Difference between revisions of "1961 AHSME Problems"

Latest revision as of 17:33, 5 April 2023

1961 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.
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

Problem 1

When simplified, $(-\frac{1}{125})^{-2/3}$ becomes:

$\textbf{(A)}\ \frac{1}{25} \qquad \textbf{(B)}\ -\frac{1}{25} \qquad \textbf{(C)}\ 25\qquad \textbf{(D)}\ -25\qquad \textbf{(E)}\ 25\sqrt{-1}$

Solution

Problem 2

An automobile travels $a/6$ feet in $r$ seconds. If this rate is maintained for $3$ minutes, how many yards does it travel in $3$ minutes?

$\textbf{(A)}\ \frac{a}{1080r}\qquad \textbf{(B)}\ \frac{30r}{a}\qquad \textbf{(C)}\ \frac{30a}{r}\qquad \textbf{(D)}\ \frac{10r}{a}\qquad \textbf{(E)}\ \frac{10a}{r}$

Solution

Problem 3

If the graphs of $2y+x+3=0$ and $3y+ax+2=0$ are to meet at right angles, the value of $a$ is:

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

Solution

Problem 4

Let the set consisting of the squares of the positive integers be called $u$ ; thus $u$ is the set $1, 4, 9, 16 \ldots$ . If a certain operation on one or more members of the set always yields a member of the set, we say that the set is closed under that operation. Then $u$ is closed under:

$\textbf{(A)}\ \text{Addition}\qquad \textbf{(B)}\ \text{Multiplication} \qquad \textbf{(C)}\ \text{Division} \qquad\\ \textbf{(D)}\ \text{Extraction of a positive integral root}\qquad \textbf{(E)}\ \text{None of these}$

Solution

Problem 5

Let $S=(x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)+1$ . Then $S$ equals:

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

Solution

Problem 6

When simplified, $\log{8} \div \log{\frac{1}{8}}$ becomes:

$\textbf{(A)}\ 6\log{2} \qquad \textbf{(B)}\ \log{2} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 0\qquad \textbf{(E)}\ -1$

Solution

Problem 7

When simplified, the third term in the expansion of $(\frac{a}{\sqrt{x}}-\frac{\sqrt{x}}{a^2})^6$ is:

$\textbf{(A)}\ \frac{15}{x}\qquad \textbf{(B)}\ -\frac{15}{x}\qquad \textbf{(C)}\ -\frac{6x^2}{a^9} \qquad \textbf{(D)}\ \frac{20}{a^3}\qquad \textbf{(E)}\ -\frac{20}{a^3}$

Solution

Problem 8

Let the two base angles of a triangle be $A$ and $B$ , with $B$ larger than $A$ . The altitude to the base divides the vertex angle $C$ into two parts, $C_1$ and $C_2$ , with $C_2$ adjacent to side $a$ . Then:

$\textbf{(A)}\ C_1+C_2=A+B \qquad \textbf{(B)}\ C_1-C_2=B-A \qquad\\ \textbf{(C)}\ C_1-C_2=A-B \qquad \textbf{(D)}\ C_1+C_2=B-A\qquad \textbf{(E)}\ C_1-C_2=A+B$

Solution

Problem 9

Let $r$ be the result of doubling both the base and exponent of $a^b$ , and $b$ does not equal to $0$ . If $r$ equals the product of $a^b$ by $x^b$ , then $x$ equals:

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

Solution

Problem 10

Each side of $\triangle ABC$ is $12$ units. $D$ is the foot of the perpendicular dropped from $A$ on $BC$ , and $E$ is the midpoint of $AD$ . The length of $BE$ , in the same unit, is:

$\textbf{(A)}\ \sqrt{18} \qquad \textbf{(B)}\ \sqrt{28} \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ \sqrt{63} \qquad \textbf{(E)}\ \sqrt{98}$

Solution

Problem 11

Two tangents are drawn to a circle from an exterior point $A$ ; they touch the circle at points $B$ and $C$ respectively. A third tangent intersects segment $AB$ in $P$ and $AC$ in $R$ , and touches the circle at $Q$ . If $AB=20$ , then the perimeter of $\triangle APR$ is

$\textbf{(A)}\ 42\qquad \textbf{(B)}\ 40.5 \qquad \textbf{(C)}\ 40\qquad \textbf{(D)}\ 39\frac{7}{8} \qquad \textbf{(E)}\ \text{not determined by the given information}$

Solution

Problem 12

The first three terms of a geometric progression are $\sqrt{2}, \sqrt[3]{2}, \sqrt[6]{2}$ . Find the fourth term.

$\textbf{(A)}\ 1\qquad \textbf{(B)}\ \sqrt[7]{2}\qquad \textbf{(C)}\ \sqrt[8]{2}\qquad \textbf{(D)}\ \sqrt[9]{2}\qquad \textbf{(E)}\ \sqrt[10]{2}$

Solution

Problem 13

The symbol $|a|$ means $a$ is a positive number or zero, and $-a$ if $a$ is a negative number. For all real values of $t$ the expression $\sqrt{t^4+t^2}$ is equal to?

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

Solution

Problem 14

A rhombus is given with one diagonal twice the length of the other diagonal. Express the side of the rhombus is terms of $K$ , where $K$ is the area of the rhombus in square inches.

