952 AHSME Problems/Problem 8|Solution]] | [[1952 AHSME Problems/Problem 8|Solution]] | ||
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== Problem 9 == | == Problem 9 == | ||
− | <math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math> | + | If <math> m=\frac{cab}{a-b} </math>, then <math> b </math> equals: |
+ | |||
+ | <math> \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} </math> | ||
[[1952 AHSME Problems/Problem 9|Solution]] | [[1952 AHSME Problems/Problem 9|Solution]] | ||
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== Problem 10 == | == Problem 10 == | ||
− | <math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math> | + | An automobile went up a hill at a speed of <math> 10 </math> miles an hour and down the same distance at a speed of <math> 20 </math> miles an hour. The average speed for the round trip was: |
+ | |||
+ | <math> \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) \ }15\text{mph} \qquad \textbf{(E) \ }\text{none of these} </math> | ||
[[1952 AHSME Problems/Problem 10|Solution]] | [[1952 AHSME Problems/Problem 10|Solution]] | ||
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== Problem 11 == | == Problem 11 == | ||
− | <math> \textbf{(A) \ } | + | If <math> y=f(x)=\frac{x+2}{x-1} </math>, then it is incorrect to say: |
+ | |||
+ | <math> \textbf{(A)\ }x=\frac{y+2}{y-1}\qquad\textbf{(B)\ }f(0)=-2\qquad\textbf{(C)\ }f(1)=0\qquad </math> | ||
+ | |||
+ | <math> \textbf{(D)\ }f(-2)=0\qquad\textbf{(E)\ }f(y)=x </math> | ||
[[1952 AHSME Problems/Problem 11|Solution]] | [[1952 AHSME Problems/Problem 11|Solution]] | ||
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== Problem 12 == | == Problem 12 == | ||
− | <math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math> | + | The sum to infinity of the terms of an infinite geometric progression is <math> 6 </math>. The sum of the first two terms is <math> 4\frac{1}{2} </math>. The first term of the progression is: |
+ | |||
+ | <math> \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\text{ or }3 </math> | ||
[[1952 AHSME Problems/Problem 12|Solution]] | [[1952 AHSME Problems/Problem 12|Solution]] | ||
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== Problem 13 == | == Problem 13 == | ||
− | <math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math> | + | The function <math> x^2+px+q </math> with <math> p </math> and <math> q </math> greater than zero has its minimum value when: |
+ | |||
+ | <math> \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 </math> | ||
+ | |||
+ | <math> \textbf{(E) \ }x=\frac{-p}{2} </math> | ||
[[1952 AHSME Problems/Problem 13|Solution]] | [[1952 AHSME Problems/Problem 13|Solution]] | ||
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== Problem 14 == | == Problem 14 == | ||
− | <math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math> | + | A house and store were sold for <math> \textdollar 12,000 </math> each. The house was sold at a loss of <math> 20\% </math> of the cost, and the store at a gain of <math> 20\% </math> of the cost. The entire transaction resulted in: |
+ | |||
+ | <math> \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} </math> | ||
[[1952 AHSME Problems/Problem 14|Solution]] | [[1952 AHSME Problems/Problem 14|Solution]] | ||
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== Problem 15 == | == Problem 15 == | ||
− | <math> \textbf{(A) \ } | + | The sides of a triangle are in the ratio <math> 6:8:9 </math>. Then: |
+ | |||
+ | <math> \textbf{(A) \ }\text{the triangle is obtuse} </math> | ||
+ | |||
+ | <math> \textbf{(B) \ }\text{the angles are in the ratio }6:8:9 </math> | ||
+ | |||
+ | <math> \textbf{(C) \ }\text{the triangle is acute} </math> | ||
+ | |||
+ | <math> \textbf{(D) \ }\text{the angle opposite the largest side is double the angle opposite the smallest side} </math> | ||
