Difference between revisions of "1970 AHSME Problems"

Latest revision as of 06:52, 26 December 2023

1970 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.
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

Problem 1

The fourth power of $\sqrt{1+\sqrt{1+\sqrt{1}}}$ is:

$\textbf{(A) }\sqrt{2}+\sqrt{3}\qquad \textbf{(B) }\frac{1}{2}(7+3\sqrt{5})\qquad \textbf{(C) }1+2\sqrt{3}\qquad \textbf{(D) }3\qquad \textbf{(E) }3+2\sqrt{2}$

Solution

Problem 2

A square and a circle have equal perimeters. The ratio of the area of the circle to the area of the square is:

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

Solution

Problem 3

If $x=1+2^p$ and $y=1+2^{-p}$ , then $y$ in terms of $x$ is:

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

Solution

Problem 4

Let $S$ be the set of all numbers which are the sum of the squares of three consecutive integers. Then we can say that:

$\textbf{(A) }\text{No member of }S\text{ is divisible by }2\qquad\\ \textbf{(B) }\text{No member of }S\text{ is divisible by }3\text{ but some member is divisible by }11\qquad\\ \textbf{(C) }\text{No member of }S\text{ is divisible by }3\text{ or }5\qquad\\ \textbf{(D) }\text{No member of }S\text{ is divisible by }3\text{ or }7\qquad\\ \textbf{(E) }\text{None of these}$

Solution

Problem 5

If $f(x)=\dfrac{x^4+x^2}{x+1}$ , then $f(i)$ , where $i=\sqrt{-1}$ , is equal to:

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

Solution

Problem 6

The smallest value of $x^2+8x$ for real values of $x$ is:

$\textbf{(A) }-16.25\qquad \textbf{(B) }-16\qquad \textbf{(C) }-15\qquad \textbf{(D) }-8\qquad \textbf{(E) }\text{None of these}$

Solution

Problem 7

Inside square $ABCD$ with side $s$ , quarter-circle arcs with radii $s$ and centers at $A$ and $B$ are drawn. These arcs intersect at point $X$ inside the square. How far is $X$ from side $CD$ ?

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

Solution

Problem 8

If $a=\log_8{225}$ and $b=\log_2{15}$ , then

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

Solution

Problem 9

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

$\textbf{(A) }12\qquad \textbf{(B) }28\qquad \textbf{(C) }70\qquad \textbf{(D) }75\qquad \textbf{(E) }105$

Solution

Problem 10

Let $F=.48181\cdots$ be an infinite repeating decimal with the digits $8$ and $1$ repeating. When $F$ is written as a fraction in lowest terms, the denominator exceeds the numerator by

$\textbf{(A) }13\qquad \textbf{(B) }14\qquad \textbf{(C) }29\qquad \textbf{(D) }57\qquad \textbf{(E) }126$

Solution

Problem 11

If two factors of $2x^3-hx+k$ are $x+2$ and $x-1$ , the value of $|2h-3k|$ is

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

Solution

Problem 12

A circle with radius $r$ is tangent to sides $AB$ , $AD$ , and $CD$ of rectangle $ABCD$ and passes through the midpoint of diagonal $AC$ . The area of the rectangle in terms of $r$ , is

$\textbf{(A) }4r^2\qquad \textbf{(B) }6r^2\qquad \textbf{(C) }8r^2\qquad \textbf{(D) }12r^2\qquad \textbf{(E) }20r^2$

Solution

Problem 13

Given the binary operation $\ast$ defined by $a\ast b=a^b$ for all positive numbers $a$ and $b$ . The for all positive $a,b,c,n$ , we have

$\textbf{(A) }a\ast b=b\ast a\qquad \textbf{(B) }a\ast (b\ast c)=(a\ast b)\ast c\qquad\\ \textbf{(C) }(a\ast b^n)=(a\ast n)\ast b\qquad \textbf{(D) }(a\ast b)^n=a\ast (bn)\qquad \textbf{(E) }\text{None of these}$

