Home » 国际竞赛 » 数学国际竞赛 » AMC美国数学竞赛 » Details

1987 AMC 8 真题及答案详细解析

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月4日上午11:20

1987 AMC 8 真题

答案详细解析请参考文末

Problem 1

$.4+.02+.006=$

$text{(A)} .012 qquad text{(B)} .066 qquad text{(C)} .12 qquad text{(D)} .24 qquad text{(E)} .426$

Problem 2

$frac{2}{25}=$

$text{(A)} .008 qquad text{(B)} .08 qquad text{(C)} .8 qquad text{(D)} 1.25 qquad text{(E)} 12.5$

Problem 3

$2(81+83+85+87+89+91+93+95+97+99)=$

$text{(A)} 1600 qquad text{(B)} 1650 qquad text{(C)} 1700 qquad text{(D)} 1750 qquad text{(E)} 1800$

Problem 4

Martians measure angles in clerts. There are $500$ clerts in a full circle. How many clerts are there in a right angle?

$text{(A)} 90 qquad text{(B)} 100 qquad text{(C)} 125 qquad text{(D)} 180 qquad text{(E)} 250$

Problem 5

The area of the rectangular region is

$[asy] draw((0,0)--(4,0)--(4,2.2)--(0,2.2)--cycle,linewidth(.5 mm)); label(".22 m",(4,1.1),E); label(".4 m",(2,0),S); [/asy]$

$text{(A)} text{.088 m}^2 qquad text{(B)} text{.62 m}^2 qquad text{(C)} text{.88 m}^2 qquad text{(D)} text{1.24 m}^2 qquad text{(E)} text{4.22 m}^2$

Problem 6

The smallest product one could obtain by multiplying two numbers in the set ${ -7, -5, -1, 1, 3 }$ is

$text{(A)} -35 qquad text{(B)} -21 qquad text{(C)} -15 qquad text{(D)} -1 qquad text{(E)} 3$

Problem 7

The large cube shown is made up of $27$ identical sized smaller cubes. For each face of the large cube, the opposite face is shaded the same way. The total number of smaller cubes that must have at least one face shaded is

$text{(A)} 10 qquad text{(B)} 16 qquad text{(C)} 20 qquad text{(D)} 22 qquad text{(E)} 24$

$[asy] unitsize(36); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5.2,1.4)--(5.2,4.4)--(3,3)); draw((0,3)--(2.2,4.4)--(5.2,4.4)); fill((0,0)--(0,1)--(1,1)--(1,0)--cycle,black); fill((0,2)--(0,3)--(1,3)--(1,2)--cycle,black); fill((1,1)--(1,2)--(2,2)--(2,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); draw((1,3)--(3.2,4.4)); draw((2,3)--(4.2,4.4)); draw((.733333333,3.4666666666)--(3.73333333333,3.466666666666)); draw((1.466666666,3.9333333333)--(4.466666666,3.9333333333)); fill((1.73333333,3.46666666666)--(2.7333333333,3.46666666666)--(3.46666666666,3.93333333333)--(2.46666666666,3.93333333333)--cycle,black); fill((3,1)--(3.733333333333,1.466666666666)--(3.73333333333,2.46666666666)--(3,2)--cycle,black); fill((3.73333333333,.466666666666)--(4.466666666666,.93333333333)--(4.46666666666,1.93333333333)--(3.733333333333,1.46666666666)--cycle,black); fill((3.73333333333,2.466666666666)--(4.466666666666,2.93333333333)--(4.46666666666,3.93333333333)--(3.733333333333,3.46666666666)--cycle,black); fill((4.466666666666,1.9333333333333)--(5.2,2.4)--(5.2,3.4)--(4.4666666666666,2.9333333333333)--cycle,black); [/asy]$

Problem 8

If $text{A}$ and $text{B}$ are nonzero digits, then the number of digits (not necessarily different) in the sum of the three whole numbers is

$[begin{tabular}[t]{cccc} 9 & 8 & 7 & 6 \ & A & 3 & 2 \ & & B & 1 \ hline end{tabular}]$

