1963 AMC 12 第 19 题

Question

清点

n

个彩球，其中一些是红球，另一些是黑球。最先清点的

50

个球中有

49

个红球；此后每清点

8

个球，就有

7

个红球。如果清点的所有球中至少有

90\%

是红球，那么

n

的最大值是：

In counting

n

colored balls, some red and some black, it was found that

49

of the first

50

counted were red. Thereafter,

7

out of every

8

counted were red. If, in all,

90\%

or more of the balls counted were red, the maximum value of

n

is:

225

210

200

180

175

把红球数写成

49+\frac78(n-50)

Write the number of red balls as

49+\frac78(n-50)

令这个数量至少为

0.9n

Require that quantity to be at least

0.9n

条件为

49+\frac78(n-50)\geq\frac9{10}n\text{。}

化简得

\frac{21}{4}\geq \frac{n}{40}

，所以

n\leq210

。最大值为

210

。

所以正确答案是 B。

The condition is

49+\frac78(n-50)\geq\frac9{10}n.

Simplifying gives

\frac{21}{4}\geq \frac{n}{40},

so

n\leq210.

The maximum is

210.

Thus, the correct answer is B.

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