经验教程|AMC8 -- 1990年真题解析( 三 )

Problem 22经验教程|AMC8 -- 1990年真题解析
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Answer: B
Solution:

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 B.
Problem 23经验教程|AMC8 -- 1990年真题解析
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经验教程|AMC8 -- 1990年真题解析
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Answer: B
Solution:
The time when the average speed is greatest is when the slope of the graph is steepest. This is in the second hour.
Problem 24经验教程|AMC8 -- 1990年真题解析
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Answer: C
Solution:
For simplicity, suppose triangle=a, trapezoid =b, and dot =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=6c => 2b=3c. Thus, we need 3 dots.
Problem 25经验教程|AMC8 -- 1990年真题解析
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Answer: C
Solution:
Case 1: At least one square is a vertex: 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.
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 3 squares.
In total, Its 5+3=8.
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【经验教程|AMC8 -- 1990年真题解析】


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