$text{(A)} 4 qquad text{(B)} 5 qquad text{(C)} 6 qquad text{(D)} 9 qquad text{(E)} text{depends on the values of A and B}$

Problem 9

When finding the sum $frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}+frac{1}{7}$ , the least common denominator used is

$text{(A)} 120 qquad text{(B)} 210 qquad text{(C)} 420 qquad text{(D)} 840 qquad text{(E)} 5040$

Problem 10

$4(299)+3(299)+2(299)+298=$

$text{(A)} 2889 qquad text{(B)} 2989 qquad text{(C)} 2991 qquad text{(D)} 2999 qquad text{(E)} 3009$

Problem 11

The sum $2frac17+3frac12+5frac{1}{19}$ is between

$text{(A)} 10text{ and }10frac12 qquad text{(B)} 10frac12 text{ and } 11 qquad text{(C)} 11text{ and }11frac12 qquad text{(D)} 11frac12 text{ and }12 qquad text{(E)} 12text{ and }12frac12$

Problem 12

What fraction of the large $12$ by $18$ rectangular region is shaded?

$[asy] draw((0,0)--(18,0)--(18,12)--(0,12)--cycle); draw((0,6)--(18,6)); for(int a=6; a<12; ++a) { draw((1.5a,0)--(1.5a,6)); } fill((15,0)--(18,0)--(18,6)--(15,6)--cycle,black); label("0",(0,0),W); label("9",(9,0),S); label("18",(18,0),S); label("6",(0,6),W); label("12",(0,12),W); [/asy]$

$text{(A)} frac{1}{108} qquad text{(B)} frac{1}{18} qquad text{(C)} frac{1}{12} qquad text{(D)} frac29 qquad text{(E)} frac13$

Problem 13

Which of the following fractions has the largest value?

$text{(A)} frac{3}{7} qquad text{(B)} frac{4}{9} qquad text{(C)} frac{17}{35} qquad text{(D)} frac{100}{201} qquad text{(E)} frac{151}{301}$

Problem 14

A computer can do $10,000$ additions per second. How many additions can it do in one hour?

$text{(A)} 6text{ million} qquad text{(B)} 36text{ million} qquad text{(C)} 60text{ million} qquad text{(D)} 216text{ million} qquad text{(E)} 360text{ million}$

Problem 15

The sale ad read: "Buy three tires at the regular price and get the fourth tire for three dollars;." Sam paid $240text{ dollars}$ for a set of four tires at the sale. What was the regular price of one tire?

$text{(A)} 59.25text{ dollars} qquad text{(B)} 60text{ dollars} qquad text{(C)} 70text{ dollars} qquad text{(D)} 79text{ dollars} qquad text{(E)} 80text{ dollars}$

Problem 16

Joyce made $12$ of her first $30$ shots in the first three games of this basketball game, so her seasonal shooting average was $40%$ . In her next game, she took $10$ shots and raised her seasonal shooting average to $50%$ . How many of these $10$ shots did she make?

$text{(A)} 2 qquad text{(B)} 3 qquad text{(C)} 5 qquad text{(D)} 6 qquad text{(E)} 8$

Problem 17

Abby, Bret, Carl, and Dana are seated in a row of four seats numbered #1 to #4. Joe looks at them and says:

"Bret is next to Carl."
"Abby is between Bret and Carl."

However each one of Joe's statements is false. Bret is actually sitting in seat #3. Who is sitting in seat #2?

$text{(A)} text{Abby} qquad text{(B)} text{Bret} qquad text{(C)} text{Carl} qquad text{(D)} text{Dana} qquad text{(E)} text{There is not enough information to be sure.}$

Problem 18

Half the people in a room left. One third of those remaining started to dance. There were then $12$ people who were not dancing. The original number of people in the room was

$text{(A)} 24 qquad text{(B)} 30 qquad text{(C)} 36 qquad text{(D)} 42 qquad text{(E)} 72$

Problem 19

A calculator has a squaring key $boxed{x^2}$ which replaces the current number displayed with its square. For example, if the display is $boxed{000003}$ and the $boxed{x^2}$ key is depressed, then the display becomes $boxed{000009}$ . If the display reads $boxed{000002}$ , how many times must you depress the $boxed{x^2}$ key to produce a displayed number greater than $500$ ?