$\textbf{(A)}\ \sqrt{K}\qquad \textbf{(B)}\ \frac{1}{2}\sqrt{2K}\qquad \textbf{(C)}\ \frac{1}{3}\sqrt{3K}\qquad \textbf{(D)}\ \frac{1}{4}\sqrt{4K}\qquad \textbf{(E)}\ \text{None of these are correct}$

Solution

Problem 15

If $x$ men working $x$ hours a day for $x$ days produce $x$ articles, then the number of articles (not necessarily an integer) produced by $y$ men working $y$ hours a day for $y$ days is:

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

Solution

Problem 16

An altitude $h$ of a triangle is increased by a length $m$ . How much must be taken from the corresponding base $b$ so that the area of the new triangle is one-half that of the original triangle?

$\textbf{(A)}\ \frac{bm}{h+m}\qquad \textbf{(B)}\ \frac{bh}{2h+2m}\qquad \textbf{(C)}\ \frac{b(2m+h)}{m+h}\qquad \textbf{(D)}\ \frac{b(m+h)}{2m+h}\qquad \textbf{(E)}\ \frac{b(2m+h)}{2(h+m)}$

Solution

Problem 17

In the base ten number system the number $526$ means $5 \times 10^2+2 \times 10 + 6$ . In the Land of Mathesis, however, numbers are written in the base $r$ . Jones purchases an automobile there for $440$ monetary units (abbreviated m.u). He gives the salesman a $1000$ m.u bill, and receives, in change, $340$ m.u. The base $r$ is:

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 12$

Solution

Problem 18

The yearly changes in the population census of a town for four consecutive years are, respectively, 25% increase, 25% increase, 25% decrease, 25% decrease. The net change over the four years, to the nearest percent, is:

$\textbf{(A)}\ -12 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 12$

Solution

Problem 19

Consider the graphs of $y=2\log{x}$ and $y=\log{2x}$ . We may say that:

$\textbf{(A)}\ \text{They do not intersect}\qquad \\ \textbf{(B)}\ \text{They intersect at 1 point only}\qquad \\ \textbf{(C)}\ \text{They intersect at 2 points only} \qquad \\ \textbf{(D)}\ \text{They intersect at a finite number of points but greater than 2} \qquad \\ \textbf{(E)}\ \text{They coincide}$

Solution

Problem 20

The set of points satisfying the pair of inequalities $y>2x$ and $y>4-x$ is contained entirely in quadrants:

$\textbf{(A)}\ \text{I and II}\qquad \textbf{(B)}\ \text{II and III}\qquad \textbf{(C)}\ \text{I and III}\qquad \textbf{(D)}\ \text{III and IV}\qquad \textbf{(E)}\ \text{I and IV}$

Solution

Problem 21

Medians $AD$ and $CE$ of $\triangle ABC$ intersect in $M$ . The midpoint of $AE$ is $N$ . Let the area of $\triangle MNE$ be $k$ times the area of $\triangle ABC$ . Then $k$ equals:

$\textbf{(A)}\ \frac{1}{6}\qquad \textbf{(B)}\ \frac{1}{8}\qquad \textbf{(C)}\ \frac{1}{9}\qquad \textbf{(D)}\ \frac{1}{12}\qquad \textbf{(E)}\ \frac{1}{16}$

Solution

Problem 22

If $3x^3-9x^2+kx-12$ is divisible by $x-3$ , then it is also divisible by:

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

Solution

Problem 23

Points $P$ and $Q$ are both in the line segment $AB$ and on the same side of its midpoint. $P$ divides $AB$ in the ratio $2:3$ , and $Q$ divides $AB$ in the ratio $3:4$ . If $PQ=2$ , then the length of $AB$ is:

$\textbf{(A)}\ 60\qquad \textbf{(B)}\ 70\qquad \textbf{(C)}\ 75\qquad \textbf{(D)}\ 80\qquad \textbf{(E)}\ 85$

Solution

Problem 24

Thirty-one books are arranged from left to right in order of increasing prices. The price of each book differs by $\textdollar{2}$ from that of each adjacent book. For the price of the book at the extreme right a customer can buy the middle book and the adjacent one. Then:

$\textbf{(A)}\ \text{The adjacent book referred to is at the left of the middle book}\qquad \\ \textbf{(B)}\ \text{The middle book sells for \textdollar 36} \qquad \\ \textbf{(C)}\ \text{The cheapest book sells for \textdollar4} \qquad \\ \textbf{(D)}\ \text{The most expensive book sells for \textdollar64 } \qquad \\ \textbf{(E)}\ \text{None of these is correct }$

Solution

Problem 25

$\triangle ABC$ is isosceles with base $AC$ . Points $P$ and $Q$ are respectively in $CB$ and $AB$ and such that $AC=AP=PQ=QB$ . The number of degrees in $\angle B$ is:

$\textbf{(A)}\ 25\frac{5}{7}\qquad \textbf{(B)}\ 26\frac{1}{3}\qquad \textbf{(C)}\ 30\qquad \textbf{(D)}\ 40\qquad \textbf{(E)}\ \text{Not determined by the information given}$

Solution

Problem 26

For a given arithmetic series the sum of the first $50$ terms is $200$ , and the sum of the next $50$ terms is $2700$ . The first term in the series is:

$\textbf{(A)}\ -1221 \qquad \textbf{(B)}\ -21.5 \qquad \textbf{(C)}\ -20.5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 3.5$

Solution

Problem 27

Given two equiangular polygons $P_1$ and $P_2$ with different numbers of sides; each angle of $P_1$ is $x$ degrees and each angle of $P_2$ is $kx$ degrees, where $k$ is an integer greater than $1$ . The number of possibilities for the pair $(x, k)$ is:

$\textbf{(A)}\ \infty\qquad \textbf{(B)}\ \text{finite, but greater than 2}\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 0$

Solution

Problem 28

If $2137^{753}$ is multiplied out, the units' digit in the final product is:

$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$

Solution

Problem 29

Let the roots of $ax^2+bx+c=0$ be $r$ and $s$ . The equation with roots $ar+b$ and $as+b$ is:

$\textbf{(A)}\ x^2-bx-ac=0\qquad \textbf{(B)}\ x^2-bx+ac=0 \qquad\\ \textbf{(C)}\ x^2+3bx+ca+2b^2=0 \qquad \textbf{(D)}\ x^2+3bx-ca+2b^2=0 \qquad\\ \textbf{(E)}\ x^2+bx(2-a)+a^2c+b^2(a+1)=0$

Solution

Problem 30

If $\log_{10}2=a$ and $\log_{10}3=b$ , then $\log_{5}12=$ ?

