Difference between revisions of "1953 AHSME Problems"

Latest revision as of 13:24, 20 February 2020

1953 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

A boy buys oranges at $3$ for $10$ cents. He will sell them at $5$ for $20$ cents. In order to make a profit of $ $1.00$ , he must sell:

$\textbf{(A)}\ 67 \text{ oranges} \qquad \textbf{(B)}\ 150 \text{ oranges} \qquad \textbf{(C)}\ 200\text{ oranges}\\ \textbf{(D)}\ \text{an infinite number of oranges}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 2

A refrigerator is offered at sale at $250.00 less successive discounts of 20% and 15%. The sale price of the refrigerator is:

$\textbf{(A) } \text{35\% less than 250.00} \qquad \textbf{(B) } \text{65\% of 250.00} \qquad \textbf{(C) } \text{77\% of 250.00} \qquad \textbf{(D) } \text{68\% of 250.00} \qquad \textbf{(E) } \text{none of these}$

Solution

Problem 3

The factors of the expression $x^2+y^2$ are:

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

Solution

Problem 4

The roots of $x(x^2+8x+16)(4-x)=0$ are:

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 0,4 \qquad \textbf{(C)}\ 0,4,-4 \qquad \textbf{(D)}\ 0,4,-4,-4 \qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 5

If $\log_6 x=2.5$ , the value of $x$ is:

$\textbf{(A)}\ 90 \qquad \textbf{(B)}\ 36 \qquad \textbf{(C)}\ 36\sqrt{6} \qquad \textbf{(D)}\ 0.5 \qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 6

Charles has $5q + 1$ quarters and Richard has $q + 5$ quarters. The difference in their money in dimes is:

$\textbf{(A)}\ 10(q - 1) \qquad \textbf{(B)}\ \frac {2}{5}(4q - 4) \qquad \textbf{(C)}\ \frac {2}{5}(q - 1) \\ \textbf{(D)}\ \frac{5}{2}(q-1)\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 7

The fraction $\frac{\sqrt{a^2+x^2}-\frac{x^2-a^2}{\sqrt{a^2+x^2}}}{a^2+x^2}$ reduces to:

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{2a^2}{a^2+x^2} \qquad \textbf{(C)}\ \frac{2x^2}{(a^2+x^2)^{\frac{3}{2}}}\qquad \textbf{(D)}\ \frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}\qquad \textbf{(E)}\ \frac{2x^2}{a^2+x^2}$

Solution

Problem 8

The value of $x$ at the intersection of $y=\frac{8}{x^2+4}$ and $x+y=2$ is:

$\textbf{(A)}\ -2+\sqrt{5} \qquad \textbf{(B)}\ -2-\sqrt{5} \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 9

The number of ounces of water needed to reduce $9$ ounces of shaving lotion containing $50$ % alcohol to a lotion containing $30$ % alcohol is:

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$

Solution

Problem 10

The number of revolutions of a wheel, with fixed center and with an outside diameter of $6$ feet, required to cause a point on the rim to go one mile is:

$\textbf{(A)}\ 880 \qquad \textbf{(B)}\ \frac{440}{\pi} \qquad \textbf{(C)}\ \frac{880}{\pi} \qquad \textbf{(D)}\ 440\pi\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 11

A running track is the ring formed by two concentric circles. It is $10$ feet wide. The circumference of the two circles differ by about:

$\textbf{(A)}\ 10\text{ feet} \qquad \textbf{(B)}\ 30\text{ feet} \qquad \textbf{(C)}\ 60\text{ feet} \qquad \textbf{(D)}\ 100\text{ feet}\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 12

The diameters of two circles are $8$ inches and $12$ inches respectively. The ratio of the area of the smaller to the area of the larger circle is:

$\textbf{(A)}\ \frac{2}{3} \qquad \textbf{(B)}\ \frac{4}{9} \qquad \textbf{(C)}\ \frac{9}{4} \qquad \textbf{(D)}\ \frac{1}{2}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 13

A triangle and a trapezoid are equal in area. They also have the same altitude. If the base of the triangle is 18 inches, the median of the trapezoid is:

$\textbf{(A)}\ 36\text{ inches} \qquad \textbf{(B)}\ 9\text{ inches} \qquad \textbf{(C)}\ 18\text{ inches}\\ \textbf{(D)}\ \text{not obtainable from these data}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 14

Given the larger of two circles with center $P$ and radius $p$ and the smaller with center $Q$ and radius $q$ . Draw $PQ$ . Which of the following statements is false?

$\textbf{(A)}\ p-q\text{ can be equal to }\overline{PQ}\\ \textbf{(B)}\ p+q\text{ can be equal to }\overline{PQ}\\ \textbf{(C)}\ p+q\text{ can be less than }\overline{PQ}\\ \textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 15

A circular piece of metal of maximum size is cut out of a square piece and then a square piece of maximum size is cut out of the circular piece. The total amount of metal wasted is:

$\textbf{(A)}\ \frac{1}{4} \text{ the area of the original square}\\ \textbf{(B)}\ \frac{1}{2}\text{ the area of the original square}\\ \textbf{(C)}\ \frac{1}{2}\text{ the area of the circular piece}\\ \textbf{(D)}\ \frac{1}{4}\text{ the area of the circular piece}\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 16

Adams plans a profit of $10$ % on the selling price of an article and his expenses are $15$ % of sales. The rate of markup on an article that sells for $ $5.00$ is:

$\textbf{(A)}\ 20\% \qquad \textbf{(B)}\ 25\% \qquad \textbf{(C)}\ 30\% \qquad \textbf{(D)}\ 33\frac {1}{3}\% \qquad \textbf{(E)}\ 35\%$

Solution

Problem 17

A man has part of $ $4500$ invested at $4$ % and the rest at $6$ %. If his annual return on each investment is the same, the average rate of interest which he realizes of the $4500 is:

$\textbf{(A)}\ 5\% \qquad \textbf{(B)}\ 4.8\% \qquad \textbf{(C)}\ 5.2\% \qquad \textbf{(D)}\ 4.6\% \qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 18

One of the factors of $x^4+4$ is:

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

Solution

Problem 19

In the expression $xy^2$ , the values of $x$ and $y$ are each decreased $25$ %; the value of the expression is:

$\textbf{(A)}\ \text{decreased } 50\% \qquad \textbf{(B)}\ \text{decreased }75\%\\ \textbf{(C)}\ \text{decreased }\frac{37}{64}\text{ of its value}\qquad \textbf{(D)}\ \text{decreased }\frac{27}{64}\text{ of its value}\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 20

If $y=x+\frac{1}{x}$ , then $x^4+x^3-4x^2+x+1=0$ becomes:

$\textbf{(A)}\ x^2(y^2+y-2)=0 \qquad \textbf{(B)}\ x^2(y^2+y-3)=0\\ \textbf{(C)}\ x^2(y^2+y-4)=0 \qquad \textbf{(D)}\ x^2(y^2+y-6)=0\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 21

If $\log_{10} (x^2-3x+6)=1$ , the value of $x$ is:

$\textbf{(A)}\ 10\text{ or }2 \qquad \textbf{(B)}\ 4\text{ or }-2 \qquad \textbf{(C)}\ 3\text{ or }-1 \qquad \textbf{(D)}\ 4\text{ or }-1\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 22

The logarithm of $27\sqrt[4]{9}\sqrt[3]{9}$ to the base $3$ is:

$\textbf{(A)}\ 8\frac{1}{2} \qquad \textbf{(B)}\ 4\frac{1}{6} \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 23

The equation $\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5$ has:

$\textbf{(A)}\ \text{an extraneous root between } - 5\text{ and } - 1 \\ \textbf{(B)}\ \text{an extraneous root between }-10\text{ and }-6\\ \textbf{(C)}\ \text{a true root between }20\text{ and }25\qquad \textbf{(D)}\ \text{two true roots}\\ \textbf{(E)}\ \text{two extraneous roots}$

Solution

Problem 24

If $a,b,c$ are positive integers less than $10$ , then $(10a + b)(10a + c) = 100a(a + 1) + bc$ if:

$\textbf{(A)}\ b + c = 10 \qquad \textbf{(B)}\ b = c \qquad \textbf{(C)}\ a + b = 10 \qquad \textbf{(D)}\ a = b \\ \textbf{(E)}\ a+b+c = 10$

Solution

Problem 25

In a geometric progression whose terms are positive, any term is equal to the sum of the next two following terms. then the common ratio is:

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ \text{about }\frac{\sqrt{5}}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2}\qquad \textbf{(D)}\ \frac{1-\sqrt{5}}{2}\qquad \textbf{(E)}\ \frac{2}{\sqrt{5}}$

Solution

Problem 26

The base of a triangle is $15$ inches. Two lines are drawn parallel to the base, terminating in the other two sides, and dividing the triangle into three equal areas. The length of the parallel closer to the base is:

$\textbf{(A)}\ 5\sqrt{6}\text{ inches} \qquad \textbf{(B)}\ 10\text{ inches} \qquad \textbf{(C)}\ 4\sqrt{3}\text{ inches}\qquad \textbf{(D)}\ 7.5\text{ inches}\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 27

The radius of the first circle is $1$ inch, that of the second $\frac{1}{2}$ inch, that of the third $\frac{1}{4}$ inch and so on indefinitely. The sum of the areas of the circles is:

$\textbf{(A)}\ \frac{3\pi}{4} \qquad \textbf{(B)}\ 1.3\pi \qquad \textbf{(C)}\ 2\pi \qquad \textbf{(D)}\ \frac{4\pi}{3}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 28

In $\triangle ABC$ , sides $a,b$ and $c$ are opposite $\angle{A},\angle{B}$ and $\angle{C}$ respectively. $AD$ bisects $\angle{A}$ and meets $BC$ at $D$ . Then if $x = \overline{CD}$ and $y = \overline{BD}$ the correct proportion is:

$\textbf{(A)}\ \frac {x}{a} = \frac {a}{b + c} \qquad \textbf{(B)}\ \frac {x}{b} = \frac {a}{a + c} \qquad \textbf{(C)}\ \frac{y}{c}=\frac{c}{b+c}\\ \textbf{(D)}\ \frac{y}{c}=\frac{a}{b+c}\qquad \textbf{(E)}\ \frac{x}{y}=\frac{c}{b}$

Solution

Problem 29

The number of significant digits in the measurement of the side of a square whose computed area is $1.1025$ square inches to the nearest ten-thousandth of a square inch is:

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 1$

Solution

Problem 30

A house worth $ $9000$ is sold by Mr. A to Mr. B at a $10$ % loss. Mr. B sells the house back to Mr. A at a $10$ % gain. The result of the two transactions is:

$\textbf{(A)}\ \text{Mr. A breaks even} \qquad \textbf{(B)}\ \text{Mr. B gains }$900 \qquad \textbf{(C)}\ \text{Mr. A loses }$900\\ \textbf{(D)}\ \text{Mr. A loses }$810\qquad \textbf{(E)}\ \text{Mr. B gains }$1710$

Solution

Problem 31

The rails on a railroad are $30$ feet long. As the train passes over the point where the rails are joined, there is an audible click. The speed of the train in miles per hour is approximately the number of clicks heard in:

$\textbf{(A)}\ 20\text{ seconds} \qquad \textbf{(B)}\ 2\text{ minutes} \qquad \textbf{(C)}\ 1\frac{1}{2}\text{ minutes}\qquad \textbf{(D)}\ 5\text{ minutes}\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 32

Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form:

$\textbf{(A)}\ \text{a square} \qquad \textbf{(B)}\ \text{a rectangle} \qquad \textbf{(C)}\ \text{a parallelogram with unequal sides}\\ \textbf{(D)}\ \text{a rhombus}\qquad \textbf{(E)}\ \text{a quadrilateral with no special properties}$

Solution

Problem 33

The perimeter of an isosceles right triangle is $2p$ . Its area is:

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

Solution

Problem 34

If one side of a triangle is $12$ inches and the opposite angle is $30^{\circ}$ , then the diameter of the circumscribed circle is:

$\textbf{(A)}\ 18\text{ inches} \qquad \textbf{(B)}\ 30\text{ inches} \qquad \textbf{(C)}\ 24\text{ inches} \qquad \textbf{(D)}\ 20\text{ inches}\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 35

If $f(x)=\frac{x(x-1)}{2}$ , then $f(x+2)$ equals:

