Difference between revisions of "1971 AHSME Problems"

Latest revision as of 18:08, 6 August 2024

1971 AHSC (Answer Key) Printable versions: Wiki • AoPS Resources • PDF
Instructions This is a 35-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 number of digits in the number $N=2^{12}\times 5^8$ is

$\textbf{(A) }9\qquad \textbf{(B) }10\qquad \textbf{(C) }11\qquad \textbf{(D) }12\qquad \textbf{(E) }20$

Solution

Problem 2

If $b$ men take $c$ days to lay $f$ bricks, then the number of days it will take $c$ men working at the same rate to lay $b$ bricks, is

$\textbf{(A) }fb^2\qquad \textbf{(B) }b/f^2\qquad \textbf{(C) }f^2/b\qquad \textbf{(D) }b^2/f\qquad \textbf{(E) }f/b^2$

Solution

Problem 3

If the point $(x,-4)$ lies on the straight line joining the points $(0,8)$ and $(-4,0)$ in the $xy$ -plane, then $x$ is equal to

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

Solution

Problem 4

After simple interest for two months at $5$ % per annum was credited, a Boy Scout Troop had a total of $\textdollar 255.31$ in the Council Treasury. The interest credited was a number of dollars plus the following number of cents

$\textbf{(A) }11\qquad \textbf{(B) }12\qquad \textbf{(C) }13\qquad \textbf{(D) }21\qquad \textbf{(E) }31$

Solution

Problem 5

Points $A,B,Q,D$ , and $C$ lie on the circle shown and the measures of arcs $\widehat{BQ}$ and $\widehat{QD}$ are $42^\circ$ and $38^\circ$ respectively. The sum of the measures of angles $P$ and $Q$ is

$\textbf{(A) }80^\circ\qquad \textbf{(B) }62^\circ\qquad \textbf{(C) }40^\circ\qquad \textbf{(D) }46^\circ\qquad \textbf{(E) }\text{None of these}$

$[asy] size(3inch); draw(Circle((1,0),1)); pair A, B, C, D, P, Q; P = (-2,0); B=(sqrt(2)/2+1,sqrt(2)/2); D=(sqrt(2)/2+1,-sqrt(2)/2); Q = (2,0); A = intersectionpoints(Circle((1,0),1),B--P)[1]; C = intersectionpoints(Circle((1,0),1),D--P)[0]; draw(B--P--D); draw(A--Q--C); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SW); label("$D$",D,SE); label("$P$",P,W); label("$Q$",Q,E); //Credit to chezbgone2 for the diagram[/asy]$

Solution

Problem 6

Let $\ast$ be the symbol denoting the binary operation on the set $S$ of all non-zero real numbers as follows: For any two numbers $a$ and $b$ of $S$ , $a\ast b=2ab$ . Then the one of the following statements which is not true, is

$\textbf{(A) }\ast\text{ is commutative over }S \qquad \textbf{(B) }\ast\text{ is associative over }S\qquad \\ \textbf{(C) }\frac{1}{2}\text{ is an identity element for }\ast\text{ in }S\qquad \textbf{(D) }\text{Every element of }S\text{ has an inverse for }\ast\qquad \textbf{(E) }\dfrac{1}{2a}\text{ is an inverse for }\ast\text{ of the element }a\text{ of }S$

Solution

Problem 7

$2^{-(2k+1)}-2^{-(2k-1)}+2^{-2k}$ is equal to

$\textbf{(A) }2^{-2k}\qquad \textbf{(B) }2^{-(2k-1)}\qquad \textbf{(C) }-2^{-(2k+1)}\qquad \textbf{(D) }0\qquad \textbf{(E) }2$

Solution

Problem 8

The solution set of $6x^2+5x<4$ is the set of all values of $x$ such that

$\textbf{(A) }\textstyle -2<x<1\qquad \textbf{(B) }-\frac{4}{3}<x<\frac{1}{2}\qquad \textbf{(C) }-\frac{1}{2}<x<\frac{4}{3}\qquad \\ \textbf{(D) }x<\textstyle\frac{1}{2}\text{ or }x>-\frac{4}{3}\qquad \textbf{(E) }x<-\frac{4}{3}\text{ or }x>\frac{1}{2}$

