1994 AMC 8 第 20 题

先试着解答 1994 AMC 8 第 20 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1994 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

20.

从下列集合中选出四个不同数字 WW,XX,YY 和 ZZ:
{1,2,3,4,5,6,7,8,9}。\{1, 2, 3, 4, 5, 6, 7, 8, 9\}\text{。}
如果要使 WX+YZ\dfrac{W}{X} + \dfrac{Y}{Z} 尽可能小,那么 WX+YZ\dfrac{W}{X} + \dfrac{Y}{Z} 必须等于
Let W,W, X,X, Y,Y, and ZZ be four different digits selected from the set
{1,2,3,4,5,6,7,8,9}.\{1, 2, 3, 4, 5, 6, 7, 8, 9\}.
If the sum WX+YZ\dfrac{W}{X} + \dfrac{Y}{Z} is to be as small as possible, then WX+YZ\dfrac{W}{X} + \dfrac{Y}{Z} must equal
217\dfrac{2}{17}
317\dfrac{3}{17}
1772\dfrac{17}{72}
2572\dfrac{25}{72}
1336\dfrac{13}{36}
答案:D
知识点:分数最优化
小提示:
要让每个分数小,应使用最小的数字作分子、最大的数字作分母
To make each fraction small, use the smallest digits as numerators and the largest as denominators
大提示:
比较 18+29\dfrac18 + \dfrac29 和 19+28\dfrac19 + \dfrac28,看哪一个更小
Compare 18+29\dfrac18 + \dfrac29 with 19+28\dfrac19 + \dfrac28 to see which is smaller
解答:
小分子和大分母会产生较小分数,所以用 11 和 22 作分子,用 88 和 99 作分母。
把较大的分子配给较大的分母,得到 18+29=9+1672=2572\dfrac18 + \dfrac29 = \dfrac{9 + 16}{72} = \dfrac{25}{72},它小于 19+28=2672\dfrac19 + \dfrac28 = \dfrac{26}{72}。所以最小和是 2572\dfrac{25}{72}。
所以正确答案是 D。
Small numerators and large denominators make small fractions, so use 11 and 22 as numerators and 88 and 99 as denominators.
Pairing the larger numerator with the larger denominator gives 18+29=9+1672=2572,\dfrac18 + \dfrac29 = \dfrac{9 + 16}{72} = \dfrac{25}{72}, which is smaller than 19+28=2672.\dfrac19 + \dfrac28 = \dfrac{26}{72}. So the minimum sum is 2572.\dfrac{25}{72}.
Thus, the correct answer is D .
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