Difference between revisions of "1952 AHSME Problems"

Latest revision as of 14:35, 30 December 2023

1952 AHSC (Answer Key) Printable version: Wiki \| AoPS Resources • PDF
Instructions This is a 50-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. You will receive ? points for each correct answer, ? points for each problem left unanswered, and ? points for each incorrect answer. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers. Figures are not necessarily drawn to scale. You will have ? minutes working time to complete the test.
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Problem 1

If the radius of a circle is a rational number, its area is given by a number which is:

$\textbf{(A)\ } \text{rational} \qquad \textbf{(B)\ } \text{irrational} \qquad \textbf{(C)\ } \text{integral} \qquad \textbf{(D)\ } \text{a perfect square }\qquad \textbf{(E)\ } \text{none of these}$

Solution

Problem 2

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:

$\textbf{(A)}\ 75\%\qquad \textbf{(B)}\ 74\%\qquad \textbf{(C)}\ 72\%\qquad \textbf{(D)}\ 77\%\qquad \textbf{(E)\ }\text{none of these}$

Solution

Problem 3

The expression $a^3-a^{-3}$ equals:

$\textbf{(A) \ }\left(a-\frac{1}{a}\right)\left(a^2+1+\frac{1}{a^2}\right) \qquad \textbf{(B) \ }\left(\frac{1}{a}-a\right)\left(a^2-1+\frac{1}{a^2}\right) \qquad \textbf{(C) \ }\left(a-\frac{1}{a}\right)\left(a^2-2+\frac{1}{a^2}\right) \qquad$ $\textbf{(D) \ }\left(\frac{1}{a}-a\right)\left(\frac{1}{a^2}+1+a^2\right) \qquad \textbf{(E) \ }\text{none of these}$

Solution

Problem 4

The cost $C$ of sending a parcel post package weighing $P$ pounds, $P$ an integer, is $10$ cents for the first pound and $3$ cents for each additional pound. The formula for the cost is:

$\textbf{(A) \ }C=10+3P \qquad \textbf{(B) \ }C=10P+3 \qquad \textbf{(C) \ }C=10+3(P-1) \qquad$

$\textbf{(D) \ }C=9+3P \qquad \textbf{(E) \ }C=10P-7$

Solution

Problem 5

The points $(6,12)$ and $(0,-6)$ are connected by a straight line. Another point on this line is:

$\textbf{(A) \ }(3,3) \qquad \textbf{(B) \ }(2,1) \qquad \textbf{(C) \ }(7,16) \qquad \textbf{(D) \ }(-1,-4) \qquad \textbf{(E) \ }(-3,-8)$

Solution

Problem 6

The difference of the roots of $x^2-7x-9=0$ is:

$\textbf{(A) \ }+7 \qquad \textbf{(B) \ }+\frac{7}{2} \qquad \textbf{(C) \ }+9 \qquad \textbf{(D) \ }2\sqrt{85} \qquad \textbf{(E) \ }\sqrt{85}$

Solution

Problem 7

When simplified, $(x^{-1}+y^{-1})^{-1}$ is equal to:

$\textbf{(A) \ }x+y \qquad \textbf{(B) \ }\frac{xy}{x+y} \qquad \textbf{(C) \ }xy \qquad \textbf{(D) \ }\frac{1}{xy} \qquad \textbf{(E) \ }\frac{x+y}{xy}$

Solution

Problem 8

Two equal circles in the same plane cannot have the following number of common tangents.

$\textbf{(A) \ }1 \qquad \textbf{(B) \ }2 \qquad \textbf{(C) \ }3 \qquad \textbf{(D) \ }4 \qquad \textbf{(E) \ }\text{none of these}$

Solution

Problem 9

If $m=\frac{cab}{a-b}$ , then $b$ equals:

$\textbf{(A) \ }\frac{m(a-b)}{ca} \qquad \textbf{(B) \ }\frac{cab-ma}{-m} \qquad \textbf{(C) \ }\frac{1}{1+c} \qquad \textbf{(D) \ }\frac{ma}{m+ca} \qquad \textbf{(E) \ }\frac{m+ca}{ma}$

Solution

Problem 10

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:

$\textbf{(A) \ }12\frac{1}{2}\text{mph} \qquad \textbf{(B) \ }13\frac{1}{3}\text{mph} \qquad \textbf{(C) \ }14\frac{1}{2}\text{mph} \qquad \textbf{(D) \ }15\text{mph} \qquad \textbf{(E) \ }\text{none of these}$

Solution

Problem 11

If $y=f(x)=\frac{x+2}{x-1}$ , then it is incorrect to say:

$\textbf{(A)\ }x=\frac{y+2}{y-1}\qquad\textbf{(B)\ }f(0)=-2\qquad\textbf{(C)\ }f(1)=0\qquad$

$\textbf{(D)\ }f(-2)=0\qquad\textbf{(E)\ }f(y)=x$

Solution

Problem 12

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:

$\textbf{(A) \ }3 \text{ or } 1\frac{1}{2} \qquad \textbf{(B) \ }1 \qquad \textbf{(C) \ }2\frac{1}{2} \qquad \textbf{(D) \ }6 \qquad \textbf{(E) \ }9\text{ or }3$

Solution

Problem 13

The function $x^2+px+q$ with $p$ and $q$ greater than zero has its minimum value when:

$\textbf{(A) \ }x=-p \qquad \textbf{(B) \ }x=\frac{p}{2} \qquad \textbf{(C) \ }x=-2p \qquad \textbf{(D) \ }x=\frac{p^2}{4q} \qquad$

$\textbf{(E) \ }x=\frac{-p}{2}$

Solution

Problem 14

A house and store were sold for $\textdollar 12,000$ each. The house was sold at a loss of $20\%$ of the cost, and the store at a gain of $20\%$ of the cost. The entire transaction resulted in:

$\textbf{(A) \ }\text{no loss or gain} \qquad \textbf{(B) \ }\text{loss of }\textdollar 1000 \qquad \textbf{(C) \ }\text{gain of }\textdollar 1000 \qquad \textbf{(D) \ }\text{gain of }\textdollar 2000 \qquad \textbf{(E) \ }\text{none of these}$

Solution

Problem 15

The sides of a triangle are in the ratio $6:8:9$ . Then:

$\textbf{(A) \ }\text{the triangle is obtuse}$

$\textbf{(B) \ }\text{the angles are in the ratio }6:8:9$

$\textbf{(C) \ }\text{the triangle is acute}$

$\textbf{(D) \ }\text{the angle opposite the largest side is double the angle opposite the smallest side}$

$\textbf{(E) \ }\text{none of these}$

Solution

Problem 16

If the base of a rectangle is increased by $10\%$ and the area is unchanged, then the altitude is decreased by:

$\textbf{(A) \ }9\% \qquad \textbf{(B) \ }10\% \qquad \textbf{(C) \ }11\% \qquad \textbf{(D) \ }11\frac{1}{9}\% \qquad \textbf{(E) \ }9\frac{1}{11}\%$

Solution

Problem 17

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 percent of the list price at which he should mark them is:

$\textbf{(A) \ }20 \qquad \textbf{(B) \ }100 \qquad \textbf{(C) \ }125 \qquad \textbf{(D) \ }80 \qquad \textbf{(E) \ }120$

Solution

Problem 18

$\log p+\log q=\log(p+q)$ only if:

$\textbf{(A) \ }p=q=\text{zero} \qquad \textbf{(B) \ }p=\frac{q^2}{1-q} \qquad \textbf{(C) \ }p=q=1 \qquad$

$\textbf{(D) \ }p=\frac{q}{q-1} \qquad \textbf{(E) \ }p=\frac{q}{q+1}$

Solution

Problem 19

Angle $B$ of triangle $ABC$ is trisected by $BD$ and $BE$ which meet $AC$ at $D$ and $E$ respectively. Then:

$\textbf{(A) \ }\frac{AD}{EC}=\frac{AE}{DC} \qquad \textbf{(B) \ }\frac{AD}{EC}=\frac{AB}{BC} \qquad \textbf{(C) \ }\frac{AD}{EC}=\frac{BD}{BE} \qquad$

$\textbf{(D) \ }\frac{AD}{EC}=\frac{(AB)(BD)}{(BE)(BC)} \qquad \textbf{(E) \ }\frac{AD}{EC}=\frac{(AE)(BD)}{(DC)(BE)}$

Solution

Problem 20

If $\frac{x}{y}=\frac{3}{4}$ , then the incorrect expression in the following is:

$\textbf{(A) \ }\frac{x+y}{y}=\frac{7}{4} \qquad \textbf{(B) \ }\frac{y}{y-x}=\frac{4}{1} \qquad \textbf{(C) \ }\frac{x+2y}{x}=\frac{11}{3} \qquad$

$\textbf{(D) \ }\frac{x}{2y}=\frac{3}{8} \qquad \textbf{(E) \ }\frac{x-y}{y}=\frac{1}{4}$

Solution

Problem 21

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:

$\textbf{(A) \ }\frac{360}{n} \qquad \textbf{(B) \ }\frac{(n-4)180}{n} \qquad \textbf{(C) \ }\frac{(n-2)180}{n} \qquad$

$\textbf{(D) \ }180-\frac{90}{n} \qquad \textbf{(E) \ }\frac{180}{n}$

Solution

Problem 22

On hypotenuse $AB$ of a right triangle $ABC$ a second right triangle $ABD$ is constructed with hypotenuse $AB$ . If $\overline{BC}=1$ , $\overline{AC}=b$ , and $\overline{AD}=2$ , then $\overline{BD}$ equals:

$\textbf{(A) \ }\sqrt{b^2+1} \qquad \textbf{(B) \ }\sqrt{b^2-3} \qquad \textbf{(C) \ }\sqrt{b^2+1}+2 \qquad$

$\textbf{(D) \ }b^2+5 \qquad \textbf{(E) \ }\sqrt{b^2+3}$

Solution

Problem 23

If $\frac{x^2-bx}{ax-c}=\frac{m-1}{m+1}$ has roots which are numerically equal but of opposite signs, the value of $m$ must be:

$\textbf{(A) \ }\frac{a-b}{a+b} \qquad \textbf{(B) \ }\frac{a+b}{a-b} \qquad \textbf{(C) \ }c \qquad \textbf{(D) \ }\frac{1}{c} \qquad \textbf{(E) \ }1$

Solution

Problem 24

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: $[asy] unitsize(7); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A,B,C,D,E; A=(0,0); B=(20,0); C=(36/5,48/5); D=(10,0); E=(10,75/10); draw(A--B--C--cycle); draw(D--E); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NE); draw(rightanglemark(B,D,E,30)); [/asy]$

$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 58\frac{1}{2}\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 37\frac{1}{2}\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 25

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:

$\textbf{(A)}\ \text{200 yd.}\qquad\textbf{(B)}\ \text{352 yd.}\qquad\textbf{(C)}\ \text{300 yd.}\qquad\textbf{(D)}\ \text{245 yd.}\qquad\textbf{(E)}\ \text{512 yd.}$

Solution

Problem 26

If $\left(r+\frac1r\right)^2=3$ , then $r^3+\frac1{r^3}$ equals

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 6$

Solution

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:

$\textbf{(A)}\ 1:2\qquad\textbf{(B)}\ 1:3\qquad\textbf{(C)}\ 1:\sqrt3\qquad\textbf{(D)}\ \sqrt3:2 \qquad\textbf{(E)}\ 2:3$

Solution

Problem 28

In the table shown, the formula relating $x$ and $y$ is:

$\begin{tabular}{|l|l|l|l|l|l|} \hline x & 1 & 2 & 3 & 4 & 5 \\ \hline y & 3 & 7 & 13 & 21 & 31 \\ \hline \end{tabular}$

$\textbf{(A)}\ y=4x-1 \qquad \textbf{(B)}\ y=x^{3}-x^{2}+x+2 \qquad \textbf{(C)}\ y=x^2+x+1 \\ \qquad \textbf{(D)}\ y=(x^2+x+1)(x-1) \qquad \textbf{(E)}\ \text{None of these}$

Solution

Problem 29

In a circle of radius $5$ units, $CD$ and $AB$ are perpendicular diameters. A chord $CH$ cutting $AB$ at $K$ is $8$ units long. The diameter $AB$ is divided into two segments whose dimensions are:

$\textbf{(A)}\ 1.25, 8.75 \qquad \textbf{(B)}\ 2.75, 7.25\qquad \textbf{(C)}\ 2,8 \qquad \textbf{(D)} \ 4,6 \qquad \textbf{(E)} \text{None of these}$

Solution

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:

$\textbf{(A)}\ 1: 2 \qquad \textbf{(B)}\ 2: 1 \qquad \textbf{(C)}\ 1: 4 \qquad \textbf{(D)}\ 4: 1 \qquad \textbf{(E)}\ 1: 1$

Solution

Problem 31

Given $12$ points in a plane no three of which are collinear, the number of lines they determine is:

$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 54 \qquad \textbf{(C)}\ 120 \qquad \textbf{(D)}\ 66 \qquad \textbf{(E)}\ \text{none of these}$ Solution

Problem 32

$K$ 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:

$\textbf{(A)}\ \dfrac{x + \frac {1}{3}}{30} \qquad \textbf{(B)}\ \dfrac{x - \frac {1}{3}}{30} \qquad \textbf{(C)}\ \frac{30}{x+\frac{1}{3}}\qquad \textbf{(D)}\ \frac{30}{x}\qquad \textbf{(E)}\ \frac{x}{30}$

Solution

Problem 33

A circle and a square have the same perimeter. Then:

$\text{(A) their areas are equal}\qquad\\ \text{(B) the area of the circle is the greater} \qquad\\ \text{(C) the area of the square is the greater} \qquad\\ \text{(D) the area of the circle is } \pi \text{ times the area of the square}\qquad\\ \text{(E) none of these}$

Solution

Problem 34

The price of an article was increased $p\%$ . Later the new price was decreased $p\%$ . If the last price was one dollar, the original price was:

$\textbf{(A)}\ \frac{1-p^2}{200}\qquad \textbf{(B)}\ \frac{\sqrt{1-p^2}}{100}\qquad \textbf{(C)}\ \text{one dollar}\qquad\\ \textbf{(D)}\ 1-\frac{p^2}{10000-p^2}\qquad \textbf{(E)}\ \frac{10000}{10000-p^2}$

Solution

Problem 35

With a rational denominator, the expression $\frac {\sqrt {2}}{\sqrt {2} + \sqrt {3} - \sqrt {5}}$ is equivalent to:

$\textbf{(A)}\ \frac {3 + \sqrt {6} + \sqrt {15}}{6} \qquad \textbf{(B)}\ \frac {\sqrt {6} - 2 + \sqrt {10}}{6} \qquad \textbf{(C)}\ \frac{2+\sqrt{6}+\sqrt{10}}{10} \qquad\\ \textbf{(D)}\ \frac {2 + \sqrt {6} - \sqrt {10}}{6} \qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 36

To be continuous at $x = - 1$ , the value of $\frac {x^3 + 1}{x^2 - 1}$ is taken to be:

$\textbf{(A)}\ - 2 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ \frac {3}{2} \qquad \textbf{(D)}\ \infty \qquad \textbf{(E)}\ -\frac{3}{2}$

Solution

Problem 37

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:

$\textbf{(A)}\ 21\frac{1}{3}\pi-32\sqrt{3}\qquad \textbf{(B)}\ 32\sqrt{3}+21\frac{1}{3}\pi\qquad \textbf{(C)}\ 32\sqrt{3}+42\frac{2}{3}\pi \qquad\\ \textbf{(D)}\ 16\sqrt {3} + 42\frac {2}{3}\pi \qquad \textbf{(E)}\ 42\frac {2}{3}\pi$

Solution

Problem 38

The area of a trapezoidal field is $1400$ square yards. Its altitude is $50$ yards. Find the two bases, if the number of yards in each base is an integer divisible by $8$ . The number of solutions to this problem is:

$\textbf{(A)}\ \text{none} \qquad \textbf{(B)}\ \text{one} \qquad \textbf{(C)}\ \text{two} \qquad \textbf{(D)}\ \text{three} \qquad \textbf{(E)}\ \text{more than three}$

Solution

Problem 39

If the perimeter of a rectangle is $p$ and its diagonal is $d$ , the difference between the length and width of the rectangle is:

$\textbf{(A)}\ \frac {\sqrt {8d^2 - p^2}}{2} \qquad \textbf{(B)}\ \frac {\sqrt {8d^2 + p^2}}{2} \qquad \textbf{(C)}\ \frac{\sqrt{6d^2-p^2}}{2}\qquad\\ \textbf{(D)}\ \frac {\sqrt {6d^2 + p^2}}{2} \qquad \textbf{(E)}\ \frac {8d^2 - p^2}{4}$

