Difference between revisions of "1960 AHSME Problems"

Latest revision as of 20:27, 30 December 2020

1960 AHSC (Answer Key) Printable versions: Wiki • AoPS Resources • PDF
Instructions This is a 40-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. You will receive ? points for each correct answer, ? points for each problem left unanswered, and ? points for each incorrect answer. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers. Figures are not necessarily drawn to scale. You will have ? minutes working time to complete the test.
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Problem 1

If $2$ is a solution (root) of $x^3+hx+10=0$ , then $h$ equals:

$\textbf{(A)}10\qquad \textbf{(B)}9 \qquad \textbf{(C)}2\qquad \textbf{(D)}-2\qquad \textbf{(E)}-9$

Solution

Problem 2

It takes $5$ seconds for a clock to strike $6$ o'clock beginning at $6:00$ o'clock precisely. If the strikings are uniformly spaced, how long, in seconds, does it take to strike $12$ o'clock?

$\textbf{(A)}9\frac{1}{5}\qquad \textbf{(B )}10\qquad \textbf{(C )}11\qquad \textbf{(D )}14\frac{2}{5}\qquad \textbf{(E )}\text{none of these}$

Solution

Problem 3

Applied to a bill for $\textdollar{10,000}$ the difference between a discount of $40$ % and two successive discounts of $36$ % and $4$ %, expressed in dollars, is:

$\textbf{(A) }0\qquad \textbf{(B) }144\qquad \textbf{(C) }256\qquad \textbf{(D) }400\qquad \textbf{(E) }416$

Solution

Problem 4

Each of two angles of a triangle is $60^{\circ}$ and the included side is $4$ inches. The area of the triangle, in square inches, is:

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

Solution

Problem 5

The number of distinct points common to the graphs of $x^2+y^2=9$ and $y^2=9$ is:

$\textbf{(A) }\text{infinitely many}\qquad \textbf{(B) }\text{four}\qquad \textbf{(C) }\text{two}\qquad \textbf{(D) }\text{one}\qquad \textbf{(E) }\text{none}$

Solution

Problem 6

The circumference of a circle is $100$ inches. The side of a square inscribed in this circle, expressed in inches, is:

$\textbf{(A) }\frac{25\sqrt{2}}{\pi}\qquad \textbf{(B) }\frac{50\sqrt{2}}{\pi}\qquad \textbf{(C) }\frac{100}{\pi}\qquad \textbf{(D) }\frac{100\sqrt{2}}{\pi}\qquad \textbf{(E) }50\sqrt{2}$

Solution

Problem 7

Circle $I$ passes through the center of, and is tangent to, circle $II$ . The area of circle $I$ is $4$ square inches. Then the area of circle $II$ , in square inches, is:

$\textbf{(A) }8\qquad \textbf{(B) }8\sqrt{2}\qquad \textbf{(C) }8\sqrt{\pi}\qquad \textbf{(D) }16\qquad \textbf{(E) }16\sqrt{2}$

Solution

Problem 8

The number $2.5252525\ldots$ can be written as a fraction. When reduced to lowest terms the sum of the numerator and denominator of this fraction is:

$\textbf{(A) }7\qquad \textbf{(B) } 29\qquad \textbf{(C) }141\qquad \textbf{(D) }349\qquad \textbf{(E) }\text{none of these}$

Solution

Problem 9

The fraction $\frac{a^2+b^2-c^2+2ab}{a^2+c^2-b^2+2ac}$ is (with suitable restrictions of the values of a, b, and c):

$\text{(A) irreducible}\qquad$

$\text{(B) reducible to negative 1}\qquad$

$\text{(C) reducible to a polynomial of three terms}\qquad$

$\text{(D) reducible to} \frac{a-b+c}{a+b-c}\qquad$

$\text{(E) reducible to} \frac{a+b-c}{a-b+c}$

Solution

Problem 10

Given the following six statements: $\text{(1) All women are good drivers}$ $\text{(2) Some women are good drivers}$ $\text{(3) No men are good drivers}$ $\text{(4) All men are bad drivers}$ $\text{(5) At least one man is a bad driver}$ $\text{(6) All men are good drivers.}$

The statement that negates statement $(6)$ is:

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

Solution

Problem 11

For a given value of $k$ the product of the roots of $x^2-3kx+2k^2-1=0$ is $7$ . The roots may be characterized as:

$\textbf{(A) }\text{integral and positive} \qquad\textbf{(B) }\text{integral and negative} \qquad \\ \textbf{(C) }\text{rational, but not integral} \qquad\textbf{(D) }\text{irrational} \qquad\textbf{(E) } \text{imaginary}$

Solution

Problem 12

The locus of the centers of all circles of given radius $a$ , in the same plane, passing through a fixed point, is:

$\textbf{(A) }\text{a point}\qquad \textbf{(B) }\text{ a straight line}\qquad \textbf{(C) }\text{two straight lines}\qquad \textbf{(D) }\text{a circle}\qquad \textbf{(E) }\text{two circles}$

Solution

Problem 13

The polygon(s) formed by $y=3x+2, y=-3x+2$ , and $y=-2$ , is (are):

$\textbf{(A) }\text{an equilateral triangle}\qquad\textbf{(B) }\text{an isosceles triangle} \qquad\textbf{(C) }\text{a right triangle} \qquad \\ \textbf{(D) }\text{a triangle and a trapezoid}\qquad\textbf{(E) }\text{a quadrilateral}$

Solution

Problem 14

If $a$ and $b$ are real numbers, the equation $3x-5+a=bx+1$ has a unique solution $x$ [The symbol $a \neq 0$ means that $a$ is different from zero]:

$\text{(A) for all a and b} \qquad \text{(B) if a }\neq\text{2b}\qquad \text{(C) if a }\neq 6\qquad \\ \text{(D) if b }\neq 0\qquad \text{(E) if b }\neq 3$

Solution

Problem 15

Triangle $I$ is equilateral with side $A$ , perimeter $P$ , area $K$ , and circumradius $R$ (radius of the circumscribed circle). Triangle $II$ is equilateral with side $a$ , perimeter $p$ , area $k$ , and circumradius $r$ . If $A$ is different from $a$ , then:

$\textbf{(A)}\ P:p = R:r \text{ } \text{only sometimes} \qquad \textbf{(B)}\ P:p = R:r \text{ } \text{always}\qquad \\ \textbf{(C)}\ P:p = K:k \text{ } \text{only sometimes} \qquad \textbf{(D)}\ R:r = K:k \text{ } \text{always}\qquad \textbf{(E)}\ R:r = K:k \text{ } \text{only sometimes}$

Solution

Problem 16

In the numeration system with base $5$ , counting is as follows: $1, 2, 3, 4, 10, 11, 12, 13, 14, 20,\ldots$ . The number whose description in the decimal system is $69$ , when described in the base $5$ system, is a number with:

$\textbf{(A)}\ \text{two consecutive digits} \qquad\textbf{(B)}\ \text{two non-consecutive digits} \qquad \\ \textbf{(C)}\ \text{three consecutive digits} \qquad\textbf{(D)}\ \text{three non-consecutive digits} \qquad \\ \textbf{(E)}\ \text{four digits}$

Solution

Problem 17

The formula $N=8 \times 10^{8} \times x^{-3/2}$ gives, for a certain group, the number of individuals whose income exceeds $x$ dollars. The lowest income, in dollars, of the wealthiest $800$ individuals is at least:

$\textbf{(A)}\ 10^4\qquad \textbf{(B)}\ 10^6\qquad \textbf{(C)}\ 10^8\qquad \textbf{(D)}\ 10^{12} \qquad \textbf{(E)}\ 10^{16}$

Solution

Problem 18

The pair of equations $3^{x+y}=81$ and $81^{x-y}=3$ has:

$\textbf{(A)}\ \text{no common solution} \qquad \\ \textbf{(B)}\ \text{the solution} \text{ } x=2, y=2\qquad \\ \textbf{(C)}\ \text{the solution} \text{ } x=2\frac{1}{2}, y=1\frac{1}{2} \qquad \\ \textbf{(D)}\text{ a common solution in positive and negative integers} \qquad \\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 19

Consider equation $I: x+y+z=46$ where $x, y$ , and $z$ are positive integers, and equation $II: x+y+z+w=46$ , where $x, y, z$ , and $w$ are positive integers. Then