Solution

Problem 14

Consider $x^2+px+q=0$ where $p$ and $q$ are positive numbers. If the roots of this equation differ by $1$ , then $p$ equals

$\textbf{(A) }\sqrt{4q+1}\qquad \textbf{(B) }q-1\qquad \textbf{(C) }-\sqrt{4q+1}\qquad\\ \textbf{(D) }q+1\qquad \textbf{(E) }\sqrt{4q-1}$

Solution

Problem 15

Lines in the $xy$ -plane are drawn through the point $(3,4)$ and the trisection points of the line segment joining the points $(-4,5)$ and $(5,-1)$ . One of these lines has the equation

$\textbf{(A) }3x-2y-1=0\qquad \textbf{(B) }4x-5y+8=0\qquad \textbf{(C) }5x+2y-23=0\qquad\\ \textbf{(D) }x+7y-31=0\qquad \textbf{(E) }x-4y+13=0$

Solution

Problem 16

If $F(n)$ is a function such that $F(1)=F(2)=F(3)=1$ , and such that $F(n+1)=\dfrac{F(n)\cdot F(n-1)+1}{F(n-2)}$ for $n\ge 3$ , then $F(6)$ is equal to

$\textbf{(A) }2\qquad \textbf{(B) }3\qquad \textbf{(C) }7\qquad \textbf{(D) }11\qquad \textbf{(E) }26$

Solution

Problem 17

If $r\ge 0$ , then for all $p$ and $q$ such that $pq\neq 0$ and $pr>qr$ , we have

$\textbf{(A) }-p>-q\qquad \textbf{(B) }-p>q\qquad \textbf{(C) }1>-q/p\qquad\\ \textbf{(D) }1<q/p\qquad \textbf{(E) }\text{None of These}$

Solution

Problem 18

$\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}$ is equal to

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

Solution

Problem 19

The sum of an infinite geometric series with common ratio $r$ such that $|r|<1$ , is $15$ , and the sum of the squares of the terms of this series is $45$ . The first term of the series is

$\textbf{(A) }12\qquad \textbf{(B) }10\qquad \textbf{(C) }5\qquad \textbf{(D) }3\qquad \textbf{(E) }2$

Solution

Problem 20

Lines $HK$ and $BC$ lie in a plane. $M$ is the midpoint of line segment $BC$ , and $BH$ and $CK$ are perpendicular to $HK$ . Then we

$\textbf{(A) }\text{always have }MH=MK\qquad\\ \textbf{(B) }\text{always have }MH>BK\qquad\\ \textbf{(C) }\text{sometimes have }MH=MK\text{ but not always}\qquad\\ \textbf{(D) }\text{always have }MH>MB\qquad \\ \textbf{(E) }\text{always have }BH<BC$

Solution

Problem 21

On an auto trip, the distance read from the instrument panel was $450$ miles. With snow tires on for the return trip over the same route, the reading was $440$ miles. Find, to the nearest hundredth of an inch, the increase in radius of the wheels if the original radius was $15$ inches.

$\textbf{(A) }.33\qquad \textbf{(B) }.34\qquad \textbf{(C) }.35\qquad \textbf{(D) }.38\qquad \textbf{(E) }.66$

Solution

Problem 22

If the sum of the first $3n$ positive integers is $150$ more than the sum of the first $n$ positive integers, then the sum of the first $4n$ positive integers is

$\textbf{(A) }300\qquad \textbf{(B) }350\qquad \textbf{(C) }400\qquad \textbf{(D) }450\qquad \textbf{(E) }600$

Solution

Problem 23

The number $10!$ ( $10$ is written in base $10$ ), when written in the base $12$ system, ends in exactly $k$ zeroes. The value of $k$ is

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

Solution

Problem 24

An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is $2$ , then the area of the hexagon is