$text{(A)} 4 qquad text{(B)} 5 qquad text{(C)} 8 qquad text{(D)} 9 qquad text{(E)} 250$

Problem 20

"If a whole number $n$ is not prime, then the whole number $n-2$ is not prime." A value of $n$ which shows this statement to be false is

$text{(A)} 9 qquad text{(B)} 12 qquad text{(C)} 13 qquad text{(D)} 16 qquad text{(E)} 23$

Problem 21

Suppose $n^{*}$ means $frac{1}{n}$ , the reciprocal of $n$ . For example, $5^{*}=frac{1}{5}$ . How many of the following statements are true?

i) 
ii) 
iii) 
iv)

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

Problem 22

$text{ABCD}$ is a rectangle, $text{D}$ is the center of the circle, and $text{B}$ is on the circle. If $text{AD}=4$ and $text{CD}=3$ , then the area of the shaded region is between

$[asy] pair A,B,C,D; A=(0,4); B=(3,4); C=(3,0); D=origin; draw(circle(D,5)); fill((0,5)..(1.5,4.7697)..B--A--cycle,black); fill(B..(4,3)..(5,0)--C--cycle,black); draw((0,5)--D--(5,0)); label("A",A,NW); label("B",B,NE); label("C",C,S); label("D",D,SW); [/asy]$

$text{(A)} 4text{ and }5 qquad text{(B)} 5text{ and }6 qquad text{(C)} 6text{ and }7 qquad text{(D)} 7text{ and }8 qquad text{(E)} 8text{ and }9$

Problem 23

Assume the adjoining chart shows the $1980$ U.S. population, in millions, for each region by ethnic group. To the nearest percent, what percent of the U.S. Black population lived in the South?

$[begin{tabular}[t]{c|cccc} & NE & MW & South & West \ hline White & 42 & 52 & 57 & 35 \ Black & 5 & 5 & 15 & 2 \ Asian & 1 & 1 & 1 & 3 \ Other & 1 & 1 & 2 & 4 end{tabular}]$

$text{(A)} 20% qquad text{(B)} 25% qquad text{(C)} 40% qquad text{(D)} 56% qquad text{(E)} 80%$

Problem 24

A multiple choice examination consists of $20$ questions. The scoring is $+5$ for each correct answer, $-2$ for each incorrect answer, and $0$ for each unanswered question. John's score on the examination is $48$ . What is the maximum number of questions he could have answered correctly?

$text{(A)} 9 qquad text{(B)} 10 qquad text{(C)} 11 qquad text{(D)} 12 qquad text{(E)} 16$

Problem 25

Ten balls numbered $1$ to $10$ are in a jar. Jack reaches into the jar and randomly removes one of the balls. Then Jill reaches into the jar and randomly removes a different ball. The probability that the sum of the two numbers on the balls removed is even is

$text{(A)} frac{4}{9} qquad text{(B)} frac{9}{19} qquad text{(C)} frac{1}{2} qquad text{(D)} frac{10}{19} qquad text{(E)} frac{5}{9}$

1. $frac{2}{25}=frac{2cdot 4}{25cdot 4} = frac{8}{100} = 0.08rightarrow boxed{text{B}}$

Find that $[(81+83+85+87+89+91+93+95+97+99) = 5 cdot 180]$ Which gives us $begin{align*} 2(5 cdot 180) &= 10 cdot 180\ &= 1800 & text{ Thus boxed{text{E}} is the correct answer} end{align*}$

Solution 2

$2(81+83+85+87+89+91+93+95+97+99)$ Pair the least with the greatest, second least with the second greatest, etc, until you have five pairs, each adding up to $81+99$ = $83+97$ = $85+95$ = $87+93$ = $89+91$ = $180$ . Since we have $5$ pairs, we multiply $180$ by $5$ to get $900$ . But since we have to multiply by 2 (remember the 2 at the beginning of the parentheses!), we get $1800$ , which is $boxed{text{E}}$ .