$\textbf{(A)}\ \text{I can be solved in consecutive integers} \qquad \\ \textbf{(B)}\ \text{I can be solved in consecutive even integers} \qquad \\ \textbf{(C)}\ \text{II can be solved in consecutive integers} \qquad \\ \textbf{(D)}\ \text{II can be solved in consecutive even integers} \qquad \\ \textbf{(E)}\ \text{II can be solved in consecutive odd integers}$

Solution

Problem 20

The coefficient of $x^7$ in the expansion of $(\frac{x^2}{2}-\frac{2}{x})^8$ is:

$\textbf{(A)}\ 56\qquad \textbf{(B)}\ -56\qquad \textbf{(C)}\ 14\qquad \textbf{(D)}\ -14\qquad \textbf{(E)}\ 0$

Solution

Problem 21

The diagonal of square $I$ is $a+b$ . The perimeter of square $II$ with twice the area of $I$ is:

$\textbf{(A)}\ (a+b)^2\qquad \textbf{(B)}\ \sqrt{2}(a+b)^2\qquad \textbf{(C)}\ 2(a+b)\qquad \textbf{(D)}\ \sqrt{8}(a+b) \qquad \textbf{(E)}\ 4(a+b)$

Solution

Problem 22

The equality $(x+m)^2-(x+n)^2=(m-n)^2$ , where $m$ and $n$ are unequal non-zero constants, is satisfied by $x=am+bn$ , where:

$\textbf{(A)}\ a = 0, b \text{ } \text{has a unique non-zero value}\qquad \\ \textbf{(B)}\ a = 0, b \text{ } \text{has two non-zero values}\qquad \\ \textbf{(C)}\ b = 0, a \text{ } \text{has a unique non-zero value}\qquad \\ \textbf{(D)}\ b = 0, a \text{ } \text{has two non-zero values}\qquad \\ \textbf{(E)}\ a \text{ } \text{and} \text{ } b \text{ } \text{each have a unique non-zero value}$

Solution

Problem 23

The radius $R$ of a cylindrical box is $8$ inches, the height $H$ is $3$ inches. The volume $V = \pi R^2H$ is to be increased by the same fixed positive amount when $R$ is increased by $x$ inches as when $H$ is increased by $x$ inches. This condition is satisfied by:

$\textbf{(A)}\ \text{no real value of x} \qquad \\ \textbf{(B)}\ \text{one integral value of x} \qquad \\ \textbf{(C)}\ \text{one rational, but not integral, value of x} \qquad \\ \textbf{(D)}\ \text{one irrational value of x}\qquad \\ \textbf{(E)}\ \text{two real values of x}$

Solution

Problem 24

If $\log_{2x}216 = x$ , where $x$ is real, then $x$ is:

$\textbf{(A)}\ \text{A non-square, non-cube integer}\qquad$ $\textbf{(B)}\ \text{A non-square, non-cube, non-integral rational number}\qquad$ $\textbf{(C)}\ \text{An irrational number}\qquad$ $\textbf{(D)}\ \text{A perfect square}\qquad$ $\textbf{(E)}\ \text{A perfect cube}$

Solution

Problem 25

Let $m$ and $n$ be any two odd numbers, with $n$ less than $m$ . The largest integer which divides all possible numbers of the form $m^2-n^2$ is:

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 16$

Solution

Problem 26

Find the set of $x$ -values satisfying the inequality $|\frac{5-x}{3}|<2$ . [The symbol $|a|$ means $+a$ if $a$ is positive, $-a$ if $a$ is negative, $0$ if $a$ is zero. The notation $1<a<2$ means that a can have any value between $1$ and $2$ , excluding $1$ and $2$ . ]

$\textbf{(A)}\ 1 < x < 11\qquad \textbf{(B)}\ -1 < x < 11\qquad \textbf{(C)}\ x< 11\qquad \textbf{(D)}\ x>11\qquad \textbf{(E)}\ |x| < 6$

Solution

Problem 27

Let $S$ be the sum of the interior angles of a polygon $P$ for which each interior angle is $7\frac{1}{2}$ times the exterior angle at the same vertex. Then