$\textbf{(A) }2\qquad \textbf{(B) }3\qquad \textbf{(C) }4\qquad \textbf{(D) }6\qquad \textbf{(E) }12$

Solution

Problem 25

For every real number $x$ , let $[x]$ be the greatest integer less than or equal to $x$ . If the postal rate for first class mail is six cents for every ounce or portion thereof, then the cost in cents of first-class postage on a letter weighing $W$ ounces is always

$\textbf{(A) }6W\qquad \textbf{(B) }6[W]\qquad \textbf{(C) }6([W]-1)\qquad \textbf{(D) }6([W]+1)\qquad \textbf{(E) }-6[-W]$

Solution

Problem 26

The number of distinct points in the $xy$ -plane common to the graphs of $(x+y-5)(2x-3y+5)=0$ and $(x-y+1)(3x+2y-12)=0$ is

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

Solution

Problem 27

In a triangle, the area is numerically equal to the perimeter. What is the radius of the inscribed circle?

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

Solution

Problem 28

In triangle $ABC$ , the median from vertex $A$ is perpendicular to the median from vertex $B$ . If the lengths of sides $AC$ and $BC$ are $6$ and $7$ respectively, then the length of side $AB$ is

$\textbf{(A) }\sqrt{17}\qquad \textbf{(B) }4\qquad \textbf{(C) }4\dfrac{1}{2}\qquad \textbf{(D) }2\sqrt{5}\qquad \textbf{(E) }4\frac{1}{4}$

Solution

Problem 29

It is now between $10:00$ and $11:00$ o'clock, and six minutes from now, the minute hand of the watch will be exactly opposite the place where the hour hand was three minutes ago. What is the exact time now?

$\textbf{(A) }10:05\dfrac{5}{11}\qquad \textbf{(B) }10:07\dfrac{1}{2}\qquad \textbf{(C) }10:10\qquad\\ \textbf{(D) }10:15\qquad \textbf{(E) }10:17\dfrac{1}{2}$

Solution

Problem 30

$[asy] size(175); defaultpen(linewidth(0.8)); real r=50, a=4,b=2.5,c=6.25; pair A=origin,B=c*dir(r),D=(a,0),C=shift(b*dir(r))*D; draw(A--B--C--D--cycle); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,E); label("$D$",D,S); label("$a$",D/2,N); label("$b$",(C+D)/2,NW); [/asy]$

In the accompanying figure, segments $AB$ and $CD$ are parallel, the measure of $\angle{D}$ is twice the measure of $\angle{B}$ , and the measures of segments $AD$ and $CD$ are $a$ and $b$ respectively. Then the measure of $AB$ is equal to

$\textbf{(A) }\dfrac{1}{2}a+2b\qquad \textbf{(B) }\dfrac{3}{2}b+\dfrac{3}{4}a\qquad \textbf{(C) }2a-b\qquad \textbf{(D) }4b-\frac{1}{2}a\qquad \textbf{(E) }a+b$

Solution

Problem 31

If a number is selected at random from the set of all five-digit numbers in which the sum of the digits is equal to $43$ , what is the probability that this number is divisible by $11$ ?

$\textbf{(A) }2/5\qquad \textbf{(B) }1/5\qquad \textbf{(C) }1/6\qquad \textbf{(D) }1/11\qquad \textbf{(E) }1/15$

Solution

Problem 32

$A$ and $B$ travel around a circular track at uniform speeds in opposite directions, starting from diametrically opposite points. If they start at the same time, meet first after $B$ has travelled $100$ yards, and meet a second time $60$ yards before $A$ completes one lap, then the circumference of the track in yards is

$\textbf{(A) }400\qquad \textbf{(B) }440\qquad \textbf{(C) }480\qquad \textbf{(D) }560\qquad \textbf{(E) }880$

Solution

Problem 33

Find the sum of the digits of all numerals in the sequence $1,2,3,4,\cdots ,10000$ .

