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1988 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11:48

1988 AMC 8 真题

答案详细解析请参考文末

Problem 1

The diagram shows part of a scale of a measuring device. The arrow indicates an approximate reading of

$[asy] draw((-3,0)..(0,3)..(3,0)); draw((-3.5,0)--(-2.5,0)); draw((0,2.5)--(0,3.5)); draw((2.5,0)--(3.5,0)); draw((1.8,1.8)--(2.5,2.5)); draw((-1.8,1.8)--(-2.5,2.5)); draw((0,0)--3*dir(120),EndArrow); label("$10$",(-2.6,0),E); label("$11$",(2.6,0),W); [/asy]$

$text{(A)} 10.05 qquad text{(B)} 10.15 qquad text{(C)} 10.25 qquad text{(D)} 10.3 qquad text{(E)} 10.6$

Problem 2

The product $8times .25times 2times .125 =$

$text{(A)} frac18 qquad text{(B)} frac14 qquad text{(C)} frac12 qquad text{(D)} 1 qquad text{(E)} 2$

Problem 3

$frac{1}{10}+frac{2}{20}+frac{3}{30} =$

$text{(A)} .1 qquad text{(B)} .123 qquad text{(C)} .2 qquad text{(D)} .3 qquad text{(E)} .6$

Problem 4

The figure consists of alternating light and dark squares. The number of dark squares exceeds the number of light squares by

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

$[asy] unitsize(12); for(int a=0; a<7; ++a) { fill((2a,0)--(2a+1,0)--(2a+1,1)--(2a,1)--cycle,black); draw((2a+1,0)--(2a+2,0)); } for(int b=7; b<15; ++b) { fill((b,14-b)--(b+1,14-b)--(b+1,15-b)--(b,15-b)--cycle,black); } for(int c=1; c<7; ++c) { fill((c,c)--(c+1,c)--(c+1,c+1)--(c,c+1)--cycle,black); } for(int d=1; d<6; ++d) { draw((2d+1,1)--(2d+2,1)); } fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); draw((5,4)--(6,4)); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); draw((7,4)--(8,4)); fill((8,4)--(9,4)--(9,5)--(8,5)--cycle,black); draw((9,4)--(10,4)); label("same",(6.3,2.45),N); label("pattern here",(7.5,1.4),N); [/asy]$

Problem 5

If $angle text{CBD}$ is a right angle, then this protractor indicates that the measure of $angle text{ABC}$ is approximately

$[asy] unitsize(36); pair A,B,C,D; A=3*dir(160); B=origin; C=3*dir(110); D=3*dir(20); draw((1.5,0)..(0,1.5)..(-1.5,0)); draw((2.5,0)..(0,2.5)..(-2.5,0)--cycle); draw(A--B); draw(C--B); draw(D--B); label("O",(-2.5,0),W); label("A",A,W); label("B",B,S); label("C",C,W); label("D",D,E); label("0",(-1.8,0),W); label("20",(-1.7,.5),NW); label("160",(1.6,.5),NE); label("180",(1.7,0),E); [/asy]$

$text{(A)} 20^circ qquad text{(B)} 40^circ qquad text{(C)} 50^circ qquad text{(D)} 70^circ qquad text{(E)} 120^circ$

Problem 6

$frac{(.2)^3}{(.02)^2} =$

$text{(A)} .2 qquad text{(B)} 2 qquad text{(C)} 10 qquad text{(D)} 15 qquad text{(E)} 20$

Problem 7

$2.46times 8.163times (5.17+4.829)$ is closest to

$text{(A)} 100 qquad text{(B)} 200 qquad text{(C)} 300 qquad text{(D)} 400 qquad text{(E)} 500$

Problem 8

Betty used a calculator to find the product $0.075 times 2.56$ . She forgot to enter the decimal points. The calculator showed $19200$ . If Betty had entered the decimal points correctly, the answer would have been

$text{(A)} .0192 qquad text{(B)} .192 qquad text{(C)} 1.92 qquad text{(D)} 19.2 qquad text{(E)} 192$

Problem 9

An isosceles triangle is a triangle with two sides of equal length. How many of the five triangles on the square grid below are isosceles?

$[asy] for(int a=0; a<12; ++a) { draw((a,0)--(a,6)); } for(int b=0; b<7; ++b) { draw((0,b)--(11,b)); } draw((0,6)--(2,6)--(1,4)--cycle,linewidth(1)); draw((3,4)--(3,6)--(5,4)--cycle,linewidth(1)); draw((0,1)--(3,2)--(6,1)--cycle,linewidth(1)); draw((7,4)--(6,6)--(9,4)--cycle,linewidth(1)); draw((8,1)--(9,3)--(10,0)--cycle,linewidth(1)); [/asy]$

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

Problem 10

Chris's birthday is on a Thursday this year. What day of the week will it be $60$ days after her birthday?

