Difference between revisions of "1951 AHSME Problems"

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1951 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

The percent that $M$ is greater than $N$ is:

$\textbf{(A) \ } \frac {100(M - N)}{M} \qquad \textbf{(B) \ } \frac {100(M - N)}{N} \qquad \textbf{(C) \ } \frac {M - N}{N} \qquad \textbf{(D) \ } \frac {M - N}{M} \qquad \textbf{(E) \ } \frac {100(M + N)}{N}$

Solution

Problem 2

A rectangular field is half as wide as it is long and is completely enclosed by $x$ yards of fencing. The area in terms of $x$ is:

$(\textbf{A})\ \frac{x^2}2 \qquad (\textbf{B})\ 2x^2 \qquad (\textbf{C})\ \frac{2x^2}9 \qquad (\textbf{D})\ \frac{x^2}{18} \qquad (\textbf{E})\ \frac{x^2}{72}$

Solution

Problem 3

If the length of a diagonal of a square is $a + b$ , then the area of the square is:

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

Solution

Problem 4

A barn with a flat roof is rectangular in shape, $10$ yd. wide, $13$ yd. long and $5$ yd. high. It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of sq. yd. to be painted is:

$\textbf{(A) \ } 360 \qquad \textbf{(B) \ } 460 \qquad \textbf{(C) \ } 490 \qquad \textbf{(D) \ } 590 \qquad \textbf{(E) \ } 720$

Solution

Problem 5

Mr. A owns a home worth $10,000$ dollars. He sells it to Mr. B at a $10 \%$ profit based on the worth of the house. Mr. B sells the house back to Mr. A at a $10 \%$ loss. Then:

$\textbf{(A)}\ \text{A comes out even} \qquad\textbf{(B)}\ \text{A makes 1100 on the deal} \qquad\textbf{(C)}\ \text{A makes 1000 on the deal}$ $\textbf{(D)}\ \text{A loses 900 on the deal} \qquad\textbf{(E)}\ \text{A loses 1000 on the deal}$

Solution

Problem 6

The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to:

$\textbf{(A)}\ \text{the volume of the box} \qquad\textbf{(B)}\ \text{the square root of the volume} \qquad\textbf{(C)}\ \text{twice the volume}$ $\textbf{(D)}\ \text{the square of the volume} \qquad\textbf{(E)}\ \text{the cube of the volume}$

Solution

Problem 7

An error of $.02"$ is made in the measurement of a line $10"$ long, while an error of only $.2"$ is made in a measurement of a line $100"$ long. In comparison with the relative error of the first measurement, the relative error of the second measurement is:

$\textbf{(A) \ } \text{greater by }.18 \qquad\textbf{(B) \ } \text{the same} \qquad \textbf{(C) \ } \text{less} \qquad\textbf{(D) \ } 10\text{ times as great} \qquad\textbf{(E) \ } \text{correctly described by both}$

Solution

Problem 8

The price of an article is cut $10 \%.$ To restore it to its former value, the new price must be increased by:

$\textbf{(A) \ } 10 \% \qquad\textbf{(B) \ } 9 \% \qquad \textbf{(C) \ } 11\frac{1}{9} \% \qquad\textbf{(D) \ } 11 \% \qquad\textbf{(E) \ } \text{none of these answers}$

Solution

Problem 9

An equilateral triangle is drawn with a side length of $a.$ A new equilateral triangle is formed by joining the midpoints of the sides of the first one. then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. the limit of the sum of the perimeters of all the triangles thus drawn is:

$\textbf{(A) \ } \text{Infinite} \qquad\textbf{(B) \ } 5\frac{1}{4}a \qquad \textbf{(C) \ } 2a \qquad\textbf{(D) \ } 6a \qquad\textbf{(E) \ } 4\frac{1}{2}a$

Solution

Problem 10

Of the following statements, the one that is incorrect is:

$\textbf{(A)}\ \text{Doubling the base of a given rectangle doubles the area.}\qquad\textbf{(B)}\ \text{Doubling the altitude of a triangle doubles the area.}$ $\textbf{(C)}\ \text{Doubling the radius of a given circle doubles the area.}$ $\textbf{(D)}\ \text{Doubling the divisor of a fraction and dividing its numerator by 2 changes the quotient.}$ $\textbf{(E)}\ \text{Doubling a given quantity may make it less than it originally was.}$

Solution

Problem 11

The limit of the sum of an infinite number of terms in a geometric progression is $\frac {a}{1- r}$ where $a$ denotes the first term and $-1 < r < 1$ denotes the common ratio. The limit of the sum of their squares is:

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

Solution

Problem 12

At $2: 15$ o'clock, the hour and minute hands of a clock form an angle of:

$\textbf{(A)}\ 30^{\circ} \qquad\textbf{(B)}\ 5^{\circ} \qquad\textbf{(C)}\ 22\frac {1}{2}^{\circ} \qquad\textbf{(D)}\ 7\frac {1}{2} ^{\circ} \qquad\textbf{(E)}\ 28^{\circ}$

Solution

Problem 13

$A$ can do a piece of work in $9$ days. $B$ is $50\%$ more efficient than $A$ . The number of days it takes $B$ to do the same piece of work is:

$\textbf{(A)}\ 13\frac {1}{2} \qquad\textbf{(B)}\ 4\frac {1}{2} \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ \text{none of these answers}$

Solution

Problem 14

In connection with proof in geometry, indicate which one of the following statements is incorrect:

$\textbf{(A)}\ \text{Some statements are accepted without being proved.}$ $\textbf{(B)}\ \text{In some cases there is more than one correct order in proving certain propositions.}$ $\textbf{(C)}\ \text{Every term used in a proof must have been defined previously.}$ $\textbf{(D)}\ \text{It is not possible to arrive by correct reasoning at a true conclusion if, in the given, there is an untrue proposition.}$ $\textbf{(E)}\ \text{Indirect proof can be used whenever there are two or more contrary propositions.}$

Solution

Problem 15

The largest number by which the expression $n^3-n$ is divisible for all possible integer values of $n$ , is:

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

Solution

Problem 16

If in applying the quadratic formula to a quadratic equation

$f(x) \equiv ax^2 + bx + c = 0,$

it happens that $c = \frac{b^2}{4a}$ , then the graph of $y = f(x)$ will certainly:

$\textbf{(A)}\ \text{have a maximum}\qquad\textbf{(B)}\ \text{have a minimum}\qquad\textbf{(C)}\ \text{be tangent to the x-axis}\\ \qquad\textbf{(D)}\ \text{be tangent to the y-axis}\qquad\textbf{(E)}\ \text{lie in one quadrant only}$

Solution

Problem 17

Indicate in which one of the following equations $y$ is neither directly nor inversely proportional to $x$ :

$\textbf{(A)}\ x+y = 0\qquad\textbf{(B)}\ 3xy = 10\qquad\textbf{(C)}\ x = 5y \qquad\textbf{(D)}\ 3x+y = 10 \qquad\textbf{(E)}\ x/y = \sqrt{3}\qquad$