Solution

Problem 9

An uncrossed belt is fitted without slack around two circular pulleys with radii of $14$ inches and $4$ inches. If the distance between the points of contact of the belt with the pulleys is $24$ inches, then the distance between the centers of the pulleys in inches is

$\textbf{(A) }24\qquad \textbf{(B) }2\sqrt{119}\qquad \textbf{(C) }25\qquad \textbf{(D) }26\qquad \textbf{(E) }4\sqrt{35}$

Solution

Problem 10

Each of a group of $50$ girls is blonde or brunette and is blue eyed of brown eyed. If $14$ are blue-eyed blondes, $31$ are brunettes, and $18$ are brown-eyed, then the number of brown-eyed brunettes is

$\textbf{(A) }5\qquad \textbf{(B) }7\qquad \textbf{(C) }9\qquad \textbf{(D) }11\qquad \textbf{(E) }13$

Solution

Problem 11

The numeral $47$ in base a represents the same number as $74$ in base $b$ . Assuming that both bases are positive integers, the least possible value of $a+b$ written as a Roman numeral, is

$\textbf{(A) }\mathrm{XIII}\qquad \textbf{(B) }\mathrm{XV}\qquad \textbf{(C) }\mathrm{XXI}\qquad \textbf{(D) }\mathrm{XXIV}\qquad \textbf{(E) }\mathrm{XVI}$

Solution

Problem 12

For each integer $N>1$ , there is a mathematical system in which two or more positive integers are defined to be congruent if they leave the same non-negative remainder when divided by $N.$ If $69,90$ , and $125$ are congruent in one such system, then in that same system, $81$ is congruent to

$\textbf{(A) }3\qquad \textbf{(B) }4\qquad \textbf{(C) }5\qquad \textbf{(D) }7\qquad \textbf{(E) }8$

Solution

Problem 13

If $(1.0025)^{10}$ is evaluated correct to $5$ decimal places, then the digit in the fifth decimal place is

$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }5\qquad \textbf{(E) }8$

Solution

Problem 14

The number $(2^{48}-1)$ is exactly divisible by two numbers between $60$ and $70$ . These numbers are

$\textbf{(A) }61,63\qquad \textbf{(B) }61,65\qquad \textbf{(C) }63,65\qquad \textbf{(D) }63,67\qquad \textbf{(E) }67,69$

Solution

Problem 15

An aquarium on a level table has rectangular faces and is $10$ inches wide and $8$ inches high. When it was tilted, the water in it covered an $8\times 10$ end but only $\tfrac{3}{4}$ of the rectangular bottom. The depth of the water when the bottom was again made level, was

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

Solution

Problem 16

After finding the average of $35$ scores, a student carelessly included the average with the $35$ scores and found the average of these $36$ numbers. The ratio of the second average to the true average was

$\textbf{(A) }1:1\qquad \textbf{(B) }35:36\qquad \textbf{(C) }36:35\qquad \textbf{(D) }2:1\qquad \textbf{(E) }\text{None of these}$

Solution

Problem 17

A circular disk is divided by $2n$ equally spaced radii( $n>0$ ) and one secant line. The maximum number of non-overlapping areas into which the disk can be divided is

$\textbf{(A) }2n+1\qquad \textbf{(B) }2n+2\qquad \textbf{(C) }3n-1\qquad \textbf{(D) }3n\qquad \textbf{(E) }3n+1$

Solution

Problem 18

The current in a river is flowing steadily at $3$ miles per hour. A motor boat which travels at a constant rate in still water goes downstream $4$ miles and then returns to its starting point. The trip takes one hour, excluding the time spent in turning the boat around. The ratio of the downstream to the upstream rate is

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

Solution

Problem 19

If the line $y=mx+1$ intersects the ellipse $x^2+4y^2=1$ exactly once, then the value of $m^2$ is

$\textbf{(A) }\textstyle\frac{1}{2}\qquad \textbf{(B) }\frac{2}{3}\qquad \textbf{(C) }\frac{3}{4}\qquad \textbf{(D) }\frac{4}{5}\qquad \textbf{(E) }\frac{5}{6}$

Solution

Problem 20

The sum of the squares of the roots of the equation $x^2+2hx=3$ is $10$ . The absolute value of $h$ is equal to

