Question

In a certain office, $\frac{1}{3}$ of the workers are women, $\frac{1}{2}$ of the women are married and $\frac{1}{3}$ of the married women have children. If $\frac{3}{4}$ of the man are married and $\frac{2}{3}$ of the married men have children, what part of workers are without children?

• A. $\frac{5}{18}$
• B. $\frac{7}{17}$
• C. $\frac{10}{21}$
• D. $\frac{11}{18}$

Answer 1

The correct option is D $\frac{11}{18}$

Lets work this out piece by piece.

$\frac{1}{3}$ of workers are women. So,$\frac{2}{3}$ of workers are men.

$\frac{1}{2}$ of women are married. So, $\frac{1}{2}$ of women are not married.

$\frac{3}{4}$ Men are married. $\frac{1}{4}$ men are not married.

$\frac{1}{3}$ married women have children. $\frac{2}{3}$ married men have children.

Let total workers be $x$

We know, $\frac{1}{3}$ married women have children.

Married women=$\frac{1}{2}\times \frac{1}{3}\times x=\frac{x}{6}$

Thus child bearers=$\frac{1}{3}\times \frac{x}{6}=\frac{x}{18}$

We know, $\frac{2}{3}$ married men have children.

Married men=$\frac{3}{4}\times \frac{2}{3}\times x=\frac{x}{2}$

Thus child bearers=$\frac{2}{3}\times \frac{x}{2}=\frac{x}{3}$

Sp, total childbearers=$\frac{x}{18}+\frac{x}{3}=\frac{7x}{18}$

Which is $\frac{7}{18}$ part of the total workers.

So, on child bearers $=$ $1-\frac{7}{18}=\frac{11}{18}$ part of the total workers.