Solution

Problem 18

The expression $21x^2+ax+21$ is to be factored into prime binomial factors and without a numerical monomial factor. This can be done if the value ascribed to $a$ is:

$\textbf{(A)}\ \text{any odd number}\qquad\textbf{(B)}\ \text{some odd number}\qquad\textbf{(C)}\ \text{any even number}\qquad\textbf{(D)}\ \text{some even number}\qquad\textbf{(E)}\ \text{zero}$

Solution

Problem 19

A six place number is formed by repeating a three place number; for example, $256256$ or $678678$ , etc. Any number of this form is always exactly divisible by:

$\textbf{(A)}\ 7\text{ only}\qquad\textbf{(B)}\ 11\text{ only}\qquad\textbf{(C)}\ 13\text{ only}\qquad\textbf{(D)}\ 101\qquad\textbf{(E)}\ 1001$

Solution

Problem 20

When simplified and expressed with negative exponents, the expression $(x+y)^{-1}(x^{-1}+y^{-1})$ is equal to:

$\textbf{(A)}\ x^{-2}+2x^{-1}y^{-1}+y^{-2}\qquad\textbf{(B)}\ x^{-2}+2^{-1}x^{-1}y^{-1}+y^{-2}\qquad\textbf{(C)}\ x^{-1}y^{-1}$ $\textbf{(D)}\ \text{some even number}\qquad\textbf{(E)}\ \text{zero}$

Solution

Problem 21

Given: $x > 0, y > 0, x > y$ and $z\not = 0$ . The inequality which is not always correct is:

$\textbf{(A)}\ x+z > y+z\qquad\textbf{(B)}\ x-z > y-z\qquad\textbf{(C)}\ xz > yz$ $\textbf{(D)}\ \frac{x}{z^{2}}>\frac{y}{z^{2}}\qquad\textbf{(E)}\ xz^{2}> yz^{2}$

Solution

Problem 22

The values of $a$ in the equation: $\log_{10}(a^{2}-15a) = 2$ are:

$\textbf{(A)}\ \frac{15\pm\sqrt{233}}{2}\qquad\textbf{(B)}\ 20,-5\qquad\textbf{(C)}\ \frac{15\pm\sqrt{305}}{2}\qquad\textbf{(D)}\ \pm20$ $\textbf{(E)}\ \text{none of these}$

Solution

Problem 23

The radius of a cylindrical box is $8$ inches and the height is $3$ inches. The number of inches that may be added to either the radius or the height to give the same nonzero increase in volume is:

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 5\frac{1}{3}\qquad\textbf{(C)}\ \text{any number}\qquad\textbf{(D)}\ \text{non-existent}\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 24

$\frac{2^{n+4}-2(2^{n})}{2(2^{n+3})}$ when simplified is:

$\textbf{(A)}\ 2^{n+1}-\frac{1}{8}\qquad\textbf{(B)}\ -2^{n+1}\qquad\textbf{(C)}\ 1-2^{n}\qquad\textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{7}{4}$

Solution

Problem 25

The apothem of a square having its area numerically equal to its perimeter is compared with the apothem of an equilateral triangle having its area numerically equal to its perimeter. The first apothem will be:

$\textbf{(A)}\ \text{equal to the second}\qquad\textbf{(B)}\ \frac{4}{3}\text{ times the second}\qquad\textbf{(C)}\ \frac{2}{\sqrt{3}}\text{ times the second}\\ \textbf{(D)}\ \frac{\sqrt{2}}{\sqrt{3}}\text{ times the second}\qquad\textbf{(E)}\ \text{indeterminately related to the second}$

Solution

Problem 26

In the equation $\frac{x(x-1)-(m+1)}{(x-1)(m-1)}=\frac{x}{m}$ the roots are equal when:

$\textbf{(A)}\ m = 1\qquad\textbf{(B)}\ m =\frac{1}{2}\qquad\textbf{(C)}\ m = 0\qquad\textbf{(D)}\ m =-1\qquad\textbf{(E)}\ m =-\frac{1}{2}$

Solution

Problem 27

Through a point inside a triangle, three lines are drawn from the vertices to the opposite sides forming six triangular sections. Then:

$\textbf{(A)}\ \text{the triangles are similar in opposite pairs}\qquad\textbf{(B)}\ \text{the triangles are congruent in opposite pairs}$ $\textbf{(C)}\ \text{the triangles are equal in area in opposite pairs}\qquad\textbf{(D)}\ \text{three similar quadrilaterals are formed}$ $\textbf{(E)}\ \text{none of the above relations are true}$

Solution

Problem 28

The pressure $(P)$ of wind on a sail varies jointly as the area $(A)$ of the sail and the square of the velocity $(V)$ of the wind. The pressure on a square foot is $1$ pound when the velocity is $16$ miles per hour. The velocity of the wind when the pressure on a square yard is $36$ pounds is:

$\textbf{(A)}\ 10\frac{2}{3}\text{ mph}\qquad\textbf{(B)}\ 96\text{ mph}\qquad\textbf{(C)}\ 32\text{ mph}\qquad\textbf{(D)}\ 1\frac{2}{3}\text{ mph}\qquad\textbf{(E)}\ 16\text{ mph}$

Solution

Problem 29

Of the following sets of data the only one that does not determine the shape of a triangle is:

$\textbf{(A)}\ \text{the ratio of two sides and the inc{}luded angle}\\ \qquad\textbf{(B)}\ \text{the ratios of the three altitudes}\\ \qquad\textbf{(C)}\ \text{the ratios of the three medians}\\ \qquad\textbf{(D)}\ \text{the ratio of the altitude to the corresponding base}\\ \qquad\textbf{(E)}\ \text{two angles}$

Solution

Problem 30

If two poles $20''$ and $80''$ high are $100''$ apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:

$\textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 31

A total of $28$ handshakes were exchanged at the conclusion of a party. Assuming that each participant was equally polite toward all the others, the number of people present was:

$\textbf{(A)}\ 14\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 56\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 7$

Solution

Problem 32

If $\triangle ABC$ is inscribed in a semicircle whose diameter is $AB$ , then $AC+BC$ must be

$\textbf{(A)}\ \text{equal to }AB\qquad\textbf{(B)}\ \text{equal to }AB\sqrt{2}\qquad\textbf{(C)}\ \geq AB\sqrt{2}\qquad\textbf{(D)}\ \leq AB\sqrt{2} \qquad \textbf{(E)}\ AB^{2}$

Solution

Problem 33

The roots of the equation $x^{2}-2x = 0$ can be obtained graphically by finding the abscissas of the points of intersection of each of the following pairs of equations except the pair:

$\textbf{(A)}\ y = x^{2}, y = 2x\qquad\textbf{(B)}\ y = x^{2}-2x, y = 0\qquad\textbf{(C)}\ y = x, y = x-2\qquad\textbf{(D)}\ y = x^{2}-2x+1, y = 1$ $\textbf{(E)}\ y = x^{2}-1, y = 2x-1$

[Note: Abscissa means x-coordinate.]