+ | |||
+ | <math> \textbf{(E) \ }\text{none of these} </math> | ||
[[1952 AHSME Problems/Problem 15|Solution]] | [[1952 AHSME Problems/Problem 15|Solution]] | ||
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== Problem 16 == | == Problem 16 == | ||
− | <math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math> | + | If the base of a rectangle is increased by <math> 10\% </math> and the area is unchanged, then the altitude is decreased by: |
+ | |||
+ | <math> \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}\% </math> | ||
[[1952 AHSME Problems/Problem 16|Solution]] | [[1952 AHSME Problems/Problem 16|Solution]] | ||
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== Problem 17 == | == Problem 17 == | ||
− | <math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math> | + | A merchant bought some goods at a discount of <math> 20\% </math> of the list price. He wants to mark them at such a price that he can give a discount of <math> 20\% </math> of the marked price and still make a profit of <math> 20\% </math> of the selling price.. The percent of the list price at which he should mark them is: |
+ | |||
+ | <math> \textbf{(A) \ }20 \qquad \textbf{(B) \ }100 \qquad \textbf{(C) \ }125 \qquad \textbf{(D) \ }80 \qquad \textbf{(E) \ }120 </math> | ||
[[1952 AHSME Problems/Problem 17|Solution]] | [[1952 AHSME Problems/Problem 17|Solution]] | ||
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== Problem 18 == | == Problem 18 == | ||
− | <math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math> | + | <math> \log p+\log q=\log(p+q) </math> only if: |
+ | |||
+ | <math> \textbf{(A) \ }p=q=\text{zero} \qquad \textbf{(B) \ }p=\frac{q^2}{1-q} \qquad \textbf{(C) \ }p=q=1 \qquad </math> | ||
+ | |||
+ | <math> \textbf{(D) \ }p=\frac{q}{q-1} \qquad \textbf{(E) \ }p=\frac{q}{q+1} </math> | ||
[[1952 AHSME Problems/Problem 18|Solution]] | [[1952 AHSME Problems/Problem 18|Solution]] | ||
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== Problem 19 == | == Problem 19 == | ||
− | <math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math> | + | Angle <math> B </math> of triangle <math> ABC </math> is trisected by <math> BD </math> and <math> BE </math> which meet <math> AC </math> at <math> D </math> and <math> E </math> respectively. Then: |
+ | |||
+ | <math> \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 </math> | ||
+ | |||
+ | <math> \textbf{(D) \ }\frac{AD}{EC}=\frac{(AB)(BD)}{(BE)(BC)} \qquad \textbf{(E) \ }\frac{AD}{EC}=\frac{(AE)(BD)}{(DC)(BE)} </math> | ||
[[1952 AHSME Problems/Problem 19|Solution]] | [[1952 AHSME Problems/Problem 19|Solution]] | ||
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== Problem 20 == | == Problem 20 == | ||
− | <math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math> | + | If <math> \frac{x}{y}=\frac{3}{4} </math>, then the incorrect expression in the following is: |
+ | |||
+ | <math> \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 </math> | ||
+ | |||
+ | <math> \textbf{(D) \ }\frac{x}{2y}=\frac{3}{8} \qquad \textbf{(E) \ }\frac{x-y}{y}=\frac{1}{4} </math> | ||
[[1952 AHSME Problems/Problem 20|Solution]] | [[1952 AHSME Problems/Problem 20|Solution]] | ||
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== Problem 21 == | == Problem 21 == | ||
− | <math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math> | + | The sides of a regular polygon of <math> n </math> sides, <math> n>4 </math>, are extended to form a star. The number of degrees at each point of the star is: |
+ | |||
+ | <math> \textbf{(A) \ }\frac{360}{n} \qquad \textbf{(B) \ }\frac{(n-4)180}{n} \qquad \textbf{(C) \ }\frac{(n-2)180}{n} \qquad </math> | ||
+ | |||
+ | <math> \textbf{(D) \ }180-\frac{90}{n} \qquad \textbf{(E) \ }\frac{180}{n} </math> | ||
[[1952 AHSME Problems/Problem 21|Solution]] | [[1952 AHSME Problems/Problem 21|Solution]] | ||