$\textbf{(A)}\ \frac{a+b}{a+1}\qquad \textbf{(B)}\ \frac{2a+b}{a+1}\qquad \textbf{(C)}\ \frac{a+2b}{1+a}\qquad \textbf{(D)}\ \frac{2a+b}{1-a}\qquad \textbf{(E)}\ \frac{a+2b}{1-a}$

Solution

Problem 31

In $\triangle ABC$ the ratio $AC:CB$ is $3:4$ . The bisector of the exterior angle at $C$ intersects $BA$ extended at $P$ ( $A$ is between $P$ and $B$ ). The ratio $PA:AB$ is:

$\textbf{(A)}\ 1:3 \qquad \textbf{(B)}\ 3:4 \qquad \textbf{(C)}\ 4:3 \qquad \textbf{(D)}\ 3:1 \qquad \textbf{(E)}\ 7:1$

Solution

Problem 32

A regular polygon of $n$ sides is inscribed in a circle of radius $R$ . The area of the polygon is $3R^2$ . Then $n$ equals:

$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 15\qquad \textbf{(E)}\ 18$

Solution

Problem 33

The number of solutions of $2^{2x}-3^{2y}=55$ , in which $x$ and $y$ are integers, is:

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ \text{More than three, but finite}$

Solution

Problem 34

Let S be the set of values assumed by the fraction $\frac{2x+3}{x+2}$ . When $x$ is any member of the interval $x \ge 0$ . If there exists a number $M$ such that no number of the set $S$ is greater than $M$ , then $M$ is an upper bound of $S$ . If there exists a number $m$ such that such that no number of the set $S$ is less than $m$ , then $m$ is a lower bound of $S$ . We may then say:

$\textbf{(A)}\ \text{m is in S, but M is not in S}\qquad\\ \textbf{(B)}\ \text{M is in S, but m is not in S}\qquad\\ \textbf{(C)}\ \text{Both m and M are in S}\qquad\\ \textbf{(D)}\ \text{Neither m nor M are in S}\qquad\\ \textbf{(E)}\ \text{M does not exist either in or outside S}$

Solution

Problem 35

The number $695$ is to be written with a factorial base of numeration, that is, $695=a_1+a_2\times2!+a_3\times3!+ \ldots a_n \times n!$ where $a_1, a_2, a_3 ... a_n$ are integers such that $0 \le a_k \le k,$ and $n!$ means $n(n-1)(n-2)...2 \times 1$ . Find $a_4$

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4$

Solution

Problem 36

In $\triangle ABC$ the median from $A$ is given perpendicular to the median from $B$ . If $BC=7$ and $AC=6$ , find the length of $AB$ .

$\textbf{(A)}\ 4\qquad \textbf{(B)}\ \sqrt{17} \qquad \textbf{(C)}\ 4.25\qquad \textbf{(D)}\ 2\sqrt{5} \qquad \textbf{(E)}\ 4.5$

Solution

Problem 37

In racing over a distance $d$ at uniform speed, $A$ can beat $B$ by $20$ yards, $B$ can beat $C$ by $10$ yards, and $A$ can beat $C$ by $28$ yards. Then $d$ , in yards, equals:

$\textbf{(A)}\ \text{Not determined by the given information}\qquad \textbf{(B)}\ 58\qquad \textbf{(C)}\ 100\qquad \textbf{(D)}\ 116\qquad \textbf{(E)}\ 120$

Solution

Problem 38

$\triangle ABC$ is inscribed in a semicircle of radius $r$ so that its base $AB$ coincides with diameter $AB$ . Point $C$ does not coincide with either $A$ or $B$ . Let $s=AC+BC$ . Then, for all permissible positions of $C$ :

$\textbf{(A)}\ s^2\le8r^2\qquad \textbf{(B)}\ s^2=8r^2 \qquad \textbf{(C)}\ s^2 \ge 8r^2 \qquad\\ \textbf{(D)}\ s^2\le4r^2 \qquad \textbf{(E)}\ s^2=4r^2$

Solution

Problem 39

Any five points are taken inside or on a square with side length $1$ . Let a be the smallest possible number with the property that it is always possible to select one pair of points from these five such that the distance between them is equal to or less than $a$ . Then $a$ is:

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

Solution

Problem 40

Find the minimum value of $\sqrt{x^2+y^2}$ if $5x+12y=60$ .