$\textbf{(A)}\ f(x)+f(2) \qquad \textbf{(B)}\ (x+2)f(x) \qquad \textbf{(C)}\ x(x+2)f(x) \qquad \textbf{(D)}\ \frac{xf(x)}{x+2}\\ \textbf{(E)}\ \frac{(x+2)f(x+1)}{x}$

Solution

Problem 36

Determine $m$ so that $4x^2-6x+m$ is divisible by $x-3$ . The obtained value, $m$ , is an exact divisor of:

$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 64$

Solution

Problem 37

The base of an isosceles triangle is $6$ inches and one of the equal sides is $12$ inches. The radius of the circle through the vertices of the triangle is:

$\textbf{(A)}\ \frac{7\sqrt{15}}{5} \qquad \textbf{(B)}\ 4\sqrt{3} \qquad \textbf{(C)}\ 3\sqrt{5} \qquad \textbf{(D)}\ 6\sqrt{3}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 38

If $f(a)=a-2$ and $F(a,b)=b^2+a$ , then $F(3,f(4))$ is:

$\textbf{(A)}\ a^2-4a+7 \qquad \textbf{(B)}\ 28 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 11$

Solution

Problem 39

The product, $\log_a b \cdot \log_b a$ is equal to:

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ a \qquad \textbf{(C)}\ b \qquad \textbf{(D)}\ ab \qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 40

The negation of the statement "all men are honest," is:

$\textbf{(A)}\ \text{no men are honest} \qquad \textbf{(B)}\ \text{all men are dishonest} \\ \textbf{(C)}\ \text{some men are dishonest}\qquad \textbf{(D)}\ \text{no men are dishonest}\\ \textbf{(E)}\ \text{some men are honest}$

Solution

Problem 41

A girls' camp is located $300$ rods from a straight road. On this road, a boys' camp is located $500$ rods from the girls' camp. It is desired to build a canteen on the road which shall be exactly the same distance from each camp. The distance of the canteen from each of the camps is:

$\textbf{(A)}\ 400\text{ rods} \qquad \textbf{(B)}\ 250\text{ rods} \qquad \textbf{(C)}\ 87.5\text{ rods} \qquad \textbf{(D)}\ 200\text{ rods}\\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 42

The centers of two circles are $41$ inches apart. The smaller circle has a radius of $4$ inches and the larger one has a radius of $5$ inches. The length of the common internal tangent is:

$\textbf{(A)}\ 41\text{ inches} \qquad \textbf{(B)}\ 39\text{ inches} \qquad \textbf{(C)}\ 39.8\text{ inches} \qquad \textbf{(D)}\ 40.1\text{ inches}\\ \textbf{(E)}\ 40\text{ inches}$

Solution

Problem 43

If the price of an article is increased by percent $p$ , then the decrease in percent of sales must not exceed $d$ in order to yield the same income. The value of $d$ is:

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

Solution

Problem 44

In solving a problem that reduces to a quadratic equation one student makes a mistake only in the constant term of the equation and obtains $8$ and $2$ for the roots. Another student makes a mistake only in the coefficient of the first degree term and find $-9$ and $-1$ for the roots. The correct equation was:

$\textbf{(A)}\ x^2-10x+9=0 \qquad \textbf{(B)}\ x^2+10x+9=0 \qquad \textbf{(C)}\ x^2-10x+16=0\\ \textbf{(D)}\ x^2-8x-9=0\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 45

The lengths of two line segments are $a$ units and $b$ units respectively. Then the correct relation between them is:

$\textbf{(A)}\ \frac{a+b}{2} > \sqrt{ab} \qquad \textbf{(B)}\ \frac{a+b}{2} < \sqrt{ab} \qquad \textbf{(C)}\ \frac{a+b}{2}=\sqrt{ab}\\ \textbf{(D)}\ \frac{a+b}{2}\leq\sqrt{ab}\qquad \textbf{(E)}\ \frac{a+b}{2}\geq\sqrt{ab}$

Solution

Problem 46

Instead of walking along two adjacent sides of a rectangular field, a boy took a shortcut along the diagonal of the field and saved a distance equal to $\frac{1}{2}$ the longer side. The ratio of the shorter side of the rectangle to the longer side was:

$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{1}{4} \qquad \textbf{(D)}\ \frac{3}{4}\qquad \textbf{(E)}\ \frac{2}{5}$

Solution

Problem 47

If $x>0$ , then the correct relationship is:

$\textbf{(A)}\ \log (1+x) = \frac{x}{1+x} \qquad \textbf{(B)}\ \log (1+x) < \frac{x}{1+x} \\ \textbf{(C)}\ \log(1+x) > x\qquad \textbf{(D)}\ \log (1+x) < x\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 48

If the larger base of an isosceles trapezoid equals a diagonal and the smaller base equals the altitude, then the ratio of the smaller base to the larger base is:

$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{3}{4} \qquad \textbf{(D)}\ \frac{3}{5}\qquad \textbf{(E)}\ \frac{2}{5}$

Solution

Problem 49

The coordinates of $A,B$ and $C$ are $(5,5),(2,1)$ and $(0,k)$ respectively. The value of $k$ that makes $\overline{AC}+\overline{BC}$ as small as possible is:

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4\frac{1}{2} \qquad \textbf{(C)}\ 3\frac{6}{7} \qquad \textbf{(D)}\ 4\frac{5}{6}\qquad \textbf{(E)}\ 2\frac{1}{7}$

Solution

Problem 50

One of the sides of a triangle is divided into segments of $6$ and $8$ units by the point of tangency of the inscribed circle. If the radius of the circle is $4$ , then the length of the shortest side of the triangle is:

$\textbf{(A)}\ 12\text{ units} \qquad \textbf{(B)}\ 13\text{ units} \qquad \textbf{(C)}\ 14\text{ units} \qquad \textbf{(D)}\ 15\text{ units}\qquad \textbf{(E)}\ 16\text{ units}$