Solution

Problem 40

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:

$\text{(A) } 4096 \qquad \text{(B) } 4356 \qquad \text{(C) } 4489 \qquad \text{(D) } 4761 \qquad \text{(E) } \text{none of these}$

Solution

Problem 41

Increasing the radius of a cylinder by $6$ units increased the volume by $y$ cubic units. Increasing the altitude of the cylinder by $6$ units also increases the volume by $y$ cubic units. If the original altitude is $2$ , then the original radius is:

$\text{(A) } 2 \qquad \text{(B) } 4 \qquad \text{(C) } 6 \qquad \text{(D) } 6\pi \qquad \text{(E) } 8$

Solution

Problem 42

Let $D$ represent a repeating decimal. If $P$ denotes the $r$ figures of $D$ which do not repeat themselves, and $Q$ denotes the $s$ figures of $D$ which do repeat themselves, then the incorrect expression is:

$\textbf{(A) } D = .PQQQ\ldots \qquad\\ \textbf{(B) } 10^rD = P.QQQ\ldots \\ \textbf{(C) } 10^{r + s}D = PQ.QQQ\ldots \qquad\\ \textbf{(D) } 10^r(10^s - 1)D = Q(P - 1) \\ \textbf{(E) } 10^r\cdot10^{2s}D = PQQ.QQQ\ldots$

Solution

Problem 43

The diameter of a circle is divided into $n$ equal parts. On each part a semicircle is constructed. As $n$ becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length:

$\textbf{(A) } \qquad$ equal to the semi-circumference of the original circle

$\textbf{(B) } \qquad$ equal to the diameter of the original circle

$\textbf{(C) }$ greater than the diameter, but less than the semi-circumference of the original circle

$\textbf{(D) } \qquad$ that is infinite

$\textbf{(E) }$ greater than the semi-circumference

Solution

Problem 44

If an integer of two digits is $k$ times the sum of its digits, the number formed by interchanging the the digits is the sum of the digits multiplied by

$\textbf{(A) \ } 9-k \qquad \textbf{(B) \ } 10-k \qquad \textbf{(C) \ } 11-k \qquad \textbf{(D) \ } k-1 \qquad \textbf{(E) \ } k+1$

Solution

Problem 45

If $a$ and $b$ are two unequal positive numbers, then:

$\text{(A) } \frac{2ab}{a+b}>\sqrt{ab}>\frac{a+b}{2}\qquad \text{(B) } \sqrt{ab}>\frac{2ab}{a+b}>\frac{a+b}{2} \\ \text{(C) } \frac{2ab}{a+b}>\frac{a+b}{2}>\sqrt{ab}\qquad \text{(D) } \frac{a+b}{2}>\frac{2ab}{a+b}>\sqrt{ab} \\ \text{(E) } \frac {a + b}{2} > \sqrt {ab} > \frac {2ab}{a + b}$

Solution

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:

$\text{(A) greater than the area of the given rectangle} \quad\\ \text{(B) equal to the area of the given rectangle} \quad\\ \text{(C) equal to the area of a square with its side equal to the smaller side of the given rectangle} \quad\\ \text{(D) equal to the area of a square with its side equal to the greater side of the given rectangle} \quad\\ \text{(E) equal to the area of a rectangle whose dimensions are the diagonal and the shorter side of the given rectangle}$

Solution

Problem 47

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:

$\textbf{(A) } 3,4,9 \qquad \textbf{(B) } 9,-5,-12 \qquad \textbf{(C) } 12,-5,9 \qquad \textbf{(D) } 4,3,9 \qquad \textbf{(E) } 4,9,3$

Solution

Problem 48

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:

$\text{(A) } \frac {r + t}{r - t} \qquad \text{(B) } \frac {r}{r - t} \qquad \text{(C) } \frac {r + t}{r} \qquad \text{(D) } \frac{r}{t}\qquad \text{(E) } \frac{r+k}{t-k}$