$text{(A)} text{Monday} qquad text{(B)} text{Wednesday} qquad text{(C)} text{Thursday} qquad text{(D)} text{Friday} qquad text{(E)} text{Saturday}$

Problem 11

$sqrt{164}$ is

$text{(A)} 42 qquad text{(B)} text{less than }10 qquad text{(C)} text{between }10text{ and }11 qquad text{(D)} text{between }11text{ and }12 qquad text{(E)} text{between }12text{ and }13$

Problem 12

Suppose the estimated $20$ billion dollar cost to send a person to the planet Mars is shared equally by the $250$ million people in the U.S. Then each person's share is

$text{(A)} 40text{ dollars} qquad text{(B)} 50text{ dollars} qquad text{(C)} 80text{ dollars} qquad text{(D)} 100text{ dollars} qquad text{(E)} 125text{ dollars}$

Problem 13

If rose bushes are spaced about $1$ foot apart, approximately how many bushes are needed to surround a circular patio whose radius is $12$ feet?

$text{(A)} 12 qquad text{(B)} 38 qquad text{(C)} 48 qquad text{(D)} 75 qquad text{(E)} 450$

Problem 14

$diamondsuit$ and $Delta$ are whole numbers and $diamondsuit times Delta =36$ . The largest possible value of $diamondsuit + Delta$ is

$text{(A)} 12 qquad text{(B)} 13 qquad text{(C)} 15 qquad text{(D)} 20 qquad text{(E)} 37$

Problem 15

The reciprocal of $left( frac{1}{2}+frac{1}{3}right)$ is

$text{(A)} frac{1}{6} qquad text{(B)} frac{2}{5} qquad text{(C)} frac{6}{5} qquad text{(D)} frac{5}{2} qquad text{(E)} 5$

Problem 16

Placing no more than one $text{X}$ in each small square, what is the greatest number of $text{X}$ 's that can be put on the grid shown without getting three $text{X}$ 's in a row vertically, horizontally, or diagonally?

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

$[asy] for(int a=0; a<4; ++a) { draw((a,0)--(a,3)); } for(int b=0; b<4; ++b) { draw((0,b)--(3,b)); } [/asy]$

Problem 17

The shaded region formed by the two intersecting perpendicular rectangles, in square units, is

$[asy] fill((0,0)--(6,0)--(6,-3.5)--(9,-3.5)--(9,0)--(10,0)--(10,2)--(9,2)--(9,4.5)--(6,4.5)--(6,2)--(0,2)--cycle,black); label("2",(0,.9),W); label("3",(7.3,4.5),N); draw((0,-3.3)--(0,-5.3),linewidth(1)); draw((0,-4.3)--(3.7,-4.3),linewidth(1)); label("10",(4.7,-3.7),S); draw((5.7,-4.3)--(10,-4.3),linewidth(1)); draw((10,-3.3)--(10,-5.3),linewidth(1)); draw((11,4.5)--(13,4.5),linewidth(1)); draw((12,4.5)--(12,2),linewidth(1)); label("8",(11.3,1),E); draw((12,0)--(12,-3.5),linewidth(1)); draw((11,-3.5)--(13,-3.5),linewidth(1)); [/asy]$

$text{(A)} 23 qquad text{(B)} 38 qquad text{(C)} 44 qquad text{(D)} 46 qquad text{(E)} text{unable to be determined from the information given}$

Problem 18

The average weight of $6$ boys is $150$ pounds and the average weight of $4$ girls is $120$ pounds. The average weight of the $10$ children is

$text{(A)} 135text{ pounds} qquad text{(B)} 137text{ pounds} qquad text{(C)} 138text{ pounds} qquad text{(D)} 140text{ pounds} qquad text{(E)} 141text{ pounds}$

Problem 19

What is the $100text{th}$ number in the arithmetic sequence: $1,5,9,13,17,21,25,...$ ?

$text{(A)} 397 qquad text{(B)} 399 qquad text{(C)} 401 qquad text{(D)} 403 qquad text{(E)} 405$

Problem 20

The glass gauge on a cylindrical coffee maker shows that there are $45$ cups left when the coffee maker is $36%$ full. How many cups of coffee does it hold when it is full?