$\textbf{(A) }-1\qquad \textbf{(B) }\textstyle\frac{1}{2}\qquad \textbf{(C) }\textstyle\frac{3}{2}\qquad \textbf{(D) }2\qquad \textbf{(E) }\text{None of these}$

Solution

Problem 21

If $\log_2(\log_3(\log_4 x))=\log_3(\log_4(\log_2 y))=\log_4(\log_2(\log_3 z))=0$ , then the sum $x+y+z$ is equal to

$\textbf{(A) }50\qquad \textbf{(B) }58\qquad \textbf{(C) }89\qquad \textbf{(D) }111\qquad \textbf{(E) }1296$

Solution

Problem 22

If $w$ is one of the imaginary roots of the equation $x^3=1$ , then the product $(1-w+w^2)(1+w-w^2)$ is equal to

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

Solution

Problem 23

Teams $A$ and $B$ are playing a series of games. If the odds for either to win any game are even and Team $A$ must win two or Team $B$ three games to win the series, then the odds favoring Team $A$ to win the series are

$\textbf{(A) }11\text{ to }5\qquad \textbf{(B) }5\text{ to }2\qquad \textbf{(C) }8\text{ to }3\qquad \textbf{(D) }3\text{ to }2\qquad \textbf{(E) }13\text{ to }6$

Solution

Problem 24

$[asy] label("$1$",(0,0),S); label("$1$",(-1,-1),S); label("$1$",(-2,-2),S); label("$1$",(-3,-3),S); label("$1$",(-4,-4),S); label("$1$",(1,-1),S); label("$1$",(2,-2),S); label("$1$",(3,-3),S); label("$1$",(4,-4),S); label("$2$",(0,-2),S); label("$3$",(-1,-3),S); label("$3$",(1,-3),S); label("$4$",(-2,-4),S); label("$4$",(2,-4),S); label("$6$",(0,-4),S); label("etc.",(0,-5),S); //Credit to chezbgone2 for the diagram[/asy]$

Pascal's triangle is an array of positive integers(See figure), in which the first row is $1$ , the second row is two $1$ 's, each row begins and ends with $1$ , and the $k^\text{th}$ number in any row when it is not $1$ , is the sum of the $k^\text{th}$ and $(k-1)^\text{th}$ numbers in the immediately preceding row. The quotient of the number of numbers in the first $n$ rows which are not $1$ 's and the number of $1$ 's is

$\textbf{(A) }\dfrac{n^2-n}{2n-1}\qquad \textbf{(B) }\dfrac{n^2-n}{4n-2}\qquad \textbf{(C) }\dfrac{n^2-2n}{2n-1}\qquad \textbf{(D) }\frac{n^2-3n+2}{4n-2}\qquad \textbf{(E) }\text{None of these}$

Solution

Problem 25

A teen age boy wrote his own age after his father's. From this new four place number, he subtracted the absolute value of the difference of their ages to get $4,289$ . The sum of their ages was

$\textbf{(A) }48\qquad \textbf{(B) }52\qquad \textbf{(C) }56\qquad \textbf{(D) }59\qquad \textbf{(E) }64$

Solution

Problem 26

$[asy] size(2.5inch); pair A, B, C, E, F, G; A = (0,3); B = (-1,0); C = (3,0); E = (0,0); F = (1,2); G = intersectionpoint(B--F,A--E); draw(A--B--C--cycle); draw(A--E); draw(B--F); label("$A$",A,N); label("$B$",B,W); label("$C$",C,dir(0)); label("$E$",E,S); label("$F$",F,NE); label("$G$",G,SE); //Credit to chezbgone2 for the diagram[/asy]$

In $\triangle ABC$ , point $F$ divides side $AC$ in the ratio $1:2$ . Let $E$ be the point of intersection of side $BC$ and $AG$ where $G$ is the midpoint of $BF$ . The point $E$ divides side $BC$ in the ratio

$\textbf{(A) }1:4\qquad \textbf{(B) }1:3\qquad \textbf{(C) }2:5\qquad \textbf{(D) }4:11\qquad \textbf{(E) }3:8$

Solution

Problem 27

A box contains chips, each of which is red, white, or blue. The number of blue chips is at least half the number of white chips, and at most one third the number of red chips. The number which are white or blue is at least $55$ . The minimum number of red chips is