$\textbf{(A)}\ \frac{60}{13}\qquad \textbf{(B)}\ \frac{13}{5}\qquad \textbf{(C)}\ \frac{13}{12}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 0$

Solution

1961 AHSC (Problems • Answer Key • Resources)
Preceded by 1960 AHSC	Followed by 1962 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
All AHSME Problems and Solutions

@@ Line 1: / Line 1: @@
+{{AHSC 40 Problems
+|year = 1961
+}}
 == Problem 1==
@@ Line 6: / Line 9: @@
 \textbf{(B)}\ -\frac{1}{25} \qquad
 \textbf{(C)}\ 25\qquad
-\textbf{(D)}\ -25}\qquad
+\textbf{(D)}\ -25\qquad
-\textbf{(E)}\ 25\sqrt{-1}} </math>
+\textbf{(E)}\ 25\sqrt{-1}</math>
 [[1961 AHSME Problems/Problem 1|Solution]]
@@ Line 18: / Line 21: @@
 \textbf{(B)}\ \frac{30r}{a}\qquad
 \textbf{(C)}\ \frac{30a}{r}\qquad
-\textbf{(D)}\ \frac{10r}{a}}\qquad
+\textbf{(D)}\ \frac{10r}{a}\qquad
-\textbf{(E)}\ \frac{10a}{r}}     </math>
+\textbf{(E)}\ \frac{10a}{r}     </math>
 [[1961 AHSME Problems/Problem 2|Solution]]
@@ Line 30: / Line 33: @@
 \textbf{(B)}\ -\frac{2}{3}\qquad
 \textbf{(C)}\ -\frac{3}{2} \qquad
-\textbf{(D)}\ 6}\qquad
+\textbf{(D)}\ 6\qquad
-\textbf{(E)}\ -6}  </math>
+\textbf{(E)}\ -6 </math>
 [[1961 AHSME Problems/Problem 3|Solution]]
@@ Line 44: / Line 47: @@
 \textbf{(B)}\ \text{Multiplication} \qquad
 \textbf{(C)}\ \text{Division} \qquad\\
-\textbf{(D)}\ \text{Extraction of a positive integral root}}\qquad
+\textbf{(D)}\ \text{Extraction of a positive integral root}\qquad
-\textbf{(E)}\text{None of these}}    </math>
+\textbf{(E)}\ \text{None of these}  </math>
 [[1961 AHSME Problems/Problem 4|Solution]]
@@ Line 56: / Line 59: @@
 \textbf{(B)}\ (x-1)^4 \qquad
 \textbf{(C)}\ x^4 \qquad
-\textbf{(D)}\ (x+1)^4 }\qquad
+\textbf{(D)}\ (x+1)^4 \qquad
-\textbf{(E)}\ x^4+1}   </math>
+\textbf{(E)}\ x^4+1 </math>
 [[1961 AHSME Problems/Problem 5|Solution]]
@@ Line 68: / Line 71: @@
 \textbf{(B)}\ \log{2} \qquad
 \textbf{(C)}\ 1 \qquad
-\textbf{(D)}\ 0}\qquad
+\textbf{(D)}\ 0\qquad
-\textbf{(E)}\ -1}}  </math>
+\textbf{(E)}\ -1 </math>
 [[1961 AHSME Problems/Problem 6|Solution]]
@@ Line 80: / Line 83: @@
 \textbf{(B)}\ -\frac{15}{x}\qquad
 \textbf{(C)}\ -\frac{6x^2}{a^9} \qquad
-\textbf{(D)}\ \frac{20}{a^3}}\qquad
+\textbf{(D)}\ \frac{20}{a^3}\qquad
-\textbf{(E)}\ -\frac{20}{a^3}} </math>
+\textbf{(E)}\ -\frac{20}{a^3}</math>
 [[1961 AHSME Problems/Problem 7|Solution]]
@@ Line 91: / Line 94: @@
 <math>\textbf{(A)}\ C_1+C_2=A+B \qquad
-\textbf{(B)}\ C_1-C_2=B-A \qquad
+\textbf{(B)}\ C_1-C_2=B-A \qquad\\
-\textbf{(C)}\ C_1-C_2=A-B} \qquad
+\textbf{(C)}\ C_1-C_2=A-B \qquad
-\textbf{(D)}\ C_1+C_2=B-A}\qquad
+\textbf{(D)}\ C_1+C_2=B-A\qquad
-\textbf{(E)}\ C_1-C_2=A+B} </math>
+\textbf{(E)}\ C_1-C_2=A+B</math>
 [[1961 AHSME Problems/Problem 8|Solution]]
@@ Line 103: / Line 106: @@
 If <math>r</math> equals the product of <math>a^b</math> by <math>x^b</math>, then <math>x</math> equals:
-<math>\textbf{(A)} a\qquad
+<math>\textbf{(A)}\ a \qquad
 \textbf{(B)}\ 2a \qquad
 \textbf{(C)}\ 4a \qquad
-\textbf{(D)}\ 2}\qquad
+\textbf{(D)}\ 2\qquad
-\textbf{(E)}\ 4}   </math>
+\textbf{(E)}\ 4   </math>
 [[1961 AHSME Problems/Problem 9|Solution]]
@@ Line 119: / Line 122: @@
 \textbf{(B)}\ \sqrt{28} \qquad
 \textbf{(C)}\ 6 \qquad
-\textbf{(D)}\ \sqrt{63} }\qquad
+\textbf{(D)}\ \sqrt{63} \qquad
-\textbf{(E)}\ \sqrt{98}} </math>
+\textbf{(E)}\ \sqrt{98}</math>
 [[1961 AHSME Problems/Problem 10|Solution]]
@@ Line 132: / Line 135: @@
 \textbf{(B)}\ 40.5 \qquad
 \textbf{(C)}\ 40\qquad
-\textbf{(D)}\ 39\frac{7}{8} }\qquad
+\textbf{(D)}\ 39\frac{7}{8} \qquad
-\textbf{(E)}\ \text{not determined by the given information}} </math>
+\textbf{(E)}\ \text{not determined by the given information}</math>
 [[1961 AHSME Problems/Problem 11|Solution]]
@@ Line 144: / Line 147: @@
 \textbf{(B)}\ \sqrt[7]{2}\qquad
 \textbf{(C)}\ \sqrt[8]{2}\qquad
-\textbf{(D)}\ \sqrt[9]{2}}\qquad
+\textbf{(D)}\ \sqrt[9]{2}\qquad
-\textbf{(E)}\ \sqrt[10]{2}} </math>
