Question

$4$ men and $10$ women were put on a work. They completed $\frac{1}{3}$ of the work in $4$ days. After this, $2$ men and $2$ women were increased. They completed $\frac{2}{9}$ more of the work in $2$ days. If the remaining work is to be completed in $3$ days, then how many more women must be increased?

• A. $8$
• B. $32$
• C. $50$
• D. $55$

Answer 1

The correct option is A $8$

Let one man's $1$ day's work $=x$
and one woman's $1$ day's work $=y.$
Then, $4x+10y=\frac{1}{3}\times \frac{1}{4}=\frac{1}{12}$
$\Rightarrow 2x+5y=\frac{1}{24}\cdots \left(i\right)$
And, $6x+12y=\frac{2}{9}\times \frac{1}{2}=\frac{1}{9}$
$\Rightarrow 2x+4y=\frac{1}{27}\cdots \left(ii\right)$
Subtracting $\left(ii\right)$ from $\left(i\right),$ we get
$y=\frac{1}{24}-\frac{1}{27}=\frac{1}{216}$

Now, $(6$ men $+$ $12$ women$)$'s $3$ day's work $=(\frac{1}{9}\times 3)=\frac{1}{3}$
Work completed $=(\frac{1}{3}+\frac{2}{9}+\frac{1}{3})=\frac{8}{9}$
$\therefore$ Remaining work $=(1-\frac{8}{9})=\frac{1}{9}$
One women's $3$ day's work
$=(\frac{1}{216}\times 3)=\frac{1}{72}$
In $3$ days, $\frac{1}{72}$ work is done by one woman.
$\therefore$ In $3$ days, $\frac{1}{9}$ work is done by $=(\frac{1}{9}\times 72)=8$ woman.