1987 AMC 8 第 11 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

11.

217+312+51192\dfrac17 + 3\dfrac12 + 5\dfrac{1}{19} 位于哪两个值之间?

The sum 217+312+51192\dfrac17 + 3\dfrac12 + 5\dfrac{1}{19} is between which two values?

1010101210\dfrac12

1010 and 101210\dfrac12

101210\dfrac121111

101210\dfrac12 and 1111

1111111211\dfrac12

1111 and 111211\dfrac12

111211\dfrac121212

111211\dfrac12 and 1212

1212121212\dfrac12

1212 and 121212\dfrac12

答案:B
知识点:分数估算
难度评级:860
解答:

整数部分的和是 1010,分数部分是 12+17+119\dfrac12 + \dfrac17 + \dfrac{1}{19}

因为 17+119\dfrac17 + \dfrac{1}{19} 是一个较小的正数,所以分数部分总和大于 12\dfrac12,但远小于 11。因此原和位于 101210\dfrac121111 之间。

所以正确答案是 B

The whole-number parts sum to 10.10. The fractional parts are 12+17+119.\dfrac12 + \dfrac17 + \dfrac{1}{19}.

Since 17+119\dfrac17 + \dfrac{1}{19} is a small positive amount, the fractional total is more than 12\dfrac12 but well under 1.1. So the sum lies between 101210\dfrac12 and 11.11.

Thus, the correct answer is B .

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