Difference between revisions of "1954 AHSME Problems"

Latest revision as of 18:16, 8 November 2024

1954 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 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:

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

Solution

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 $12$ 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

1954 AHSC (Problems • Answer Key • Resources)
Preceded by 1953 AHSC	Followed by 1955 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=1954
+}}
+== Problem 1==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The square of <math>5-\sqrt{y^2-25}</math> is:
-[[1953 AHSME Problems/Problem 1|Solution]]
+<math>\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}  </math>
+[[1954 AHSME Problems/Problem 1|Solution]]
-== Problem 2 ==
+== Problem 2==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
-[[1953 AHSME Problems/Problem 2|Solution]]
+The equation <math>\frac{2x^2}{x-1}-\frac{2x+7}{3}+\frac{4-6x}{x-1}+1=0</math> can be transformed by eliminating fractions to the equation <math>x^2-5x+4=0</math>.
+The roots of the latter equation are <math>4</math> and <math>1</math>. Then the roots of the first equation are:
-== Problem 3 ==
+<math>\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}  </math>
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+[[1954 AHSME Problems/Problem 2|Solution]]
-[[1953 AHSME Problems/Problem 3|Solution]]
+== Problem 3==
-== Problem 4 ==
+If <math>x</math> varies as the cube of <math>y</math>, and <math>y</math> varies as the fifth root of <math>z</math>, then <math>x</math> varies as the nth power of <math>z</math>, where n is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
-[[1953 AHSME Problems/Problem 4|Solution]]
+<math>\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  </math>
+[[1954 AHSME Problems/Problem 3|Solution]]
-== Problem 5 ==
+== Problem 4==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
-[[1953 AHSME Problems/Problem 5|Solution]]
+If the Highest Common Divisor of <math>6432</math> and <math>132</math> is diminished by <math>8</math>, it will equal:
-== Problem 6 ==
+<math>\textbf{(A)}\ -6 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ -2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4     </math>
+[[1954 AHSME Problems/Problem 4|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 5==
-[[1953 AHSME Problems/Problem 6|Solution]]
+A regular hexagon is inscribed in a circle of radius <math>10</math> inches. Its area is:
-== Problem 7 ==
+<math>\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.}  </math>
+[[1954 AHSME Problems/Problem 5|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 6==
-[[1953 AHSME Problems/Problem 7|Solution]]
+The value of <math>\frac{1}{16}a^0+\left (\frac{1}{16a} \right )^0- \left (64^{-\frac{1}{2}} \right )- (-32)^{-\frac{4}{5}}</math> is:
-== Problem 8 ==
+<math>\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}  </math>
+[[1954 AHSME Problems/Problem 6|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 7==
-[[1953 AHSME Problems/Problem 8|Solution]]
+A housewife saved <math>\textdollar{2.50}</math> in buying a dress on sale. If she spent <math>\textdollar{25}</math> for the dress, she saved about:
-== Problem 9 ==
+<math>\textbf{(A)}\ 8 \% \qquad \textbf{(B)}\ 9 \% \qquad \textbf{(C)}\ 10 \% \qquad \textbf{(D)}\ 11 \% \qquad \textbf{(E)}\ 12\% </math>
+[[1954 AHSME Problems/Problem 7|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 8==
-[[1953 AHSME Problems/Problem 9|Solution]]
+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:
-== Problem 10 ==
+<math>\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 4  </math>
+[[1954 AHSME Problems/Problem 8|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 9==
-[[1953 AHSME Problems/Problem 10|Solution]]
+A point <math>P</math> is outside a circle and is <math>13</math> inches from the center. A secant from <math>P</math> cuts the circle at <math>Q</math> and <math>R</math>
+so that the external segment of the secant <math>PQ</math> is <math>9</math> inches and <math>QR</math> is <math>7</math> inches. The radius of the circle is:
-== Problem 11 ==
+<math>\textbf{(A)}\ 3" \qquad \textbf{(B)}\ 4" \qquad \textbf{(C)}\ 5" \qquad \textbf{(D)}\ 6"\qquad\textbf{(E)}\ 7" </math>
+[[1954 AHSME Problems/Problem 9|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 10==
-[[1953 AHSME Problems/Problem 11|Solution]]
+The sum of the numerical coefficients in the expansion of the binomial <math>(a+b)^6</math> is:
-== Problem 12 ==
+<math>\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 7 </math>
+[[1954 AHSME Problems/Problem 10|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 11==
-[[1953 AHSME Problems/Problem 12|Solution]]
+A merchant placed on display some dresses, each with a marked price. He then posted a sign “<math>\frac{1}{3}</math> off on these dresses.”
