1989 AMC 8 第 13 题

先试着解答 1989 AMC 8 第 13 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 1989 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。

13.

下列哪个式子等于
97×53\frac{9}{7 \times 53}
Which of the following is equal to
97×53?\frac{9}{7 \times 53}?
0.90.7×53\dfrac{0.9}{0.7 \times 53}
0.90.7×0.53\dfrac{0.9}{0.7 \times 0.53}
0.90.7×5.3\dfrac{0.9}{0.7 \times 5.3}
0.97×0.53\dfrac{0.9}{7 \times 0.53}
0.090.07×0.53\dfrac{0.09}{0.07 \times 0.53}
答案:A
知识点:分数位值
小提示:
如果把分子除以十,分母中的某一个因数也必须除以十,分数值才不变
Dividing the numerator and one factor of the denominator by the same number keeps the fraction unchanged
大提示:
要把 99 变成 0.90.9,需除以 1010;为保持分数值不变,也把分母中的一个因数除以 1010
To turn 99 into 0.9,0.9, divide by 10;10; to keep the value equal, divide one factor of the denominator by 1010 as well
解答:
将分子 99 除以 1010,得到 0.90.9。为了保持分数值不变,分母整体也要除以 1010,即把其中一个因数除以 1010。
把分母中的因数 77 除以 1010,而另一个因数保持不变,便得到 0.90.7×53\frac{0.9}{0.7 \times 53}。其他选项会使原式改变 1010 倍或 100100 倍。
所以正确答案是 A。
To rewrite the numerator 99 as 0.9,0.9, divide it by 10.10. To keep the fraction equal, divide the denominator by 1010 too, which means dividing one of its factors by 10.10.
Dividing the factor 77 by 1010 gives 0.90.7×53,\frac{0.9}{0.7 \times 53}, which equals the original. Each other choice changes the value by a factor of 1010 or 100.100.
Thus, the correct answer is A .
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