$\textbf{(A)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may be regular}\qquad \\ \textbf{(B)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad \\ \textbf{(C)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is regular}\qquad \\ \textbf{(D)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad \\ \textbf{(E)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may or may not be regular}$

Solution

Problem 28

The equation $x-\frac{7}{x-3}=3-\frac{7}{x-3}$ has:

$\textbf{(A)}\ \text{infinitely many integral roots}\qquad\textbf{(B)}\ \text{no root}\qquad\textbf{(C)}\ \text{one integral root}\qquad$ $\textbf{(D)}\ \text{two equal integral roots}\qquad\textbf{(E)}\ \text{two equal non-integral roots}$

Solution

Problem 29

Five times $A$ 's money added to $B$ 's money is more than $$51.00$ . Three times $A$ 's money minus $B$ 's money is $$21.00$ . If $a$ represents $A$ 's money in dollars and $b$ represents $B$ 's money in dollars, then:

$\textbf{(A)}\ a>9, b>6 \qquad \textbf{(B)}\ a>9, b<6 \qquad \textbf{(C)}\ a>9, b=6\qquad \textbf{(D)}\ a>9, \text{but we can put no bounds on} \text{ } b\qquad \textbf{(E)}\ 2a=3b$

Solution

Problem 30

Given the line $3x+5y=15$ and a point on this line equidistant from the coordinate axes. Such a point exists in:

$\textbf{(A)}\ \text{none of the quadrants}\qquad\textbf{(B)}\ \text{quadrant I only}\qquad\textbf{(C)}\ \text{quadrants I, II only}\qquad$ $\textbf{(D)}\ \text{quadrants I, II, III only}\qquad\textbf{(E)}\ \text{each of the quadrants}$

Solution

Problem 31

For $x^2+2x+5$ to be a factor of $x^4+px^2+q$ , the values of $p$ and $q$ must be, respectively:

$\textbf{(A)}\ -2, 5\qquad \textbf{(B)}\ 5, 25\qquad \textbf{(C)}\ 10, 20\qquad \textbf{(D)}\ 6, 25\qquad \textbf{(E)}\ 14, 25$

Solution

Problem 32

In this figure the center of the circle is $O$ . $AB \perp BC$ , $ADOE$ is a straight line, $AP = AD$ , and $AB$ has a length twice the radius. Then:

$[asy] size(150); defaultpen(linewidth(0.8)+fontsize(10)); real e=350,c=55; pair O=origin,E=dir(e),C=dir(c),B=dir(180+c),D=dir(180+e), rot=rotate(90,B)*O,A=extension(E,D,B,rot); path tangent=A--B; pair P=waypoint(tangent,abs(A-D)/abs(A-B)); draw(unitcircle^^C--B--A--E); dot(A^^B^^C^^D^^E^^P,linewidth(2)); label("$O$",O,dir(290)); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,NE); label("$D$",D,dir(120)); label("$E$",E,SE); label("$P$",P,SW);[/asy]$

$\textbf{(A)} AP^2 = PB \times AB\qquad \\ \textbf{(B)}\ AP \times DO = PB \times AD\qquad \\ \textbf{(C)}\ AB^2 = AD \times DE\qquad \\ \textbf{(D)}\ AB \times AD = OB \times AO\qquad \\ \textbf{(E)}\ \text{none of these}$

Solution

Problem 33

You are given a sequence of $58$ terms; each term has the form $P+n$ where $P$ stands for the product $2 \times 3 \times 5 \times\ldots \times 61$ of all prime numbers less than or equal to $61$ , and $n$ takes, successively, the values $2, 3, 4,\ldots, 59$ . Let $N$ be the number of primes appearing in this sequence. Then $N$ is:

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 57\qquad \textbf{(E)}\ 58$

Solution

Problem 34

Two swimmers, at opposite ends of a $90$ -foot pool, start to swim the length of the pool, one at the rate of $3$ feet per second, the other at $2$ feet per second. They swim back and forth for $12$ minutes. Allowing no loss of times at the turns, find the number of times they pass each other.