+The cost of the dresses was <math>\frac{3}{4}</math> of the price at which he actually sold them. Then the ratio of the cost to the marked price was:
-== Problem 13 ==
+<math>\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}   </math>
+[[1954 AHSME Problems/Problem 11|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 12==
-[[1953 AHSME Problems/Problem 13|Solution]]
+The solution of the equations
-== Problem 14 ==
+<cmath>\begin{align*}2x-3y &=7 \\ 4x-6y &=20\end{align*}</cmath>
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+is:
-[[1953 AHSME Problems/Problem 14|Solution]]
+<math>\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 </math>
+[[1954 AHSME Problems/Problem 12|Solution]]
-== Problem 15 ==
+== Problem 13==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+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:
-[[1953 AHSME Problems/Problem 15|Solution]]
+<math>\textbf{(A)}\ 180^\circ \qquad \textbf{(B)}\ 540^\circ \qquad \textbf{(C)}\ 360^\circ \qquad \textbf{(D)}\ 450^\circ\qquad\textbf{(E)}\ 1080^\circ  </math>
+[[1954 AHSME Problems/Problem 13|Solution]]
-== Problem 16 ==
+== Problem 14==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+When simplified <math>\sqrt{1+ \left (\frac{x^4-1}{2x^2} \right )^2}</math> equals:
-[[1953 AHSME Problems/Problem 16|Solution]]
+<math>\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}  </math>
+[[1954 AHSME Problems/Problem 14|Solution]]
-== Problem 17 ==
+== Problem 15==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math>\log 125</math> equals:
-[[1953 AHSME Problems/Problem 17|Solution]]
+<math>\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)    </math>
+[[1954 AHSME Problems/Problem 15|Solution]]
-== Problem 18 ==
+== Problem 16==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+If <math>f(x) = 5x^2 - 2x - 1</math>, then <math>f(x + h) - f(x)</math> equals:
-[[1953 AHSME Problems/Problem 18|Solution]]
+<math>\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 </math>
+[[1954 AHSME Problems/Problem 16|Solution]]
-== Problem 19 ==
+== Problem 17==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The graph of the function <math>f(x) = 2x^3 - 7</math> goes:
-[[1953 AHSME Problems/Problem 19|Solution]]
+<math>\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.}    </math>
+[[1954 AHSME Problems/Problem 17|Solution]]
-== Problem 20 ==
+== Problem 18==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+Of the following sets, the one that includes all values of <math>x</math> which will satisfy <math>2x - 3 > 7 - x</math> is:
-[[1953 AHSME Problems/Problem 20|Solution]]
+<math>\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 </math>
+[[1954 AHSME Problems/Problem 18|Solution]]
-== Problem 21 ==
+== Problem 19==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+If the three points of contact of a circle inscribed in a triangle are joined, the angles of the resulting triangle:
-[[1953 AHSME Problems/Problem 21|Solution]]
+<math> \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} </math>
-== Problem 22 ==
+[[1954 AHSME Problems/Problem 19|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 20==
-[[1953 AHSME Problems/Problem 22|Solution]]
+The equation <math>x^3+6x^2+11x+6=0</math> has:
-== Problem 23 ==
+<math> \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} </math>
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+[[1954 AHSME Problems/Problem 20|Solution]]
-[[1953 AHSME Problems/Problem 23|Solution]]
+== Problem 21==
-== Problem 24 ==
+The roots of the equation <math>2\sqrt {x} + 2x^{ - \frac {1}{2}} = 5</math> can be found by solving:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \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 </math>
+[[1954 AHSME Problems/Problem 21|Solution]]
-[[1953 AHSME Problems/Problem 24|Solution]]
+== Problem 22==
-== Problem 25 ==
+The expression <math>\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}</math> cannot be evaluated for <math>x=-1</math> or <math>x=2</math>,
