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1990 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12:49

1990 AMC 8 真题

答案详细解析请参考文末

Problem 1

What is the smallest sum of two $3$ -digit numbers that can be obtained by placing each of the six digits $4,5,6,7,8,9$ in one of the six boxes in this addition problem?

$[asy] unitsize(12); draw((0,0)--(10,0)); draw((-1.5,1.5)--(-1.5,2.5)); draw((-1,2)--(-2,2)); draw((1,1)--(3,1)--(3,3)--(1,3)--cycle); draw((1,4)--(3,4)--(3,6)--(1,6)--cycle); draw((4,1)--(6,1)--(6,3)--(4,3)--cycle); draw((4,4)--(6,4)--(6,6)--(4,6)--cycle); draw((7,1)--(9,1)--(9,3)--(7,3)--cycle); draw((7,4)--(9,4)--(9,6)--(7,6)--cycle); [/asy]$

$text{(A)} 947 qquad text{(B)} 1037 qquad text{(C)} 1047 qquad text{(D)} 1056 qquad text{(E)} 1245$

Problem 2

Which digit of $.12345$ , when changed to $9$ , gives the largest number?

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

Problem 3

What fraction of the square is shaded?

$[asy] draw((0,0)--(0,3)--(3,3)--(3,0)--cycle); draw((0,2)--(2,2)--(2,0)); draw((0,1)--(1,1)--(1,0)); draw((0,0)--(3,3)); fill((0,0)--(0,1)--(1,1)--cycle,grey); fill((1,0)--(1,1)--(2,2)--(2,0)--cycle,grey); fill((0,2)--(2,2)--(3,3)--(0,3)--cycle,grey); [/asy]$

$text{(A)} frac{1}{3} qquad text{(B)} frac{2}{5} qquad text{(C)} frac{5}{12} qquad text{(D)} frac{3}{7} qquad text{(E)} frac{1}{2}$

Problem 4

Which of the following could not be the unit's digit [one's digit] of the square of a whole number?

$text{(A)} 1 qquad text{(B)} 4 qquad text{(C)} 5 qquad text{(D)} 6 qquad text{(E)} 8$

Problem 5

Which of the following is closest to the product $(.48017)(.48017)(.48017)$ ?

$text{(A)} 0.011 qquad text{(B)} 0.110 qquad text{(C)} 1.10 qquad text{(D)} 11.0 qquad text{(E)} 110$

Problem 6

Which of these five numbers is the largest?

$text{(A)} 13579+frac{1}{2468} qquad text{(B)} 13579-frac{1}{2468} qquad text{(C)} 13579times frac{1}{2468}$

$text{(D)} 13579div frac{1}{2468} qquad text{(E)} 13579.2468$

Problem 7

When three different numbers from the set ${ -3, -2, -1, 4, 5 }$ are multiplied, the largest possible product is

$text{(A)} 10 qquad text{(B)} 20 qquad text{(C)} 30 qquad text{(D)} 40 qquad text{(E)} 60$

Problem 8

A dress originally priced at $80$ dollars was put on sale for $25%$ off. If $10%$ tax was added to the sale price, then the total selling price (in dollars) of the dress was

$text{(A)} text{45 dollars} qquad text{(B)} text{52 dollars} qquad text{(C)} text{54 dollars} qquad text{(D)} text{66 dollars} qquad text{(E)} text{68 dollars}$

Problem 9

The grading scale shown is used at Jones Junior High. The fifteen scores in Mr. Freeman's class were: $[begin{tabular}[t]{lllllllll} 89, & 72, & 54, & 97, & 77, & 92, & 85, & 74, & 75, \ 63, & 84, & 78, & 71, & 80, & 90. & & & \ end{tabular}]$

In Mr. Freeman's class, what percent of the students received a grade of C?