$\textbf{(A) }24\qquad \textbf{(B) }33\qquad \textbf{(C) }45\qquad \textbf{(D) }54\qquad \textbf{(E) }57$

Solution

Problem 28

Nine lines parallel to the base of a triangle divide the other sides each into $10$ equal segments and the area into $10$ distinct parts. If the area of the largest of these parts is $38$ , then the area of the original triangle is

$\textbf{(A) }180\qquad \textbf{(B) }190\qquad \textbf{(C) }200\qquad \textbf{(D) }210\qquad \textbf{(E) }240$

Solution

Problem 29

Given the progression $10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}$ . The least positive integer $n$ such that the product of the first $n$ terms of the progression exceeds $100,000$ is

$\textbf{(A) }7\qquad \textbf{(B) }8\qquad \textbf{(C) }9\qquad \textbf{(D) }10\qquad \textbf{(E) }11$

Solution

Problem 30

Given the linear fractional transformation of $x$ into $f_1(x)=\dfrac{2x-1}{x+1}$ . Define $f_{n+1}(x)=f_1(f_n(x))$ for $n=1,2,3,\cdots$ . Assuming that $f_{35}(x)=f_5(x)$ , it follows that $f_{28}(x)$ is equal to

$\textbf{(A) }x\qquad \textbf{(B) }\frac{1}{x}\qquad \textbf{(C) }\frac{x-1}{x}\qquad \textbf{(D) }\frac{1}{1-x}\qquad \textbf{(E) }\text{None of these}$

Solution

Problem 31

$[asy] size(2.5inch); pair A = (-2,0), B = 2dir(150), D = (2,0), C; draw(A..(0,2)..D--cycle); C = intersectionpoint(A..(0,2)..D,Circle(B,arclength(A--B))); draw(A--B--C--D--cycle); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,N); label("$D$",D,E); label("$4$",A--D,S); label("$1$",A--B,E); label("$1$",B--C,SE); //Credit to chezbgone2 for the diagram[/asy]$

Quadrilateral $ABCD$ is inscribed in a circle with side $AD$ , a diameter of length $4$ . If sides $AB$ and $BC$ each have length $1$ , then side $CD$ has length

$\textbf{(A) }\frac{7}{2}\qquad \textbf{(B) }\frac{5\sqrt{2}}{2}\qquad \textbf{(C) }\sqrt{11}\qquad \textbf{(D) }\sqrt{13}\qquad \textbf{(E) }2\sqrt{3}$

Solution

Problem 32

If $s=(1+2^{-\frac{1}{32}})(1+2^{-\frac{1}{16}})(1+2^{-\frac{1}{8}})(1+2^{-\frac{1}{4}})(1+2^{-\frac{1}{2}})$ , then $s$ is equal to

$\textbf{(A) }\textstyle{\frac{1}{2}}(1-2^{-\frac{1}{32}})^{-1}\qquad \textbf{(B) }(1-2^{-\frac{1}{32}})^{-1}\qquad \textbf{(C) }1-2^{-\frac{1}{32}}\qquad \\ \textbf{(D) }\textstyle{\frac{1}{2}}(1-2^{-\frac{1}{32}})\qquad \textbf{(E) }\frac{1}{2}$

Solution

Problem 33

If $P$ is the product of $n$ quantities in Geometric Progression, $S$ their sum, and $S'$ the sum of their reciprocals, then $P$ in terms of $S, S'$ , and $n$ is

$\textbf{(A) }(SS')^{\frac{1}{2}n}\qquad \textbf{(B) }(S/S')^{\frac{1}{2}n}\qquad \textbf{(C) }(SS')^{n-2}\qquad \textbf{(D) }(S/S')^n\qquad \textbf{(E) }(S/S')^{\frac{1}{2}(n-1)}$

Solution

Problem 34

An ordinary clock in a factory is running slow so that the minute hand passes the hour hand at the usual dial position( $12$ o'clock, etc.) but only every $69$ minutes. At time and one-half for overtime, the extra pay to which a $\textdollar 4.00$ per hour worker should be entitled after working a normal $8$ hour day by that slow running clock, is