+\textbf{(E)}\ \sqrt[10]{2}</math>
 [[1961 AHSME Problems/Problem 12|Solution]]
@@ Line 157: / Line 160: @@
 \textbf{(B)}\ t^2+t\qquad
 \textbf{(C)}\ |t^2+t|\qquad
-\textbf{(D)}\ t\sqrt{t^2+1}}\qquad
+\textbf{(D)}\ t\sqrt{t^2+1}\qquad
-\textbf{(E)}\ |t|\sqrt{1+t^2}} </math>
+\textbf{(E)}\ |t|\sqrt{1+t^2}</math>
 [[1961 AHSME Problems/Problem 13|Solution]]
@@ Line 170: / Line 173: @@
 \textbf{(B)}\ \frac{1}{2}\sqrt{2K}\qquad
 \textbf{(C)}\ \frac{1}{3}\sqrt{3K}\qquad
-\textbf{(D)}\ \frac{1}{4}\sqrt{4K}}\qquad
+\textbf{(D)}\ \frac{1}{4}\sqrt{4K}\qquad
-\textbf{(E)}\ \text{None of these are correct}} </math>
+\textbf{(E)}\ \text{None of these are correct}</math>
 [[1961 AHSME Problems/Problem 14|Solution]]
@@ Line 183: / Line 186: @@
 \textbf{(B)}\ \frac{y^3}{x^2}\qquad
 \textbf{(C)}\ \frac{x^2}{y^3}\qquad
-\textbf{(D)}\ \frac{y^2}{x^3}}\qquad
+\textbf{(D)}\ \frac{y^2}{x^3}\qquad
-\textbf{(E)}\ y} </math>
+\textbf{(E)}\ y</math>
 [[1961 AHSME Problems/Problem 15|Solution]]
@@ Line 196: / Line 199: @@
 \textbf{(B)}\ \frac{bh}{2h+2m}\qquad
 \textbf{(C)}\ \frac{b(2m+h)}{m+h}\qquad
-\textbf{(D)}\ \frac{b(m+h)}{2m+h}}\qquad
+\textbf{(D)}\ \frac{b(m+h)}{2m+h}\qquad
-\textbf{(E)}\ \frac{b(2m+h)}{2(h+m)}} </math>
+\textbf{(E)}\ \frac{b(2m+h)}{2(h+m)}</math>
 [[1961 AHSME Problems/Problem 16|Solution]]
@@ Line 205: / Line 208: @@
 In the base ten number system the number <math>526</math> means <math>5 \times 10^2+2 \times 10 + 6</math>.
 In the Land of Mathesis, however, numbers are written in the base <math>r</math>.
-King Rusczyk purchases an automobile there for <math>440</math> monetary units (abbreviated m.u).
+Jones purchases an automobile there for <math>440</math> monetary units (abbreviated m.u).
 He gives the salesman a <math>1000</math> m.u bill, and receives, in change, <math>340</math> m.u. The base <math>r</math> is:
@@ Line 211: / Line 214: @@
 \textbf{(B)}\ 5\qquad
 \textbf{(C)}\ 7\qquad
-\textbf{(D)}\ 8}\qquad
+\textbf{(D)}\ 8\qquad
-\textbf{(E)}\ 12}     </math>
+\textbf{(E)}\ 12     </math>
 [[1961 AHSME Problems/Problem 17|Solution]]
@@ Line 225: / Line 228: @@
 \textbf{(B)}\ -1 \qquad
 \textbf{(C)}\ 0 \qquad
-\textbf{(D)}\ 1}\qquad
+\textbf{(D)}\ 1\qquad
-\textbf{(E)}\ 12}     </math>
+\textbf{(E)}\ 12    </math>
 [[1961 AHSME Problems/Problem 18|Solution]]
@@ Line 234: / Line 237: @@
 Consider the graphs of <math>y=2\log{x}</math> and <math>y=\log{2x}</math>. We may say that:
-<math>\textbf{(A)}\ \text{They do not intersect}\qquad\\
+<math>\textbf{(A)}\ \text{They do not intersect}\qquad \\
-\textbf{(B)}\ \text{They intersect at 1 point only}\qquad\\
+\textbf{(B)}\ \text{They intersect at 1 point only}\qquad \\
-\textbf{(C)}\ \text{They intersect at 2 points only} \qquad\\
+\textbf{(C)}\ \text{They intersect at 2 points only} \qquad \\
-\textbf{(D)}\ \text{They intersect at a finite number of points but greater than 2 }\qquad\\
+\textbf{(D)}\ \text{They intersect at a finite number of points but greater than 2} \qquad \\
-\textbf{(E)}\ \text{They coincide} }   </math>
+\textbf{(E)}\ \text{They coincide}   </math>
 [[1961 AHSME Problems/Problem 19|Solution]]
@@ Line 249: / Line 252: @@
 \textbf{(B)}\ \text{II and III}\qquad
 \textbf{(C)}\ \text{I and III}\qquad
-\textbf{(D)}\ \text{III and IV}}\qquad
+\textbf{(D)}\ \text{III and IV}\qquad
-\textbf{(E)}\ \text{I and IV}} </math>
+\textbf{(E)}\ \text{I and IV}</math>
 [[1961 AHSME Problems/Problem 20|Solution]]
@@ Line 262: / Line 265: @@
 \textbf{(B)}\ \frac{1}{8}\qquad
 \textbf{(C)}\ \frac{1}{9}\qquad
-\textbf{(D)}\ \frac{1}{12}}\qquad
+\textbf{(D)}\ \frac{1}{12}\qquad
-\textbf{(E)}\ \frac{1}{16}} </math>
+\textbf{(E)}\ \frac{1}{16}</math>
 [[1961 AHSME Problems/Problem 21|Solution]]
@@ Line 274: / Line 277: @@
 \textbf{(B)}\ 3x^2-4\qquad
 \textbf{(C)}\ 3x^2+4\qquad
-\textbf{(D)}\ 3x-4 }\qquad
+\textbf{(D)}\ 3x-4 \qquad
-\textbf{(E)}\ 3x+4 } </math>
