「Answer」AMC8 -- 2000年真题解析( 二 )
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Answer: D
Solution:
Since the bricks are 1 foot high, there will be 7 rows. To minimize the number of blocks used, rows 1,3,5 and 7 will look like the bottom row of the picture, which takes 100/2=50 bricks to construct. Rows 2,4 and 6 will look like the upper row pictured, which has 49 2-foot bricks in the middle, and 2 1-foot bricks on each end for a total of 51 bricks. Four rows of 50 bricks and three rows of 51 bricks totals 4*50+3*51=353 bricks.Problem 13
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Answer: C
Solution:
In triangle ACT, <c=<T=(180-36)/2=72. <RTC=1/2*<ATC=36;In Triangle CRT, <c=72, <RTC=36, thus <CRT=180-72-26=72.Problem 14
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Answer: D
Solution:
even powers of 19 have a units digit of 1, and odd powers of 19 have a units digit of 9. So, 19^19 has a units digit of 9. Powers of 99 have the exact same property, so 99^99 also has a units digit of 9. 9+9=18 which has a units digit of 8.Problem 15
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Answer: C
Solution:
The large triangle ABC has sides of length 4. The medium triangle has sides of length 2. The small triangle has sides of length 1. There are 3 segment sizes, and all segments depicted are one of these lengths.Starting at A and going clockwise, the perimeter is:AB+BC+CD+DE+EF+FG+GA=4+4+2+2+1+1+1=15.Problem 16
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Answer: C
Solution:
The length L of the rectangle is 1000/25=40 meters. The perimeter is 1000/10=100 meters. So the width is 100/2-40=10 meters. Area is 40*10=400 .Problem 17
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Answer: A
Solution:
1@2=1*1/2=1/2; 1/2@3=(1/4)/3=1/12;2@3=4/3;1@4/3=1/(4/3)=3/4;the answer is 1/12-3/4=-2/3.Problem 18
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Answer: E
Solution:
Area of quadrilateral I is : 1/2*1*1+1/2*1*1=1;Area of quadrilateral II is : 1/2*1*1+1/2*1*1=1;Perimeter of quadrilateral I is : 1+sqrt(2)+1+sqrt(1);Perimeter of quadrilateral II is : sqrt(5)+sqrt(2)+1+sqrt(2)Problem 19
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Answer: C
Solution:
Draw line BD. Then draw CO, where o is the center of the semicircle. You have two quarter circles on top, and two quarter circle-sized "bites" on the bottom. Move the pieces from the top to fit in the bottom like a jigsaw puzzle. You now have a rectangle with length BD and height AO, which are equal to 10 and 5, respectively. Thus, the total area is 50.Problem 20
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Answer: A
Solution:
Problem 21
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Answer: B
Solution:
Let K(n) be the probability that Keiko gets n heads, and let E(n) be the probability that Ephriam gets n heads. E(0)=1/4; E(1)=1/2; E(2)=1/4; K(0)=1/2; K(1)=1/2; K(2)=0;The probability that Keiko gets 0 heads and Ephriam gets 0 heads is K(0)*E(0)=1/2*1/4=1/8; Similarly for 1 head and 2 heads. Thus, we have:P=K(0)*E(0)+K(1)*E(1)+K(2)*E(2)=1/8+1/2*1/2+0=3/8.Problem 22
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Answer: C
Solution:
The original cube has 6 faces, each with an area of 2*2=4 square units. Thus the original figure had a total surface area of 24 square units.The new figure has the original surface, with 6 new faces that each have an area of 1 square unit, for a total surface area of 6 additional square units added to it. But 1 square unit of the top of the bigger cube, and 1 square unit on the bottom of smaller cube, is not on the surface, and does not count towards the surface area.The total surface area is therefore 24+6-1-1=28 square units.The percent increase in surface area 28-24/24=4/24=17%Problem 23
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