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Difference between revisions of "1954 AHSME Problems"

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{{AHSC 50 Problems
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== Problem 1==
 
== Problem 1==
  

Revision as of 13:39, 19 February 2020

1954 AHSC (Answer Key)
Printable version: Wiki | AoPS Resources • PDF

Instructions

  1. 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.
  2. You will receive ? points for each correct answer, ? points for each problem left unanswered, and ? points for each incorrect answer.
  3. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers.
  4. Figures are not necessarily drawn to scale.
  5. You will have ? minutes working time to complete the test.
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

Problem 1

The square of $5-\sqrt{y^2-25}$ is:

$\textbf{(A)}\ y^2-5\sqrt{y^2-25} \qquad \textbf{(B)}\ -y^2 \qquad \textbf{(C)}\ y^2 \\ \textbf{(D)}\ (5-y)^2\qquad\textbf{(E)}\ y^2-10\sqrt{y^2-25}$

Solution

Problem 2

The equation $\frac{2x^2}{x-1}-\frac{2x+7}{3}+\frac{4-6x}{x-1}+1=0$ can be transformed by eliminating fractions to the equation $x^2-5x+4=0$. The roots of the latter equation are $4$ and $1$. Then the roots of the first equation are:

$\textbf{(A)}\ 4 \text{ and }1 \qquad \textbf{(B)}\ \text{only }1 \qquad \textbf{(C)}\ \text{only }4 \qquad \textbf{(D)}\ \text{neither 4 nor 1}\qquad\textbf{(E)}\ \text{4 and some other root}$

Solution

Problem 3

If $x$ varies as the cube of $y$, and $y$ varies as the fifth root of $z$, then $x$ varies as the nth power of $z$, where n is:

$\textbf{(A)}\ \frac{1}{15} \qquad\textbf{(B)}\ \frac{5}{3} \qquad\textbf{(C)}\ \frac{3}{5} \qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 8$

Solution

Problem 4

If the Highest Common Divisor of $6432$ and $132$ is diminished by $8$, it will equal:

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

Solution

Problem 5

A regular hexagon is inscribed in a circle of radius $10$ inches. Its area is:

$\textbf{(A)}\ 150\sqrt{3} \text{ sq. in.} \qquad \textbf{(B)}\ \text{150 sq. in.} \qquad \textbf{(C)}\ 25\sqrt{3}\text{ sq. in.}\qquad\textbf{(D)}\ \text{600 sq. in.}\qquad\textbf{(E)}\ 300\sqrt{3}\text{ sq. in.}$

Solution

Problem 6

The value of $\frac{1}{16}a^0+\left (\frac{1}{16a} \right )^0- \left (64^{-\frac{1}{2}} \right )- (-32)^{-\frac{4}{5}}$ is:

$\textbf{(A)}\ 1 \frac{13}{16} \qquad \textbf{(B)}\ 1 \frac{3}{16} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{1}{16}$

Solution

Problem 7

A housewife saved $\textdollar{2.50}$ in buying a dress on sale. If she spent $\textdollar{25}$ for the dress, she saved about:

$\textbf{(A)}\ 8 \% \qquad \textbf{(B)}\ 9 \% \qquad \textbf{(C)}\ 10 \% \qquad \textbf{(D)}\ 11 \% \qquad \textbf{(E)}\ 12\%$

Solution

Problem 8

The base of a triangle is twice as long as a side of a square and their areas are the same. Then the ratio of the altitude of the triangle to the side of the square is:

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

Solution

Problem 9

A point $P$ is outside a circle and is $13$ inches from the center. A secant from $P$ cuts the circle at $Q$ and $R$ so that the external segment of the secant $PQ$ is $9$ inches and $QR$ is $7$ inches. The radius of the circle is:

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

Solution

Problem 10

The sum of the numerical coefficients in the expansion of the binomial $(a+b)^6$ is:

$\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 7$

Solution

Problem 11

A merchant placed on display some dresses, each with a marked price. He then posted a sign “$\frac{1}{3}$ off on these dresses.” The cost of the dresses was $\frac{3}{4}$ of the price at which he actually sold them. Then the ratio of the cost to the marked price was:

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

Solution

Problem 12

The solution of the equations

\begin{align*}2x-3y &=7 \\ 4x-6y &=20\end{align*}

is:

$\textbf{(A)}\ x=18, y=12 \qquad \textbf{(B)}\ x=0, y=0 \qquad \textbf{(C)}\ \text{There is no solution} \\ \textbf{(D)}\ \text{There are an unlimited number of solutions}\qquad \textbf{(E)}\ x=8, y=5$

Solution

Problem 13

A quadrilateral is inscribed in a circle. If angles are inscribed in the four arcs cut off by the sides of the quadrilateral, their sum will be:

$\textbf{(A)}\ 180^\circ \qquad \textbf{(B)}\ 540^\circ \qquad \textbf{(C)}\ 360^\circ \qquad \textbf{(D)}\ 450^\circ\qquad\textbf{(E)}\ 1080^\circ$

Solution

Problem 14

When simplified $\sqrt{1+ \left (\frac{x^4-1}{2x^2} \right )^2}$ equals:

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

Solution

Problem 15

$\log 125$ equals:

$\textbf{(A)}\ 100 \log 1.25 \qquad \textbf{(B)}\ 5 \log 3 \qquad \textbf{(C)}\ 3 \log 25 \\ \textbf{(D)}\ 3 - 3\log 2 \qquad \textbf{(E)}\ (\log 25)(\log 5)$

Solution

Problem 16

If $f(x) = 5x^2 - 2x - 1$, then $f(x + h) - f(x)$ equals:

$\textbf{(A)}\ 5h^2 - 2h \qquad \textbf{(B)}\ 10xh - 4x + 2 \qquad \textbf{(C)}\ 10xh - 2x - 2 \\ \textbf{(D)}\ h(10x+5h-2)\qquad\textbf{(E)}\ 3h$

Solution

Problem 17

The graph of the function $f(x) = 2x^3 - 7$ goes:

$\textbf{(A)}\ \text{up to the right and down to the left} \\ \textbf{(B)}\ \text{down to the right and up to the left}\\ \textbf{(C)}\ \text{up to the right and up to the left}\\ \textbf{(D)}\ \text{down to the right and down to the left}\\ \textbf{(E)}\ \text{none of these ways.}$

Solution

Problem 18

Of the following sets, the one that includes all values of $x$ which will satisfy $2x - 3 > 7 - x$ is:

$\textbf{(A)}\ x > 4 \qquad \textbf{(B)}\ x < \frac {10}{3} \qquad \textbf{(C)}\ x = \frac {10}{3} \qquad \textbf{(D)}\ x >\frac{10}{3}\qquad\textbf{(E)}\ x < 0$

Solution

Problem 19

If the three points of contact of a circle inscribed in a triangle are joined, the angles of the resulting triangle:

$\textbf{(A)}\ \text{are always equal to }60^\circ\\ \textbf{(B)}\ \text{are always one obtuse angle and two unequal acute angles}\\ \textbf{(C)}\ \text{are always one obtuse angle and two equal acute angles}\\ \textbf{(D)}\ \text{are always acute angles}\\ \textbf{(E)}\ \text{are always unequal to each other}$

Solution

Problem 20

The equation $x^3+6x^2+11x+6=0$ has:

$\textbf{(A)}\ \text{no negative real roots}\qquad\textbf{(B)}\ \text{no positive real roots}\qquad\textbf{(C)}\ \text{no real roots}\\ \textbf{(D)}\ \text{1 positive and 2 negative roots}\qquad\textbf{(E)}\ \text{2 positive and 1 negative root}$

Solution

Problem 21

The roots of the equation $2\sqrt {x} + 2x^{ - \frac {1}{2}} = 5$ can be found by solving:

$\textbf{(A)}\ 16x^2-92x+1 = 0\qquad\textbf{(B)}\ 4x^2-25x+4 = 0\qquad\textbf{(C)}\ 4x^2-17x+4 = 0\\ \textbf{(D)}\ 2x^2-21x+2 = 0\qquad\textbf{(E)}\ 4x^2-25x-4 = 0$

Solution

Problem 22

The expression $\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}$ cannot be evaluated for $x=-1$ or $x=2$, since division by zero is not allowed. For other values of $x$:

$\textbf{(A)}\ \text{The expression takes on many different values.}\\ \textbf{(B)}\ \text{The expression has only the value 2.}\\ \textbf{(C)}\ \text{The expression has only the value 1.}\\ \textbf{(D)}\ \text{The expression always has a value between }-1\text{ and }+2.\\ \textbf{(E)}\ \text{The expression has a value greater than 2 or less than }-1.$

Solution

Problem 23

If the margin made on an article costing $C$ dollars and selling for $S$ dollars is $M=\frac{1}{n}C$, then the margin is given by:

$\textbf{(A)}\ M=\frac{1}{n-1}S\qquad\textbf{(B)}\ M=\frac{1}{n}S\qquad\textbf{(C)}\ M=\frac{n}{n+1}S\\ \textbf{(D)}\ M=\frac{1}{n+1}S\qquad\textbf{(E)}\ M=\frac{n}{n-1}S$

Solution

Problem 24

The values of $k$ for which the equation $2x^2-kx+x+8=0$ will have real and equal roots are:

$\textbf{(A)}\ 9\text{ and }-7\qquad\textbf{(B)}\ \text{only }-7\qquad\textbf{(C)}\ \text{9 and 7}\\ \textbf{(D)}\ -9\text{ and }-7\qquad\textbf{(E)}\ \text{only 9}$

Solution

Problem 25

The two roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are $1$ and:

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

Problem 26

The straight line $\overline{AB}$ is divided at $C$ so that $AC=3CB$. Circles are described on $\overline{AC}$ and $\overline{CB}$ as diameters and a common tangent meets $AB$ produced at $D$. Then $BD$ equals:

$\textbf{(A)}\ \text{diameter of the smaller circle}\\ \textbf{(B)}\ \text{radius of the smaller circle}\\ \textbf{(C)}\ \text{radius of the larger circle}\\ \textbf{(D)}\ CB\sqrt{3}\\ \textbf{(E)}\ \text{the difference of the two radii}$

Solution

Problem 27

A right circular cone has for its base a circle having the same radius as a given sphere. The volume of the cone is one-half that of the sphere. The ratio of the altitude of the cone to the radius of its base is:

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

Solution

Problem 28

If $\frac{m}{n}=\frac{4}{3}$ and $\frac{r}{t}=\frac{9}{14}$, the value of $\frac{3mr-nt}{4nt-7mr}$ is:

$\textbf{(A)}\ -5\frac{1}{2}\qquad\textbf{(B)}\ -\frac{11}{14}\qquad\textbf{(C)}\ -1\frac{1}{4}\qquad\textbf{(D)}\ \frac{11}{14}\qquad\textbf{(E)}\ -\frac{2}{3}$

Solution

Problem 29

If the ratio of the legs of a right triangle is $1: 2$, then the ratio of the corresponding segments of the hypotenuse made by a perpendicular upon it from the vertex is:

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

Solution

Problem 30

$A$ and $B$ together can do a job in $2$ days; $B$ and $C$ can do it in four days; and $A$ and $C$ in $2\frac{2}{5}$ days. The number of days required for A to do the job alone is:

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 2.8$

Solution

Problem 31

In $\triangle ABC$, $AB=AC$, $\angle A=40^\circ$. Point $O$ is within the triangle with $\angle OBC \cong \angle OCA$. The number of degrees in $\angle BOC$ is:

$\textbf{(A)}\ 110^{\circ} \qquad \textbf{(B)}\ 35^{\circ} \qquad \textbf{(C)}\ 140^{\circ} \qquad \textbf{(D)}\ 55^{\circ} \qquad \textbf{(E)}\ 70^{\circ}$

Solution

Problem 32

The factors of $x^4+64$ are:

$\textbf{(A)}\ (x^2+8)^2\qquad\textbf{(B)}\ (x^2+8)(x^2-8)\qquad\textbf{(C)}\ (x^2+2x+4)(x^2-8x+16)\\ \textbf{(D)}\ (x^2-4x+8)(x^2-4x-8)\qquad\textbf{(E)}\ (x^2-4x+8)(x^2+4x+8)$

Solution

Problem 33

A bank charges $\textdollar{6}$ for a loan of $\textdollar{120}$. The borrower receives $\textdollar{114}$ and repays the loan in $12$ easy installments of $\textdollar{10}$ a month. The interest rate is approximately:

$\textbf{(A)}\ 5 \% \qquad \textbf{(B)}\ 6 \% \qquad \textbf{(C)}\ 7 \% \qquad \textbf{(D)}\ 9\% \qquad \textbf{(E)}\ 15 \%$

Solution

Problem 34

The fraction $\frac{1}{3}$:

$\textbf{(A)}\ \text{equals 0.33333333}\qquad\textbf{(B)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10^8}\\ \textbf{(C)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10^9}\\ \textbf{(D)}\ \text{is greater than 0.33333333 by }\frac{1}{3\cdot 10^8}\\ \textbf{(E)}\ \text{is greater than 0.33333333 by }\frac{1}{3\cdot 10^9}$

Solution

Problem 35

In the right triangle shown the sum of the distances $BM$ and $MA$ is equal to the sum of the distances $BC$ and $CA$. If $MB = x, CB = h$, and $CA = d$, then $x$ equals:

[asy] defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; draw((0,0)--(8,0)--(0,5)--cycle); label("C",(0,0),SW); label("A",(8,0),SE); label("M",(0,5),N); dot((0,3.5)); label("B",(0,3.5),W); label("$x$",(0,4.25),W); label("$h$",(0,1),W); label("$d$",(4,0),S);[/asy]

$\textbf{(A)}\ \frac{hd}{2h+d}\qquad\textbf{(B)}\ d-h\qquad\textbf{(C)}\ \frac{1}{2}d\qquad\textbf{(D)}\ h+d-\sqrt{2d}\qquad\textbf{(E)}\ \sqrt{h^2+d^2}-h$

Solution

Problem 36

A boat has a speed of $15$ mph in still water. In a stream that has a current of $5$ mph it travels a certain distance downstream and returns. The ratio of the average speed for the round trip to the speed in still water is:

$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{1}{1}\qquad\textbf{(C)}\ \frac{8}{9}\qquad\textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{9}{8}$

Solution

Problem 37

Given $\triangle PQR$ with $\overline{RS}$ bisecting $\angle R$, $PQ$ extended to $D$ and $\angle n$ a right angle, then:

[asy] path anglemark2(pair A, pair B, pair C, real t=8, bool flip=false) { pair M,N; path mark; M=t*0.03*unit(A-B)+B; N=t*0.03*unit(C-B)+B; if(flip) mark=Arc(B,t*0.03,degrees(C-B)-360,degrees(A-B)); else mark=Arc(B,t*0.03,degrees(A-B),degrees(C-B)); return mark; } unitsize(1.5cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair P=(0,0), R=(3,2), Q=(4,0); pair S0=bisectorpoint(P,R,Q); pair Sp=extension(P,Q,S0,R); pair D0=bisectorpoint(R,Sp), Np=midpoint(R--Sp); pair D=extension(Np,D0,P,Q), M=extension(Np,D0,P,R); draw(P--R--Q); draw(R--Sp); draw(P--D--M); draw(anglemark2(Sp,P,R,17)); label("$p$",P+(0.35,0.1)); draw(anglemark2(R,Q,P,11)); label("$q$",Q+(-0.17,0.1)); draw(anglemark2(R,Np,D,8,true)); label("$n$",Np+(+0.12,0.07)); draw(anglemark2(R,M,D,13,true)); label("$m$",M+(+0.25,0.03)); draw(anglemark2(M,D,P,29)); label("$d$",D+(-0.75,0.095)); pen f=fontsize(10pt); label("$R$",R,N,f); label("$P$",P,S,f); label("$S$",Sp,S,f); label("$Q$",Q,S,f); label("$D$",D,S,f);[/asy]