$text{(A)} 80 qquad text{(B)} 100 qquad text{(C)} 125 qquad text{(D)} 130 qquad text{(E)} 262$

$[asy] draw((5,0)..(0,-1.3)..(-5,0)); draw((5,0)--(5,10)); draw((-5,0)--(-5,10)); draw(ellipse((0,10),5,1.3)); draw(circle((.3,1.3),.4)); draw((-.1,1.7)--(-.1,7.9)--(.7,7.9)--(.7,1.7)--cycle); fill((-.1,1.7)--(-.1,4)--(.7,4)--(.7,1.7)--cycle,black); draw((-2,11.3)--(2,11.3)..(2.6,11.9)..(2,12.2)--(-2,12.2)..(-2.6,11.9)..cycle); [/asy]$

Problem 21

A fifth number, $n$ , is added to the set ${ 3,6,9,10 }$ to make the mean of the set of five numbers equal to its median. The number of possible values of $n$ is

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

Problem 22

Tom's Hat Shoppe increased all original prices by $25%$ . Now the shoppe is having a sale where all prices are $20%$ off these increased prices. Which statement best describes the sale price of an item?

$text{(A)} text{The sale price is }5% text{ higher than the original price.}$

$text{(B)} text{The sale price is higher than the original price, but by less than }5% .$

$text{(C)} text{The sale price is higher than the original price, but by more than }5% .$

$text{(D)} text{The sale price is lower than the original price.}$

$text{(E)} text{The sale price is the same as the original price.}$

Problem 23

Maria buys computer disks at a price of $4$ for $<dollar>5$ and sells them at a price of $3$ for $<dollar>$ $5$ . How many computer disks must she sell in order to make a profit of $<dollar>100$ ?

$text{(A)} 100 qquad text{(B)} 120 qquad text{(C)} 200 qquad text{(D)} 240 qquad text{(E)} 1200$

Problem 24

$[asy] unitsize(15); for (int a=0; a<6; ++a) { draw(2*dir(60a)--2*dir(60a+60),linewidth(1)); } draw((1,1.7320508075688772935274463415059)--(1,3.7320508075688772935274463415059)--(-1,3.7320508075688772935274463415059)--(-1,1.7320508075688772935274463415059)--cycle,linewidth(1)); fill((.4,1.7320508075688772935274463415059)--(0,3.35)--(-.4,1.7320508075688772935274463415059)--cycle,black); label("1.",(0,-2),S); draw(arc((1,1.7320508075688772935274463415059),1,90,300,CW)); draw((1.5,0.86602540378443864676372317075294)--(1.75,1.7)); draw((1.5,0.86602540378443864676372317075294)--(2.2,1)); draw((7,0)--(6,1.7320508075688772935274463415059)--(4,1.7320508075688772935274463415059)--(3,0)--(4,-1.7320508075688772935274463415059)--(6,-1.7320508075688772935274463415059)--cycle,linewidth(1)); draw((7,0)--(6,1.7320508075688772935274463415059)--(7.7320508075688772935274463415059,2.7320508075688772935274463415059)--(8.7320508075688772935274463415059,1)--cycle,linewidth(1)); label("2.",(5,-2),S); draw(arc((7,0),1,30,240,CW)); draw((6.5,-0.86602540378443864676372317075294)--(7.1,-.7)); draw((6.5,-0.86602540378443864676372317075294)--(6.8,-1.5)); draw((14,0)--(13,1.7320508075688772935274463415059)--(11,1.7320508075688772935274463415059)--(10,0)--(11,-1.7320508075688772935274463415059)--(13,-1.7320508075688772935274463415059)--cycle,linewidth(1)); draw((14,0)--(13,-1.7320508075688772935274463415059)--(14.7320508075688772935274463415059,-2.7320508075688772935274463415059)--(15.7320508075688772935274463415059,-1)--cycle,linewidth(1)); label("3.",(12,-2.5),S); draw((21,0)--(20,1.7320508075688772935274463415059)--(18,1.7320508075688772935274463415059)--(17,0)--(18,-1.7320508075688772935274463415059)--(20,-1.7320508075688772935274463415059)--cycle,linewidth(1)); draw((18,-1.7320508075688772935274463415059)--(20,-1.7320508075688772935274463415059)--(20,-3.7320508075688772935274463415059)--(18,-3.7320508075688772935274463415059)--cycle,linewidth(1)); label("4.",(19,-4),S); [/asy]$

The square in the first diagram "rolls" clockwise around the fixed regular hexagon until it reaches the bottom. In which position will the solid triangle be in diagram $4$ ?