+since division by zero is not allowed. For other values of <math>x</math>:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \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. </math>
-[[1953 AHSME Problems/Problem 25|Solution]]
+[[1954 AHSME Problems/Problem 22|Solution]]
-== Problem 26 ==
+== Problem 23==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+If the margin made on an article costing <math>C</math> dollars and selling for <math>S</math> dollars is <math>M=\frac{1}{n}C</math>, then the margin is given by:
-[[1953 AHSME Problems/Problem 26|Solution]]
+<math> \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 </math>
-== Problem 27 ==
+[[1954 AHSME Problems/Problem 23|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 24==
-[[1953 AHSME Problems/Problem 27|Solution]]
+The values of <math>k</math> for which the equation <math>2x^2-kx+x+8=0</math> will have real and equal roots are:
-== Problem 28 ==
+<math> \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} </math>
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+[[1954 AHSME Problems/Problem 24|Solution]]
-[[1953 AHSME Problems/Problem 28|Solution]]
+== Problem 25==
-== Problem 29 ==
+The two roots of the equation <math>a(b-c)x^2+b(c-a)x+c(a-b)=0</math> are <math>1</math> and:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \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)} </math>
-[[1953 AHSME Problems/Problem 29|Solution]]
+[[1954 AHSME Problems/Problem 25|Solution]]
-== Problem 30 ==
+== Problem 26==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The straight line <math>\overline{AB}</math> is divided at <math>C</math> so that <math>AC=3CB</math>. Circles are described on <math>\overline{AC}</math>
+and <math>\overline{CB}</math> as diameters and a common tangent meets <math>AB</math> produced at <math>D</math>. Then <math>BD</math> equals:
-[[1953 AHSME Problems/Problem 30|Solution]]
+<math> \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} </math>
-== Problem 31 ==
+[[1954 AHSME Problems/Problem 26|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 27==
-[[1953 AHSME Problems/Problem 31|Solution]]
+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:
+<math> \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}} </math>
-== Problem 32 ==
+[[1954 AHSME Problems/Problem 27|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 28==
-[[1953 AHSME Problems/Problem 32|Solution]]
+If <math>\frac{m}{n}=\frac{4}{3}</math> and <math>\frac{r}{t}=\frac{9}{14}</math>, the value of <math>\frac{3mr-nt}{4nt-7mr}</math> is:
-== Problem 33 ==
+<math> \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} </math>
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+[[1954 AHSME Problems/Problem 28|Solution]]
-[[1953 AHSME Problems/Problem 33|Solution]]
+== Problem 29==
-== Problem 34 ==
+If the ratio of the legs of a right triangle is <math>1: 2</math>, then the ratio of the corresponding segments of
+the hypotenuse made by a perpendicular upon it from the vertex is:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \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 </math>
-[[1953 AHSME Problems/Problem 34|Solution]]
+[[1954 AHSME Problems/Problem 29|Solution]]
-== Problem 35 ==
+== Problem 30==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math>A</math> and <math>B</math> together can do a job in <math>2</math> days; <math>B</math> and <math>C</math> can do it in four days; and <math>A</math> and <math>C</math> in <math>2\frac{2}{5}</math> days.
+The number of days required for A to do the job alone is:
-[[1953 AHSME Problems/Problem 35|Solution]]
+<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 2.8  </math>
+[[1954 AHSME Problems/Problem 30|Solution]]
-== Problem 36 ==
+== Problem 31==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+In <math>\triangle ABC</math>, <math>AB=AC</math>, <math>\angle A=40^\circ</math>. Point <math>O</math> is within the triangle with <math>\angle OBC \cong \angle OCA</math>.