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== Problem 22 == | == Problem 22 == | ||
− | <math> \textbf{(A) \ } \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math> | + | On hypotenuse <math> AB </math> of a right triangle <math> ABC </math> a second right triangle <math> ABD </math> is constructed with hypotenuse <math> AB </math>. If <math> \overline{BC}=1 </math>, <math> \overline{AC}=b </math>, and <math> \overline{AD}=2 </math>, then <math> \overline{BD} </math> equals: |
+ | |||
+ | <math> \textbf{(A) \ }\sqrt{b^2+1} \qquad \textbf{(B) \ }\sqrt{b^2-3} \qquad \textbf{(C) \ }\sqrt{b^2+1}+2 \qquad </math> | ||
+ | |||
+ | <math> \textbf{(D) \ }b^2+5 \qquad \textbf{(E) \ }\sqrt{b^2+3} </math> | ||
[[1952 AHSME Problems/Problem 22|Solution]] | [[1952 AHSME Problems/Problem 22|Solution]] | ||
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== Problem 23 == | == Problem 23 == | ||
− | <math> \textbf{(A) \ } | + | If <math> \frac{x^2-bx}{ax-c}=\frac{m-1}{m+1} </math> has roots which are numerically equal but of opposite signs, the value of <math> m </math> must be: |
+ | |||
+ | <math> \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 </math> | ||
[[1952 AHSME Problems/Problem 23|Solution]] | [[1952 AHSME Problems/Problem 23|Solution]] | ||
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== Problem 24 == | == Problem 24 == | ||
− | <math> \textbf{(A) \ | + | In the figure, it is given that angle <math> C = 90^{\circ} </math>,<math> \overline{AD} = \overline{DB} </math>,<math> DE \perp AB </math>, <math> \overline{AB} = 20 </math>, and <math> \overline{AC} = 12 </math>. The area of quadrilateral <math> ADEC </math> 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=(10,75/10); | ||
+ | draw(A--B--C--cycle); draw(D--E); | ||
+ | label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NE); | ||
+ | draw(rightanglemark(B,D,E,30)); | ||
+ | </asy> | ||
+ | |||
+ | <math> \textbf{(A)}\ 75\qquad\textbf{(B)}\ 58\frac{1}{2}\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 37\frac{1}{2}\qquad\textbf{(E)}\ \text{none of these} </math> | ||
[[1952 AHSME Problems/Problem 24|Solution]] | [[1952 AHSME Problems/Problem 24|Solution]] | ||
== Problem 25 == | == Problem 25 == | ||
+ | A powderman set a fuse for a blast to take place in <math>30</math> seconds. He ran away at a rate of <math>8</math> yards per second. Sound travels at the rate of <math>1080</math> feet per second. When the powderman heard the blast, he had run approximately: | ||
− | <math> \textbf{(A) \ } | + | <math> \textbf{(A)}\ \text{200 yd.}\qquad\textbf{(B)}\ \text{352 yd.}\qquad\textbf{(C)}\ \text{300 yd.}\qquad\textbf{(D)}\ \text{245 yd.}\qquad\textbf{(E)}\ \text{512 yd.} </math> |
[[1952 AHSME Problems/Problem 25|Solution]] | [[1952 AHSME Problems/Problem 25|Solution]] | ||
== Problem 26 == | == Problem 26 == | ||
+ | If <math>\left(r+\frac1r\right)^2=3</math>, then <math>r^3+\frac1{r^3}</math> equals | ||
− | <math> \textbf{(A) \ | + | <math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 6</math> |
[[1952 AHSME Problems/Problem 26|Solution]] | [[1952 AHSME Problems/Problem 26|Solution]] | ||
== Problem 27 == | == Problem 27 == | ||
+ | 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: | ||
− | <math> \textbf{(A) \ | + | <math> \textbf{(A)}\ 1:2\qquad\textbf{(B)}\ 1:3\qquad\textbf{(C)}\ 1:\sqrt3\qquad\textbf{(D)}\ \sqrt3:2 \qquad\textbf{(E)}\ 2:3</math> |
[[1952 AHSME Problems/Problem 27|Solution]] | [[1952 AHSME Problems/Problem 27|Solution]] | ||
== Problem 28 == | == Problem 28 == | ||
+ | In the table shown, the formula relating <math>x</math> and <math>y</math> is: | ||
+ | |||
+ | <math>\begin{tabular}{|l|l|l|l|l|l|} | ||
+ | \hline | ||
+ | x & 1 & 2 & 3 & 4 & 5 \ \hline | ||
+ | y & 3 & 7 & 13 & 21 & 31 \ | ||
+ | \hline | ||
