1952 AMC 12 第 16 题

Question

若一个长方形的底边增加

10\%

，而面积不变，则高减少：

If the base of a rectangle is increased by

10\%

and the area is unchanged, then the altitude is decreased by:

9\%

10\%

11\%

11\dfrac19\%

9\dfrac1{11}\%

新的高必须除以底边所乘的同一个倍数

The new altitude must be divided by the same factor by which the base was multiplied

底边增加

10\%

，相当于将底边乘以

\frac{11}{10}

A

10\%

base increase multiplies the base by

\frac{11}{10}

要保持面积不变，高应乘以

\frac{10}{11}

。其减少的比例为

1-\frac{10}{11}=\frac1{11}\text{，}

即

\frac{100}{11}\%=9\dfrac1{11}\%

。

因此，正确答案是 E。

To keep the area fixed, the altitude is multiplied by

\frac{10}{11}.

Its fractional decrease is

1-\frac{10}{11}=\frac1{11},

which is

\frac{100}{11}\%=9\dfrac1{11}\%.

Thus, the correct answer is E.

其他年份的第 16 题