+The number of degrees in <math>\angle BOC</math> is:
-[[1953 AHSME Problems/Problem 36|Solution]]
+<math>\textbf{(A)}\ 110^{\circ} \qquad \textbf{(B)}\ 35^{\circ} \qquad \textbf{(C)}\ 140^{\circ} \qquad \textbf{(D)}\ 55^{\circ} \qquad \textbf{(E)}\ 70^{\circ}</math>
+[[1954 AHSME Problems/Problem 31|Solution]]
-== Problem 37 ==
+== Problem 32==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The factors of <math>x^4+64</math> are:
-[[1953 AHSME Problems/Problem 37|Solution]]
+<math> \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) </math>
-== Problem 38 ==
+[[1954 AHSME Problems/Problem 32|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 33==
-[[1953 AHSME Problems/Problem 38|Solution]]
+A bank charges <math>\textdollar{6}</math> for a loan of <math>\textdollar{120}</math>. The borrower receives <math>\textdollar{114}</math> and
+repays the loan in <math>12</math> installments of <math>\textdollar{10}</math> a month. The interest rate is approximately:
-== Problem 39 ==
+<math>\textbf{(A)}\ 5 \% \qquad \textbf{(B)}\ 6 \% \qquad \textbf{(C)}\ 7 \% \qquad \textbf{(D)}\ 9\% \qquad \textbf{(E)}\ 15 \%     </math>
+[[1954 AHSME Problems/Problem 33|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 34==
-[[1953 AHSME Problems/Problem 39|Solution]]
+The fraction <math>\frac{1}{3}</math>:
-== Problem 40 ==
+<math> \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} </math>
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+[[1954 AHSME Problems/Problem 34|Solution]]
-[[1953 AHSME Problems/Problem 40|Solution]]
+== Problem 35==
-== Problem 41 ==
+In the right triangle shown the sum of the distances <math>BM</math> and <math>MA</math> is equal to the sum of the distances <math>BC</math> and <math>CA</math>.
+If <math>MB = x, CB = h</math>, and <math>CA = d</math>, then <math>x</math> equals:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<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>
-[[1953 AHSME Problems/Problem 41|Solution]]
+<math> \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 </math>
-== Problem 42 ==
+[[1954 AHSME Problems/Problem 35|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 36==
-[[1953 AHSME Problems/Problem 42|Solution]]
+A boat has a speed of <math>15</math> mph in still water. In a stream that has a current of <math>5</math> 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:
-== Problem 43 ==
+<math> \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} </math>
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+[[1954 AHSME Problems/Problem 36|Solution]]
-[[1953 AHSME Problems/Problem 43|Solution]]
+== Problem 37==
-== Problem 44 ==
+Given <math>\triangle PQR</math> with <math>\overline{RS}</math> bisecting <math>\angle R</math>, <math>PQ</math> extended to <math>D</math> and <math>\angle n</math> a right angle, then:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ }  \qquad \textbf{(C) \ }  \qquad \textbf{(D) \ }  \qquad \textbf{(E) \ } </math>
+<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>
-[[1953 AHSME Problems/Problem 44|Solution]]
+<math>\textbf{(A)}\ \angle m = \frac {1}{2}(\angle p - \angle q) \qquad
+\textbf{(B)}\ \angle m = \frac {1}{2}(\angle p + \angle q) </math>
+<math> \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} </math>
-== Problem 45 ==
+[[1954 AHSME Problems/Problem 37|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 38==
-[[1953 AHSME Problems/Problem 45|Solution]]
+If <math>\log 2=.3010</math> and <math>\log 3=.4771</math>, the value of <math>x</math> when <math>3^{x+3}=135</math> is approximately:
-== Problem 46 ==
+<math>\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 1.47 \qquad \textbf{(C)}\ 1.67 \qquad \textbf{(D)}\ 1.78 \qquad \textbf{(E)}\ 1.63    </math>
+[[1954 AHSME Problems/Problem 38|Solution]]
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+== Problem 39==
-[[1953 AHSME Problems/Problem 46|Solution]]
+The locus of the midpoint of a line segment that is drawn from a given external point <math>P</math> to a given circle with center <math>O</math> and radius <math>r</math>, is:
-== Problem 47 ==
+<math> \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 </math>
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+[[1954 AHSME Problems/Problem 39|Solution]]
-[[1953 AHSME Problems/Problem 47|Solution]]
+== Problem 40==
-== Problem 48 ==
+If <math>\left (a+\frac{1}{a} \right )^2=3</math>, then <math>a^3+\frac{1}{a^3}</math> equals:
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+<math> \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} </math>
-[[1953 AHSME Problems/Problem 48|Solution]]
+[[1954 AHSME Problems/Problem 40|Solution]]
-== Problem 49 ==
+== Problem 41==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+The sum of all the roots of <math>4x^3-8x^2-63x-9=0</math> is:
-[[1953 AHSME Problems/Problem 49|Solution]]
+<math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ -8 \qquad \textbf{(D)}\ -2 \qquad \textbf{(E)}\ 0   </math>
+[[1954 AHSME Problems/Problem 41|Solution]]
-== Problem 50 ==
+== Problem 42==
-<math> \textbf{(A) \ }  \qquad \textbf{(B) \ } \qquad \textbf{(C) \ } \qquad \textbf{(D) \ }\qquad \textbf{(E) \ } </math>
+Consider the graphs of
+<cmath>(1)\qquad y=x^2-\frac{1}{2}x+2</cmath>
+and
+<cmath>(2)\qquad y=x^2+\frac{1}{2}x+2</cmath>
+on the same set of axis.