+ | \end{tabular}</math> | ||
+ | |||
− | <math> \textbf{(A) \ | + | <math> \textbf{(A)}\ y=4x-1 \qquad \textbf{(B)}\ y=x^{3}-x^{2}+x+2 \qquad \textbf{(C)}\ y=x^2+x+1 \\ \qquad \textbf{(D)}\ y=(x^2+x+1)(x-1) \qquad \textbf{(E)}\ \text{None of these}</math> |
[[1952 AHSME Problems/Problem 28|Solution]] | [[1952 AHSME Problems/Problem 28|Solution]] | ||
== Problem 29 == | == Problem 29 == | ||
+ | In a circle of radius <math>5</math> units, <math>CD</math> and <math>AB</math> are perpendicular diameters. A chord <math>CH</math> cutting <math>AB</math> at <math>K</math> is <math>8</math> units long. The diameter <math>AB</math> is divided into two segments whose dimensions are: | ||
− | <math> \textbf{(A) \ | + | <math> \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}</math> |
[[1952 AHSME Problems/Problem 29|Solution]] | [[1952 AHSME Problems/Problem 29|Solution]] | ||
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== Problem 30 == | == Problem 30 == | ||
− | <math> \textbf{(A) \ | + | 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: |
+ | |||
+ | <math>\textbf{(A)}\ 1: 2 \qquad | ||
+ | \textbf{(B)}\ 2: 1 \qquad | ||
+ | \textbf{(C)}\ 1: 4 \qquad | ||
+ | \textbf{(D)}\ 4: 1 \qquad | ||
+ | \textbf{(E)}\ 1: 1 </math> | ||
[[1952 AHSME Problems/Problem 30|Solution]] | [[1952 AHSME Problems/Problem 30|Solution]] | ||
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== Problem 31 == | == Problem 31 == | ||
− | <math> | + | Given <math>12</math> points in a plane no three of which are collinear, the number of lines they determine is: |
+ | <math>\textbf{(A)}\ 24 \qquad | ||
+ | \textbf{(B)}\ 54 \qquad | ||
+ | \textbf{(C)}\ 120 \qquad | ||
+ | \textbf{(D)}\ 66 \qquad | ||
+ | \textbf{(E)}\ \text{none of these}</math> | ||
[[1952 AHSME Problems/Problem 31|Solution]] | [[1952 AHSME Problems/Problem 31|Solution]] | ||
== Problem 32 == | == Problem 32 == | ||
− | <math> \textbf{(A) \ } | + | <math>K</math> takes <math>30</math> minutes less time than <math>M</math> to travel a distance of <math>30</math> miles. <math>K</math> travels <math>\frac {1}{3}</math> mile per hour faster than <math>M</math>. If <math>x</math> is <math>K</math>'s rate of speed in miles per hours, then <math>K</math>'s time for the distance is: |
+ | |||
+ | <math>\textbf{(A)}\ \dfrac{x + \frac {1}{3}}{30} \qquad | ||
+ | \textbf{(B)}\ \dfrac{x - \frac {1}{3}}{30} \qquad | ||
+ | \textbf{(C)}\ \frac{30}{x+\frac{1}{3}}\qquad | ||
+ | \textbf{(D)}\ \frac{30}{x}\qquad | ||
+ | \textbf{(E)}\ \frac{x}{30} </math> | ||
[[1952 AHSME Problems/Problem 32|Solution]] | [[1952 AHSME Problems/Problem 32|Solution]] | ||
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== Problem 33 == | == Problem 33 == | ||
− | <math> \ | + | A circle and a square have the same perimeter. Then: |
+ | |||
+ | <math>\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 these}</math> | ||
[[1952 AHSME Problems/Problem 33|Solution]] | [[1952 AHSME Problems/Problem 33|Solution]] | ||
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== Problem 34 == | == Problem 34 == | ||
− | <math> \textbf{(A) \ } | + | The price of an article was increased <math>p\%</math>. Later the new price was decreased <math>p\%</math>. If the last price was one dollar, the original price was: |
+ | |||
+ | <math>\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} </math> | ||
+ | |||
[[1952 AHSME Problems/Problem 34|Solution]] | [[1952 AHSME Problems/Problem 34|Solution]] | ||
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== Problem 35 == | == Problem 35 == | ||
− | <math> \textbf{(A) \ } | + | With a rational denominator, the expression <math>\frac {\sqrt {2}}{\sqrt {2} + \sqrt {3} - \sqrt {5}}</math> is equivalent to: |
+ | |||
+ | <math>\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} </math> | ||