$\textbf{(A)}\ 24\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 20\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 18$

Solution

Problem 35

From point $P$ outside a circle, with a circumference of $10$ units, a tangent is drawn. Also from $P$ a secant is drawn dividing the circle into unequal arcs with lengths $m$ and $n$ . It is found that $t_1$ , the length of the tangent, is the mean proportional between $m$ and $n$ . If $m$ and $t$ are integers, then $t$ may have the following number of values:

$\textbf{(A)}\ \text{zero}\qquad \textbf{(B)}\ \text{one}\qquad \textbf{(C)}\ \text{two}\qquad \textbf{(D)}\ \text{three}\qquad \textbf{(E)}\ \text{infinitely many}$

Solution

Problem 36

Let $s_1, s_2, s_3$ be the respective sums of $n, 2n, 3n$ terms of the same arithmetic progression with $a$ as the first term and $d$ as the common difference. Let $R=s_3-s_2-s_1$ . Then $R$ is dependent on:

$\textbf{(A)}\ a\text{ }\text{and}\text{ }d\qquad \textbf{(B)}\ d\text{ }\text{and}\text{ }n\qquad \textbf{(C)}\ a\text{ }\text{and}\text{ }n\qquad \textbf{(D)}\ a, d,\text{ }\text{and}\text{ }n\qquad \textbf{(E)}\ \text{neither} \text{ } a \text{ } \text{nor} \text{ } d \text{ } \text{nor} \text{ } n$

Solution

Problem 37

The base of a triangle is of length $b$ , and the altitude is of length $h$ . A rectangle of height $x$ is inscribed in the triangle with the base of the rectangle in the base of the triangle. The area of the rectangle is:

$\textbf{(A)}\ \frac{bx}{h}(h-x)\qquad \textbf{(B)}\ \frac{hx}{b}(b-x)\qquad \textbf{(C)}\ \frac{bx}{h}(h-2x)\qquad \textbf{(D)}\ x(b-x)\qquad \textbf{(E)}\ x(h-x)$

Solution

Problem 38

In this diagram $AB$ and $AC$ are the equal sides of an isosceles $\triangle ABC$ , in which is inscribed equilateral $\triangle DEF$ . Designate $\angle BFD$ by $a$ , $\angle ADE$ by $b$ , and $\angle FEC$ by $c$ . Then:

$[asy] size(150); defaultpen(linewidth(0.8)+fontsize(10)); pair A=(5,12),B=origin,C=(10,0),D=(5/3,4),E=(10-5*.45,12*.45),F=(6,0); draw(A--B--C--cycle^^D--E--F--cycle); draw(anglemark(E,D,A,1,45)^^anglemark(F,E,C,1,45)^^anglemark(D,F,B,1,45)); label("$b$",(D.x+.2,D.y+.25),dir(30)); label("$c$",(E.x,E.y-.4),S); label("$a$",(F.x-.4,F.y+.1),dir(150)); label("$A$",A,N); label("$B$",B,S); label("$C$",C,S); label("$D$",D,dir(150)); label("$E$",E,dir(60)); label("$F$",F,S);[/asy]$

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

Solution

Problem 39

To satisfy the equation $\frac{a+b}{a}=\frac{b}{a+b}$ , $a$ and $b$ must be:

$\textbf{(A)}\ \text{both rational}\qquad\textbf{(B)}\ \text{both real but not rational}\qquad\textbf{(C)}\ \text{both not real}\qquad$ $\textbf{(D)}\ \text{one real, one not real}\qquad\textbf{(E)}\ \text{one real, one not real or both not real}$

Solution

Problem 40

Given right $\triangle ABC$ with legs $BC=3, AC=4$ . Find the length of the shorter angle trisector from $C$ to the hypotenuse:

$\textbf{(A)}\ \frac{32\sqrt{3}-24}{13}\qquad\textbf{(B)}\ \frac{12\sqrt{3}-9}{13}\qquad\textbf{(C)}\ 6\sqrt{3}-8\qquad\textbf{(D)}\ \frac{5\sqrt{10}}{6}\qquad\textbf{(E)}\ \frac{25}{12}\qquad$