3. $(.4) cdot (.22) = frac{4}{10} cdot frac{22}{100} = frac{4cdot 22}{10cdot 100} = frac{88}{1000} = 0.088rightarrow boxed{text{A}}$

4.The right angle is $1/4$ of the circle, hence it contains $500/4=125rightarrow boxed{text{C}}$ clerts.

5. $(.4) cdot (.22) = frac{4}{10} cdot frac{22}{100} = frac{4cdot 22}{10cdot 100} = frac{88}{1000} = 0.088rightarrow boxed{text{A}}$

6.To get the smallest possible product, we want to multiply the smallest negative number by the largest positive number. These are $-7$ and $3$ , respectively, and their product is $-21$ , which is $boxed{text{B}}$

7.Clearly no cube has more than one face painted. Therefore, the number of cubes with at least one face painted is equal to the number of painted unit squares.

There are $10$ painted unit squares on the half of the cube shown, so there are $10cdot 2=20$ cubes with at least one face painted.

$boxed{text{C}}$

8.The minimum possible value of this sum is when $A=B=1$ , which is $[9876+132+11=10019]$

The largest possible value of the sum is when $A=B=9$ , making the sum $[9876+932+91=10899]$

Since all the possible sums are between $10019$ and $10899$ , they must have $5$ digits.

$boxed{text{B}}$

9.We want the least common multiple of $2,3,4,5,6,7$ , which is $420$ , or choice $boxed{text{C}}$ .

10.

We can make use of the distributive property as follows: $begin{align*} 4(299)+3(299)+2(299)+298 &= 4(299)+3(299)+2(299)+1(299)-1 \ &= (4+3+2+1)(299)-1 \ &= 10(299)-1 \ &= 2989 \ end{align*}$

$boxed{text{B}}$

11.

Since $frac{1}{7}<frac14$ and $frac{1}{19}<frac14$ , $[2frac17+3frac12+5frac{1}{19}<2frac14+3frac12+5frac14 =11]$

Clearly, $[2frac17+3frac12+5frac{1}{19}>2+3frac12+5=10frac12]$

Thus, the sum is between $10frac12$ and $11$ .

$boxed{text{B}}$

12.The shaded region makes up $frac13$ of the quarter rectangle. The quarter rectangle is then $frac14$ of the large rectangle, so the shaded region takes up $frac{1}{3}times frac{1}{4}=frac{1}{12}$ of the large rectangle.

$boxed{text{C}}$

13.Note that the first four choices are a little less than $frac{1}{2}$ , but the last choice is just above $frac{1}{2}$ . Thus, the largest fraction is clearly $boxed{text{E}}$

14.There are $3600$ seconds per hour, so we have $begin{align*} frac{3600text{ seconds}}{text{hour}}cdot frac{10,000text{ additions}}{text{second}} &= frac{36,000,000text{ additions}}{text{hour}} \ &= 36text{ million additions per hour} end{align*}$

15.

Let the regular price of one tire be $x$ . We have $begin{align*} 3x+3=240 &Rightarrow 3x=237 \ &Rightarrow x=79 end{align*}$

$boxed{text{D}}$ Good Job!

16.After the fourth game, she took $40$ shots, $50%$ of which she made, so she made $40times .5=20$ shots. Twelve of them were made in the first three games, so in the last game she made $20-12=8$ shots.

$boxed{text{E}}$

17.We know that Carl does not sit next to Bret, so he must sit in seat #1. Since Abby is not between Bret and Carl, she must sit in seat #4. Finally, Dana has to take the last seat available, which is #2.

$boxed{text{D}}$

18.Let the original number of people in the room be $x$ . Half of them left, so $frac{x}{2}$ of them are left in the room.

After that, one third of this group is dancing, so $frac{x}{2}-frac{1}{3}left( frac{x}{2}right) =frac{x}{3}$ people are not dancing.

This is given to be $12$ , so $[frac{x}{3}=12Rightarrow x=36]$

$boxed{text{C}}$

19.

