「艾蕾特教育」AMC8 -- 2006年真题解析( 二 )
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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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Answer: D
Solution:
If Cassie leaves half an hour earlier then Brian, when Brian starts, the distance between them will be 62-12/2=56. Every hour, they will get 12+16=28 miles closer. 56/28=2, so 2 hours from 9:00 AM is when they meet, which is 11:00.
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Problem 14
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Answer: B
Solution:
760*45-760*30=760*15=11400.Problem 15
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Answer: C
Solution:
we set up an equation and solve.Let x be the number of pages that Chandra reads. 30*x=45*(760-x), thus x=456.Problem 16
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Answer: E
Solution:
The amount of pages Bob, Chandra, and Alice will read is in the ratio 4:6:9. Therefore, Bob, Chandra, and Alice read 160, 240, and 360 pages respectively. They would also be reading for the same amount of time because the ratio of the pages read was based on the time it takes each of them to read a page. Therefore, the amount of seconds each person reads is simply 160*45=7200.Problem 17
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Answer: B
Solution:
In order for Jeff to have an odd number sum, the numbers must either be Odd + Odd + Odd or Even + Even + Odd. We easily notice that we cannot obtain Odd + Odd + Odd because spinner Q contains only even numbers. Therefore we must work with Even + Even + Odd and spinner Q will give us one of our even numbers. We also see that spinner R only contains odd, so spinner R must give us our odd number. We still need one even number from spinner P. There is only 1 even number: 2. Since spinning the required numbers are automatic on the other spinners, we only have to find the probability of spinning a 2 in spinner P, which is clearly 1/3.Problem 18
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