Solution

@@ Line 1: / Line 1: @@
+{{AHSC 40 Problems
+|year = 1960
+}}
 == Problem 1==
 If <math>2</math> is a solution (root) of <math>x^3+hx+10=0</math>, then <math>h</math> equals:
 <math>\textbf{(A)}10\qquad
-\textbf{(B )}9 \qquad
+\textbf{(B)}9 \qquad
-\textbf{(C )}2\qquad
+\textbf{(C)}2\qquad
-\textbf{(D )}-2\qquad
+\textbf{(D)}-2\qquad
-\textbf{(E )}-9   </math>
+\textbf{(E)}-9   </math>
 [[1960 AHSME Problems/Problem 1|Solution]]
@@ Line 29: / Line 32: @@
 expressed in dollars, is:
-<math>\textbf{(A)}0\qquad
+<math>\textbf{(A) }0\qquad
-\textbf{(B )}144\qquad
+\textbf{(B) }144\qquad
-\textbf{(C )}256\qquad
+\textbf{(C) }256\qquad
-\textbf{(D )}400\qquad
+\textbf{(D) }400\qquad
-\textbf{(E )}416     </math>
+\textbf{(E) }416     </math>
@@ Line 42: / Line 45: @@
 Each of two angles of a triangle is <math>60^{\circ}</math> and the included side is <math>4</math> inches. The area of the triangle, in square inches, is:
-<math>\textbf{(A)}8\sqrt{3}\qquad
+<math>\textbf{(A) }8\sqrt{3}\qquad
-\textbf{(B )}8\qquad
+\textbf{(B) }8\qquad
-\textbf{(C )}4\sqrt{3}\qquad
+\textbf{(C) }4\sqrt{3}\qquad
-\textbf{(D )}4\qquad
+\textbf{(D) }4\qquad
-\textbf{(E )}2\sqrt{3}     </math>
+\textbf{(E) }2\sqrt{3}     </math>
@@ Line 55: / Line 58: @@
 The number of distinct points common to the graphs of <math>x^2+y^2=9</math> and <math>y^2=9</math> is:
-<math>\textbf{(A)}\text{infinitely many}\qquad
+<math>\textbf{(A) }\text{infinitely many}\qquad
-\textbf{(B )}\text{four}\qquad
+\textbf{(B) }\text{four}\qquad
-\textbf{(C )}\text{two}\qquad
+\textbf{(C) }\text{two}\qquad
-\textbf{(D )}\text{one}\qquad
+\textbf{(D) }\text{one}\qquad
-\textbf{(E )}\text{none} </math>
+\textbf{(E) }\text{none} </math>
@@ Line 68: / Line 71: @@
 The circumference of a circle is <math>100</math> inches. The side of a square inscribed in this circle, expressed in inches, is:
-<math>\textbf{(A)}\frac{25\sqrt{2}}{\pi}\qquad
+<math>\textbf{(A) }\frac{25\sqrt{2}}{\pi}\qquad
-\textbf{(B )}\frac{50\sqrt{2}}{\pi}\qquad
+\textbf{(B) }\frac{50\sqrt{2}}{\pi}\qquad
-\textbf{(C )}\frac{100}{\pi}\qquad
+\textbf{(C) }\frac{100}{\pi}\qquad
-\textbf{(D )}\frac{100\sqrt{2}}{\pi}\qquad
+\textbf{(D) }\frac{100\sqrt{2}}{\pi}\qquad
-\textbf{(E )}50\sqrt{2} </math>
+\textbf{(E) }50\sqrt{2} </math>
@@ Line 82: / Line 85: @@
 Then the area of circle <math>II</math>, in square inches, is:
-<math>\textbf{(A)}8\qquad
+<math>\textbf{(A) }8\qquad
-\textbf{(B )}8\sqrt{2}\qquad
+\textbf{(B) }8\sqrt{2}\qquad
-\textbf{(C )}8\sqrt{\pi}\qquad
+\textbf{(C) }8\sqrt{\pi}\qquad
-\textbf{(D )}16\qquad
+\textbf{(D) }16\qquad
-\textbf{(E )}16\sqrt{2}     </math>
+\textbf{(E) }16\sqrt{2}     </math>
@@ Line 96: / Line 99: @@
 When reduced to lowest terms the sum of the numerator and denominator of this fraction is:
-<math>\textbf{(A)}7\qquad
+<math>\textbf{(A) }7\qquad
-\textbf{(B)} 29\qquad
+\textbf{(B) } 29\qquad
-\textbf{(C )}141\qquad
+\textbf{(C) }141\qquad
-\textbf{(D )}349\qquad
+\textbf{(D) }349\qquad
-\textbf{(E )}\text{none of these}     </math>
+\textbf{(E) }\text{none of these}     </math>
@@ Line 124: / Line 127: @@
 Given the following six statements:
-<cmath>\text{(1) All women are good drivers} \\
+<cmath>\text{(1) All women are good drivers}</cmath>