Solution

Problem 34

The value of $10^{\log_{10}7}$ is:

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ \log_{10}7\qquad\textbf{(E)}\ \log_{7}10$

Solution

Problem 35

If $a^{x}= c^{q}= b$ and $c^{y}= a^{z}= d$ , then

$\textbf{(A)}\ xy = qz\qquad\textbf{(B)}\ \frac{x}{y}=\frac{q}{z}\qquad\textbf{(C)}\ x+y = q+z\qquad\textbf{(D)}\ x-y = q-z$ $\textbf{(E)}\ x^{y}= q^{z}$

Solution

Problem 36

Which of the following methods of proving a geometric figure a locus is not correct?

$\textbf{(A)}\ \text{Every point of the locus satisfies the conditions and every point not on the locus does not satisfy the conditions.}$ $\textbf{(B)}\ \text{Every point not satisfying the conditions is not on the locus and every point on the locus does satisfy the conditions.}$ $\textbf{(C)}\ \text{Every point satisfying the conditions is on the locus and every point on the locus satisfies the conditions.}$ $\textbf{(D)}\ \text{Every point not on the locus does not satisfy the conditions and every point not satisfying}\\ \text{the conditions is not on the locus.}$ $\textbf{(E)}\ \text{Every point satisfying the conditions is on the locus and every point not satisfying the conditions is not on the locus.}$

Solution

Problem 37

A number which when divided by $10$ leaves a remainder of $9$ , when divided by $9$ leaves a remainder of $8$ , by $8$ leaves a remainder of $7$ , etc., down to where, when divided by $2$ , it leaves a remainder of $1$ , is:

$\textbf{(A)}\ 59\qquad\textbf{(B)}\ 419\qquad\textbf{(C)}\ 1259\qquad\textbf{(D)}\ 2519\qquad\textbf{(E)}\ \text{none of these answers}$

Solution

Problem 38

A rise of $600$ feet is required to get a railroad line over a mountain. The grade can be kept down by lengthening the track and curving it around the mountain peak. The additional length of track required to reduce the grade from $3\%$ to $2\%$ is approximately:

$\textbf{(A)}\ 10000\text{ ft.}\qquad\textbf{(B)}\ 20000\text{ ft.}\qquad\textbf{(C)}\ 30000\text{ ft.}\qquad\textbf{(D)}\ 12000\text{ ft.}\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 39

A stone is dropped into a well and the report of the stone striking the bottom is heard $7.7$ seconds after it is dropped. Assume that the stone falls $16t^2$ feet in t seconds and that the velocity of sound is $1120$ feet per second. The depth of the well is:

$\textbf{(A)}\ 784\text{ ft.}\qquad\textbf{(B)}\ 342\text{ ft.}\qquad\textbf{(C)}\ 1568\text{ ft.}\qquad\textbf{(D)}\ 156.8\text{ ft.}\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 40

$\left(\frac{(x+1)^{2}(x^{2}-x+1)^{2}}{(x^{3}+1)^{2}}\right)^{2}\cdot\left(\frac{(x-1)^{2}(x^{2}+x+1)^{2}}{(x^{3}-1)^{2}}\right)^{2}$ equals:

$\textbf{(A)}\ (x+1)^{4}\qquad\textbf{(B)}\ (x^{3}+1)^{4}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ [(x^{3}+1)(x^{3}-1)]^{2}$ $\textbf{(E)}\ [(x^{3}-1)^{2}]^{2}$

Solution

Problem 41

The formula expressing the relationship between $x$ and $y$ in the table is: $\begin{tabular}{|c|c|c|c|c|c|}\hline x & 2 & 3 & 4 & 5 & 6\\ \hline y & 0 & 2 & 6 & 12 & 20\\ \hline\end{tabular}$

$\textbf{(A)}\ y = 2x-4\qquad\textbf{(B)}\ y = x^{2}-3x+2\qquad\textbf{(C)}\ y = x^{3}-3x^{2}+2x$ $\textbf{(D)}\ y = x^{2}-4x\qquad\textbf{(E)}\ y = x^{2}-4$

Solution

Problem 42

If $x =\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}$ , then:

$\textbf{(A)}\ x = 1\qquad\textbf{(B)}\ 0 < x < 1\qquad\textbf{(C)}\ 1 < x < 2\qquad\textbf{(D)}\ x\text{ is infinite}$ $\textbf{(E)}\ x > 2\text{ but finite}$

Solution

Problem 43

Of the following statements, the only one that is incorrect is:

$\textbf{(A)}\ \text{An inequality will remain true after each side is increased,}$ $\text{ decreased, multiplied or divided (zero excluded) by the same positive quantity.}$

$\textbf{(B)}\ \text{The arithmetic mean of two unequal positive quantities is greater than their geometric mean.}$

$\textbf{(C)}\ \text{If the sum of two positive quantities is given, ther product is largest when they are equal.}$

$\textbf{(D)}\ \text{If }a\text{ and }b\text{ are positive and unequal, }\frac{1}{2}(a^{2}+b^{2})\text{ is greater than }[\frac{1}{2}(a+b)]^{2}.$

$\textbf{(E)}\ \text{If the product of two positive quantities is given, their sum is greatest when they are equal.}$

Solution

Problem 44

If $\frac{xy}{x+y}= a,\frac{xz}{x+z}= b,\frac{yz}{y+z}= c$ , where $a, b, c$ are other than zero, then $x$ equals:

$\textbf{(A)}\ \frac{abc}{ab+ac+bc}\qquad\textbf{(B)}\ \frac{2abc}{ab+bc+ac}\qquad\textbf{(C)}\ \frac{2abc}{ab+ac-bc}$ $\textbf{(D)}\ \frac{2abc}{ab+bc-ac}\qquad\textbf{(E)}\ \frac{2abc}{ac+bc-ab}$

Solution

Problem 45

If you are given $\log 8\approx .9031$ and $\log 9\approx .9542$ , then the only logarithm that cannot be found without the use of tables is:

$\textbf{(A)}\ \log 17\qquad\textbf{(B)}\ \log\frac{5}{4}\qquad\textbf{(C)}\ \log 15\qquad\textbf{(D)}\ \log 600\qquad\textbf{(E)}\ \log .4$

Solution

Problem 46

$AB$ is a fixed diameter of a circle whose center is $O$ . From $C$ , any point on the circle, a chord $CD$ is drawn perpendicular to $AB$ . Then, as $C$ moves over a semicircle, the bisector of angle $OCD$ cuts the circle in a point that always:

$\textbf{(A)}\ \text{bisects the arc }AB\qquad\textbf{(B)}\ \text{trisects the arc }AB\qquad\textbf{(C)}\ \text{varies}$ $\textbf{(D)}\ \text{is as far from }AB\text{ as from }D\qquad\textbf{(E)}\ \text{is equidistant from }B\text{ and }C$