Solution

Problem 49

$[asy] unitsize(27); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A,B,C,D,E,F,X,Y,Z; A=(3,3); B=(0,0); C=(6,0); D=(4,0); E=(4,2); F=(1,1); draw(A--B--C--cycle); draw(A--D); draw(B--E); draw(C--F); X=intersectionpoint(A--D,C--F); Y=intersectionpoint(B--E,A--D); Z=intersectionpoint(B--E,C--F); label("$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, $\overline{CD}$ , $\overline{AE}$ and $\overline{BF}$ are one-third of their respective sides. It follows that $\overline{AN_2}: \overline{N_2N_1}: \overline{N_1D} = 3: 3: 1$ , and similarly for lines BE and CF. Then the area of triangle $N_1N_2N_3$ is:

$\text{(A) } \frac {1}{10} \triangle ABC \qquad \text{(B) } \frac {1}{9} \triangle ABC \qquad \text{(C) } \frac{1}{7}\triangle ABC\qquad \text{(D) } \frac{1}{6}\triangle ABC\qquad \text{(E) } \text{none of these}$

Solution

Problem 50

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:

$\textbf{(A) } \infty\qquad \textbf{(B) } \frac{4}{3}\qquad \textbf{(C) } \frac{8}{3}\qquad \textbf{(D) } \frac{1}{3}(4+\sqrt{2})\qquad \textbf{(E) } \frac{2}{3}(4+\sqrt{2})$

Solution

1952 AHSC (Problems • Answer Key • Resources)
Preceded by 1951 AHSC	Followed by 1953 AHSC
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

@@ Line 1: / Line 1: @@
+{{AHSC 50 Problems
+|year=1952
+}}
 == Problem 1 ==
@@ Line 144: / Line 147: @@
 == Problem 17 ==
-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 selling price. The per cent of the list price at which he should mark them is:
+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>
@@ Line 152: / Line 155: @@
 == 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]]
@@ Line 158: / Line 165: @@
 == 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]]
@@ Line 164: / Line 175: @@
 == 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]]
@@ Line 170: / Line 185: @@
 == 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]]
@@ Line 176: / Line 195: @@
 == 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]]
@@ Line 182: / Line 205: @@
 == Problem 23 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+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]]
@@ Line 188: / Line 213: @@
 == Problem 24 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+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]]
 == 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) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<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]]
 == Problem 26 ==
+If <math>\left(r+\frac1r\right)^2=3</math>, then <math>r^3+\frac1{r^3}</math> equals
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<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]]
 == 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) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<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]]
 == 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) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<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]]
 == 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) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<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]]
@@ Line 224: / Line 273: @@
 == Problem 30 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+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]]
@@ Line 230: / Line 285: @@
 == Problem 31 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </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]]
 == Problem 32 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<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]]
@@ Line 242: / Line 308: @@
 == Problem 33 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </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]]
@@ Line 248: / Line 320: @@
 == Problem 34 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+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]]
@@ Line 254: / Line 333: @@
 == Problem 35 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+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]]
@@ Line 260: / Line 345: @@
 == Problem 36 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+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]]
@@ Line 266: / Line 357: @@
 == Problem 37 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+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]]
@@ Line 272: / Line 369: @@
 == Problem 38 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+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]]
@@ Line 278: / Line 382: @@
 == Problem 39 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+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]]
@@ Line 284: / Line 394: @@
 == Problem 40 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </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]]
@@ Line 290: / Line 406: @@
 == Problem 41 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </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]]
 == 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) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<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]]
@@ Line 302: / Line 425: @@
 == Problem 43 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+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]]
@@ Line 316: / Line 449: @@
 == Problem 45 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </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]]
@@ Line 322: / Line 461: @@
 == Problem 46 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </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]]
@@ Line 328: / Line 475: @@
 == Problem 47 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+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]]
@@ Line 334: / Line 487: @@
 == Problem 48 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </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]]
@@ Line 340: / Line 500: @@
 == Problem 49 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </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]]
@@ Line 346: / Line 526: @@
 == Problem 50 ==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+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]]
 == See also ==
-* [[AHSME]]
-* [[AHSME Problems and Solutions]]
+* [[AMC 12 Problems and Solutions]]
 * [[Mathematics competition resources]]
+{{AHSME 50p box|year=1952|before=[[1951 AHSME|1951 AHSC]]|after=[[1953 AHSME|1953 AHSC]]}}
 {{MAA Notice}}

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Difference between revisions of "1952 AHSME Problems"

Latest revision as of 14:35, 30 December 2023

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

See also