+\textbf{(E)}\ 3x+4 </math>
 [[1961 AHSME Problems/Problem 22|Solution]]
@@ Line 287: / Line 290: @@
 \textbf{(B)}\ 70\qquad
 \textbf{(C)}\ 75\qquad
-\textbf{(D)}\ 80}\qquad
+\textbf{(D)}\ 80\qquad
-\textbf{(E)}\ 85}     </math>
+\textbf{(E)}\ 85    </math>
 [[1961 AHSME Problems/Problem 23|Solution]]
@@ Line 295: / Line 298: @@
 Thirty-one books are arranged from left to right in order of increasing prices.
-The price of each book differs by &#036;2 from that of each adjacent book.
+The price of each book differs by <math>\textdollar{2}</math> from that of each adjacent book.
 For the price of the book at the extreme right a customer can buy the middle book and the adjacent one. Then:
-<math>\textbf{(A)}\ \text{The adjacent book referred to is at the left of the middle book}\qquad
+<math>\textbf{(A)}\ \text{The adjacent book referred to is at the left of the middle book}\qquad \\
-\textbf{(B)}\ \text{The middle book sells for &#036;36 }\qquad
+\textbf{(B)}\ \text{The middle book sells for \textdollar 36} \qquad \\
-\textbf{(C)}\ \text{The cheapest book sells for &#036;4 }\qquad
+\textbf{(C)}\ \text{The cheapest book sells for \textdollar4} \qquad \\
-\textbf{(D)}\ \text{The most expensive book sells for &#036;64 }\qquad
+\textbf{(D)}\ \text{The most expensive book sells for \textdollar64 } \qquad \\
 \textbf{(E)}\ \text{None of these is correct }  </math>
@@ Line 314: / Line 317: @@
 \textbf{(B)}\ 26\frac{1}{3}\qquad
 \textbf{(C)}\ 30\qquad
-\textbf{(D)}\ 40}\qquad
+\textbf{(D)}\ 40\qquad
-\textbf{(E)}\ \text{Not determined by the information given}} </math>
+\textbf{(E)}\ \text{Not determined by the information given}</math>
 [[1961 AHSME Problems/Problem 25|Solution]]
@@ Line 327: / Line 330: @@
 \textbf{(B)}\ -21.5 \qquad
 \textbf{(C)}\ -20.5 \qquad
-\textbf{(D)}\ 3 }\qquad
+\textbf{(D)}\ 3 \qquad
-\textbf{(E)}\ 3.5 }    </math>
+\textbf{(E)}\ 3.5    </math>
 [[1961 AHSME Problems/Problem 26|Solution]]
@@ Line 349: / Line 352: @@
 == Problem 28==
-If <math>2137^{753}</math> is multiplied out, the units' digit in the final product in the final product is:
+If <math>2137^{753}</math> is multiplied out, the units' digit in the final product is:
 <math>\textbf{(A)}\ 1\qquad
 \textbf{(B)}\ 3\qquad
 \textbf{(C)}\ 5\qquad
-\textbf{(D)}\ 7}\qquad
+\textbf{(D)}\ 7\qquad
-\textbf{(E)}\ 9} </math>
+\textbf{(E)}\ 9</math>
 [[1961 AHSME Problems/Problem 28|Solution]]
@@ Line 364: / Line 367: @@
 <math>\textbf{(A)}\ x^2-bx-ac=0\qquad
-\textbf{(B)}\ x^2-bx+ac=0 \qquad
+\textbf{(B)}\ x^2-bx+ac=0 \qquad\\
 \textbf{(C)}\ x^2+3bx+ca+2b^2=0 \qquad
-\textbf{(D)}\ x^2+3bx-ca+2b^2=0 }\qquad
+\textbf{(D)}\ x^2+3bx-ca+2b^2=0 \qquad\\
-\textbf{(E)}\ x^2+bx(2-a)+a^2c+b^2(a+1)=0} </math>
+\textbf{(E)}\ x^2+bx(2-a)+a^2c+b^2(a+1)=0</math>
 [[1961 AHSME Problems/Problem 29|Solution]]
@@ Line 378: / Line 381: @@
 \textbf{(B)}\ \frac{2a+b}{a+1}\qquad
 \textbf{(C)}\ \frac{a+2b}{1+a}\qquad
-\textbf{(D)}\ \frac{2a+b}{1-a}}\qquad
+\textbf{(D)}\ \frac{2a+b}{1-a}\qquad
-\textbf{(E)}\ \frac{a+2b}{1-a}} </math>
+\textbf{(E)}\ \frac{a+2b}{1-a}</math>
 [[1961 AHSME Problems/Problem 30|Solution]]
@@ Line 391: / Line 394: @@
 \textbf{(B)}\ 3:4 \qquad
 \textbf{(C)}\ 4:3 \qquad
-\textbf{(D)}\ 3:1 }\qquad
+\textbf{(D)}\ 3:1 \qquad
-\textbf{(E)}\ 7:1 }
+\textbf{(E)}\ 7:1      </math>
 [[1961 AHSME Problems/Problem 31|Solution]]
@@ Line 398: / Line 401: @@
 == Problem 32==
-A regular polygon of </math>n<math> sides is inscribed in a circle of radius </math>R<math>. The area of the polygon is </math>3R^2<math>. Then </math>n<math> equals:
+A regular polygon of <math>n</math> sides is inscribed in a circle of radius <math>R</math>. The area of the polygon is <math>3R^2</math>. Then <math>n</math> equals:
-</math>\textbf{(A)}\ 8\qquad
+<math>\textbf{(A)}\ 8\qquad
 \textbf{(B)}\ 10\qquad
 \textbf{(C)}\ 12\qquad
-\textbf{(D)}\ 15}\qquad
+\textbf{(D)}\ 15\qquad
-\textbf{(E)}\ 18}  <math>