$[boxed{begin{tabular}[t]{cc} A: & 93 - 100 \ B: & 85 - 92 \ C: & 75 - 84 \ D: & 70 - 74 \ F: & 0 - 69 end{tabular}}]$

$text{(A)} 20% qquad text{(B)} 25% qquad text{(C)} 30% qquad text{(D)} 33frac{1}{3}% qquad text{(E)} 40%$

Problem 10

On this monthly calendar, the date behind one of the letters is added to the date behind $text{C}$ . If this sum equals the sum of the dates behind $text{A}$ and $text{B}$ , then the letter is

$[asy] unitsize(12); draw((1,1)--(23,1)); draw((0,5)--(23,5)); draw((0,9)--(23,9)); draw((0,13)--(23,13)); for(int a=0; a<6; ++a) { draw((4a+2,0)--(4a+2,14)); } label("Tues.",(4,14),N); label("Wed.",(8,14),N); label("Thurs.",(12,14),N); label("Fri.",(16,14),N); label("Sat.",(20,14),N); label("C",(12,10.3),N); label("$textbf{A}$",(16,10.3),N); label("Q",(12,6.3),N); label("S",(4,2.3),N); label("$textbf{B}$",(8,2.3),N); label("P",(12,2.3),N); label("T",(16,2.3),N); label("R",(20,2.3),N); [/asy]$

$text{(A)} text{P} qquad text{(B)} text{Q} qquad text{(C)} text{R} qquad text{(D)} text{S} qquad text{(E)} text{T}$

Problem 11

The numbers on the faces of this cube are consecutive whole numbers. The sums of the two numbers on each of the three pairs of opposite faces are equal. The sum of the six numbers on this cube is

$[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5,2)--(5,5)--(2,5)--(0,3)); draw((3,3)--(5,5)); label("$15$",(1.5,1.2),N); label("$11$",(4,2.3),N); label("$14$",(2.5,3.7),N); [/asy]$

$text{(A)} 75 qquad text{(B)} 76 qquad text{(C)} 78 qquad text{(D)} 80 qquad text{(E)} 81$

Problem 12

There are twenty-four $4$ -digit numbers that use each of the four digits $2$ , $4$ , $5$ , and $7$ exactly once. Listed in numerical order from smallest to largest, the number in the $17text{th}$ position in the list is

$text{(A)} 4527 qquad text{(B)} 5724 qquad text{(C)} 5742 qquad text{(D)} 7245 qquad text{(E)} 7524$

Problem 13

One proposal for new postage rates for a letter was $30$ cents for the first ounce and $22$ cents for each additional ounce (or fraction of an ounce). The postage for a letter weighing $4.5$ ounces was

$text{(A)} text{96 cents} qquad text{(B)} text{1.07 dollars} qquad text{(C)} text{1.18 dollars} qquad text{(D)} text{1.20 dollars} qquad text{(E)} text{1.40 dollars}$

Problem 14

A bag contains only blue balls and green balls. There are $6$ blue balls. If the probability of drawing a blue ball at random from this bag is $frac{1}{4}$ , then the number of green balls in the bag is

$text{(A)} 12 qquad text{(B)} 18 qquad text{(C)} 24 qquad text{(D)} 30 qquad text{(E)} 36$

Problem 15

The area of this figure is $100text{ cm}^2$ . Its perimeter is

$[asy] draw((0,2)--(2,2)--(2,1)--(3,1)--(3,0)--(1,0)--(1,1)--(0,1)--cycle,linewidth(1)); draw((1,2)--(1,1)--(2,1)--(2,0),dashed); [/asy]$

[figure consists of four identical squares] $text{(A)} text{20 cm} qquad text{(B)} text{25 cm} qquad text{(C)} text{30 cm} qquad text{(D)} text{40 cm} qquad text{(E)} text{50 cm}$

Problem 16

$1990-1980+1970-1960+cdots -20+10 =$

$text{(A)} -990 qquad text{(B)} -10 qquad text{(C)} 990 qquad text{(D)} 1000 qquad text{(E)} 1990$

Problem 17

A straight concrete sidewalk is to be $3$ feet wide, $60$ feet long, and $3$ inches thick. How many cubic yards of concrete must a contractor order for the sidewalk if concrete must be ordered in a whole number of cubic yards?