Solution

1953 AHSC (Problems • Answer Key • Resources)
Preceded by 1952 AHSC	Followed by 1954 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: @@
-== Problem 1 ==
+{{AHSC 50 Problems
+|year=1953
+}}
+==Problem 1==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+A boy buys oranges at <math>3</math> for <math>10</math> cents. He will sell them at <math>5</math> for <math>20</math> cents. In order to make a profit of \$ <math>1.00</math>, he must sell:
+<math>\textbf{(A)}\ 67 \text{ oranges} \qquad
+\textbf{(B)}\ 150 \text{ oranges} \qquad
+\textbf{(C)}\ 200\text{ oranges}\\
+\textbf{(D)}\ \text{an infinite number of oranges}\qquad
+\textbf{(E)}\ \text{none of these} </math>
 [[1953 AHSME Problems/Problem 1|Solution]]
+==Problem 2==
+A refrigerator is offered at sale at \$250.00 less successive discounts of 20% and 15%. The sale price of the refrigerator is:
-== Problem 2 ==
+<math>\textbf{(A) } \text{35\% less than 250.00} \qquad
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+\textbf{(B) } \text{65\% of 250.00} \qquad
+\textbf{(C) } \text{77\% of 250.00} \qquad
+\textbf{(D) } \text{68\% of 250.00} \qquad
+\textbf{(E) } \text{none of these} </math>
+[[1953 AHSME Problems/Problem 2|Solution]]
-[[1953 AHSME Problems/Problem 2|Solution]]
+==Problem 3==
-== Problem 3 ==
+The factors of the expression <math>x^2+y^2</math> are:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math>\textbf{(A)}\ (x+y)(x-y) \qquad
+\textbf{(B)}\ (x+y)^2 \qquad
+\textbf{(C)}\ (x^{\frac{2}{3}}+y^{\frac{2}{3}})(x^{\frac{4}{3}}+y^{\frac{4}{3}})\\
+\textbf{(D)}\ (x+iy)(x-iy)\qquad
+\textbf{(E)}\ \text{none of these}  </math>
 [[1953 AHSME Problems/Problem 3|Solution]]
+==Problem 4==
-== Problem 4 ==
+The roots of <math>x(x^2+8x+16)(4-x)=0</math> are:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math>\textbf{(A)}\ 0 \qquad
+\textbf{(B)}\ 0,4 \qquad
+\textbf{(C)}\ 0,4,-4 \qquad
+\textbf{(D)}\ 0,4,-4,-4 \qquad
+\textbf{(E)}\ \text{none of these}  </math>
 [[1953 AHSME Problems/Problem 4|Solution]]
-== Problem 5 ==
+==Problem 5==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+If <math>\log_6 x=2.5</math>, the value of <math>x</math> is:
+<math>\textbf{(A)}\ 90 \qquad
+\textbf{(B)}\ 36 \qquad
+\textbf{(C)}\ 36\sqrt{6} \qquad
+\textbf{(D)}\ 0.5 \qquad
+\textbf{(E)}\ \text{none of these} </math>
 [[1953 AHSME Problems/Problem 5|Solution]]
-== Problem 6 ==
+==Problem 6==
+Charles has <math>5q + 1</math> quarters and Richard has <math>q + 5</math> quarters. The difference in their money in dimes is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math>\textbf{(A)}\ 10(q - 1) \qquad
+\textbf{(B)}\ \frac {2}{5}(4q - 4) \qquad
+\textbf{(C)}\ \frac {2}{5}(q - 1) \\
+\textbf{(D)}\ \frac{5}{2}(q-1)\qquad
+\textbf{(E)}\ \text{none of these}  </math>
 [[1953 AHSME Problems/Problem 6|Solution]]
-== Problem 7 ==
+==Problem 7==
+The fraction <math>\frac{\sqrt{a^2+x^2}-\frac{x^2-a^2}{\sqrt{a^2+x^2}}}{a^2+x^2}</math> reduces to:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math>\textbf{(A)}\ 0 \qquad
+\textbf{(B)}\ \frac{2a^2}{a^2+x^2} \qquad
+\textbf{(C)}\ \frac{2x^2}{(a^2+x^2)^{\frac{3}{2}}}\qquad
+\textbf{(D)}\ \frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}\qquad
+\textbf{(E)}\ \frac{2x^2}{a^2+x^2}   </math>
 [[1953 AHSME Problems/Problem 7|Solution]]
-== Problem 8 ==
+==Problem 8==
+The value of <math>x</math> at the intersection of <math>y=\frac{8}{x^2+4}</math> and <math>x+y=2</math> is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math>\textbf{(A)}\ -2+\sqrt{5} \qquad
+\textbf{(B)}\ -2-\sqrt{5} \qquad
+\textbf{(C)}\ 0 \qquad
+\textbf{(D)}\ 2 \qquad
+\textbf{(E)}\ \text{none of these}     </math>
 [[1953 AHSME Problems/Problem 8|Solution]]
-== Problem 9 ==
+==Problem 9==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The number of ounces of water needed to reduce <math>9</math> ounces of shaving lotion containing <math>50</math> % alcohol to a lotion containing <math>30</math> % alcohol is:
+<math>\textbf{(A)}\ 3 \qquad
+\textbf{(B)}\ 4 \qquad
+\textbf{(C)}\ 5 \qquad
+\textbf{(D)}\ 6 \qquad
+\textbf{(E)}\ 7     </math>
 [[1953 AHSME Problems/Problem 9|Solution]]
-== Problem 10 ==
+==Problem 10==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The number of revolutions of a wheel, with fixed center and with an outside diameter of <math>6</math> feet, required to cause a point on the rim to go one mile is:
+<math>\textbf{(A)}\ 880 \qquad
+\textbf{(B)}\ \frac{440}{\pi} \qquad
+\textbf{(C)}\ \frac{880}{\pi} \qquad
+\textbf{(D)}\ 440\pi\qquad
+\textbf{(E)}\ \text{none of these}     </math>
 [[1953 AHSME Problems/Problem 10|Solution]]
-== Problem 11 ==
+==Problem 11==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+A running track is the ring formed by two concentric circles. It is <math>10</math> feet wide. The circumference of the two circles differ by about:
+<math>\textbf{(A)}\ 10\text{ feet} \qquad
+\textbf{(B)}\ 30\text{ feet} \qquad
+\textbf{(C)}\ 60\text{ feet} \qquad
+\textbf{(D)}\ 100\text{ feet}\\ \textbf{(E)}\ \text{none of these}    </math>
 [[1953 AHSME Problems/Problem 11|Solution]]