$\textbf{(A)}\ \angle m = \frac {1}{2}(\angle p - \angle q) \qquad \textbf{(B)}\ \angle m = \frac {1}{2}(\angle p + \angle q)$ $\textbf{(C)}\ \angle d =\frac{1}{2}(\angle q+\angle p)\qquad \textbf{(D)}\ \angle d =\frac{1}{2}\angle m\qquad \textbf{(E)}\ \text{none of these is correct}$

Solution

Problem 38

If $\log 2=.3010$ and $\log 3=.4771$, the value of $x$ when $3^{x+3}=135$ is approximately:

$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 1.47 \qquad \textbf{(C)}\ 1.67 \qquad \textbf{(D)}\ 1.78 \qquad \textbf{(E)}\ 1.63$

Solution

Problem 39

The locus of the midpoint of a line segment that is drawn from a given external point $P$ to a given circle with center $O$ and radius $r$, is:

$\textbf{(A)}\ \text{a straight line perpendicular to }\overline{PO}\\ \textbf{(B)}\ \text{a straight line parallel to }\overline{PO}\\ \textbf{(C)}\ \text{a circle with center }P\text{ and radius }r\\ \textbf{(D)}\ \text{a circle with center at the midpoint of }\overline{PO}\text{ and radius }2r\\ \textbf{(E)}\ \text{a circle with center at the midpoint }\overline{PO}\text{ and radius }\frac{1}{2}r$

Solution

Problem 40

If $\left (a+\frac{1}{a} \right )^2=3$, then $a^3+\frac{1}{a^3}$ equals:

$\textbf{(A)}\ \frac{10\sqrt{3}}{3}\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 7\sqrt{7}\qquad\textbf{(E)}\ 6\sqrt{3}$

Solution

Problem 41

The sum of all the roots of $4x^3-8x^2-63x-9=0$ is:

$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ -8 \qquad \textbf{(D)}\ -2 \qquad \textbf{(E)}\ 0$

Solution

Problem 42

Consider the graphs of \[(1)\qquad y=x^2-\frac{1}{2}x+2\] and \[(2)\qquad y=x^2+\frac{1}{2}x+2\] on the same set of axis. These parabolas are exactly the same shape. Then:

$\textbf{(A)}\ \text{the graphs coincide.}\\ \textbf{(B)}\ \text{the graph of (1) is lower than the graph of (2).}\\ \textbf{(C)}\ \text{the graph of (1) is to the left of the graph of (2).}\\ \textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}\\ \textbf{(E)}\ \text{the graph of (1) is higher than the graph of (2).}$

Solution

Problem 43

The hypotenuse of a right triangle is $10$ inches and the radius of the inscribed circle is $1$ inch. The perimeter of the triangle in inches is:

$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 30$

Solution

Problem 44

A man born in the first half of the nineteenth century was $x$ years old in the year $x^2$. He was born in:

$\textbf{(A)}\ 1849 \qquad \textbf{(B)}\ 1825 \qquad \textbf{(C)}\ 1812 \qquad \textbf{(D)}\ 1836 \qquad \textbf{(E)}\ 1806$

Solution

Problem 45

In a rhombus, $ABCD$, line segments are drawn within the rhombus, parallel to diagonal $BD$, and terminated in the sides of the rhombus. A graph is drawn showing the length of a segment as a function of its distance from vertex $A$. The graph is:

$\textbf{(A)}\ \text{A straight line passing through the origin.}\\ \textbf{(B)}\ \text{A straight line cutting across the upper right quadrant.}\\ \textbf{(C)}\ \text{Two line segments forming an upright V.}\\ \textbf{(D)}\ \text{Two line segments forming an inverted V.}\\ \textbf{(E)}\ \text{None of these.}$

Solution

Problem 46

In the diagram, if points $A, B$ and $C$ are points of tangency, then $x$ equals:

[asy] unitsize(5cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=3; pair A=(-3*sqrt(3)/32,9/32), B=(3*sqrt(3)/32, 9/32), C=(0,9/16); pair O=(0,3/8); draw((-2/3,9/16)--(2/3,9/16)); draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2)); draw(Circle(O,3/16)); draw((-2/3,0)--(2/3,0)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$\frac{3}{8}$",O); draw(O+.07*dir(60)--O+3/16*dir(60),EndArrow(3)); draw(O+.07*dir(240)--O+3/16*dir(240),EndArrow(3)); label("$\frac{1}{2}$",(.5,.25)); draw((.5,.33)--(.5,.5),EndArrow(3)); draw((.5,.17)--(.5,0),EndArrow(3)); label("$x$",midpoint((.5,.5)--(.5,9/16))); draw((.5,5/8)--(.5,9/16),EndArrow(3)); label("$60^{\circ}$",(0.01,0.12)); dot(A); dot(B); dot(C);[/asy]

$\textbf{(A)}\ \frac{3}{16}"\qquad\textbf{(B)}\ \frac{1}{8}"\qquad\textbf{(C)}\ \frac{1}{32}"\qquad\textbf{(D)}\ \frac{3}{32}"\qquad\textbf{(E)}\ \frac{1}{16}"$

Solution

Problem 47

At the midpoint of line segment $AB$ which is $p$ units long, a perpendicular $MR$ is erected with length $q$ units. An arc is described from $R$ with a radius equal to $\frac{1}{2}AB$, meeting $AB$ at $T$. Then $AT$ and $TB$ are the roots of:

$\textbf{(A)}\ x^2+px+q^2=0\\ \textbf{(B)}\ x^2-px+q^2=0\\ \textbf{(C)}\ x^2+px-q^2=0\\ \textbf{(D)}\ x^2-px-q^2=0\\ \textbf{(E)}\ x^2-px+q=0$

Solution

Problem 48

A train, an hour after starting, meets with an accident which detains it a half hour, after which it proceeds at $\frac{3}{4}$ of its former rate and arrives $3\tfrac{1}{2}$ hours late. Had the accident happened $90$ miles farther along the line, it would have arrived only $3$ hours late. The length of the trip in miles was:

$\textbf{(A)}\ 400 \qquad \textbf{(B)}\ 465 \qquad \textbf{(C)}\ 600 \qquad \textbf{(D)}\ 640 \qquad \textbf{(E)}\ 550$

Solution

Problem 49

The difference of the squares of two odd numbers is always divisible by $8$. If $a>b$, and $2a+1$ and $2b+1$ are the odd numbers, to prove the given statement we put the difference of the squares in the form:

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

Solution

Problem 50

The times between $7$ and $8$ o'clock, correct to the nearest minute, when the hands of a clock will form an angle of $84^{\circ}$ are:

$\textbf{(A)}\ \text{7: 23 and 7: 53}\qquad \textbf{(B)}\ \text{7: 20 and 7: 50}\qquad \textbf{(C)}\ \text{7: 22 and 7: 53}\\ \textbf{(D)}\ \text{7: 23 and 7: 52}\qquad \textbf{(E)}\ \text{7: 21 and 7: 49}$

Solution

See also

1954 AHSC (Problems • Answer Key • Resources)
Preceded by
1953 AHSME 		Followed by
1955 AHSME
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All AHSME Problems and Solutions


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