We just brute force this: $begin{align*} 2 &rightarrow 4 \ &rightarrow 16 \ &rightarrow 256 \ &rightarrow 65536>500 end{align*}$

Clearly we need to press the button $4$ times, so $boxed{text{A}}$

20.To show this statement to be false, we need a non-prime value of $n$ such that $n-2$ is prime. Since $13$ and $23$ are prime, they won't prove anything relating to the truth of the statement.

Now we just check the statement for $n=9,12,16$ . If $n=12$ or $n=16$ , then $n-2$ is $10$ or $14$ , which aren't prime. However, $n=9$ makes $n-2=7$ , which is prime, so $n=9$ proves the statement false.

$boxed{text{A}}$

21.

We can just test all of these statements: $begin{align*} 3^*+6^* &= frac{1}{3}+frac{1}{6} \ &= frac{1}{2} neq 9^* \ 6^*-4^* &= frac{1}{6}-frac{1}{4} \ &= frac{-1}{12} neq 2^* \ 2^*cdot 6^* &= frac{1}{2}cdot frac{1}{6} \ &= frac{1}{12} = 12^* \ 10^* div 2^* &= frac{1}{10}div frac{1}{2} \ &= frac{1}{5} = 5^* end{align*}$

The last two statements are true and the first two aren't, so $boxed{text{C}}$

22.

The area of the shaded region is equal to the area of the quarter circle with the area of the rectangle taken away. The area of the rectangle is $4cdot 3=12$ , so we just need the quarter circle.

Applying the Pythagorean Theorem to $triangle ADC$ , we have $[(AC)^2=4^2+3^2Rightarrow AC=5]$ Since $ABCD$ is a rectangle, $[BD=AC=5]$

Clearly $BD$ is a radius of the circle, so the area of the whole circle is $5^2pi =25pi$ and the area of the quarter circle is $frac{25pi }{4}$ .

Finally, the shaded region is $[frac{25pi }{4}-12 approx 7.6]$ so the answer is $boxed{text{D}}$

23.There are $5+5+15+2=27$ million Blacks living in the U.S. Out of these, $15$ of them live in the South, so the percentage is $frac{15}{27}approx 56%$ .

$boxed{text{D}}$

24.

Solution 1

Let $c$ be the number of questions correct, $w$ be the number of questions wrong, and $b$ be the number of questions left blank. We are given that $begin{align} c+w+b &= 20 \ 5c-2w &= 48 end{align}$

Adding equation $(2)$ to double equation $(1)$ , we get $[7c+2b=88]$

Since we want to maximize the value of $c$ , we try to find the largest multiple of $7$ less than $88$ . This is $84=7times 12$ , so let $c=12$ . Then we have $[7(12)+2b=88Rightarrow b=2]$

Finally, we have $w=20-12-2=6$ . We want $c$ , so the answer is $12$ , or $boxed{text{D}}$ .

Solution 2

If John answered 16 questions correctly, then he answered at most 4 questions incorrectly, giving him at least $16 cdot 5 - 4 cdot 2 = 72$ points. Therefore, John did not answer 16 questions correctly. If he answered 12 questions correctly and 6 questions incorrectly (leaving 2 questions unanswered), then he scored $12 cdot 5 - 6 cdot 2 = 48$ points. As all other options are less than 12, we conclude that 12 is the most questions John could have answered correctly, and the answer is $boxed{text{D}}$ .

25.For the sum of the two numbers removed to be even, they must be of the same parity. There are five even values and five odd values.

No matter what Jack chooses, the number of numbers with the same parity is four. There are nine numbers total, so the probability Jill chooses a number with the same parity as Jack's is $frac49$

$boxed{text{A}}$

Solution 2

We find that it is only possible for the sum to be even if the numbers added are both even or odd. We will get an odd number when we add an even and an odd. We can use complementary counting to help solve the problem. There are a total of $90$ possibilities since Jack can chose $10$ numbers and Jill can pick $9$ . There are $50$ possibilities for the two numbers to be different since Jack can pick any of the $10$ numbers and Jill has to pick from $5$ numbers in the set with a different parity than the one that Jack picks. So the probability that the sum will be odd is $frac{50}{90}=frac{5}{9}$ . Subtracting this by one gets the answer $frac{4}{9}$