$\textbf{(A) }180,001\qquad \textbf{(B) }154,756\qquad \textbf{(C) }45,001\qquad \textbf{(D) }154,755\qquad \textbf{(E) }270,001$

Solution

Problem 34

The greatest integer that will divide $13511, 13903$ , and $14589$ and leave the same remainder is

$\textbf{(A) }28\qquad \textbf{(B) }49\qquad \textbf{(C) }98\qquad\\ \textbf{(D) }\text{an odd multiple of }7\text{ greater than }49\qquad\\ \textbf{(E) }\text{an even multiple of }7\text{ greater than }98$

Solution

Problem 35

A retiring employee receives an annual pension proportional to the square root of the number of years of his service. Had he served $a$ years more, his pension would have been $p$ dollars greater, whereas, had he served $b$ years more $b\neq a$ , his pension would have been $q$ dollars greater than the original annual pension. Find his annual pension in terms of $a,b,p$ , and $q$ .

$\textbf{(A) }\dfrac{p^2-q^2}{2(a-b)}\qquad \textbf{(B) }\dfrac{(p-q)^2}{2\sqrt{ab}}\qquad \textbf{(C) }\frac{ap^2-bq^2}{2(ap-bq)}\qquad \textbf{(D) }\frac{aq^2-bp^2}{2(bp-aq)}\qquad \textbf{(E) }\sqrt{(a-b)(p-q)}$