[[1952 AHSME Problems/Problem 35|Solution]] | [[1952 AHSME Problems/Problem 35|Solution]] | ||
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== Problem 36 == | == Problem 36 == | ||
− | <math> \textbf{(A) \ | + | To be continuous at <math>x = - 1</math>, the value of <math>\frac {x^3 + 1}{x^2 - 1}</math> is taken to be: |
+ | |||
+ | <math>\textbf{(A)}\ - 2 \qquad | ||
+ | \textbf{(B)}\ 0 \qquad | ||
+ | \textbf{(C)}\ \frac {3}{2} \qquad | ||
+ | \textbf{(D)}\ \infty \qquad | ||
+ | \textbf{(E)}\ -\frac{3}{2} </math> | ||
[[1952 AHSME Problems/Problem 36|Solution]] | [[1952 AHSME Problems/Problem 36|Solution]] | ||
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== Problem 37 == | == Problem 37 == | ||
− | <math> \textbf{(A) \ } | + | Two equal parallel chords are drawn <math>8</math> inches apart in a circle of radius <math>8</math> inches. The area of that part of the circle that lies between the chords is: |
+ | |||
+ | <math>\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)}\ 16\sqrt {3} + 42\frac {2}{3}\pi \qquad | ||
+ | \textbf{(E)}\ 42\frac {2}{3}\pi </math> | ||
[[1952 AHSME Problems/Problem 37|Solution]] | [[1952 AHSME Problems/Problem 37|Solution]] | ||
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== Problem 38 == | == Problem 38 == | ||
− | <math> \textbf{(A) \ } | + | The area of a trapezoidal field is <math>1400</math> square yards. Its altitude is <math>50</math> yards. Find the two bases, if the number of yards in each base is an integer divisible by <math>8</math>. The number of solutions to this problem is: |
+ | |||
+ | <math>\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}</math> | ||
+ | |||
[[1952 AHSME Problems/Problem 38|Solution]] | [[1952 AHSME Problems/Problem 38|Solution]] | ||
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== Problem 39 == | == Problem 39 == | ||
− | <math> \textbf{(A) \ } | + | If the perimeter of a rectangle is <math>p</math> and its diagonal is <math>d</math>, the difference between the length and width of the rectangle is: |
+ | |||
+ | <math>\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} </math> | ||
[[1952 AHSME Problems/Problem 39|Solution]] | [[1952 AHSME Problems/Problem 39|Solution]] | ||
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== Problem 40 == | == Problem 40 == | ||
− | <math> \ | + | In order to draw a graph of <math>ax^2+bx+c</math>, a table of values was constructed. These values of the function for a set of equally spaced increasing values of <math>x</math> were <math>3844, 3969, 4096, 4227, 4356, 4489, 4624</math>, and <math>4761</math>. The one which is incorrect is: |
+ | |||
+ | <math>\text{(A) } 4096 \qquad | ||
+ | \text{(B) } 4356 \qquad | ||
+ | \text{(C) } 4489 \qquad | ||
+ | \text{(D) } 4761 \qquad | ||
+ | \text{(E) } \text{none of these}</math> | ||
[[1952 AHSME Problems/Problem 40|Solution]] | [[1952 AHSME Problems/Problem 40|Solution]] | ||
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== Problem 41 == | == Problem 41 == | ||
− | <math> \ | + | Increasing the radius of a cylinder by <math>6</math> units increased the volume by <math>y</math> cubic units. Increasing the altitude of the cylinder by <math>6</math> units also increases the volume by <math>y</math> cubic units. If the original altitude is <math>2</math>, then the original radius is: |
+ | |||
+ | <math>\text{(A) } 2 \qquad | ||
+ | \text{(B) } 4 \qquad | ||
+ | \text{(C) } 6 \qquad | ||
+ | \text{(D) } 6\pi \qquad | ||
+ | \text{(E) } 8 </math> | ||
[[1952 AHSME Problems/Problem 41|Solution]] | [[1952 AHSME Problems/Problem 41|Solution]] | ||
== Problem 42 == | == Problem 42 == | ||
+ | Let <math>D</math> represent a repeating decimal. If <math>P</math> denotes the <math>r</math> figures of <math>D</math> which do not repeat themselves, and <math>Q</math> denotes the <math>s</math> figures of <math>D</math> which do repeat themselves, then the incorrect expression is: | ||
