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1989 AMC8真题及答案详解

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月4日下午12:20

1989 AMC 8 真题

答案详细解析请参考文末

Problem 1

$(1+11+21+31+41)+(9+19+29+39+49)=$

$text{(A)} 150 qquad text{(B)} 199 qquad text{(C)} 200 qquad text{(D)} 249 qquad text{(E)} 250$

Problem 2

$frac{2}{10}+frac{4}{100}+frac{6}{1000} =$

$text{(A)} .012 qquad text{(B)} .0246 qquad text{(C)} .12 qquad text{(D)} .246 qquad text{(E)} 246$

Problem 3

Which of the following numbers is the largest?

$text{(A)} .99 qquad text{(B)} .9099 qquad text{(C)} .9 qquad text{(D)} .909 qquad text{(E)} .9009$

Problem 4

Estimate to determine which of the following numbers is closest to $frac{401}{.205}$ .

$text{(A)} .2 qquad text{(B)} 2 qquad text{(C)} 20 qquad text{(D)} 200 qquad text{(E)} 2000$

Problem 5

$-15+9times (6div 3) =$

$text{(A)} -48 qquad text{(B)} -12 qquad text{(C)} -3 qquad text{(D)} 3 qquad text{(E)} 12$

Problem 6

If the markings on the number line are equally spaced, what is the number $text{y}$ ?

$[asy] draw((-4,0)--(26,0),Arrows); for(int a=0; a<6; ++a) { draw((4a,-1)--(4a,1)); } label("0",(0,-1),S); label("20",(20,-1),S); label("y",(12,-1),S); [/asy]$

$text{(A)} 3 qquad text{(B)} 10 qquad text{(C)} 12 qquad text{(D)} 15 qquad text{(E)} 16$

Problem 7

If the value of $20$ quarters and $10$ dimes equals the value of $10$ quarters and $n$ dimes, then $n=$

$text{(A)} 10 qquad text{(B)} 20 qquad text{(C)} 30 qquad text{(D)} 35 qquad text{(E)} 45$

Problem 8

$(2times 3times 4)left(frac{1}{2}+frac{1}{3}+frac{1}{4}right) =$

$text{(A)} 1 qquad text{(B)} 3 qquad text{(C)} 9 qquad text{(D)} 24 qquad text{(E)} 26$

Problem 9

There are $2$ boys for every $3$ girls in Ms. Johnson's math class. If there are $30$ students in her class, what percent of them are boys?

$text{(A)} 12% qquad text{(B)} 20% qquad text{(C)} 40% qquad text{(D)} 60% qquad text{(E)} 66frac{2}{3}%$

Problem 10

What is the number of degrees in the smaller angle between the hour hand and the minute hand on a clock that reads seven o'clock?

$text{(A)} 50^circ qquad text{(B)} 120^circ qquad text{(C)} 135^circ qquad text{(D)} 150^circ qquad text{(E)} 165^circ$

Problem 11

Which of the five "T-like shapes" would be symmetric to the one shown with respect to the dashed line?

$[asy] unitsize(48); for (int a=0; a<3; ++a) { fill((2a+1,1)--(2a+.8,1)--(2a+.8,.8)--(2a+1,.8)--cycle,black); } draw((.8,1)--(0,1)--(0,0)--(1,0)--(1,.8)); draw((2.8,1)--(2,1)--(2,0)--(3,0)--(3,.8)); draw((4.8,1)--(4,1)--(4,0)--(5,0)--(5,.8)); draw((.2,.4)--(.6,.8),linewidth(1)); draw((.4,.6)--(.8,.2),linewidth(1)); draw((2.4,.8)--(2.8,.4),linewidth(1)); draw((2.6,.6)--(2.2,.2),linewidth(1)); draw((4.4,.2)--(4.8,.6),linewidth(1)); draw((4.6,.4)--(4.2,.8),linewidth(1)); draw((7,.2)--(7,1)--(6,1)--(6,0)--(6.8,0)); fill((6.8,0)--(7,0)--(7,.2)--(6.8,.2)--cycle,black); draw((6.2,.6)--(6.6,.2),linewidth(1)); draw((6.4,.4)--(6.8,.8),linewidth(1)); draw((8,.8)--(8,0)--(9,0)--(9,1)--(8.2,1)); fill((8,1)--(8,.8)--(8.2,.8)--(8.2,1)--cycle,black); draw((8.4,.8)--(8.8,.8),linewidth(1)); draw((8.6,.8)--(8.6,.2),linewidth(1)); draw((6,1.2)--(6,1.4)); draw((6,1.6)--(6,1.8)); draw((6,2)--(6,2.2)); draw((6,2.4)--(6,2.6)); draw((6.4,2.2)--(6.4,1.4)--(7.4,1.4)--(7.4,2.4)--(6.6,2.4)); fill((6.4,2.4)--(6.4,2.2)--(6.6,2.2)--(6.6,2.4)--cycle,black); draw((6.6,1.8)--(7,2.2),linewidth(1)); draw((6.8,2)--(7.2,1.6),linewidth(1)); label("(A)",(0,1),W); label("(B)",(2,1),W); label("(C)",(4,1),W); label("(D)",(6,1),W); label("(E)",(8,1),W); [/asy]$