$\textbf{(A) }\textdollar 2.30\qquad \textbf{(B) }\textdollar 2.60\qquad \textbf{(C) }\textdollar 2.80\qquad \textbf{(D) }\textdollar 3.00\qquad \textbf{(E) }\textdollar 3.30$

Solution

Problem 35

Each circle in an infinite sequence with decreasing radii is tangent externally to the one following it and to both sides of a given right angle. The ratio of the area of the first circle to the sum of areas of all other circles in the sequence, is

$\textbf{(A) }(4+3\sqrt{2}):4\qquad \textbf{(B) }9\sqrt{2}:2\qquad \textbf{(C) }(16+12\sqrt{2}):1\qquad \\ \textbf{(D) }(2+2\sqrt{2}):1\qquad \textbf{(E) }(3+2\sqrt{2}):1$

Solution

@@ Line 1: / Line 1: @@
+{{AHSC 35 Problems
+|year = 1971
+}}
 == Problem 1 ==
 The number of digits in the number <math>N=2^{12}\times 5^8</math> is
@@ Line 160: / Line 163: @@
 For each integer <math>N>1</math>, there is a mathematical system in which two or more positive integers are defined
-to be congruent if they leave the same non-negative remainder when divided by N. If <math>69,90</math>, and <math>125</math> are
+to be congruent if they leave the same non-negative remainder when divided by <math>N.</math> If <math>69,90</math>, and <math>125</math> are
-congruent in one such system, then in that same system, 8<math>1</math> is congruent to
+congruent in one such system, then in that same system, <math>81</math> is congruent to
 <math>\textbf{(A) }3\qquad
@@ Line 170: / Line 173: @@
 [[1971 AHSME Problems/Problem 12|Solution]]
 == Problem 13 ==
@@ Line 198: / Line 201: @@
 An aquarium on a level table has rectangular faces and is <math>10</math> inches wide and <math>8</math> inches high. When it was tilted,
-the water in it covered an <math>8\times 10</math> end but only <math>\tfrac{3}{4}</math> of the rectangular room. The depth of the
+the water in it covered an <math>8\times 10</math> end but only <math>\tfrac{3}{4}</math> of the rectangular bottom. The depth of the
 water when the bottom was again made level, was
@@ Line 208: / Line 211: @@
 [[1971 AHSME Problems/Problem 15|Solution]]
 == Problem 16 ==
@@ Line 379: / Line 382: @@
 In <math>\triangle ABC</math>, point <math>F</math> divides side <math>AC</math> in the ratio <math>1:2</math>. Let <math>E</math> be the point of intersection of side <math>BC</math> and <math>AG</math> where <math>G</math> is the
-midpoints of <math>BF</math>. The point <math>E</math> divides side <math>BC</math> in the ratio
+midpoint of <math>BF</math>. The point <math>E</math> divides side <math>BC</math> in the ratio
 <math>\textbf{(A) }1:4\qquad
@@ Line 388: / Line 391: @@
 [[1971 AHSME Problems/Problem 26|Solution]]
 == Problem 27 ==
@@ Line 468: / Line 471: @@
 [[1971 AHSME Problems/Problem 31|Solution]]
 == Problem 32 ==
@@ Line 517: / Line 520: @@
 \textbf{(C) }(16+12\sqrt{2}):1\qquad \\
 \textbf{(D) }(2+2\sqrt{2}):1\qquad
-\textbf{(E) }3+2\sqrt{2}):1  </math>
+\textbf{(E) }(3+2\sqrt{2}):1  </math>
 [[1971 AHSME Problems/Problem 35|Solution]]
 == See also ==
@@ Line 527: / Line 529: @@
 * [[Mathematics competition resources]]
-{{AHSME box|year=1971|before=[[1970 AHSME]]|after=[[1972 AHSME]]}}
+{{AHSME 35p box|year=1971|before=[[1970 AHSME|1970 AHSC]]|after=[[1972 AHSME|1972 AHSC]]}}
 {{MAA Notice}}

1971 AHSC (Problems • Answer Key • Resources)
Preceded by 1970 AHSC	Followed by 1972 AHSC
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All AHSME Problems and Solutions

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Difference between revisions of "1971 AHSME Problems"

Latest revision as of 18:08, 6 August 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

See also