− | <math> \textbf{(A) \ | + | <math>\textbf{(A) } D = .PQQQ\ldots \qquad\ \textbf{(B) } 10^rD = P.QQQ\ldots \ \textbf{(C) } 10^{r + s}D = PQ.QQQ\ldots \qquad\ \textbf{(D) } 10^r(10^s - 1)D = Q(P - 1) \ \textbf{(E) } 10^r\cdot10^{2s}D = PQQ.QQQ\ldots</math> |
[[1952 AHSME Problems/Problem 42|Solution]] | [[1952 AHSME Problems/Problem 42|Solution]] | ||
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== Problem 43 == | == Problem 43 == | ||
− | <math> \textbf{(A) | + | The diameter of a circle is divided into <math>n</math> equal parts. On each part a semicircle is constructed. As <math>n</math> becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length: |
+ | |||
+ | <math>\textbf{(A) } \qquad</math> equal to the semi-circumference of the original circle | ||
+ | |||
+ | <math>\textbf{(B) } \qquad</math> equal to the diameter of the original circle | ||
+ | |||
+ | <math>\textbf{(C) } </math> greater than the diameter, but less than the semi-circumference of the original circle | ||
+ | |||
+ | <math>\textbf{(D) } \qquad</math> that is infinite | ||
+ | |||
+ | <math>\textbf{(E) } </math> greater than the semi-circumference | ||
[[1952 AHSME Problems/Problem 43|Solution]] | [[1952 AHSME Problems/Problem 43|Solution]] | ||
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== Problem 45 == | == Problem 45 == | ||
− | <math> \ | + | If <math>a</math> and <math>b</math> are two unequal positive numbers, then: |
+ | |||
+ | <math>\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} > \sqrt {ab} > \frac {2ab}{a + b} </math> | ||
[[1952 AHSME Problems/Problem 45|Solution]] | [[1952 AHSME Problems/Problem 45|Solution]] | ||
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== Problem 46 == | == Problem 46 == | ||
− | <math> \ | + | |
+ | 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: | ||
+ | |||
+ | <math>\text{(A) greater than the area of the given rectangle} \quad\ | ||
+ | \text{(B) equal to the area of the given rectangle} \quad\ | ||
+ | \text{(C) equal to the area of a square with its side equal to the smaller side of the given rectangle} \quad\ | ||
+ | \text{(D) equal to the area of a square with its side equal to the greater side of the given rectangle} \quad\ | ||
+ | \text{(E) equal to the area of a rectangle whose dimensions are the diagonal and the shorter side of the given rectangle}</math> | ||
+ | |||
[[1952 AHSME Problems/Problem 46|Solution]] | [[1952 AHSME Problems/Problem 46|Solution]] | ||
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== Problem 47 == | == Problem 47 == | ||
− | <math> \textbf{(A) | + | In the set of equations <math>z^x = y^{2x},\quad 2^z = 2\cdot4^x, \quad x + y + z = 16</math>, the integral roots in the order <math>x,y,z</math> are: |
+ | |||
+ | <math>\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 </math> | ||
[[1952 AHSME Problems/Problem 47|Solution]] | [[1952 AHSME Problems/Problem 47|Solution]] | ||
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== Problem 48 == | == Problem 48 == | ||
− | <math> \ | + | Two cyclists, <math>k</math> miles apart, and starting at the same time, would be together in <math>r</math> hours if they traveled in the same direction, but would pass each other in <math>t</math> hours if they traveled in opposite directions. The ratio of the speed of the faster cyclist to that of the slower is: |
+ | |||
+ | <math>\text{(A) } \frac {r + t}{r - t} \qquad | ||
+ | \text{(B) } \frac {r}{r - t} \qquad | ||
+ | \text{(C) } \frac {r + t}{r} \qquad | ||
+ | \text{(D) } \frac{r}{t}\qquad | ||
+ | \text{(E) } \frac{r+k}{t-k}</math> | ||
+ | |||
[[1952 AHSME Problems/Problem 48|Solution]] | [[1952 AHSME Problems/Problem 48|Solution]] | ||
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== Problem 49 == | == Problem 49 == | ||