Problem 12

$frac{1-frac{1}{3}}{1-frac{1}{2}} =$

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

Problem 13

$frac{9}{7times 53} =$

$text{(A)} frac{.9}{.7times 53} qquad text{(B)} frac{.9}{.7times .53} qquad text{(C)} frac{.9}{.7times 5.3} qquad text{(D)} frac{.9}{7times .53} qquad text{(E)} frac{.09}{.07times .53}$

Problem 14

When placing each of the digits $2,4,5,6,9$ in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?

$text{(A)} 58 qquad text{(B)} 123 qquad text{(C)} 149 qquad text{(D)} 171 qquad text{(E)} 176$

$[begin{tabular}[t]{cccc} & boxed{} & boxed{} & boxed{} \ - & & boxed{} & boxed{} \ hline end{tabular}]$

Problem 15

The area of the shaded region $text{BEDC}$ in parallelogram $text{ABCD}$ is

$[asy] unitsize(10); pair A,B,C,D,E; A=origin; B=(4,8); C=(14,8); D=(10,0); E=(4,0); draw(A--B--C--D--cycle); fill(B--E--D--C--cycle,gray); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,S); label("$10$",(9,8),N); label("$6$",(7,0),S); label("$8$",(4,4),W); draw((3,0)--(3,1)--(4,1)); [/asy]$

$text{(A)} 24 qquad text{(B)} 48 qquad text{(C)} 60 qquad text{(D)} 64 qquad text{(E)} 80$

Problem 16

In how many ways can $47$ be written as the sum of two primes?

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

Problem 17

The number $text{N}$ is between $9$ and $17$ . The average of $6$ , $10$ , and $text{N}$ could be

$text{(A)} 8 qquad text{(B)} 10 qquad text{(C)} 12 qquad text{(D)} 14 qquad text{(E)} 16$

Problem 18

Many calculators have a reciprocal key $boxed{frac{1}{x}}$ that replaces the current number displayed with its reciprocal. For example, if the display is $boxed{00004}$ and the $boxed{frac{1}{x}}$ key is pressed, then the display becomes $boxed{000.25}$ . If $boxed{00032}$ is currently displayed, what is the fewest positive number of times you must depress the $boxed{frac{1}{x}}$ key so the display again reads $boxed{00032}$ ?

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

Problem 19

The graph below shows the total accumulated dollars (in millions) spent by the Surf City government during $1988$ . For example, about $.5$ million had been spent by the beginning of February and approximately $2$ million by the end of April. Approximately how many millions of dollars were spent during the summer months of June, July, and August?

$text{(A)} 1.5 qquad text{(B)} 2.5 qquad text{(C)} 3.5 qquad text{(D)} 4.5 qquad text{(E)} 5.5$

$[asy] unitsize(18); for (int a=1; a<13; ++a) { draw((a,0)--(a,.5)); } for (int b=1; b<6; ++b) { draw((-.5,2b)--(0,2b)); } draw((0,0)--(0,12)); draw((0,0)--(14,0)); draw((0,0)--(1,.9)--(2,1.9)--(3,2.6)--(4,4.3)--(5,4.5)--(6,5.7)--(7,8.2)--(8,9.4)--(9,9.8)--(10,10.1)--(11,10.2)--(12,10.5)); label("J",(.5,0),S); label("F",(1.5,0),S); label("M",(2.5,0),S); label("A",(3.5,0),S); label("M",(4.5,0),S); label("J",(5.5,0),S); label("J",(6.5,0),S); label("A",(7.5,0),S); label("S",(8.5,0),S); label("O",(9.5,0),S); label("N",(10.5,0),S); label("D",(11.5,0),S); label("month F=February",(16,0),S); label("$1$",(-.6,2),W); label("$2$",(-.6,4),W); label("$3$",(-.6,6),W); label("$4$",(-.6,8),W); label("$5$",(-.6,10),W); label("dollars in millions",(0,11.9),N); [/asy]$

Problem 20

The figure may be folded along the lines shown to form a number cube. Three number faces come together at each corner of the cube. What is the largest sum of three numbers whose faces come together at a corner?