$text{(A)} 2 qquad text{(B)} 5 qquad text{(C)} 12 qquad text{(D)} 20 qquad text{(E)} text{more than 20}$

Problem 18

Each corner of a rectangular prism is cut off. Two (of the eight) cuts are shown. How many edges does the new figure have?

$[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5,2)--(5,5)--(2,5)--(0,3)); draw((3,3)--(5,5)); draw((2,0)--(3,1.8)--(4,1)--cycle,linewidth(1)); draw((2,3)--(4,4)--(3,2)--cycle,linewidth(1)); [/asy]$

$text{(A)} 24 qquad text{(B)} 30 qquad text{(C)} 36 qquad text{(D)} 42 qquad text{(E)} 48$

Assume that the planes cutting the prism do not intersect anywhere in or on the prism.

Problem 19

There are $120$ seats in a row. What is the fewest number of seats that must be occupied so the next person to be seated must sit next to someone?

$text{(A)} 30 qquad text{(B)} 40 qquad text{(C)} 41 qquad text{(D)} 60 qquad text{(E)} 119$

Problem 20

The annual incomes of $1,000$ families range from $8200$ dollars to $98,000$ dollars. In error, the largest income was entered on the computer as $980,000$ dollars. The difference between the mean of the incorrect data and the mean of the actual data is

$text{(A)} text{882 dollars} qquad text{(B)} text{980 dollars} qquad text{(C)} text{1078 dollars} qquad text{(D)} text{482,000 dollars} qquad text{(E)} text{882,000 dollars}$

Problem 21

A list of $8$ numbers is formed by beginning with two given numbers. Each new number in the list is the product of the two previous numbers. Find the first number if the last three are shown: $[text{underline{hspace{3 mm}?hspace{3 mm}}hspace{1 mm},hspace{1 mm} underline{hspace{7 mm}}hspace{1 mm},hspace{1 mm} underline{hspace{7 mm}}hspace{1 mm},hspace{1 mm} underline{hspace{7 mm}}hspace{1 mm},hspace{1 mm} underline{hspace{7 mm}}hspace{1 mm},hspace{1 mm}underline{hspace{2 mm}16hspace{2 mm}}hspace{1 mm},hspace{1 mm}underline{hspace{2 mm}64hspace{2 mm}}hspace{1 mm},hspace{1 mm}underline{hspace{1 mm}1024hspace{1 mm}}}]$ $text{(A)} frac{1}{64} qquad text{(B)} frac{1}{4} qquad text{(C)} 1 qquad text{(D)} 2 qquad text{(E)} 4$

Problem 22

Several students are seated at a large circular table. They pass around a bag containing $100$ pieces of candy. Each person receives the bag, takes one piece of candy and then passes the bag to the next person. If Chris takes the first and last piece of candy, then the number of students at the table could be

$text{(A)} 10 qquad text{(B)} 11 qquad text{(C)} 19 qquad text{(D)} 20 qquad text{(E)} 25$

Problem 23

The graph relates the distance traveled [in miles] to the time elapsed [in hours] on a trip taken by an experimental airplane. During which hour was the average speed of this airplane the largest?

