1959 AMC 12 第 24 题

Question

一位化学家有

m

盎司含盐量为

m\%

的盐水。他必须加入多少盎司盐，才能使溶液的含盐量变为

2m\%

？

A chemist has

m

ounces of salt water that is

m\%

salt. How many ounces of salt must he add to make a solution that is

2m\%

salt?

\dfrac{m}{100+m}

\dfrac{2m}{100-2m}

\dfrac{m^2}{100-2m}

\dfrac{m^2}{100+2m}

\dfrac{2m}{100+m}

原有盐的质量为

\frac{m^2}{100}

盎司

The original amount of salt is

\frac{m^2}{100}

ounces

若加入

x

盎司纯盐，令

\frac{\frac{m^2}{100}+x}{m+x}

等于

\frac{2m}{100}

If

x

ounces of pure salt are added, equate

\frac{\frac{m^2}{100}+x}{m+x}

to

\frac{2m}{100}

设加入

x

盎司盐。浓度方程为

\frac{\frac{m^2}{100}+x}{m+x}=\frac{2m}{100}\text{。}

两边同乘分母并合并含

x

的项，得

x(100-2m)=m^2\text{，}

所以

x=\frac{m^2}{100-2m}

。

因此，正确答案是 C。

Let

x

ounces of salt be added. The concentration equation is

\frac{\frac{m^2}{100}+x}{m+x}=\frac{2m}{100}.

Multiplying through and collecting the

x

-terms gives

x(100-2m)=m^2,

so

x=\frac{m^2}{100-2m}.

Therefore, the correct answer is C.

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