$[asy] draw((0,0)--(0,1)--(1,1)--(1,2)--(2,2)--(2,1)--(4,1)--(4,0)--(2,0)--(2,-1)--(1,-1)--(1,0)--cycle); draw((1,0)--(1,1)--(2,1)--(2,0)--cycle); draw((3,1)--(3,0)); label("$1$",(1.5,1.25),N); label("$2$",(1.5,.25),N); label("$3$",(1.5,-.75),N); label("$4$",(2.5,.25),N); label("$5$",(3.5,.25),N); label("$6$",(.5,.25),N); [/asy]$

$text{(A)} 11 qquad text{(B)} 12 qquad text{(C)} 13 qquad text{(D)} 14 qquad text{(E)} 15$

Problem 21

Jack had a bag of $128$ apples. He sold $25%$ of them to Jill. Next he sold $25%$ of those remaining to June. Of those apples still in his bag, he gave the shiniest one to his teacher. How many apples did Jack have then?

$text{(A)} 7 qquad text{(B)} 63 qquad text{(C)} 65 qquad text{(D)} 71 qquad text{(E)} 111$

Problem 22

The letters $text{A}$ , $text{J}$ , $text{H}$ , $text{S}$ , $text{M}$ , $text{E}$ and the digits $1$ , $9$ , $8$ , $9$ are "cycled" separately as follows and put together in a numbered list: $[begin{tabular}[t]{lccc} & & AJHSME & 1989 \ & & & \ 1. & & JHSMEA & 9891 \ 2. & & HSMEAJ & 8919 \ 3. & & SMEAJH & 9198 \ & & ........ & end{tabular}]$

What is the number of the line on which $text{AJHSME 1989}$ will appear for the first time?

$text{(A)} 6 qquad text{(B)} 10 qquad text{(C)} 12 qquad text{(D)} 18 qquad text{(E)} 24$

Problem 23

An artist has $14$ cubes, each with an edge of $1$ meter. She stands them on the ground to form a sculpture as shown. She then paints the exposed surface of the sculpture. How many square meters does she paint?

$text{(A)} 21 qquad text{(B)} 24 qquad text{(C)} 33 qquad text{(D)} 37 qquad text{(E)} 42$

$[asy] draw((0,0)--(2.35,-.15)--(2.44,.81)--(.09,.96)--cycle); draw((.783333333,-.05)--(.873333333,.91)--(1.135,1.135)); draw((1.566666667,-.1)--(1.656666667,.86)--(1.89,1.1)); draw((2.35,-.15)--(4.3,1.5)--(4.39,2.46)--(2.44,.81)); draw((3,.4)--(3.09,1.36)--(2.61,1.4)); draw((3.65,.95)--(3.74,1.91)--(3.29,1.94)); draw((.09,.96)--(.76,1.49)--(.71,1.17)--(2.2,1.1)--(3.6,2.2)--(3.62,2.52)--(4.39,2.46)); draw((.76,1.49)--(.82,1.96)--(2.28,1.89)--(2.2,1.1)); draw((2.28,1.89)--(3.68,2.99)--(3.62,2.52)); draw((1.455,1.135)--(1.55,1.925)--(1.89,2.26)); draw((2.5,2.48)--(2.98,2.44)--(2.9,1.65)); draw((.82,1.96)--(1.55,2.6)--(1.51,2.3)--(2.2,2.26)--(2.9,2.8)--(2.93,3.05)--(3.68,2.99)); draw((1.55,2.6)--(1.59,3.09)--(2.28,3.05)--(2.2,2.26)); draw((2.28,3.05)--(2.98,3.59)--(2.93,3.05)); draw((1.59,3.09)--(2.29,3.63)--(2.98,3.59)); [/asy]$

Problem 24

Suppose a square piece of paper is folded in half vertically. The folded paper is then cut in half along the dashed line. Three rectangles are formed-a large one and two small ones. What is the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle?

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

$[asy] draw((0,0)--(0,8)--(6,8)--(6,0)--cycle); draw((0,8)--(5,9)--(5,8)); draw((3,-1.5)--(3,10.3),dashed); draw((0,5.5)..(-.75,4.75)..(0,4)); draw((0,4)--(1.5,4),EndArrow); [/asy]$

Problem 25

Every time these two wheels are spun, two numbers are selected by the pointers. What is the probability that the sum of the two selected numbers is even?