$[asy] unitsize(12); for(int a=1; a<13; ++a) { draw((2a,-1)--(2a,1)); } draw((-1,4)--(1,4)); draw((-1,8)--(1,8)); draw((-1,12)--(1,12)); draw((-1,16)--(1,16)); draw((0,0)--(0,17)); draw((-5,0)--(33,0)); label("$0$",(0,-1),S); label("$1$",(2,-1),S); label("$2$",(4,-1),S); label("$3$",(6,-1),S); label("$4$",(8,-1),S); label("$5$",(10,-1),S); label("$6$",(12,-1),S); label("$7$",(14,-1),S); label("$8$",(16,-1),S); label("$9$",(18,-1),S); label("$10$",(20,-1),S); label("$11$",(22,-1),S); label("$12$",(24,-1),S); label("Time in hours",(11,-2),S); label("$500$",(-1,4),W); label("$1000$",(-1,8),W); label("$1500$",(-1,12),W); label("$2000$",(-1,16),W); label(rotate(90)*"Distance traveled in miles",(-4,10),W); draw((0,0)--(2,3)--(4,7.2)--(6,8.5)); draw((6,8.5)--(16,12.5)--(18,14)--(24,15)); [/asy]$

$text{(A)} text{first (0-1)} qquad text{(B)} text{second (1-2)} qquad text{(C)} text{third (2-3)} qquad text{(D)} text{ninth (8-9)} qquad text{(E)} text{last (11-12)}$

Problem 24

Three $Delta$ 's and a $diamondsuit$ will balance nine $bullet$ 's. One $Delta$ will balance a $diamondsuit$ and a $bullet$ .

$[asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,2)--(-12,0)--(12,0)--(12,2)); draw(ellipse((-12,5),8,3)); draw(ellipse((12,5),8,3)); label("$Delta hspace{2 mm}Delta hspace{2 mm}Delta hspace{2 mm}diamondsuit $",(-12,6.5),S); label("$bullet hspace{2 mm}bullet hspace{2 mm}bullet hspace{2 mm} bullet $",(12,5.2),N); label("$bullet hspace{2 mm}bullet hspace{2 mm}bullet hspace{2 mm}bullet hspace{2 mm}bullet $",(12,5.2),S); fill((44,0)--(40,-2)--(48,-2)--cycle,black); draw((34,2)--(34,0)--(54,0)--(54,2)); draw(ellipse((34,5),6,3)); draw(ellipse((54,5),6,3)); label("$Delta $",(34,6.5),S); label("$bullet hspace{2 mm}diamondsuit $",(54,6.5),S); [/asy]$

How many $bullet$ 's will balance the two $diamondsuit$ 's in this balance?

$[asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,4)--(-12,2)--(12,-2)--(12,0)); draw(ellipse((-12,7),6.5,3)); draw(ellipse((12,3),6.5,3)); label("$?$",(-12,8.5),S); label("$diamondsuit hspace{2 mm}diamondsuit $",(12,4.5),S); [/asy]$

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

Problem 25

How many different patterns can be made by shading exactly two of the nine squares? Patterns that can be matched by flips and/or turns are not considered different. For example, the patterns shown below are not considered different.

$[asy] fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,gray); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle,gray); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,linewidth(1)); draw((2,0)--(2,3),linewidth(1)); draw((0,1)--(3,1),linewidth(1)); draw((1,0)--(1,3),linewidth(1)); draw((0,2)--(3,2),linewidth(1)); fill((6,0)--(8,0)--(8,1)--(6,1)--cycle,gray); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle,linewidth(1)); draw((8,0)--(8,3),linewidth(1)); draw((6,1)--(9,1),linewidth(1)); draw((7,0)--(7,3),linewidth(1)); draw((6,2)--(9,2),linewidth(1)); fill((14,1)--(15,1)--(15,3)--(14,3)--cycle,gray); draw((12,0)--(15,0)--(15,3)--(12,3)--cycle,linewidth(1)); draw((14,0)--(14,3),linewidth(1)); draw((12,1)--(15,1),linewidth(1)); draw((13,0)--(13,3),linewidth(1)); draw((12,2)--(15,2),linewidth(1)); fill((18,1)--(19,1)--(19,3)--(18,3)--cycle,gray); draw((18,0)--(21,0)--(21,3)--(18,3)--cycle,linewidth(1)); draw((20,0)--(20,3),linewidth(1)); draw((18,1)--(21,1),linewidth(1)); draw((19,0)--(19,3),linewidth(1)); draw((18,2)--(21,2),linewidth(1)); [/asy]$

$text{(A)} 3 qquad text{(B)} 6 qquad text{(C)} 8 qquad text{(D)} 12 qquad text{(E)} 18$

1990 AMC8真答案详细解析

1.Let the two three-digit numbers be $overline{abc}$ and $overline{def}$ . Their sum is equal to $100(a+d)+10(b+e)+(c+f)$ .