Solution

Problem 47

If $r$ and $s$ are the roots of the equation $ax^2+bx+c=0$ , the value of $\frac{1}{r^{2}}+\frac{1}{s^{2}}$ is:

$\textbf{(A)}\ b^{2}-4ac\qquad\textbf{(B)}\ \frac{b^{2}-4ac}{2a}\qquad\textbf{(C)}\ \frac{b^{2}-4ac}{c^{2}}\qquad\textbf{(D)}\ \frac{b^{2}-2ac}{c^{2}}$ $\textbf{(E)}\ \text{none of these}$

Solution

Problem 48

The area of a square inscribed in a semicircle is to the area of the square inscribed in the entire circle as:

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

Solution

Problem 49

The medians of a right triangle which are drawn from the vertices of the acute angles are $5$ and $\sqrt{40}$ . The value of the hypotenuse is:

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 2\sqrt{40}\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 2\sqrt{13}\qquad\textbf{(E)}\ \text{none of these}$

Solution

Problem 50

Tom, Dick and Harry started out on a $100$ -mile journey. Tom and Harry went by automobile at the rate of $25$ mph, while Dick walked at the rate of $5$ mph. After a certain distance, Harry got off and walked on at $5$ mph, while Tom went back for Dick and got him to the destination at the same time that Harry arrived. The number of hours required for the trip was:

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ \text{none of these answers}$