$text{(A)} frac{1}{6} qquad text{(B)} frac{3}{7} qquad text{(C)} frac{1}{2} qquad text{(D)} frac{2}{3} qquad text{(E)} frac{5}{7}$

$[asy] unitsize(36); draw(circle((-3,0),1)); draw(circle((0,0),1)); draw((0,0)--dir(30)); draw((0,0)--(0,-1)); draw((0,0)--dir(150)); draw((-2.293,.707)--(-3.707,-.707)); draw((-2.293,-.707)--(-3.707,.707)); fill((-2.9,1)--(-2.65,1.25)--(-2.65,1.6)--(-3.35,1.6)--(-3.35,1.25)--(-3.1,1)--cycle,black); fill((.1,1)--(.35,1.25)--(.35,1.6)--(-.35,1.6)--(-.35,1.25)--(-.1,1)--cycle,black); label("$5$",(-3,.2),N); label("$3$",(-3.2,0),W); label("$4$",(-3,-.2),S); label("$8$",(-2.8,0),E); label("$6$",(0,.2),N); label("$9$",(-.2,.1),SW); label("$7$",(.2,.1),SE); [/asy]$

1989 AMC8真答案详细解析

1.We make use of the associative and commutative properties of addition to rearrange the sum as $begin{align*} (1+49)+(11+39)+(21+29)+(31+19)+(41+9) &= 50+50+50+50+50 \ &= 250 Longrightarrow boxed{text{E}} end{align*}$

2. $begin{align*} frac{2}{10}+frac{4}{100}+frac{6}{1000} &= frac{200}{1000}+frac{40}{1000}+frac{6}{1000} \ &= frac{246}{1000} \ &= .246 rightarrow boxed{text{D}} end{align*}$

3.We have $.99>.9099>.909>.9009>.9$ , so choice $boxed{text{A}}$ is the largest.

4. $401$ is around $400$ and $.205$ is around $.2$ so the fraction is approximately $[frac{400}{.2}=2000rightarrow boxed{text{E}}]$

5.We use the order of operations here to get

$begin{align*} -15+9times (6div 3) &= -15+9times 2 \ &= -15+18 \ &= 3 rightarrow boxed{text{D}} end{align*}$

6.Five steps are taken to get from $0$ to $20$ . Each step is of equal size, so each step is $4$ . Three steps are taken from $0$ to $y$ , so $y=3times 4=12rightarrow boxed{text{C}}$ .

7. $begin{align*} 20(25)+10(10) &= 10(25)+n(10) \ 600 &= 250+10n \ 35 &= n implies boxed{text{D}} end{align*}$

Solution 1

We use the distributive property to get $[3times 4+2times 4+2times 3 = 26 rightarrow boxed{text{E}}]$

Solution 2

Since $frac12+frac13+frac14 > frac12+frac14+frac14 = 1$ , we have $[(2times 3times 4)left(frac{1}{2}+frac{1}{3}+frac{1}{4}right) > 2times 3times 4 times 1 = 24]$ The only answer choice greater than $24$ is $boxed{text{E}}$ .

9.Besides ensuring the situation is possible, the $30$ students information is irrelevant.

From the first statement, we can deduce that $2$ of every $2+3=5$ students are boys. Thus, $2/5=40% rightarrow boxed{text{C}}$ of the students are boys.

10.The smaller angle makes up $5/12$ of the circle which is the clock. A circle is $360^circ$ , so the measure of the smaller angle is $[frac{5}{12}cdot 360^circ = 150^circ rightarrow boxed{text{D}}]$

11.Drawing the reflection, we see that it is $boxed{text{B}}$ . Imagine it as if it were a mirror reflection or if you were to flip it over the dashed line.

12. $begin{align*} frac{1-frac{1}{3}}{1-frac{1}{2}} &= frac{frac{2}{3}}{frac{1}{2}} \ &= frac{2}{3} times 2 \ &= frac{4}{3} rightarrow boxed{text{E}} end{align*}$

13. $begin{align*} frac{1-frac{1}{3}}{1-frac{1}{2}} &= frac{frac{2}{3}}{frac{1}{2}} \ &= frac{2}{3} times 2 \ &= frac{4}{3} rightarrow boxed{text{E}} end{align*}$

14.When trying to minimize $a-b$ , we minimize $a$ and maximize $b$ . Since in this problem, $a$ is three digit and $b$ is two digit, we set $a=245$ and $b=96$ . Their difference is $149rightarrow boxed{text{C}}$ .