-== Problem 12 ==
+==Problem 12==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The diameters of two circles are <math>8</math> inches and <math>12</math> inches respectively. The ratio of the area of the smaller to the area of the larger circle is:
+<math>\textbf{(A)}\ \frac{2}{3} \qquad
+\textbf{(B)}\ \frac{4}{9} \qquad
+\textbf{(C)}\ \frac{9}{4} \qquad
+\textbf{(D)}\ \frac{1}{2}\qquad
+\textbf{(E)}\ \text{none of these}     </math>
 [[1953 AHSME Problems/Problem 12|Solution]]
-== Problem 13 ==
+==Problem 13==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+A triangle and a trapezoid are equal in area. They also have the same altitude. If the base of the triangle is 18 inches, the median of the trapezoid is:
+<math>\textbf{(A)}\ 36\text{ inches} \qquad
+\textbf{(B)}\ 9\text{ inches} \qquad
+\textbf{(C)}\ 18\text{ inches}\\
+\textbf{(D)}\ \text{not obtainable from these data}\qquad
+\textbf{(E)}\ \text{none of these}    </math>
 [[1953 AHSME Problems/Problem 13|Solution]]
-== Problem 14 ==
+==Problem 14==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+Given the larger of two circles with center <math>P</math> and radius <math>p</math> and the smaller with center <math>Q</math> and radius <math>q</math>. Draw <math>PQ</math>. Which of the following statements is false?
+<math>\textbf{(A)}\ p-q\text{ can be equal to }\overline{PQ}\\
+\textbf{(B)}\ p+q\text{ can be equal to }\overline{PQ}\\
+\textbf{(C)}\ p+q\text{ can be less than }\overline{PQ}\\
+\textbf{(D)}\ p-q\text{ can be less than }\overline{PQ}\\ \textbf{(E)}\ \text{none of these}     </math>
 [[1953 AHSME Problems/Problem 14|Solution]]
-== Problem 15 ==
+==Problem 15==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+A circular piece of metal of maximum size is cut out of a square piece and then a square piece of maximum size is cut out of the circular piece. The total amount of metal wasted is:
+<math>\textbf{(A)}\ \frac{1}{4} \text{ the area of the original square}\\
+\textbf{(B)}\ \frac{1}{2}\text{ the area of the original square}\\ \textbf{(C)}\ \frac{1}{2}\text{ the area of the circular piece}\\ \textbf{(D)}\ \frac{1}{4}\text{ the area of the circular piece}\\ \textbf{(E)}\ \text{none of these}     </math>
 [[1953 AHSME Problems/Problem 15|Solution]]
-== Problem 16 ==
+==Problem 16==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+Adams plans a profit of <math>10</math> % on the selling price of an article and his expenses are <math>15</math> % of sales. The rate of markup on an article that sells for \$ <math>5.00</math> is:
+<math>\textbf{(A)}\ 20\% \qquad
+\textbf{(B)}\ 25\% \qquad
+\textbf{(C)}\ 30\% \qquad
+\textbf{(D)}\ 33\frac {1}{3}\% \qquad
+\textbf{(E)}\ 35\%   </math>
 [[1953 AHSME Problems/Problem 16|Solution]]
-== Problem 17 ==
+==Problem 17==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+A man has part of \$ <math>4500</math> invested at <math>4</math> % and the rest at <math>6</math> %. If his annual return on each investment is the same, the average rate of interest which he realizes of the \$4500 is:
+<math>\textbf{(A)}\ 5\% \qquad
+\textbf{(B)}\ 4.8\% \qquad
+\textbf{(C)}\ 5.2\% \qquad
+\textbf{(D)}\ 4.6\% \qquad
+\textbf{(E)}\ \text{none of these}   </math>
 [[1953 AHSME Problems/Problem 17|Solution]]
-== Problem 18 ==
+==Problem 18==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+One of the factors of <math>x^4+4</math> is:
+<math>\textbf{(A)}\ x^2+2 \qquad
+\textbf{(B)}\ x+1 \qquad
+\textbf{(C)}\ x^2-2x+2 \qquad
+\textbf{(D)}\ x^2-4\\
+\textbf{(E)}\ \text{none of these}   </math>
 [[1953 AHSME Problems/Problem 18|Solution]]
-== Problem 19 ==
+==Problem 19==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+In the expression <math>xy^2</math>, the values of <math>x</math> and <math>y</math> are each decreased <math>25</math> %; the value of the expression is:
+<math>\textbf{(A)}\ \text{decreased } 50\% \qquad
+\textbf{(B)}\ \text{decreased }75\%\\
+\textbf{(C)}\ \text{decreased }\frac{37}{64}\text{ of its value}\qquad
+\textbf{(D)}\ \text{decreased }\frac{27}{64}\text{ of its value}\\ \textbf{(E)}\ \text{none of these}   </math>
 [[1953 AHSME Problems/Problem 19|Solution]]
-== Problem 20 ==
+==Problem 20==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+If <math>y=x+\frac{1}{x}</math>, then <math>x^4+x^3-4x^2+x+1=0</math> becomes:
+<math>\textbf{(A)}\ x^2(y^2+y-2)=0 \qquad
+\textbf{(B)}\ x^2(y^2+y-3)=0\\
+\textbf{(C)}\ x^2(y^2+y-4)=0 \qquad
+\textbf{(D)}\ x^2(y^2+y-6)=0\\ \textbf{(E)}\ \text{none of these}  </math>
 [[1953 AHSME Problems/Problem 20|Solution]]
-== Problem 21 ==
+==Problem 21==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+If <math>\log_{10} (x^2-3x+6)=1</math>, the value of <math>x</math> is:
+<math>\textbf{(A)}\ 10\text{ or }2 \qquad
+\textbf{(B)}\ 4\text{ or }-2 \qquad
+\textbf{(C)}\ 3\text{ or }-1 \qquad
+\textbf{(D)}\ 4\text{ or }-1\\ \textbf{(E)}\ \text{none of these}   </math>
 [[1953 AHSME Problems/Problem 21|Solution]]
-== Problem 22 ==
+==Problem 22==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The logarithm of <math>27\sqrt[4]{9}\sqrt[3]{9}</math> to the base <math>3</math> is:
+<math>\textbf{(A)}\ 8\frac{1}{2} \qquad
+\textbf{(B)}\ 4\frac{1}{6} \qquad
+\textbf{(C)}\ 5 \qquad
+\textbf{(D)}\ 3 \qquad
+\textbf{(E)}\ \text{none of these}  </math>
 [[1953 AHSME Problems/Problem 22|Solution]]
-== Problem 23 ==
+==Problem 23==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The equation <math>\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5</math> has:
+<math>\textbf{(A)}\ \text{an extraneous root between } - 5\text{ and } - 1 \\
+\textbf{(B)}\ \text{an extraneous root between }-10\text{ and }-6\\ \textbf{(C)}\ \text{a true root between }20\text{ and }25\qquad
+\textbf{(D)}\ \text{two true roots}\\ \textbf{(E)}\ \text{two extraneous roots}  </math>
 [[1953 AHSME Problems/Problem 23|Solution]]
-== Problem 24 ==
+==Problem 24==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+If <math>a,b,c</math> are positive integers less than <math>10</math>, then <math>(10a + b)(10a + c) = 100a(a + 1) + bc</math> if:
+<math>\textbf{(A)}\ b + c = 10 \qquad
+\textbf{(B)}\ b = c \qquad
+\textbf{(C)}\ a + b = 10 \qquad
+\textbf{(D)}\ a = b \\
+\textbf{(E)}\ a+b+c = 10   </math>
 [[1953 AHSME Problems/Problem 24|Solution]]
-== Problem 25 ==
+==Problem 25==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+In a geometric progression whose terms are positive, any term is equal to the sum of the next two following terms. then the common ratio is:
+<math>\textbf{(A)}\ 1 \qquad
+\textbf{(B)}\ \text{about }\frac{\sqrt{5}}{2} \qquad
+\textbf{(C)}\ \frac{\sqrt{5}-1}{2}\qquad
+\textbf{(D)}\ \frac{1-\sqrt{5}}{2}\qquad
+\textbf{(E)}\ \frac{2}{\sqrt{5}}     </math>
 [[1953 AHSME Problems/Problem 25|Solution]]
-== Problem 26 ==
+==Problem 26==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The base of a triangle is <math>15</math> inches. Two lines are drawn parallel to the base, terminating in the other two sides, and dividing the triangle into three equal areas. The length of the parallel closer to the base is:
+<math>\textbf{(A)}\ 5\sqrt{6}\text{ inches} \qquad
+\textbf{(B)}\ 10\text{ inches} \qquad
+\textbf{(C)}\ 4\sqrt{3}\text{ inches}\qquad
+\textbf{(D)}\ 7.5\text{ inches}\\
+\textbf{(E)}\ \text{none of these}    </math>
 [[1953 AHSME Problems/Problem 26|Solution]]
-== Problem 27 ==
+==Problem 27==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The radius of the first circle is <math>1</math> inch, that of the second <math>\frac{1}{2}</math> inch, that of the third <math>\frac{1}{4}</math> inch and so on indefinitely.
+The sum of the areas of the circles is:
+<math>\textbf{(A)}\ \frac{3\pi}{4} \qquad
+\textbf{(B)}\ 1.3\pi \qquad
+\textbf{(C)}\ 2\pi \qquad
+\textbf{(D)}\ \frac{4\pi}{3}\qquad
+\textbf{(E)}\ \text{none of these}     </math>
 [[1953 AHSME Problems/Problem 27|Solution]]
+==Problem 28==
-== Problem 28 ==
+In <math>\triangle ABC</math>, sides <math>a,b</math> and <math>c</math> are opposite <math>\angle{A},\angle{B}</math> and <math>\angle{C}</math> respectively. <math>AD</math> bisects <math>\angle{A}</math> and meets <math>BC</math> at <math>D</math>.
+Then if <math>x = \overline{CD}</math> and <math>y = \overline{BD}</math> the correct proportion is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math>\textbf{(A)}\ \frac {x}{a} = \frac {a}{b + c} \qquad
+\textbf{(B)}\ \frac {x}{b} = \frac {a}{a + c} \qquad
+\textbf{(C)}\ \frac{y}{c}=\frac{c}{b+c}\\ \textbf{(D)}\ \frac{y}{c}=\frac{a}{b+c}\qquad
+\textbf{(E)}\ \frac{x}{y}=\frac{c}{b}  </math>
 [[1953 AHSME Problems/Problem 28|Solution]]
-== Problem 29 ==
+==Problem 29==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The number of significant digits in the measurement of the side of a square whose computed area is <math>1.1025</math> square inches to
+the nearest ten-thousandth of a square inch is:
+<math>\textbf{(A)}\ 2 \qquad
+\textbf{(B)}\ 3 \qquad
+\textbf{(C)}\ 4 \qquad
+\textbf{(D)}\ 5 \qquad
+\textbf{(E)}\ 1     </math>
 [[1953 AHSME Problems/Problem 29|Solution]]
-== Problem 30 ==
+==Problem 30==
+A house worth \$ <math>9000</math> is sold by Mr. A to Mr. B at a <math>10</math> % loss. Mr. B sells the house back to Mr. A at a <math>10</math> % gain.
+The result of the two transactions is:
+<math>\textbf{(A)}\ \text{Mr. A breaks even} \qquad
+\textbf{(B)}\ \text{Mr. B gains }\$900 \qquad
+\textbf{(C)}\ \text{Mr. A loses }\$900\\ \textbf{(D)}\ \text{Mr. A loses }\$810\qquad
+\textbf{(E)}\ \text{Mr. B gains }\$1710  </math>
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
 [[1953 AHSME Problems/Problem 30|Solution]]
-== Problem 31 ==
+==Problem 31==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The rails on a railroad are <math>30</math> feet long. As the train passes over the point where the rails are joined, there is an audible click.
+The speed of the train in miles per hour is approximately the number of clicks heard in:
+<math>\textbf{(A)}\ 20\text{ seconds} \qquad
+\textbf{(B)}\ 2\text{ minutes} \qquad
+\textbf{(C)}\ 1\frac{1}{2}\text{ minutes}\qquad
+\textbf{(D)}\ 5\text{ minutes}\\ \textbf{(E)}\ \text{none of these}     </math>
 [[1953 AHSME Problems/Problem 31|Solution]]