Solution

1951 AHSC (Problems • Answer Key • Resources)
Preceded by 1950 AHSC	Followed by 1952 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=1951
+}}
 == Problem 1 ==
 The percent that <math>M</math> is greater than <math>N</math> is:
-<math> \mathrm{(A) \ } \frac {100(M - N)}{M} \qquad \mathrm{(B) \ } \frac {100(M - N)}{N} \qquad \mathrm{(C) \ } \frac {M - N}{N} \qquad \mathrm{(D) \ } \frac {M - N}{M} \qquad \mathrm{(E) \ } \frac {100(M + N)}{N} </math>
+<math> \textbf{(A) \ } \frac {100(M - N)}{M} \qquad \textbf{(B) \ } \frac {100(M - N)}{N} \qquad \textbf{(C) \ } \frac {M - N}{N} \qquad \textbf{(D) \ } \frac {M - N}{M} \qquad \textbf{(E) \ } \frac {100(M + N)}{N} </math>
 [[1951 AHSME Problems/Problem 1|Solution]]
@@ Line 9: / Line 12: @@
 A rectangular field is half as wide as it is long and is completely enclosed by <math>x</math> yards of fencing. The area in terms of <math>x</math> is:
-<math>(\mathrm{A})\ \frac{x^2}2 \qquad (\mathrm{B})\ 2x^2 \qquad (\mathrm{C})\ \frac{2x^2}9 \qquad (\mathrm{D})\ \frac{x^2}{18} \qquad (\mathrm{E})\ \frac{x^2}{72}</math>
+<math>(\textbf{A})\ \frac{x^2}2 \qquad (\textbf{B})\ 2x^2 \qquad (\textbf{C})\ \frac{2x^2}9 \qquad (\textbf{D})\ \frac{x^2}{18} \qquad (\textbf{E})\ \frac{x^2}{72}</math>
 [[1951 AHSME Problems/Problem 2|Solution]]
@@ Line 16: / Line 19: @@
 If the length of a diagonal of a square is <math>a + b</math>, then the area of the square is:
-<math> \mathrm{(A) \ (a+b)^2 } \qquad \mathrm{(B) \ \frac{1}{2}(a+b)^2 } \qquad \mathrm{(C) \ a^2+b^2 } \qquad \mathrm{(D) \ \frac {1}{2}(a^2+b^2) } \qquad \mathrm{(E) \ \text{none of these} }  </math>
+<math> \textbf{(A)}\ (a+b)^{2}\qquad\textbf{(B)}\ \frac{1}{2}(a+b)^{2}\qquad\textbf{(C)}\ a^{2}+b^{2} </math>
+<math> \textbf{(D)}\ \frac{1}{2}(a^{2}+b^{2})\qquad\textbf{(E)}\ \text{none of these} </math>
 [[1951 AHSME Problems/Problem 3|Solution]]
@@ Line 23: / Line 27: @@
 A barn with a flat roof is rectangular in shape, <math>10</math> yd. wide, <math>13</math> yd. long and <math>5</math> yd. high.  It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of sq. yd. to be painted is:
-<math> \mathrm{(A) \ } 360 \qquad \mathrm{(B) \ } 460 \qquad \mathrm{(C) \ } 490 \qquad \mathrm{(D) \ } 590 \qquad \mathrm{(E) \ } 720 </math>
+<math> \textbf{(A) \ } 360 \qquad \textbf{(B) \ } 460 \qquad \textbf{(C) \ } 490 \qquad \textbf{(D) \ } 590 \qquad \textbf{(E) \ } 720 </math>
 [[1951 AHSME Problems/Problem 4|Solution]]
@@ Line 30: / Line 34: @@
 Mr. A owns a home worth <math>10,000</math> dollars.  He sells it to Mr. B at a <math>10 \%</math> profit based on the worth of the house. Mr. B sells the house back to Mr. A at a <math>10 \%</math> loss.  Then:
-<math> \textrm{(A)}\ \text{A comes out even} \qquad\textrm{(B)}\ \text{A makes 1100 on the deal} \qquad\textrm{(C)}\ \text{A makes 1000 on the deal}</math>
+<math> \textbf{(A)}\ \text{A comes out even} \qquad\textbf{(B)}\ \text{A makes 1100 on the deal} \qquad\textbf{(C)}\ \text{A makes 1000 on the deal}</math>
-<math>\textrm{(D)}\ \text{A loses 900 on the deal} \qquad\textrm{(E)}\ \text{A loses 1000 on the deal}</math>
+<math>\textbf{(D)}\ \text{A loses 900 on the deal} \qquad\textbf{(E)}\ \text{A loses 1000 on the deal}</math>
 [[1951 AHSME Problems/Problem 5|Solution]]
@@ Line 39: / Line 43: @@
 The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to:
-<math> \textrm{(A)}\ \text{the volume of the box} \qquad\textrm{(B)}\ \text{the square root of the volume} \qquad\textrm{(C)}\ \text{twice the volume}</math>
+<math> \textbf{(A)}\ \text{the volume of the box} \qquad\textbf{(B)}\ \text{the square root of the volume} \qquad\textbf{(C)}\ \text{twice the volume}</math>
-<math> \textrm{(D)}\ \text{the square of the volume} \qquad\textrm{(E)}\ \text{the cube of the volume}</math>
+<math> \textbf{(D)}\ \text{the square of the volume} \qquad\textbf{(E)}\ \text{the cube of the volume}</math>
 [[1951 AHSME Problems/Problem 6|Solution]]
@@ Line 48: / Line 52: @@
 An error of <math>.02"</math> is made in the measurement of a line <math>10"</math> long, while an error of only <math>.2"</math> is made in a measurement of a line <math>100"</math> long. In comparison with the relative error of the first measurement, the relative error of the second measurement is:
-<math> \mathrm{(A) \ } \text{greater by }.18 \qquad\mathrm{(B) \ } \text{the same} \qquad \mathrm{(C) \ } \text{less} \qquad\mathrm{(D) \ } 10\text{ times as great} \qquad\mathrm{(E) \ } \text{correctly described by both} </math>
+<math> \textbf{(A) \ } \text{greater by }.18 \qquad\textbf{(B) \ } \text{the same} \qquad \textbf{(C) \ } \text{less} \qquad\textbf{(D) \ } 10\text{ times as great} \qquad\textbf{(E) \ } \text{correctly described by both} </math>
 [[1951 AHSME Problems/Problem 7|Solution]]
@@ Line 56: / Line 60: @@
 The price of an article is cut <math>10 \%.</math> To restore it to its former value, the new price must be increased by:
-<math> \mathrm{(A) \ } 10 \% \qquad\mathrm{(B) \ } 9 \% \qquad \mathrm{(C) \ } 11\frac{1}{9} \% \qquad\mathrm{(D) \ } 11 \% \qquad\mathrm{(E) \ } \text{none of these answers} </math>
+<math> \textbf{(A) \ } 10 \% \qquad\textbf{(B) \ } 9 \% \qquad \textbf{(C) \ } 11\frac{1}{9} \% \qquad\textbf{(D) \ } 11 \% \qquad\textbf{(E) \ } \text{none of these answers} </math>
 [[1951 AHSME Problems/Problem 8|Solution]]
@@ Line 64: / Line 68: @@
 An equilateral triangle is drawn with a side length of <math>a.</math> A new equilateral triangle is formed by joining the midpoints of the sides of the first one. then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. the limit of the sum of the perimeters of all the triangles thus drawn is:
-<math> \mathrm{(A) \ } \text{Infinite} \qquad\mathrm{(B) \ } 5\frac{1}{4}a \qquad \mathrm{(C) \ } 2a \qquad\mathrm{(D) \ } 6a \qquad\mathrm{(E) \ } 4\frac{1}{2}a </math>
+<math> \textbf{(A) \ } \text{Infinite} \qquad\textbf{(B) \ } 5\frac{1}{4}a \qquad \textbf{(C) \ } 2a \qquad\textbf{(D) \ } 6a \qquad\textbf{(E) \ } 4\frac{1}{2}a </math>
 [[1951 AHSME Problems/Problem 9|Solution]]
@@ Line 71: / Line 75: @@
 Of the following statements, the one that is incorrect is:
-<math> \textrm{(A)}\ \text{Doubling the base of a given rectangle doubles the area.}</math>
+<math> \textbf{(A)}\ \text{Doubling the base of a given rectangle doubles the area.}\qquad\textbf{(B)}\ \text{Doubling the altitude of a triangle doubles the area.}</math>
-<math> \textrm{(B)}\ \text{Doubling the altitude of a triangle doubles the area.}</math>
+<math> \textbf{(C)}\ \text{Doubling the radius of a given circle doubles the area.}</math>