To minimize this, we need to minimize the contribution of the $100$ factor, so we let $a=4$ and $d=5$ . Similarly, we let $b=6$ , $e=7$ , and then $c=8$ and $f=9$ . The sum is $[100(9)+10(13)+(17)=1047 rightarrow boxed{text{C}}]$

2.When dealing with positive decimals, the leftmost digits affect the change in value more. Thus, to get the largest number, we change the $1$ to a $9 rightarrow boxed{text{A}}$ .

3.Reflecting the square across the diagonal drawn, we see that the shaded region covers exactly the unshaded region in the original square, and thus makes up $1/2$ of the square $rightarrow boxed{text{E}}$ .

4.We see that $1^2=1$ , $2^2=4$ , $5^2=25$ , and $4^2=16$ , so already we know that either $text{E}$ is the answer or the problem has some issues.

For integers, only the units digit affects the units digit of the final result, so we only need to test the squares of the integers from $0$ through $9$ inclusive. Testing shows that $8$ is unachievable, so the answer is $boxed{text{E}}$ .

Solution 1

Clearly, $[.4<.48017<.5]$ Since the function $f(x)=x^3$ is strictly increasing, we can say that $[.4^3<.48017^3<.5^3]$ from which it follows that $text{A}$ is much too small and $text{C}$ is much too large, so $boxed{text{B}}$ is the answer.

Solution 2

Since $0.48017$ is quite close to $0.5$ , or $dfrac{1}{2}$ , we can look for the answer choice that is just below $(dfrac{1}{2})^3=dfrac{1}{8}=0.125$ , which would be $boxed{textbf{(B)}}$ .

6.The options given in choices $text{A}$ , $text{B}$ and $text{E}$ don't change the initial value (13579) much, the option in choice $text{C}$ decreases 13579 by a lot, and the one given in choice $text{D}$ increases 13579 by a lot.

After ensuring that no trivial error was made, we have $boxed{text{D}}$ as the answer.

7.First we try for a positive product, meaning we either pick three positive numbers or one positive number and two negative numbers.

It is clearly impossible to pick three positive numbers. If we try the second case, we want to pick the numbers with the largest absolute values, so we choose $5$ , $-3$ and $-2$ . Their product is $30rightarrow boxed{text{C}}$ .

8.After the price reduction, the sale price is $80-.25times 80 = 60$ dollars. The tax makes the final price $60+.1times 60 = 66$ dollars $rightarrow boxed{text{D}}$ .

We just count to find that there are $5$ students in the $text{C}$ range.

There are $15$ total, so the percentage is $frac{5}{15}=33frac{1}{3}%$ $rightarrow boxed{text{D}}$ .

10.Let the date behind $C$ be $x$ . Now the date behind $A$ is $x+1$ , and after looking at the calendar, the date behind $B$ is $x+13$ . Now we have $x+1+x+13=x+y$ for some date $y$ , and we desire for $y$ to be $x+14$ . Now we find that $y$ is the date behind $P$ , so the answer is $boxed{(text{A})}$ ~motorfinn

11.The only possibilities for the numbers are $11,12,13,14,15,16$ and $10,11,12,13,14,15$ .

In the second case, the common sum would be $(10+11+12+13+14+15)/3=25$ , so $11$ must be paired with $14$ , which it isn't.

Thus, the only possibility is the first case and the sum of the six numbers is $81rightarrow boxed{text{E}}$ .

12.