− | <math> \ | + | |
+ | <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("$B$",B,SW); label("$C$",C,SE); | ||
+ | label("$D$",D,S); label("$E$",E,NE); label("$F$",F,NW); | ||
+ | label("$N_1$",X,NE); label("$N_2$",Y,WNW); label("$N_3$",Z,S); | ||
+ | </asy> | ||
+ | |||
+ | In the figure, <math>\overline{CD}</math>, <math>\overline{AE}</math> and <math>\overline{BF}</math> are one-third of their respective sides. It follows that <math>\overline{AN_2}: \overline{N_2N_1}: \overline{N_1D} = 3: 3: 1</math>, and similarly for lines BE and CF. Then the area of triangle <math>N_1N_2N_3</math> is: | ||
+ | |||
+ | <math>\text{(A) } \frac {1}{10} \triangle ABC \qquad | ||
+ | \text{(B) } \frac {1}{9} \triangle ABC \qquad | ||
+ | \text{(C) } \frac{1}{7}\triangle ABC\qquad | ||
+ | \text{(D) } \frac{1}{6}\triangle ABC\qquad | ||
+ | \text{(E) } \text{none of these}</math> | ||
[[1952 AHSME Problems/Problem 49|Solution]] | [[1952 AHSME Problems/Problem 49|Solution]] | ||
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== Problem 50 == | == Problem 50 == | ||
− | <math> \textbf{(A) \ | + | A line initially 1 inch long grows according to the following law, where the first term is the initial length. |
+ | <math> 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 </math> | ||
+ | |||
+ | If the growth process continues forever, the limit of the length of the line is: | ||
+ | |||
+ | <math>\textbf{(A) } \infty\qquad | ||
+ | \textbf{(B) } \frac{4}{3}\qquad | ||
+ | \textbf{(C) } \frac{8}{3}\qquad | ||
+ | \textbf{(D) } \frac{1}{3}(4+\sqrt{2})\qquad | ||
+ | \textbf{(E) } \frac{2}{3}(4+\sqrt{2})</math> | ||
[[1952 AHSME Problems/Problem 50|Solution]] | [[1952 AHSME Problems/Problem 50|Solution]] | ||
== See also == | == See also == | ||
− | + | ||
− | * [[ | + | * [[AMC 12 Problems and Solutions]] |
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
+ | |||
+ | {{AHSME 50p box|year=1952|before=[[1951 AHSME|1951 AHSC]]|after=[[1953 AHSME|1953 AHSC]]}} | ||
+ | |||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:35, 30 December 2023
1952 AHSC (Answer Key) Printable version: | AoPS Resources • PDF | ||
Instructions
| ||
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Contents
[hide]- 1 Problem 1
- 2 Problem 2
- 3 Problem 3
- 4 Problem 4
- 5 Problem 5
- 6 Problem 6
- 7 Problem 7
- 8 Problem 8
- 9 Problem 9
- 10 Problem 10
- 11 Problem 11
- 12 Problem 12
- 13 Problem 13
- 14 Problem 14
- 15 Problem 15
- 16 Problem 16
- 17 Problem 17
- 18 Problem 18
- 19 Problem 19
- 20 Problem 20
- 21 Problem 21
- 22 Problem 22
- 23 Problem 23
- 24 Problem 24
- 25 Problem 25
- 26 Problem 26
- 27 Problem 27
- 28 Problem 28
- 29 Problem 29
- 30 Problem 30
- 31 Problem 31
- 32 Problem 32
- 33 Problem 33
- 34 Problem 34
- 35 Problem 35
- 36 Problem 36
- 37 Problem 37
- 38 Problem 38
- 39 Problem 39
- 40 Problem 40
- 41 Problem 41
- 42 Problem 42
- 43 Problem 43
- 44 Problem 44
- 45 Problem 45
- 46 Problem 46
- 47 Problem 47
- 48 Problem 48
- 49 Problem 49
- 50 Problem 50
- 51 See also
Problem 1
If the radius of a circle is a rational number, its area is given by a number which is:
Problem 2
Two high school classes took the same test. One class of students made an average grade of ; the other class of students made an average grade of . The average grade for all students in both classes is:
Problem 3
The expression equals:
Problem 4
The cost of sending a parcel post package weighing pounds, an integer, is cents for the first pound and cents for each additional pound. The formula for the cost is:
Problem 5
The points and are connected by a straight line. Another point on this line is:
Problem 6
The difference of the roots of is:
Problem 7
When simplified, is equal to:
Problem 8
Two equal circles in the same plane cannot have the following number of common tangents.