-<math> \textrm{(C)}\ \text{Doubling the radius of a given circle doubles the area.}</math>
+<math> \textbf{(D)}\ \text{Doubling the divisor of a fraction and dividing its numerator by 2 changes the quotient.}</math>
-<math> \textrm{(D)}\ \text{Doubling the divisor of a fraction and dividing its numerator by 2 changes the quotient.}</math>
+<math> \textbf{(E)}\ \text{Doubling a given quantity may make it less than it originally was.}</math>
-<math> \textrm{(E)}\ \text{Doubling a given quantity may make it less than it originally was.}</math>
 [[1951 AHSME Problems/Problem 10|Solution]]
@@ Line 81: / Line 84: @@
 == Problem 11 ==
-The limit of the sum of an infinite number of terms in a geometric progression is <math> \frac {a}{1 \minus{} r}</math> where <math> a</math> denotes the first term and <math> \minus{} 1 < r < 1</math> denotes the common ratio. The limit of the sum of their squares is:
+The limit of the sum of an infinite number of terms in a geometric progression is <math> \frac {a}{1- r}</math> where <math> a</math> denotes the first term and <math> -1 < r < 1</math> denotes the common ratio. The limit of the sum of their squares is:
-<math> \textrm{(A)}\ \frac {a^2}{(1 \minus{} r)^2} \qquad\textrm{(B)}\ \frac {a^2}{1 \plus{} r^2} \qquad\textrm{(C)}\ \frac {a^2}{1 \minus{} r^2} \qquad\textrm{(D)}\ \frac {4a^2}{1 \plus{} r^2} \qquad\textrm{(E)}\ \text{none of these}</math>
+<math> \textbf{(A)}\ \frac {a^2}{(1 -r)^2} \qquad\textbf{(B)}\ \frac {a^2}{1 + r^2} \qquad\textbf{(C)}\ \frac {a^2}{1 - r^2} \qquad\textbf{(D)}\ \frac {4a^2}{1+ r^2} \qquad\textbf{(E)}\ \text{none of these}</math>
 [[1951 AHSME Problems/Problem 11|Solution]]
@@ Line 91: / Line 94: @@
 At <math> 2: 15</math> o'clock, the hour and minute hands of a clock form an angle of:
-<math> \textrm{(A)}\ 30^{\circ} \qquad\textrm{(B)}\ 5^{\circ} \qquad\textrm{(C)}\ 22\frac {1}{2}^{\circ} \qquad\textrm{(D)}\ 7\frac {1}{2} ^{\circ} \qquad\textrm{(E)}\ 28^{\circ}</math>
+<math> \textbf{(A)}\ 30^{\circ} \qquad\textbf{(B)}\ 5^{\circ} \qquad\textbf{(C)}\ 22\frac {1}{2}^{\circ} \qquad\textbf{(D)}\ 7\frac {1}{2} ^{\circ} \qquad\textbf{(E)}\ 28^{\circ}</math>
 [[1951 AHSME Problems/Problem 12|Solution]]
@@ Line 99: / Line 102: @@
 <math> A</math> can do a piece of work in <math> 9</math> days. <math> B</math> is <math> 50\%</math> more efficient than <math> A</math>. The number of days it takes <math> B</math> to do the same piece of work is:
-<math> \textrm{(A)}\ 13\frac {1}{2} \qquad\textrm{(B)}\ 4\frac {1}{2} \qquad\textrm{(C)}\ 6 \qquad\textrm{(D)}\ 3 \qquad\textrm{(E)}\ \text{none of these answers}</math>
+<math> \textbf{(A)}\ 13\frac {1}{2} \qquad\textbf{(B)}\ 4\frac {1}{2} \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ \text{none of these answers}</math>
 [[1951 AHSME Problems/Problem 13|Solution]]
@@ Line 107: / Line 110: @@
 In connection with proof in geometry, indicate which one of the following statements is ''incorrect'':
-<math> \textrm{(A)}\ \text{Some statements are accepted without being proved.} </math>
+<math> \textbf{(A)}\ \text{Some statements are accepted without being proved.} </math>
-<math> \textrm{(B)}\ \text{In some cases there is more than one correct order in proving certain propositions.} </math>
+<math> \textbf{(B)}\ \text{In some cases there is more than one correct order in proving certain propositions.} </math>
-<math> \textrm{(C)}\ \text{Every term used in a proof must have been defined previously.} </math>
+<math> \textbf{(C)}\ \text{Every term used in a proof must have been defined previously.} </math>
-<math> \textrm{(D)}\ \text{It is not possible to arrive by correct reasoning at a true conclusion if, in the given, there is an untrue proposition.} </math>
+<math> \textbf{(D)}\ \text{It is not possible to arrive by correct reasoning at a true conclusion if, in the given, there is an untrue proposition.} </math>
-<math> \textrm{(E)}\ \text{Indirect proof can be used whenever there are two or more contrary propositions.} </math>
+<math> \textbf{(E)}\ \text{Indirect proof can be used whenever there are two or more contrary propositions.} </math>
 [[1951 AHSME Problems/Problem 14|Solution]]
@@ Line 117: / Line 120: @@
 == Problem 15 ==
-The largest number by which the expression <math>n^3-n</math> is divisible for all possible integral values of <math>n</math>, is:
+The largest number by which the expression <math>n^3-n</math> is divisible for all possible integer values of <math>n</math>, is:
 <math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6 </math>
@@ Line 138: / Line 141: @@
 Indicate in which one of the following equations <math>y</math> is neither directly nor inversely proportional to <math>x</math>:
-<math> \textbf{(A)}\ x+y = 0\qquad\textbf{(B)}\ 3xy = 10\qquad\textbf{(C)}\ x = 5y\qquad\textbf{(D)}\ 3x+y = 10 </math>
+<math> \textbf{(A)}\ x+y = 0\qquad\textbf{(B)}\ 3xy = 10\qquad\textbf{(C)}\ x = 5y \qquad\textbf{(D)}\ 3x+y = 10 \qquad\textbf{(E)}\ x/y = \sqrt{3}\qquad</math>
 [[1951 AHSME Problems/Problem 17|Solution]]
@@ Line 144: / Line 147: @@
 == Problem 18 ==
-The expression <math>21x^2+ax+21</math> is to be factored into two linear prime binomial factors with integer coefficients. This can be one if <math>a</math> is:
+The expression <math>21x^2+ax+21</math> is to be factored into prime binomial factors and without a numerical monomial factor. This can be done if the value ascribed to <math>a</math> is:
 <math> \textbf{(A)}\ \text{any odd number}\qquad\textbf{(B)}\ \text{some odd number}\qquad\textbf{(C)}\ \text{any even number}\qquad\textbf{(D)}\ \text{some even number}\qquad\textbf{(E)}\ \text{zero} </math>
@@ Line 177: / Line 180: @@
 == Problem 22 ==
+The values of <math>a</math> in the equation: <math> \log_{10}(a^{2}-15a) = 2 </math> are:
+<math> \textbf{(A)}\ \frac{15\pm\sqrt{233}}{2}\qquad\textbf{(B)}\ 20,-5\qquad\textbf{(C)}\ \frac{15\pm\sqrt{305}}{2}\qquad\textbf{(D)}\ \pm20 </math>
+<math> \textbf{(E)}\ \text{none of these} </math>
 [[1951 AHSME Problems/Problem 22|Solution]]
 == Problem 23 ==
+The radius of a cylindrical box is <math>8</math> inches and the height is <math>3</math> inches. The number of inches that may be added to either the radius or the height to give the same nonzero increase in volume is:
+<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 5\frac{1}{3}\qquad\textbf{(C)}\ \text{any number}\qquad\textbf{(D)}\ \text{non-existent}\qquad\textbf{(E)}\ \text{none of these} </math>
 [[1951 AHSME Problems/Problem 23|Solution]]
 == Problem 24 ==
+<math> \frac{2^{n+4}-2(2^{n})}{2(2^{n+3})} </math> when simplified is:
+<math> \textbf{(A)}\ 2^{n+1}-\frac{1}{8}\qquad\textbf{(B)}\ -2^{n+1}\qquad\textbf{(C)}\ 1-2^{n}\qquad\textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{7}{4} </math>
 [[1951 AHSME Problems/Problem 24|Solution]]
 == Problem 25 ==
+The apothem of a square having its area numerically equal to its perimeter is compared with the apothem of an equilateral triangle having its area numerically equal to its perimeter. The first apothem will be:
+<math> \textbf{(A)}\ \text{equal to the second}\qquad\textbf{(B)}\ \frac{4}{3}\text{ times the second}\qquad\textbf{(C)}\ \frac{2}{\sqrt{3}}\text{ times the second}\\ \textbf{(D)}\ \frac{\sqrt{2}}{\sqrt{3}}\text{ times the second}\qquad\textbf{(E)}\ \text{indeterminately related to the second} </math>
 [[1951 AHSME Problems/Problem 25|Solution]]
 == Problem 26 ==
+In the equation <math> \frac{x(x-1)-(m+1)}{(x-1)(m-1)}=\frac{x}{m} </math> the roots are equal when:
+<math> \textbf{(A)}\ m = 1\qquad\textbf{(B)}\ m =\frac{1}{2}\qquad\textbf{(C)}\ m = 0\qquad\textbf{(D)}\ m =-1\qquad\textbf{(E)}\ m =-\frac{1}{2} </math>
 [[1951 AHSME Problems/Problem 26|Solution]]
 == Problem 27 ==
+Through a point inside a triangle, three lines are drawn from the vertices to the opposite sides forming six triangular sections. Then:
+<math> \textbf{(A)}\ \text{the triangles are similar in opposite pairs}\qquad\textbf{(B)}\ \text{the triangles are congruent in opposite pairs} </math>
+<math> \textbf{(C)}\ \text{the triangles are equal in area in opposite pairs}\qquad\textbf{(D)}\ \text{three similar quadrilaterals are formed} </math>
+<math> \textbf{(E)}\ \text{none of the above relations are true} </math>
 [[1951 AHSME Problems/Problem 27|Solution]]
 == Problem 28 ==
+The pressure <math>(P)</math> of wind on a sail varies jointly as the area <math>(A)</math> of the sail and the square of the velocity <math>(V)</math> of the wind. The pressure on a square foot is <math>1</math> pound when the velocity is <math>16</math> miles per hour. The velocity of the wind when the pressure on a square yard is <math>36</math> pounds is:
+<math> \textbf{(A)}\ 10\frac{2}{3}\text{ mph}\qquad\textbf{(B)}\ 96\text{ mph}\qquad\textbf{(C)}\ 32\text{ mph}\qquad\textbf{(D)}\ 1\frac{2}{3}\text{ mph}\qquad\textbf{(E)}\ 16\text{ mph} </math>
 [[1951 AHSME Problems/Problem 28|Solution]]
 == Problem 29 ==
+Of the following sets of data the only one that does not determine the shape of a triangle is:
+<math> \textbf{(A)}\ \text{the ratio of two sides and the inc{}luded angle}\\ \qquad\textbf{(B)}\ \text{the ratios of the three altitudes}\\ \qquad\textbf{(C)}\ \text{the ratios of the three medians}\\ \qquad\textbf{(D)}\ \text{the ratio of the altitude to the corresponding base}\\ \qquad\textbf{(E)}\ \text{two angles} </math>
 [[1951 AHSME Problems/Problem 29|Solution]]
 == Problem 30 ==
+If two poles <math>20''</math> and <math>80''</math> high are <math>100''</math> apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:
+<math> \textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these} </math>
 [[1951 AHSME Problems/Problem 30|Solution]]
 == Problem 31 ==
+A total of <math>28</math> handshakes were exchanged at the conclusion of a party. Assuming that each participant was equally polite toward all the others, the number of people present was:
+<math> \textbf{(A)}\ 14\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 56\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 7 </math>
 [[1951 AHSME Problems/Problem 31|Solution]]
 == Problem 32 ==
+If <math>\triangle ABC</math> is inscribed in a semicircle whose diameter is <math>AB</math>, then <math>AC+BC</math> must be
+<math> \textbf{(A)}\ \text{equal to }AB\qquad\textbf{(B)}\ \text{equal to }AB\sqrt{2}\qquad\textbf{(C)}\ \geq AB\sqrt{2}\qquad\textbf{(D)}\ \leq AB\sqrt{2} \qquad \textbf{(E)}\ AB^{2} </math>
 [[1951 AHSME Problems/Problem 32|Solution]]
 == Problem 33 ==
+The roots of the equation <math> x^{2}-2x = 0 </math> can be obtained graphically by finding the abscissas of the points of intersection of each of the following pairs of equations except the pair:
+<math> \textbf{(A)}\ y = x^{2}, y = 2x\qquad\textbf{(B)}\ y = x^{2}-2x, y = 0\qquad\textbf{(C)}\ y = x, y = x-2\qquad\textbf{(D)}\ y = x^{2}-2x+1, y = 1 </math>
+<math> \textbf{(E)}\ y = x^{2}-1, y = 2x-1 </math>
+''[Note: Abscissa means x-coordinate.]''
 [[1951 AHSME Problems/Problem 33|Solution]]
 == Problem 34 ==
+The value of <math> 10^{\log_{10}7} </math> is:
+<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ \log_{10}7\qquad\textbf{(E)}\ \log_{7}10 </math>
 [[1951 AHSME Problems/Problem 34|Solution]]
 == Problem 35 ==
+If <math> a^{x}= c^{q}= b </math> and <math> c^{y}= a^{z}= d </math>, then
+<math> \textbf{(A)}\ xy = qz\qquad\textbf{(B)}\ \frac{x}{y}=\frac{q}{z}\qquad\textbf{(C)}\ x+y = q+z\qquad\textbf{(D)}\ x-y = q-z </math>
+<math> \textbf{(E)}\ x^{y}= q^{z} </math>
 [[1951 AHSME Problems/Problem 35|Solution]]
 == Problem 36 ==
+Which of the following methods of proving a geometric figure a locus is not correct?
+<math> \textbf{(A)}\ \text{Every point of the locus satisfies the conditions and every point not on the locus does not satisfy the conditions.} </math>
+<math> \textbf{(B)}\ \text{Every point not satisfying the conditions is not on the locus and every point on the locus does satisfy the conditions.} </math>
+<math> \textbf{(C)}\ \text{Every point satisfying the conditions is on the locus and every point on the locus satisfies the conditions.} </math>
+<math> \textbf{(D)}\ \text{Every point not on the locus does not satisfy the conditions and every point not satisfying}\\ \text{the conditions is not on the locus.} </math>
+<math> \textbf{(E)}\ \text{Every point satisfying the conditions is on the locus and every point not satisfying the conditions is not on the locus.} </math>
 [[1951 AHSME Problems/Problem 36|Solution]]
 == Problem 37 ==
+A number which when divided by <math>10</math> leaves a remainder of <math>9</math>, when divided by <math>9</math> leaves a remainder of <math>8</math>, by <math>8</math> leaves a remainder of <math>7</math>, etc., down to where, when divided by <math>2</math>, it leaves a remainder of <math>1</math>, is:
+<math> \textbf{(A)}\ 59\qquad\textbf{(B)}\ 419\qquad\textbf{(C)}\ 1259\qquad\textbf{(D)}\ 2519\qquad\textbf{(E)}\ \text{none of these answers} </math>
 [[1951 AHSME Problems/Problem 37|Solution]]
 == Problem 38 ==
+A rise of <math>600</math> feet is required to get a railroad line over a mountain. The grade can be kept down by lengthening the track and curving it around the mountain peak. The additional length of track required to reduce the grade from <math>3\%</math> to <math>2\%</math> is approximately:
+<math> \textbf{(A)}\ 10000\text{ ft.}\qquad\textbf{(B)}\ 20000\text{ ft.}\qquad\textbf{(C)}\ 30000\text{ ft.}\qquad\textbf{(D)}\ 12000\text{ ft.}\qquad\textbf{(E)}\ \text{none of these} </math>
 [[1951 AHSME Problems/Problem 38|Solution]]
 == Problem 39 ==
+A stone is dropped into a well and the report of the stone striking the bottom is heard <math>7.7</math> seconds after it is dropped. Assume that the stone falls <math>16t^2</math> feet in t seconds and that the velocity of sound is <math>1120</math> feet per second. The depth of the well is:
+<math> \textbf{(A)}\ 784\text{ ft.}\qquad\textbf{(B)}\ 342\text{ ft.}\qquad\textbf{(C)}\ 1568\text{ ft.}\qquad\textbf{(D)}\ 156.8\text{ ft.}\qquad\textbf{(E)}\ \text{none of these} </math>
 [[1951 AHSME Problems/Problem 39|Solution]]
 == Problem 40 ==
+<math> \left(\frac{(x+1)^{2}(x^{2}-x+1)^{2}}{(x^{3}+1)^{2}}\right)^{2}\cdot\left(\frac{(x-1)^{2}(x^{2}+x+1)^{2}}{(x^{3}-1)^{2}}\right)^{2} </math> equals:
+<math> \textbf{(A)}\ (x+1)^{4}\qquad\textbf{(B)}\ (x^{3}+1)^{4}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ [(x^{3}+1)(x^{3}-1)]^{2} </math>
+<math> \textbf{(E)}\ [(x^{3}-1)^{2}]^{2} </math>
 [[1951 AHSME Problems/Problem 40|Solution]]
 == Problem 41 ==
+The formula expressing the relationship between <math>x</math> and <math>y</math> in the table is:
+<cmath> \begin{tabular}{|c|c|c|c|c|c|}\hline x & 2 & 3 & 4 & 5 & 6\\ \hline y & 0 & 2 & 6 & 12 & 20\\ \hline\end{tabular} </cmath>
+<math> \textbf{(A)}\ y = 2x-4\qquad\textbf{(B)}\ y = x^{2}-3x+2\qquad\textbf{(C)}\ y = x^{3}-3x^{2}+2x</math>
+<math> \textbf{(D)}\ y = x^{2}-4x\qquad\textbf{(E)}\ y = x^{2}-4 </math>
 [[1951 AHSME Problems/Problem 41|Solution]]
 == Problem 42 ==
+If <math> x =\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}} </math>, then:
+<math> \textbf{(A)}\ x = 1\qquad\textbf{(B)}\ 0 < x < 1\qquad\textbf{(C)}\ 1 < x < 2\qquad\textbf{(D)}\ x\text{ is infinite} </math>
+<math> \textbf{(E)}\ x > 2\text{ but finite} </math>
 [[1951 AHSME Problems/Problem 42|Solution]]
 == Problem 43 ==
+Of the following statements, the only one that is incorrect is:
+<math> \textbf{(A)}\ \text{An inequality will remain true after each side is increased,} </math> <math>\text{ decreased, multiplied or divided (zero excluded) by the same positive quantity.} </math>
+<math> \textbf{(B)}\ \text{The arithmetic mean of two unequal positive quantities is greater than their geometric mean.} </math>
+<math> \textbf{(C)}\ \text{If the sum of two positive quantities is given, ther product is largest when they are equal.} </math>
+<math> \textbf{(D)}\ \text{If }a\text{ and }b\text{ are positive and unequal, }\frac{1}{2}(a^{2}+b^{2})\text{ is greater than }[\frac{1}{2}(a+b)]^{2}. </math>
+<math> \textbf{(E)}\ \text{If the product of two positive quantities is given, their sum is greatest when they are equal.} </math>
 [[1951 AHSME Problems/Problem 43|Solution]]
 == Problem 44 ==
+If <math> \frac{xy}{x+y}= a,\frac{xz}{x+z}= b,\frac{yz}{y+z}= c </math>, where <math> a, b, c </math> are other than zero, then <math>x</math> equals:
+<math> \textbf{(A)}\ \frac{abc}{ab+ac+bc}\qquad\textbf{(B)}\ \frac{2abc}{ab+bc+ac}\qquad\textbf{(C)}\ \frac{2abc}{ab+ac-bc} </math>
+<math> \textbf{(D)}\ \frac{2abc}{ab+bc-ac}\qquad\textbf{(E)}\ \frac{2abc}{ac+bc-ab} </math>
 [[1951 AHSME Problems/Problem 44|Solution]]
 == Problem 45 ==
+If you are given <math> \log 8\approx .9031 </math> and <math> \log 9\approx .9542 </math>, then the only logarithm that cannot be found without the use of tables is:
+<math> \textbf{(A)}\ \log 17\qquad\textbf{(B)}\ \log\frac{5}{4}\qquad\textbf{(C)}\ \log 15\qquad\textbf{(D)}\ \log 600\qquad\textbf{(E)}\ \log .4 </math>
 [[1951 AHSME Problems/Problem 45|Solution]]
 == Problem 46 ==
+<math>AB</math> is a fixed diameter of a circle whose center is <math>O</math>. From <math>C</math>, any point on the circle, a chord <math>CD</math> is drawn perpendicular to <math>AB</math>. Then, as <math>C</math> moves over a semicircle, the bisector of angle <math>OCD</math> cuts the circle in a point that always:
+<math> \textbf{(A)}\ \text{bisects the arc }AB\qquad\textbf{(B)}\ \text{trisects the arc }AB\qquad\textbf{(C)}\ \text{varies} </math>
+<math> \textbf{(D)}\ \text{is as far from }AB\text{ as from }D\qquad\textbf{(E)}\ \text{is equidistant from }B\text{ and }C </math>
 [[1951 AHSME Problems/Problem 46|Solution]]
 == Problem 47 ==
+If <math>r</math> and <math>s</math> are the roots of the equation <math>ax^2+bx+c=0</math>, the value of <math> \frac{1}{r^{2}}+\frac{1}{s^{2}} </math> is:
+<math> \textbf{(A)}\ b^{2}-4ac\qquad\textbf{(B)}\ \frac{b^{2}-4ac}{2a}\qquad\textbf{(C)}\ \frac{b^{2}-4ac}{c^{2}}\qquad\textbf{(D)}\ \frac{b^{2}-2ac}{c^{2}} </math>
+<math> \textbf{(E)}\ \text{none of these} </math>
 [[1951 AHSME Problems/Problem 47|Solution]]
 == Problem 48 ==
+The area of a square inscribed in a semicircle is to the area of the square inscribed in the entire circle as:
+<math> \textbf{(A)}\ 1: 2\qquad\textbf{(B)}\ 2: 3\qquad\textbf{(C)}\ 2: 5\qquad\textbf{(D)}\ 3: 4\qquad\textbf{(E)}\ 3: 5 </math>
 [[1951 AHSME Problems/Problem 48|Solution]]
 == Problem 49 ==
+The medians of a right triangle which are drawn from the vertices of the acute angles are <math>5</math> and <math>\sqrt{40}</math>. The value of the hypotenuse is:
+<math> \textbf{(A)}\ 10\qquad\textbf{(B)}\ 2\sqrt{40}\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 2\sqrt{13}\qquad\textbf{(E)}\ \text{none of these} </math>
 [[1951 AHSME Problems/Problem 49|Solution]]
 == Problem 50 ==
+Tom, Dick and Harry started out on a <math>100</math>-mile journey. Tom and Harry went by automobile at the rate of <math>25</math> mph, while Dick walked at the rate of <math>5</math> mph. After a certain distance, Harry got off and walked on at <math>5</math> mph, while Tom went back for Dick and got him to the destination at the same time that Harry arrived. The number of hours required for the trip was:
+<math> \textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ \text{none of these answers} </math>
 [[1951 AHSME Problems/Problem 50|Solution]]
 == See also ==
-* [[AHSME]]
-* [[AHSME Problems and Solutions]]
+* [[AMC 12 Problems and Solutions]]
-* [[1995 AHSME]]
 * [[Mathematics competition resources]]
+{{AHSME 50p box|year=1951|before=[[1950 AHSME|1950 AHSC]]|after=[[1952 AHSME|1952 AHSC]]}}
+{{MAA Notice}}

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Difference between revisions of "1951 AHSME Problems"

Latest revision as of 20:53, 21 June 2024

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