$[asy] unitsize(12); label("(A)",(0,0),W); fill((1,-1)--(1,1)--(5,0)--cycle,black); label("(B)",(6,0),E); fill((9,-2)--(11,-2)--(10,1)--cycle,black); label("(C)",(14,0),E); fill((17,1)--(19,1)--(18,-1.8)--cycle,black); label("(D)",(22,0),E); fill((25,-1)--(27,-2)--(28,1)--cycle,black); label("(E)",(31,0),E); fill((33,0)--(37,1)--(37,-1)--cycle,black); [/asy]$

Problem 25

A palindrome is a whole number that reads the same forwards and backwards. If one neglects the colon, certain times displayed on a digital watch are palindromes. Three examples are: $boxed{1:01}$ , $boxed{4:44}$ , and $boxed{12:21}$ . How many times during a $12$ -hour period will be palindromes?

$text{(A)} 57 qquad text{(B)} 60 qquad text{(C)} 63 qquad text{(D)} 90 qquad text{(E)} 93$

1988 AMC8 真题答案详细解析

1.Clearly the arrow marks a value between $10.25$ and $10.5$ , so only $text{C}$ and $text{D}$ are possible.

Looking, we see that the arrow is closer to $10.3$ , so $boxed{text{D}}$

2.Converting the decimals to fractions, this is $begin{align*} 8times frac{1}{4} times 2times frac{1}{8} &= frac{8times 2}{4times 8} \ &= frac{16}{32} \ &= frac{1}{2} rightarrow boxed{text{C}} end{align*}$

3.Each of the fractions simplify to $frac{1}{10}$ , so this sum is $begin{align*} frac{1}{10}+frac{1}{10}+frac{1}{10} &= frac{3}{10} \ &= .3 rightarrow boxed{text{D}} end{align*}$

4.If, for a moment, we disregard the white squares, we notice that the number of black squares in each row increases by 1 continuously as we go down the pyramid. Thus, the number of black squares is $1 + 2 + cdots + 8$ .

Same goes for the white squares, except it starts a row later, making it $1 + 2 + cdots + 7$ .

Subtracting the number of white squares from the number of black squares... $[1 + 2 + cdots + 7 + 8 - (1 + 2 + cdots + 7) = 8 Rightarrow (B)]$

5.We have that $20^{circ}+angle ABD +angle CBD=160^{circ}$ , or $angle ABD +angle CBD=140^{circ}$ . Since $angle CBD$ is a right angle, we have $angle ABD=140^{circ}-90^{circ}=50^{circ}Rightarrow mathrm{(C)}$ .

6.Converting the decimals to fractions gives us $frac{(.2)^3}{(.02)^2} =frac{left( frac{1}{5}right)^3}{left(frac{1}{50}right)^2}=frac{50^2}{5^3}=frac{2500}{125}=20Rightarrow mathrm{(E)}$ .

7.We estimate the first thing to be $2.5$ , the second thing to be $8$ , and the third thing to be $10$ . We now have $2.5cdot 8cdot 10=25cdot 8=200Rightarrow mathrm{(B)}$ .

8.The decimal point of 0.075 is three away from what Betty punched in, and that of 2.56 is two away. The decimal point is therefore $3+2=5$ units to the left of where it should be, so we would want $.19200Rightarrow mathrm{(B)}$ .

9.The first triangle has two legs of length $sqrt{5}$ , the second has two legs of length 2, the leg lengths of the third triangle are $2$ , $sqrt{5}$ , and $sqrt{13}$ , two legs of the fourth triangle have length $sqrt{10}$ , and two legs of the fifth triangle have length $sqrt{5}$ . Therefore all of the triangles in the diagram except the third are isosceles, and there are $4Rightarrow mathrm{(D)}$ are isosceles.

10.

Solution 1

7 days after her birthday would be a Thursday, as would 14, 21, 28, 35, 42, 49, and 56. Therefore the 60th would be four days after a Thursday, which is a $text{Monday} Rightarrow mathrm{(A)}$ .

Solution 2

Note that $60equiv4pmod7$ . We count 4 days past Thursday, and arrive at Monday. $mathrm{(A)}$

11.Note that if $1leq a<b<c$ , then $sqrt{a}<sqrt{b}<sqrt{c}$ .

Since $1<144<164<169$ , we can say that $[sqrt{144}<sqrt{164}<sqrt{169} Rightarrow 12<sqrt{164}<13 rightarrow boxed{text{E}}]$

12.We want the cost per person, which is $begin{align*} frac{20text{ billion}}{250text{ million}} &= frac{20000text{ million}}{250text{ million}} \ &= 80 rightarrow boxed{text{C}} end{align*}$

13.The circumference of the patio is $2cdot 12cdot pi =24pi approx 75.398$ . Since the bushes are spaced $1$ foot apart, about $75rightarrow boxed{text{D}}$ are needed.