-\text{(2) Some women are good drivers} \\
+<cmath>\text{(2) Some women are good drivers}</cmath>
-\text{(3) No men are good drivers} \\
+<cmath>\text{(3) No men are good drivers}</cmath>
-\text{(4) All men are bad drivers} \\
+<cmath>\text{(4) All men are bad drivers}</cmath>
-\text{(5) At least one man is a bad driver} \\
+<cmath>\text{(5) At least one man is a bad driver}</cmath>
-\text{(6) All men are good drivers.}</cmath>
+<cmath>\text{(6) All men are good drivers.}</cmath>
@@ Line 135: / Line 138: @@
-<math>\textbf{(A )}(1)\qquad
+<math>\textbf{(A) }(1)\qquad
-\textbf{(B )}(2)\qquad
+\textbf{(B) }(2)\qquad
-\textbf{(C )}(3)\qquad
+\textbf{(C) }(3)\qquad
-\textbf{(D )}(4)\qquad
+\textbf{(D) }(4)\qquad
-\textbf{(E )}(5)</math>
+\textbf{(E) }(5)</math>
@@ Line 159: / Line 162: @@
 The locus of the centers of all circles of given radius <math>a</math>, in the same plane, passing through a fixed point, is:
-<math>\textbf{(A)}\text{a point}\qquad
+<math>\textbf{(A) }\text{a point}\qquad
-\textbf{(B )}\text{ a straight line}\qquad
+\textbf{(B) }\text{ a straight line}\qquad
-\textbf{(C )}\text{two straight lines}\qquad
+\textbf{(C) }\text{two straight lines}\qquad
-\textbf{(D )}\text{a circle}\qquad
+\textbf{(D) }\text{a circle}\qquad
-\textbf{(E )}\text{two circles}     </math>
+\textbf{(E) }\text{two circles}     </math>
@@ Line 172: / Line 175: @@
 The polygon(s) formed by <math>y=3x+2, y=-3x+2</math>, and <math>y=-2</math>, is (are):
-<math>\textbf{(A) }\text{An equilateral triangle}\qquad\textbf{(B) }\text{an isosceles triangle} \qquad\textbf{(C) }\text{a right triangle} \qquad \\
+<math>\textbf{(A) }\text{an equilateral triangle}\qquad\textbf{(B) }\text{an isosceles triangle} \qquad\textbf{(C) }\text{a right triangle} \qquad \\
 \textbf{(D) }\text{a triangle and a trapezoid}\qquad\textbf{(E) }\text{a quadrilateral} </math>
@@ Line 211: / Line 214: @@
 <math>\textbf{(A)}\ \text{two consecutive digits} \qquad\textbf{(B)}\ \text{two non-consecutive digits} \qquad \\
 \textbf{(C)}\ \text{three consecutive digits} \qquad\textbf{(D)}\ \text{three non-consecutive digits} \qquad \\
 \textbf{(E)}\ \text{four digits} </math>
@@ Line 305: / Line 306: @@
 is increased by <math>x</math> inches as when <math>H</math> is increased by <math>x</math> inches. This condition is satisfied by:
-<math>\textbf{(A)}\ \text{A non-square, non-cube integer} \qquad \\
+<math>\textbf{(A)}\ \text{no real value of x} \qquad \\
-\textbf{(B)}\ \text{A non-square, non-cube, non-integral rational number} \qquad \\
+\textbf{(B)}\ \text{one integral value of x} \qquad \\
-\textbf{(C)}\ \text{An irrational number} \qquad \\
+\textbf{(C)}\ \text{one rational, but not integral, value of x} \qquad \\
-\textbf{(D)}\ \text{A perfect square}\qquad \\
+\textbf{(D)}\ \text{one irrational value of x}\qquad \\
-\textbf{(E)}\ \text{A perfect cube} </math>
+\textbf{(E)}\ \text{two real values of x} </math>
 [[1960 AHSME Problems/Problem 23|Solution]]
@@ Line 358: / Line 359: @@
 exterior angle at the same vertex. Then
-<math>\textbf{(A)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may be regular}\qquad
+<math>\textbf{(A)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may be regular}\qquad \\