Solution

@@ Line 1: / Line 1: @@
+{{AHSC 35 Problems
+|year = 1970
+}}
 ==Problem 1==
@@ Line 4: / Line 7: @@
 <math>\textbf{(A) }\sqrt{2}+\sqrt{3}\qquad
-\textbf{(B) }\frac{1}{2}(7+3\sqrt{5}})\qquad
+\textbf{(B) }\frac{1}{2}(7+3\sqrt{5})\qquad
 \textbf{(C) }1+2\sqrt{3}\qquad
 \textbf{(D) }3\qquad
@@ Line 166: / Line 169: @@
 <math>\textbf{(A) }\sqrt{4q+1}\qquad
 \textbf{(B) }q-1\qquad
-\textbf{(C) }-\sqrt{4q+1}\qquad
+\textbf{(C) }-\sqrt{4q+1}\qquad\\
-\textbf{(D) }q+1\qquad
 \textbf{(D) }q+1\qquad
 \textbf{(E) }\sqrt{4q-1} </math>
@@ Line 180: / Line 182: @@
 <math>\textbf{(A) }3x-2y-1=0\qquad
 \textbf{(B) }4x-5y+8=0\qquad
-\textbf{(C) }5x+2y-23=0\qquad
+\textbf{(C) }5x+2y-23=0\qquad\\
 \textbf{(D) }x+7y-31=0\qquad
 \textbf{(E) }x-4y+13=0  </math>
@@ Line 205: / Line 207: @@
 <math>\textbf{(A) }-p>-q\qquad
 \textbf{(B) }-p>q\qquad
-\textbf{(C) }1>-q/p\qquad
+\textbf{(C) }1>-q/p\qquad\\
 \textbf{(D) }1<q/p\qquad
 \textbf{(E) }\text{None of These} </math>
@@ Line 219: / Line 221: @@
 \textbf{(C) }4\sqrt{2}\qquad
 \textbf{(D) }\sqrt{6}\qquad
-\textbf{(D) }\sqrt{6}\qquad
 \textbf{(E) }2\sqrt{2} </math>
@@ Line 241: / Line 242: @@
 Lines <math>HK</math> and <math>BC</math> lie in a plane. <math>M</math> is the midpoint of line segment <math>BC</math>, and <math>BH</math> and <math>CK</math> are perpendicular to <math>HK</math>. Then we
-<math>\textbf{(A) }\text{always have }MH=MK\qquad
+<math>\textbf{(A) }\text{always have }MH=MK\qquad\\
-\textbf{(B) }\text{always have }MH>BK\qquad
+\textbf{(B) }\text{always have }MH>BK\qquad\\
-\textbf{(C) }\text{sometimes have }MH=MK\text{ but not always}\qquad
+\textbf{(C) }\text{sometimes have }MH=MK\text{ but not always}\qquad\\
-\textbf{(D) }\text{always have }MH>MB\qquad
+\textbf{(D) }\text{always have }MH>MB\qquad \\
 \textbf{(E) }\text{always have }BH<BC </math>
@@ Line 354: / Line 355: @@
 ==Problem 29==
-It is now between <math>10:00</math> and <math>11:00</math> o'clock, and six minutes form now, the minute hand of the
+It is now between <math>10:00</math> and <math>11:00</math> o'clock, and six minutes from now, the minute hand of the
 watch will be exactly opposite the place where the hour hand was three minutes ago. What is the exact time now?
 <math>\textbf{(A) }10:05\dfrac{5}{11}\qquad
 \textbf{(B) }10:07\dfrac{1}{2}\qquad
-\textbf{(C) }10:10\qquad
+\textbf{(C) }10:10\qquad\\
 \textbf{(D) }10:15\qquad
 \textbf{(E) }10:17\dfrac{1}{2}   </math>
@@ Line 382: / Line 383: @@
 In the accompanying figure, segments <math>AB</math> and <math>CD</math> are parallel,
-the measure of <math>\angle{D}</math> is twice the measure of <math>\angle{B}</math>, and the measures of segments <math>AB</math> and <math>CD</math>
+the measure of <math>\angle{D}</math> is twice the measure of <math>\angle{B}</math>, and the measures of segments <math>AD</math> and <math>CD</math>
 are <math>a</math> and <math>b</math> respectively. Then the measure of <math>AB</math> is equal to
@@ Line 439: / Line 440: @@
 <math>\textbf{(A) }28\qquad
 \textbf{(B) }49\qquad
-\textbf{(C) }98\qquad
+\textbf{(C) }98\qquad\\
-\textbf{(D) }\text{an odd multiple of }7\text{ greater than }49\qquad
+\textbf{(D) }\text{an odd multiple of }7\text{ greater than }49\qquad\\
 \textbf{(E) }\text{an even multiple of }7\text{ greater than }98 </math>
@@ Line 447: / Line 448: @@
 ==Problem 35==
-A retiring employee receives and annual pension proportional to the square root of the number of years of his service.
+A retiring employee receives an annual pension proportional to the square root of the number of years of his service.
-Had he served a years more, his pension would have been <math>p</math> dollars greater, whereas, had he served <math>b</math> years more <math>b\neq a</math>,
+Had he served <math>a</math> years more, his pension would have been <math>p</math> dollars greater, whereas, had he served <math>b</math> years more <math>b\neq a</math>,
 his pension would have been <math>q</math> dollars greater than the original annual pension. Find his annual pension in terms of <math>a,b,p</math>, and <math>q</math>.
@@ Line 456: / Line 457: @@
 \textbf{(D) }\frac{aq^2-bp^2}{2(bp-aq)}\qquad
 \textbf{(E) }\sqrt{(a-b)(p-q)} </math>
+[[1970 AHSME Problems/Problem 35|Solution]]
+== See also ==
+* [[AMC 12 Problems and Solutions]]
+* [[Mathematics competition resources]]
+{{AHSME 35p box|year=1970|before=[[1969 AHSME|1969 AHSC]]|after=[[1971 AHSME|1971 AHSC]]}}
+{{MAA Notice}}

1970 AHSC (Problems • Answer Key • Resources)
Preceded by 1969 AHSC	Followed by 1971 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
All AHSME Problems and Solutions

Page

Toolbox

Search

Difference between revisions of "1970 AHSME Problems"

Latest revision as of 06:52, 26 December 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

See also