+\textbf{(E)}\ 18 </math>
 [[1961 AHSME Problems/Problem 32|Solution]]
@@ Line 410: / Line 413: @@
 == Problem 33==
-The number of solutions of </math>2^{2x}-3^{2y}=55<math>, in which </math>x<math> and </math>y<math> are integers, is:
+The number of solutions of <math>2^{2x}-3^{2y}=55</math>, in which <math>x</math> and <math>y</math> are integers, is:
-</math>\textbf{(A)}\ 0\qquad
+<math>\textbf{(A)}\ 0\qquad
 \textbf{(B)}\ 1\qquad
 \textbf{(C)}\ 2\qquad
-\textbf{(D)}\ 3}\qquad
+\textbf{(D)}\ 3\qquad
-\textbf{(E)}\ \text{More than three, but finite}} <math>
+\textbf{(E)}\ \text{More than three, but finite}</math>
 [[1961 AHSME Problems/Problem 33|Solution]]
@@ Line 422: / Line 425: @@
 == Problem 34==
-Let S be the set of values assumed by the fraction </math>\frac{2x+3}{x+2}<math>.
+Let S be the set of values assumed by the fraction <math>\frac{2x+3}{x+2}</math>.
-When </math>x<math> is any member of the interval </math>x \ge 0<math>. If there exists a number </math>M<math> such that no number of the set </math>S<math> is greater than </math>M<math>,
+When <math>x</math> is any member of the interval <math>x \ge 0</math>. If there exists a number <math>M</math> such that no number of the set <math>S</math> is greater than <math>M</math>,
-then </math>M<math> is an upper bound of </math>S<math>. If there exists a number </math>m<math> such that such that no number of the set </math>S<math> is less than </math>m<math>,
+then <math>M</math> is an upper bound of <math>S</math>. If there exists a number <math>m</math> such that such that no number of the set <math>S</math> is less than <math>m</math>,
-then </math>m<math> is a lower bound of </math>S<math>. We may then say:
+then <math>m</math> is a lower bound of <math>S</math>. We may then say:
-</math>\textbf{(A)}\ \text{m is in S, but M is not in S}\qquad\\
+<math>\textbf{(A)}\ \text{m is in S, but M is not in S}\qquad\\
 \textbf{(B)}\ \text{M is in S, but m is not in S}\qquad\\
 \textbf{(C)}\ \text{Both m and M are in S}\qquad\\
 \textbf{(D)}\ \text{Neither m nor M are in S}\qquad\\
-\textbf{(E)}\ \text{M does not exist either in or outside S}    <math>
+\textbf{(E)}\ \text{M does not exist either in or outside S}  </math>
 [[1961 AHSME Problems/Problem 34|Solution]]
@@ Line 437: / Line 440: @@
 == Problem 35==
-The number </math>695<math> is to be written with a factorial base of numeration, that is, </math>695=a_1+a_2\times2!+a_3\times3!+ \ldots a_n \times n!<math>
+The number <math>695</math> is to be written with a factorial base of numeration, that is, <math>695=a_1+a_2\times2!+a_3\times3!+ \ldots a_n \times n!</math>
-where </math>a_1, a_2, a_3 ... a_n<math> are integers such that </math>0 \le a_k \le k,<math> and </math>n!<math> means </math>n(n-1)(n-2)...2 \times 1<math>. Find </math>a_4<math>
+where <math>a_1, a_2, a_3 ... a_n</math> are integers such that <math>0 \le a_k \le k,</math> and <math>n!</math> means <math>n(n-1)(n-2)...2 \times 1</math>. Find <math>a_4</math>
-</math>\textbf{(A)}\ 0\qquad
+<math>\textbf{(A)}\ 0\qquad
 \textbf{(B)}\ 1\qquad
 \textbf{(C)}\ 2\qquad
-\textbf{(D)}\ 3}\qquad
+\textbf{(D)}\ 3\qquad
-\textbf{(E)}\ 4}     <math>
+\textbf{(E)}\ 4     </math>
 [[1961 AHSME Problems/Problem 35|Solution]]
@@ Line 450: / Line 453: @@
 == Problem 36==
-In </math>\triangle ABC<math> the median from </math>A<math> is given perpendicular to the median from </math>B<math>. If </math>BC=7<math> and </math>AC=6<math>, find the length of </math>AB<math>.
+In <math>\triangle ABC</math> the median from <math>A</math> is given perpendicular to the median from <math>B</math>. If <math>BC=7</math> and <math>AC=6</math>, find the length of <math>AB</math>.
-</math>\textbf{(A)}\ 4\qquad
+<math>\textbf{(A)}\ 4\qquad
 \textbf{(B)}\ \sqrt{17} \qquad
 \textbf{(C)}\ 4.25\qquad
-\textbf{(D)}\ 2\sqrt{5} }\qquad
+\textbf{(D)}\ 2\sqrt{5} \qquad
-\textbf{(E)}\ 4.5}  <math>
+\textbf{(E)}\ 4.5  </math>