Problem 9
If , then equals:
Problem 10
An automobile went up a hill at a speed of miles an hour and down the same distance at a speed of miles an hour. The average speed for the round trip was:
Problem 11
If , then it is incorrect to say:
Problem 12
The sum to infinity of the terms of an infinite geometric progression is . The sum of the first two terms is . The first term of the progression is:
Problem 13
The function with and greater than zero has its minimum value when:
Problem 14
A house and store were sold for each. The house was sold at a loss of of the cost, and the store at a gain of of the cost. The entire transaction resulted in:
Problem 15
The sides of a triangle are in the ratio . Then:
Problem 16
If the base of a rectangle is increased by and the area is unchanged, then the altitude is decreased by:
Problem 17
A merchant bought some goods at a discount of of the list price. He wants to mark them at such a price that he can give a discount of of the marked price and still make a profit of of the selling price.. The percent of the list price at which he should mark them is:
Problem 18
only if:
Problem 19
Angle of triangle is trisected by and which meet at and respectively. Then:
Problem 20
If , then the incorrect expression in the following is:
Problem 21
The sides of a regular polygon of sides, , are extended to form a star. The number of degrees at each point of the star is:
Problem 22
On hypotenuse of a right triangle a second right triangle is constructed with hypotenuse . If , , and , then equals:
Problem 23
If has roots which are numerically equal but of opposite signs, the value of must be:
Problem 24
In the figure, it is given that angle ,,, , and . The area of quadrilateral is:
Problem 25
A powderman set a fuse for a blast to take place in seconds. He ran away at a rate of yards per second. Sound travels at the rate of feet per second. When the powderman heard the blast, he had run approximately:
Problem 26
If , then equals
Problem 27
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:
Problem 28
In the table shown, the formula relating and is:
Problem 29
In a circle of radius units, and are perpendicular diameters. A chord cutting at is units long. The diameter is divided into two segments whose dimensions are:
Problem 30
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:
Problem 31
Given points in a plane no three of which are collinear, the number of lines they determine is:
Problem 32
takes minutes less time than to travel a distance of miles. travels mile per hour faster than . If is 's rate of speed in miles per hours, then 's time for the distance is:
Problem 33
A circle and a square have the same perimeter. Then:
Problem 34
The price of an article was increased . Later the new price was decreased . If the last price was one dollar, the original price was:
Problem 35
With a rational denominator, the expression is equivalent to:
Problem 36
To be continuous at , the value of is taken to be:
Problem 37
Two equal parallel chords are drawn inches apart in a circle of radius inches. The area of that part of the circle that lies between the chords is:
Problem 38
The area of a trapezoidal field is square yards. Its altitude is yards. Find the two bases, if the number of yards in each base is an integer divisible by . The number of solutions to this problem is:
Problem 39
If the perimeter of a rectangle is and its diagonal is , the difference between the length and width of the rectangle is:
Problem 40
In order to draw a graph of , a table of values was constructed. These values of the function for a set of equally spaced increasing values of were , and . The one which is incorrect is:
Problem 41
Increasing the radius of a cylinder by units increased the volume by cubic units. Increasing the altitude of the cylinder by units also increases the volume by cubic units. If the original altitude is , then the original radius is:
Problem 42
Let represent a repeating decimal. If denotes the figures of which do not repeat themselves, and denotes the figures of which do repeat themselves, then the incorrect expression is:
Problem 43
The diameter of a circle is divided into equal parts. On each part a semicircle is constructed. As becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length:
equal to the semi-circumference of the original circle
equal to the diameter of the original circle
greater than the diameter, but less than the semi-circumference of the original circle
that is infinite
greater than the semi-circumference
Problem 44
If an integer of two digits is times the sum of its digits, the number formed by interchanging the the digits is the sum of the digits multiplied by
Problem 45
If and are two unequal positive numbers, then:
Problem 46
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:
Problem 47
In the set of equations , the integral roots in the order are:
Problem 48
Two cyclists, miles apart, and starting at the same time, would be together in hours if they traveled in the same direction, but would pass each other in hours if they traveled in opposite directions. The ratio of the speed of the faster cyclist to that of the slower is:
Problem 49
In the figure, , and are one-third of their respective sides. It follows that , and similarly for lines BE and CF. Then the area of triangle is:
Problem 50
A line initially 1 inch long grows according to the following law, where the first term is the initial length.
If the growth process continues forever, the limit of the length of the line is:
See also
1952 AHSC (Problems • Answer Key • Resources) | ||
Preceded by 1951 AHSC |
Followed by 1953 AHSC | |
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All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.