+These parabolas are exactly the same shape. Then:
-[[1953 AHSME Problems/Problem 50|Solution]]
+<math> \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).} </math>
+[[1954 AHSME Problems/Problem 42|Solution]]
+== Problem 43==
+The hypotenuse of a right triangle is <math>10</math> inches and the radius of the inscribed circle is <math>1</math> inch. The perimeter of the triangle in inches is:
+<math>\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 30 </math>
+[[1954 AHSME Problems/Problem 43|Solution]]
+== Problem 44==
+A man born in the first half of the nineteenth century was <math>x</math> years old in the year <math>x^2</math>. He was born in:
+<math>\textbf{(A)}\ 1849 \qquad \textbf{(B)}\ 1825 \qquad \textbf{(C)}\ 1812 \qquad \textbf{(D)}\ 1836 \qquad \textbf{(E)}\ 1806 </math>
+[[1954 AHSME Problems/Problem 44|Solution]]
+== Problem 45==
+In a rhombus, <math>ABCD</math>, line segments are drawn within the rhombus, parallel to diagonal <math>BD</math>,
+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 <math>A</math>. The graph is:
+<math> \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.} </math>
+[[1954 AHSME Problems/Problem 45|Solution]]
+== Problem 46==
+In the diagram, if points <math>A, B</math> and <math>C</math> are points of tangency, then <math>x</math> 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>
+<math> \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}" </math>
+[[1954 AHSME Problems/Problem 46|Solution]]
+== Problem 47==
+At the midpoint of line segment <math>AB</math> which is <math>p</math> units long, a perpendicular <math>MR</math> is erected with length <math>q</math> units.
+An arc is described from <math>R</math> with a radius equal to <math>\frac{1}{2}AB</math>, meeting <math>AB</math> at <math>T</math>. Then <math>AT</math> and <math>TB</math> are the roots of:
+<math> \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 </math>
+[[1954 AHSME Problems/Problem 47|Solution]]
+== Problem 48==
+A train, an hour after starting, meets with an accident which detains it a half hour, after which it
+proceeds at <math>\frac{3}{4}</math> of its former rate and arrives <math>3\tfrac{1}{2}</math> hours late.
+Had the accident happened <math>90</math> miles farther along the line, it would have arrived only <math>3</math> hours late. The length of the trip in miles was:
+<math>\textbf{(A)}\ 400 \qquad \textbf{(B)}\ 465 \qquad \textbf{(C)}\ 600 \qquad \textbf{(D)}\ 640 \qquad \textbf{(E)}\ 550    </math>
+[[1954 AHSME Problems/Problem 48|Solution]]
+== Problem 49==
+The difference of the squares of two odd numbers is always divisible by <math>8</math>. If <math>a>b</math>, and <math>2a+1</math> and <math>2b+1</math> are the odd numbers,
+to prove the given statement we put the difference of the squares in the form:
+<math> \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) </math>
+[[1954 AHSME Problems/Problem 49|Solution]]
+== Problem 50==
+The times between <math>7</math> and <math>8</math> o'clock, correct to the nearest minute, when the hands of a clock will form an angle of <math>84^{\circ}</math> are:
+<math> \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} </math>
+[[1954 AHSME Problems/Problem 50|Solution]]
 == See also ==
-* [[AHSME]]
-* [[AHSME Problems and Solutions]]
+* [[AMC 12 Problems and Solutions]]
 * [[Mathematics competition resources]]
+{{AHSME 50p box|year=1954|before=[[1953 AHSME|1953 AHSC]]|after=[[1955 AHSME|1955 AHSC]]}}
+{{MAA Notice}}

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Difference between revisions of "1954 AHSME Problems"

Latest revision as of 18:16, 8 November 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