『艾蕾特教育』AMC8 -- 2006年真题解析
北京联盟_本文原题：AMC8 -- 2006年真题解析
Problem 1
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Answer: D
Solution:
The three prices round to $2, $5, and $10, which has a sum of 17.Problem 2
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Answer: C
Solution:
As the AMC 8 only rewards 1 point for each correct answer, everything is irrelevant except the number Billy answered correctly ， 13.Problem 3
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Answer: A
Solution:
When Elisa started, she finished a lap in 25/10=2.5 minutes. Now, she finishes a lap is 24/12=2 minutes. The difference is 2.5-2=0.5Problem 4
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Answer: B
Solution:
If the spinner goes clockwise 2+1/4 revolutions and then counterclockwise 3+3/4 revolutions, it ultimately goes counterclockwise 1+1/2 which brings the spinner pointing east.Problem 5
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Answer: D
Solution:
Drawing segments AC and BD, the number of triangles outside square ABCD is the same as the number of triangles inside the square. Thus areas must be equal so the area of ABCD is half the area of the larger square which is 60/2=30.Problem 6
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Answer: C
Solution:
If the two rectangles were seperate, the perimeter would be 2*2*(2+4)=24. It easy to see that their connection erases 2 from each of the rectangles, so the final perimeter is 24-2*2=20.Problem 7
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Answer: B
Solution:
Using the formulas of circles, C=2*Pi*r and A=Pi*r*r, we find that circle Y has a radius of 4 and circle Z has a radius of 3. Thus, the order from smallest to largest radius is Z, X, Y.Problem 8
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Answer: E
Solution:
Males which listen are （136-58）=78. The total males which surveyed are 78+26=104. Thus the percentage of males surveyed listen to the station is 78/104=75%.Problem 9
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Answer: C
Solution:
After looking at the problem, we immediately notice that terms cancel out, leaving us with 2006/2=1003.Problem 10
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Answer: A
Solution:
The length of the rectangle will relate invertly to the width, specifically using the theorem l=12/3. The only graph that could represent a inverted relationship is A. (The rest are linear graphs that represent direct relationships, therefore they are incorrect.)Problem 11
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Answer: C
Solution:
There is 1 integer whose digits sum to 1: 10.；There are 4 integers whose digits sum to 4： 13,22, 31, 40. ; There are 9 integers whose digits sum to 9: 18, 27, 36, 45, 54, 63, 72, 81, 90. ; There are 3 integers whose digits sum to 16: 79, 88, 97. Two digits cannot sum to 25 or any greater square since the greatest sum of digits of a two-digit number is 9+9=18. Thus, the answer is 1+4+9+3=17.Problem 12
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Answer: D
Solution:
70%*10=7; 80%*20=16； 90%*30=27. Adding them us gets 7+16+27=50. The overall percentage correct would be 50/60=83.Problem 13
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