Question

$3$ men and $4$ women do a work in $10$ days as $2$ men and $6$ women do.$4$ men and $8$ women starts the same work and after $2$ days only women remains in the work for completing it. How many days are required to complete the work?

• A. $6$ days
• B. $7.5$ days
• C. $8$ days
• D. $8.5$ days

Answer 1

The correct option is D $8.5$ days

Given,
$3$ men $+4$ women $=2$ men + $6$ women.

Therefore, 1 man = 2 women
Now,
$4$ men $+8$ women = $16$ women.

According to question,
$3$ men $+4$ women $=10$ women do the work in 10 days
So 1 day work of 1 woman $=\frac{1}{100}$ part
In 2 days work completed $=\frac{2\times 16\times 1)}{100}$
$=\frac{32}{100}$ part
Remaining work $=\frac{68}{100}$ part
and Remaining women for work $=8$
So number of days required $=\frac{68/100)}{\left(8/100\right)}=\frac{68}{8}=\frac{17}{2}$
$=8.5$