-== Problem 32 ==
+==Problem 32==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form:
+<math>\textbf{(A)}\ \text{a square} \qquad
+\textbf{(B)}\ \text{a rectangle} \qquad
+\textbf{(C)}\ \text{a parallelogram with unequal sides}\\ \textbf{(D)}\ \text{a rhombus}\qquad
+\textbf{(E)}\ \text{a quadrilateral with no special properties}    </math>
 [[1953 AHSME Problems/Problem 32|Solution]]
-== Problem 33 ==
+==Problem 33==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The perimeter of an isosceles right triangle is <math>2p</math>. Its area is:
+<math>\textbf{(A)}\ (2+\sqrt{2})p \qquad
+\textbf{(B)}\ (2-\sqrt{2})p \qquad
+\textbf{(C)}\ (3-2\sqrt{2})p^2\\
+\textbf{(D)}\ (1-2\sqrt{2})p^2\qquad
+\textbf{(E)}\ (3+2\sqrt{2})p^2    </math>
 [[1953 AHSME Problems/Problem 33|Solution]]
-== Problem 34 ==
+==Problem 34==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+If one side of a triangle is <math>12</math> inches and the opposite angle is <math>30^{\circ}</math>, then the diameter of the circumscribed circle is:
+<math>\textbf{(A)}\ 18\text{ inches} \qquad
+\textbf{(B)}\ 30\text{ inches} \qquad
+\textbf{(C)}\ 24\text{ inches} \qquad
+\textbf{(D)}\ 20\text{ inches}\\ \textbf{(E)}\ \text{none of these}    </math>
 [[1953 AHSME Problems/Problem 34|Solution]]
-== Problem 35 ==
+==Problem 35==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+If <math>f(x)=\frac{x(x-1)}{2}</math>, then <math>f(x+2)</math> equals:
+<math>\textbf{(A)}\ f(x)+f(2) \qquad
+\textbf{(B)}\ (x+2)f(x) \qquad
+\textbf{(C)}\ x(x+2)f(x) \qquad
+\textbf{(D)}\ \frac{xf(x)}{x+2}\\ \textbf{(E)}\ \frac{(x+2)f(x+1)}{x}    </math>
 [[1953 AHSME Problems/Problem 35|Solution]]
-== Problem 36 ==
+==Problem 36==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+Determine <math>m</math> so that <math>4x^2-6x+m</math> is divisible by <math>x-3</math>. The obtained value, <math>m</math>, is an exact divisor of:
+<math>\textbf{(A)}\ 12 \qquad
+\textbf{(B)}\ 20 \qquad
+\textbf{(C)}\ 36 \qquad
+\textbf{(D)}\ 48 \qquad
+\textbf{(E)}\ 64     </math>
 [[1953 AHSME Problems/Problem 36|Solution]]
-== Problem 37 ==
+==Problem 37==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The base of an isosceles triangle is <math>6</math> inches and one of the equal sides is <math>12</math> inches.
+The radius of the circle through the vertices of the triangle is:
+<math>\textbf{(A)}\ \frac{7\sqrt{15}}{5} \qquad
+\textbf{(B)}\ 4\sqrt{3} \qquad
+\textbf{(C)}\ 3\sqrt{5} \qquad
+\textbf{(D)}\ 6\sqrt{3}\qquad
+\textbf{(E)}\ \text{none of these}   </math>
 [[1953 AHSME Problems/Problem 37|Solution]]
-== Problem 38 ==
+==Problem 38==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+If <math>f(a)=a-2</math> and <math>F(a,b)=b^2+a</math>, then <math>F(3,f(4))</math> is:
+<math>\textbf{(A)}\ a^2-4a+7 \qquad
+\textbf{(B)}\ 28 \qquad
+\textbf{(C)}\ 7 \qquad
+\textbf{(D)}\ 8 \qquad
+\textbf{(E)}\ 11     </math>
 [[1953 AHSME Problems/Problem 38|Solution]]
-== Problem 39 ==
+==Problem 39==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The product, <math>\log_a b \cdot \log_b a</math> is equal to:
+<math>\textbf{(A)}\ 1 \qquad
+\textbf{(B)}\ a \qquad
+\textbf{(C)}\ b \qquad
+\textbf{(D)}\ ab \qquad
+\textbf{(E)}\ \text{none of these}     </math>
 [[1953 AHSME Problems/Problem 39|Solution]]
-== Problem 40 ==
+==Problem 40==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The negation of the statement "all men are honest," is:
+<math>\textbf{(A)}\ \text{no men are honest} \qquad
+\textbf{(B)}\ \text{all men are dishonest} \\
+\textbf{(C)}\ \text{some men are dishonest}\qquad
+\textbf{(D)}\ \text{no men are dishonest}\\ \textbf{(E)}\ \text{some men are honest}   </math>
 [[1953 AHSME Problems/Problem 40|Solution]]
-== Problem 41 ==
+==Problem 41==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+A girls' camp is located <math>300</math> rods from a straight road. On this road, a boys' camp is located <math>500</math> rods from the girls' camp.
+It is desired to build a canteen on the road which shall be exactly the same distance from each camp.
+The distance of the canteen from each of the camps is:
+<math>\textbf{(A)}\ 400\text{ rods} \qquad
+\textbf{(B)}\ 250\text{ rods} \qquad
+\textbf{(C)}\ 87.5\text{ rods} \qquad
+\textbf{(D)}\ 200\text{ rods}\\ \textbf{(E)}\ \text{none of these} </math>
 [[1953 AHSME Problems/Problem 41|Solution]]
-== Problem 42 ==
+==Problem 42==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The centers of two circles are <math>41</math> inches apart. The smaller circle has a radius of <math>4</math> inches and the larger one has a radius of <math>5</math> inches.
+The length of the common internal tangent is:
+<math>\textbf{(A)}\ 41\text{ inches} \qquad
+\textbf{(B)}\ 39\text{ inches} \qquad
+\textbf{(C)}\ 39.8\text{ inches} \qquad
+\textbf{(D)}\ 40.1\text{ inches}\\ \textbf{(E)}\ 40\text{ inches} </math>
 [[1953 AHSME Problems/Problem 42|Solution]]
-== Problem 43 ==
+==Problem 43==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+If the price of an article is increased by percent <math>p</math>, then the decrease in percent of sales must not exceed <math>d</math> in order to yield the same income.