 [[1961 AHSME Problems/Problem 36|Solution]]
@@ Line 462: / Line 465: @@
 == Problem 37==
-In racing over a distance </math>d<math> at uniform speed, </math>A<math> can beat </math>B<math> by </math>20<math> yards, </math>B<math> can beat </math>C<math> by </math>10<math> yards,
+In racing over a distance <math>d</math> at uniform speed, <math>A</math> can beat <math>B</math> by <math>20</math> yards, <math>B</math> can beat <math>C</math> by <math>10</math> yards,
-and </math>A<math> can beat </math>C<math> by </math>28<math> yards. Then </math>d<math>, in yards, equals:
+and <math>A</math> can beat <math>C</math> by <math>28</math> yards. Then <math>d</math>, in yards, equals:
-</math>\textbf{(A)}\ \text{Not determined by the given information}\qquad
+<math>\textbf{(A)}\ \text{Not determined by the given information}\qquad
 \textbf{(B)}\ 58\qquad
 \textbf{(C)}\ 100\qquad
-\textbf{(D)}\ 116}\qquad
+\textbf{(D)}\ 116\qquad
-\textbf{(E)}\ 120} <math>
+\textbf{(E)}\ 120</math>
 [[1961 AHSME Problems/Problem 37|Solution]]
@@ Line 475: / Line 478: @@
 == Problem 38==
-</math>\triangle ABC<math> is inscribed in a semicircle of radius </math>r<math> so that its base </math>AB<math> coincides with diameter </math>AB<math>.
+<math>\triangle ABC</math> is inscribed in a semicircle of radius <math>r</math> so that its base <math>AB</math> coincides with diameter <math>AB</math>.
-Point </math>C<math> does not coincide with either </math>A<math> or </math>B<math>. Let </math>s=AC+BC<math>. Then, for all permissible positions of </math>C<math>:
+Point <math>C</math> does not coincide with either <math>A</math> or <math>B</math>. Let <math>s=AC+BC</math>. Then, for all permissible positions of <math>C</math>:
-</math>\textbf{(A)}\ s^2\le8r^2\qquad
+<math>\textbf{(A)}\ s^2\le8r^2\qquad
 \textbf{(B)}\ s^2=8r^2 \qquad
-\textbf{(C)}\ s^2 \ge 8r^2 \qquad
+\textbf{(C)}\ s^2 \ge 8r^2 \qquad\\
-\textbf{(D)}\ s^2\le4r^2 }\qquad
+\textbf{(D)}\ s^2\le4r^2 \qquad
-\textbf{(E)}\ x^2=4r^2 }  <math>
+\textbf{(E)}\ s^2=4r^2   </math>
 [[1961 AHSME Problems/Problem 38|Solution]]
@@ Line 488: / Line 491: @@
 == Problem 39==
-Any five points are taken inside or on a square with side length </math>1<math>. Let a be the smallest possible number with the
+Any five points are taken inside or on a square with side length <math>1</math>. Let a be the smallest possible number with the
 property that it is always possible to select one pair of points from these five such that the distance between them
-is equal to or less than </math>a<math>. Then </math>a<math> is:
+is equal to or less than <math>a</math>. Then <math>a</math> is:
-</math>\textbf{(A)}\ \sqrt{3}/3\qquad
+<math>\textbf{(A)}\ \sqrt{3}/3\qquad
 \textbf{(B)}\ \sqrt{2}/2\qquad
 \textbf{(C)}\ 2\sqrt{2}/3\qquad
-\textbf{(D)}\ 1 }\qquad
+\textbf{(D)}\ 1 \qquad
-\textbf{(E)}\ \sqrt{2}} <math>
+\textbf{(E)}\ \sqrt{2}</math>
 [[1961 AHSME Problems/Problem 39|Solution]]
@@ Line 502: / Line 505: @@
 == Problem 40==
-Find the minimum value of </math>\sqrt{x^2+y^2}<math> if </math>5x+12y=60<math>.
+Find the minimum value of <math>\sqrt{x^2+y^2}</math> if <math>5x+12y=60</math>.
-</math>\textbf{(A)}\ \frac{60}{13}\qquad
+<math>\textbf{(A)}\ \frac{60}{13}\qquad
 \textbf{(B)}\ \frac{13}{5}\qquad
 \textbf{(C)}\ \frac{13}{12}\qquad
-\textbf{(D)}\ 1}\qquad
+\textbf{(D)}\ 1\qquad
-\textbf{(E)}\ 0 } $
+\textbf{(E)}\ 0 </math>
 [[1961 AHSME Problems/Problem 40|Solution]]
 == See also ==
-* [[AHSME]]
-* [[AHSME Problems and Solutions]]
 * [[AMC 12 Problems and Solutions]]
 * [[Mathematics competition resources]]
+{{AHSME 40p box|year=1961|before=[[1960 AHSME|1960 AHSC]]|after=[[1962 AHSME|1962 AHSC]]}}
 {{MAA Notice}}

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Difference between revisions of "1961 AHSME Problems"

Latest revision as of 17:33, 5 April 2023

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