-\textbf{(B)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad
+\textbf{(B)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad \\
-\textbf{(C)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is regular}\qquad
+\textbf{(C)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is regular}\qquad \\
-\textbf{(D)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad
+\textbf{(D)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad \\
 \textbf{(E)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may or may not be regular}     </math>
@@ Line 379: / Line 380: @@
 == Problem 29==
-Five times <math>A</math>'s money added to <math>B</math>'s money is more than <math>\texdollar{51.00}</math>. Three times <math>A</math>'s money minus <math>B</math>'s money is <math>\textdollar{21.00}</math>.
+Five times <math>A</math>'s money added to <math>B</math>'s money is more than <math>\$51.00</math>. Three times <math>A</math>'s money minus <math>B</math>'s money is <math>\$21.00</math>.
 If <math>a</math> represents <math>A</math>'s money in dollars and <math>b</math> represents <math>B</math>'s money in dollars, then:
@@ Line 434: / Line 435: @@
 label("$P$",P,SW);</asy>
-<math>\textbf{(A)} AP^2 = PB \times AB\qquad
+<math>\textbf{(A)} AP^2 = PB \times AB\qquad \\
-\textbf{(B)}\ AP \times DO = PB \times AD\qquad
+\textbf{(B)}\ AP \times DO = PB \times AD\qquad \\
-\textbf{(C)}\ AB^2 = AD \times DE\qquad
+\textbf{(C)}\ AB^2 = AD \times DE\qquad \\
-\textbf{(D)}\ AB \times AD = OB \times AO\qquad
+\textbf{(D)}\ AB \times AD = OB \times AO\qquad \\
 \textbf{(E)}\ \text{none of these}     </math>
@@ Line 505: / Line 506: @@
 == Problem 37==
-The base of a triangle is of length <math>b</math>, and the latitude is of length <math>h</math>.
+The base of a triangle is of length <math>b</math>, and the altitude is of length <math>h</math>.
 A rectangle of height <math>x</math> is inscribed in the triangle with the base of the rectangle in the base of the triangle. The area of the rectangle is:
@@ Line 561: / Line 562: @@
 Given right <math>\triangle ABC</math> with legs <math>BC=3, AC=4</math>. Find the length of the shorter angle trisector from <math>C</math> to the hypotenuse:
-<math> \textbf{(A)}\ \frac{32\sqrt{3}-24}{13}\qquad\textbf{(B)}\ \frac{12\sqrt{3}-9}{13}\qquad\textbf{(C)}\ 6\sqrt{3}-8\qquad\textbf{(D)}\ \frac{5\sqrt{10}}{6}\qquad </math>
+<math> \textbf{(A)}\ \frac{32\sqrt{3}-24}{13}\qquad\textbf{(B)}\ \frac{12\sqrt{3}-9}{13}\qquad\textbf{(C)}\ 6\sqrt{3}-8\qquad\textbf{(D)}\ \frac{5\sqrt{10}}{6}\qquad\textbf{(E)}\ \frac{25}{12}\qquad</math>
 [[1960 AHSME Problems/Problem 40|Solution]]
 == See also ==
@@ Line 571: / Line 572: @@
 * [[Mathematics competition resources]]
-{{AHSME 40p box|year=1960|before=[[1959 AHSME]]|after=[[1961 AHSME]]}}
+{{AHSME 40p box|year=1960|before=[[1959 AHSME|1959 AHSC]]|after=[[1961 AHSME|1961 AHSC]]}}
 {{MAA Notice}}

1960 AHSC (Problems • Answer Key • Resources)
Preceded by 1959 AHSC	Followed by 1961 AHSC
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Difference between revisions of "1960 AHSME Problems"

Latest revision as of 20:27, 30 December 2020

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

See also