+The value of <math>d</math> is:
+<math>\textbf{(A)}\ \frac{1}{1+p} \qquad
+\textbf{(B)}\ \frac{1}{1-p} \qquad
+\textbf{(C)}\ \frac{p}{1+p} \qquad
+\textbf{(D)}\ \frac{p}{p-1}\qquad
+\textbf{(E)}\ \frac{1-p}{1+p}    </math>
 [[1953 AHSME Problems/Problem 43|Solution]]
-== Problem 44 ==
+==Problem 44==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ }  \qquad \textbf{(C) \ }  \qquad \textbf{(D) \ }  \qquad \textbf{(E) \ } </math>
+In solving a problem that reduces to a quadratic equation one student makes a mistake only in the constant term of the equation and
+obtains <math>8</math> and <math>2</math> for the roots. Another student makes a mistake only in the coefficient of the first degree term and
+find <math>-9</math> and <math>-1</math> for the roots. The correct equation was:
+<math>\textbf{(A)}\ x^2-10x+9=0 \qquad
+\textbf{(B)}\ x^2+10x+9=0 \qquad
+\textbf{(C)}\ x^2-10x+16=0\\
+\textbf{(D)}\ x^2-8x-9=0\qquad
+\textbf{(E)}\ \text{none of these}  </math>
 [[1953 AHSME Problems/Problem 44|Solution]]
-== Problem 45 ==
+==Problem 45==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The lengths of two line segments are <math>a</math> units and <math>b</math> units respectively. Then the correct relation between them is:
+<math>\textbf{(A)}\ \frac{a+b}{2} > \sqrt{ab} \qquad
+\textbf{(B)}\ \frac{a+b}{2} < \sqrt{ab} \qquad
+\textbf{(C)}\ \frac{a+b}{2}=\sqrt{ab}\\ \textbf{(D)}\ \frac{a+b}{2}\leq\sqrt{ab}\qquad
+\textbf{(E)}\ \frac{a+b}{2}\geq\sqrt{ab}   </math>
 [[1953 AHSME Problems/Problem 45|Solution]]
-== Problem 46 ==
+==Problem 46==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+Instead of walking along two adjacent sides of a rectangular field, a boy took a shortcut along the diagonal of the field and
+saved a distance equal to <math>\frac{1}{2}</math> the longer side. The ratio of the shorter side of the rectangle to the longer side was:
+<math>\textbf{(A)}\ \frac{1}{2} \qquad
+\textbf{(B)}\ \frac{2}{3} \qquad
+\textbf{(C)}\ \frac{1}{4} \qquad
+\textbf{(D)}\ \frac{3}{4}\qquad
+\textbf{(E)}\ \frac{2}{5}     </math>
 [[1953 AHSME Problems/Problem 46|Solution]]
-== Problem 47 ==
+==Problem 47==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+If <math>x>0</math>, then the correct relationship is:
+<math>\textbf{(A)}\ \log (1+x) = \frac{x}{1+x} \qquad
+\textbf{(B)}\ \log (1+x) < \frac{x}{1+x} \\
+\textbf{(C)}\ \log(1+x) > x\qquad
+\textbf{(D)}\ \log (1+x) < x\qquad
+\textbf{(E)}\ \text{none of these}  </math>
 [[1953 AHSME Problems/Problem 47|Solution]]
-== Problem 48 ==
+==Problem 48==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+If the larger base of an isosceles trapezoid equals a diagonal and the smaller base equals the altitude,
+then the ratio of the smaller base to the larger base is:
+<math>\textbf{(A)}\ \frac{1}{2} \qquad
+\textbf{(B)}\ \frac{2}{3} \qquad
+\textbf{(C)}\ \frac{3}{4} \qquad
+\textbf{(D)}\ \frac{3}{5}\qquad
+\textbf{(E)}\ \frac{2}{5}    </math>
 [[1953 AHSME Problems/Problem 48|Solution]]
-== Problem 49 ==
+==Problem 49==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The coordinates of <math>A,B</math> and <math>C</math> are <math>(5,5),(2,1)</math> and <math>(0,k)</math> respectively.
+The value of <math>k</math> that makes <math>\overline{AC}+\overline{BC}</math> as small as possible is:
+<math>\textbf{(A)}\ 3 \qquad
+\textbf{(B)}\ 4\frac{1}{2} \qquad
+\textbf{(C)}\ 3\frac{6}{7} \qquad
+\textbf{(D)}\ 4\frac{5}{6}\qquad
+\textbf{(E)}\ 2\frac{1}{7}     </math>
 [[1953 AHSME Problems/Problem 49|Solution]]
-== Problem 50 ==
+==Problem 50==
-One of the sides of a triangle is divided into segments of <math>6</math> and <math>8</math> units by the point of tangency of the inscribed circle. If the radius of the circle is <math>4</math>, then the length of the shortest side is
+One of the sides of a triangle is divided into segments of <math>6</math> and <math>8</math> units by the point of tangency of the inscribed circle.
+If the radius of the circle is <math>4</math>, then the length of the shortest side of the triangle is:
-<math> \textbf{(A) \ } 12 \mathrm{\ units} \qquad \textbf{(B) \ } 13 \mathrm{\ units} \qquad \textbf{(C) \ } 14 \mathrm{\ units} \qquad \textbf{(D) \ } 15 \mathrm{\ units} \qquad \textbf{(E) \ } 16 \mathrm{\ units} </math>
+<math>\textbf{(A)}\ 12\text{ units} \qquad
+\textbf{(B)}\ 13\text{ units} \qquad
+\textbf{(C)}\ 14\text{ units} \qquad
+\textbf{(D)}\ 15\text{ units}\qquad
+\textbf{(E)}\ 16\text{ units}</math>
+[[1953 AHSME Problems/Problem 50|Solution]]
-[[1953 AHSME Problems/Problem 50|Solution]]
 == See also ==
-* [[AHSME]]
-* [[AHSME Problems and Solutions]]
+* [[AMC 12 Problems and Solutions]]
 * [[Mathematics competition resources]]
+{{AHSME 50p box|year=1953|before=[[1952 AHSME|1952 AHSC]]|after=[[1954 AHSME|1954 AHSC]]}}
+{{MAA Notice}}

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Difference between revisions of "1953 AHSME Problems"

Latest revision as of 13:24, 20 February 2020

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