15.

Solution 1

Let $[ABC]$ denote the area of figure $ABC$ .

Clearly, $[BEDC]=[ABCD]-[ABE]$ . Using basic area formulas,

$[ABCD]=(BC)(BE)=80$ $[ABE]=(BE)(AE)/2 = 4(AE)$ Since $AE+ED=BC=10$ and $ED=6$ , $AE=4$ and the area of $triangle ABE$ is $4(4)=16$ .

Finally, we have $[BEDC]=80-16=64rightarrow boxed{text{D}}$

Solution 2

Notice that $BEDC$ is a trapezoid. Therefore its area is $[8left(frac{6+10}{2}right)=8left(frac{16}{2}right)=8(8)=64Rightarrow mathrm{(D)}]$

16.For $47$ to be written as the sum of two integers, one must be odd and the other must be even. There is only one even prime, namely $1989 AMC8真题及答案详解$ , so one of the numbers must be $2$ , making the other $45$ .

However, $45$ is not prime, so there are no ways to write $47$ as the sum of two primes $rightarrow boxed{text{A}}$ .

17.

We know that $9<N<17$ and we wish to bound $frac{6+10+N}{3}=frac{16+N}{3}$ .

From what we know, we can deduce that $25<N+16<33$ , and thus $[8.overline{3}<frac{N+16}{3}<11]$

The only answer choice that falls in this range is choice $boxed{text{B}}$

18.Let $f(x)=frac{1}{x}$ . We have $[f(f(x))=frac{1}{frac{1}{x}}=x]$ Thus, we need to iterate the key pressing twice to get the display back to the original $rightarrow boxed{text{B}}$ .

19.Since we want to know how much money is spent in June, July and August, we need the difference between the amount of money spent by the beginning of June and the amount of money spent by the end of August.

We estimate these to be about $2.2$ million and $4.8$ million, respectively. The difference is $[4.8-2.2=2.6approx 2.5 rightarrow boxed{text{B}}]$

20.It is clear that $6$ , $5$ , and $4$ will not come together to get a sum of $15$ .

The faces $6$ , $5$ , and $3$ come together at a common vertex, making the maximal sum $6+5+3=14rightarrow boxed{text{D}}$ .

21.First he gives $128times .25 = 32$ apples to Jill, so he has $128-32=96$ apples left. Then he gives $96times .25 = 24$ apples to June, so he has $96-24=72$ left.

Finally, he gives one to the teacher, leaving $71rightarrow boxed{text{D}}$

22.Every $4text{th}$ line has $1989$ as part of it and every $6text{th}$ line has $text{AJHSME}$ as part of it. In order for both to be part of line $n$ , $n$ must be a multiple of $4$ and $6$ , the least of which is $text{lcm}(4,6)=12rightarrow boxed{text{C}}$ .

23.

We can consider the contributions of the sides of the three layers and the tops of the layers separately.

Layer $n$ (counting from the top starting at $1$ ) has $4$ side faces each with $n$ unit squares, so the sides of the pyramid contribute $4+8+12=24$ for the surface area.

The tops of the layers when combined form the same arrangement of unit cubes as the bottom of the pyramid, which is a $3times 3$ square, hence this contributes $9$ for the surface area.

Thus, the artist paints $24+9=33 rightarrow boxed{text{C}}$ square meters.

24.

From here on a blue line represents a cut, the dashed line represents the fold.

$[asy] draw((0,0)--(4,0)--(4,4)--(0,4)--cycle,linewidth(1)); draw((1,0)--(1,4),blue+linewidth(1)); draw((2,0)--(2,4),dashed); draw((3,0)--(3,4),blue+linewidth(1)); label("$x$",(0.5,4),N); label("$x$",(1.5,4),N); label("$x$",(2.5,4),N); label("$x$",(3.5,4),N); label("$4x$",(0,2),W); [/asy]$

From the diagram, we can tell the perimeter of one of the small rectangles is $2(4x+x)=10x$ and the perimeter of the large rectangle is $2(4x+2x)=12x$ . The desired ratio is $[frac{10x}{12x}=5/6rightarrow boxed{text{E}}]$

25.

For the sum to be even, the two selected numbers must have the same parity.

The first spinner has $2$ odd numbers and $2$ even, so no matter what the second spinner is, there is a $1/2$ chance the first spinner lands on a number with the same parity, so the probability of an even sum is $1/2rightarrow boxed{text{C}}$ .