14.Since it doesn't take too long, we can just make a table with all the possible values of the sum: $[begin{array}{|c|c|c|} multicolumn{3}{}{} \ hline diamondsuit & Delta & diamondsuit + Delta \ hline 36 & 1 & 37 \ hline 18 & 2 & 20 \ hline 12 & 3 & 15 \ hline 9 & 4 & 13 \ hline 6 & 6 & 12 \ hline end{array}]$

15.The reciprocal for a fraction $frac{a}{b}$ turns out to be $frac{b}{a}$ , so if we can express the expression as a single fraction, we're basically done.

The expression is equal to $frac{3}{6}+frac{2}{6}=frac{5}{6}$ , so the reciprocal is $frac{6}{5}rightarrow boxed{text{C}}$ .

16.By the Pigeonhole Principle, if there are at least $7$ $text{X}$ 's, then there will be some row with $3$ $text{X}$ 's. We can put in $6$ by leaving out the three boxes in one of the main diagonals.

$rightarrow boxed{text{E}}$

17.Looking at the diagram, the shaded region is the union of two rectangles, with a small rectangle as overlap. If we just add the areas of the two rectangles, then we'll overcount the small rectangle, so we must subtract that area to get the desired area.

The areas of the two larger rectangles are $2cdot 10=20$ and $3cdot 8=24$ , and the area of the small rectangle is $2cdot 3=6$ . The desired area is thus $20+24-6=38 rightarrow boxed{text{B}}$ .

18.Let the $6$ boys have total weight $S_B$ and let the $4$ girls have total weight $S_G$ . We are given

$begin{align*} frac{S_B}{6} &= 150 \ frac{S_G}{4} &= 120 end{align*}$

We want the average of the $10$ children, which is $[frac{S_B+S_G}{10}]$ From the first two equations, we can determine that $S_B=900$ and $S_G=480$ , so $S_B+S_G=1380$ . Therefore, the average we desire is $[frac{1380}{10}=138 rightarrow boxed{text{C}}]$

19.To get from the $1^text{st}$ term of an arithmetic sequence to the $100^text{th}$ term, we must add the common difference $99$ times. The first term is $1$ and the common difference is $5-1=9-5=13-9=cdots = 4$ , so the $100^text{th}$ term is $[1+4(99)=397 rightarrow boxed{text{A}}]$

20.Let the amount of coffee the maker will hold when full be $x$ . Then, $[.36x=45 Rightarrow x=125 rightarrow boxed{text{C}}]$

21.

The possible medians after $n$ is added are $6$ , $n$ , or $9$ . Now we use casework.

Case 1: The median is $6$

In this case, $n<6$ and $[frac{3+n+6+9+10}{5}=6 Rightarrow n=2]$ so this case contributes $1$ .

Case 2: The median is $n$

We have $6<n<9$ and $[frac{3+6+n+9+10}{5}=n Rightarrow n=7]$ so this case also contributes $1$ .

Case 3: The median is $9$

We have $9<n$ and $[frac{3+6+9+n+10}{5}=9 Rightarrow 17]$ so this case adds $1$ .

In all there are $3rightarrow boxed{text{C}}$ possible values of $n$ .

22.Let the original price of an item be $x$ . The shoppe originally increased this to $1.25x$ . The sale brings it down to $.8(1.25x)=x$ , which is the same as the original $rightarrow boxed{text{E}}$

23.This is the equivalent of saying she buys $12$ for $$15$ and sells $12$ for $$20$ , so for every dozen disks she sells, she profits $$5$ .

She needs to profit $$100$ , so she needs to sell $frac{100}{5}=20$ dozen disks, which is $240rightarrow boxed{text{D}}$

24.

Solution 1

The inner angle of the hexagon is $120^circ$ , and the inner angle of the square is $90^circ$ . Hence during each rotation the square is rotated by $360^circ - 120^circ - 90^circ = 150^circ$ clockwise. In the diagram $4$ the square is therefore rotated by $3cdot 150^circ = 450^circ$ clockwise from its original state. Rotation by $450^circ$ is identical to rotation by $450^circ - 360^circ = 90^circ$ , hence the black triangle in the diagram $4$ will be pointing to the right, and the answer is $boxed{text{(A)}}$ .

Solution 2

Alternately, we can simply keep track of the "bottom" side of the square. In the diagrams below, this bottom side is shown in red.

25.From $1$ to $9$ , the times will be of the form $boxed{a:ba}$ . There are $9$ choices for $a$ and $6$ choices for $b$ , so there are $9cdot 6 =54$ times in this period.