For each choice of the thousands digit, there are $6$ numbers with that as the thousands digit. Thus, the six smallest are in the two thousands, the next six are in the four thousands, and then we need $5$ more numbers.

We can just list from here: $5247,5274,5427,5472,5724 rightarrow boxed{text{B}}$ .

13.

After the first ounce, there are $3.5$ ounces left. Since each additional ounce or fraction of an ounce adds $22$ cents to the total cost, we need to add $4times 22$ to the cost for the first ounce.

So, the total price is $30+4times 22=118$ cents. The answer is choice $boxed{textbf{(C)}~mathbf{1.18}~textbf{dollars}}$ .

14.

Solution 1

The total number of balls in the bag must be $4times 6=24$ , so there are $24-6=18$ green balls $rightarrow boxed{text{B}}$

Solution 2

If b= number of blue balls in the bag and g = number of green balls in the bag then b/(b+g) = 1/4. Substituting b=6 and solving for g we get g=18, or B

- goldenn

15.

Since the area of the whole figure is $100$ , each square has an area of $25$ and the side length is $5$ .

There are $10$ sides of this length, so the perimeter is $10(5)=50rightarrow boxed{text{E}}$ .

16.

In the middle, we have $cdots + 1010-1000+990 -cdots$ .

If we match up the back with the front, and then do the same for the rest, we get pairs with $2000$ and $-2000$ , so these will cancel out. In the middle, we have $2000-1000$ which doesn't cancel, but gives us $1000 rightarrow boxed{text{D}}$ .

17.This is a $1$ yard by $20$ yard by $1/12$ yard sidewalk, so its volume in yards is $[1times 20times frac{1}{12} = 1.overline{6}.]$ Since concrete must be ordered in a whole number of cubic yards, we need $2rightarrow boxed{text{A}}$ .

18.

In addition to the original $12$ edges, each original vertex contributes $3$ new edges.

There are $8$ original vertices, so there are $12+3times 8=36$ edges in the new figure $rightarrow boxed{text{C}}$ .

19.

p is a person seated, o is an empty seat

The pattern of seating that results in the fewest occupied seats is opoopoopoo...po we can group the seats in 3s opo opo opo ... opo

there are a total of $boxed{B}$ groups

20.Let $S$ be the sum of all the incomes but the largest one. For the actual data, the mean is $frac{S+98000}{1000}$ , and for the incorrect data the mean is $frac{S+980000}{1000}$ . The difference is $882rightarrow boxed{text{A}}$ .

21.We just use the definition to find the first number is $1/4 rightarrow boxed{text{B}}$ .

22.If this is the case, then if there were only $99$ pieces of candy, the bag would have gone around the table a whole number of times. Thus, the number of students is a divisor of $99$ . The only choice that satisfies this is choice $boxed{text{B}}$ .

23.

The time when the average speed is greatest is when the slope of the graph is steepest. This is in the second hour $rightarrow boxed{text{B}}$ .

24.

For simplicity, suppose $Delta = a$ , $diamondsuit = b$ and $bullet = c$ . Then, $[3a+b=9c]$ $[a=b+c]$ and we want to know what $2b$ is in terms of $c$ . Substituting the second equation into the first, we have $[4b=6cRightarrow 2b=3c]$

Thus, we need $3$ $bullet$ 's $rightarrow boxed{text{C}}$ .

25.

We break this into cases.

Case 1: At least one square is a vertex: WLOG, suppose one of them is in the upper-left corner. Then, consider the diagonal through that square. The two squares on that diagonal could be the second square, or the second square is on one side of the diagonal.

The square is reflectionally symmetric about this diagonal, so we only consider the squares on one side, giving another three possibilities.

In this case, there are $2+3=5$ distinct squares.

Case 2: At least one square is on an edge, but no square is on a vertex: There are clearly two edge-edge combinations and one edge-center combination, so this case has $3$ squares.

In total, there are $3+5=8$ distinct squares $rightarrow boxed{text{C}}$