Question

$3$ men of $7$ women can do a piece of work in $32$ days. The number of days required by $7$ men and $5$ women to do a piece of work twice as large is

• A. $19$
• B. $21$
• C. $27$
• D. $36$

Answer 1

Answer: (B)

$21$

Work of 3 men $=$ work of 7 women
Work of 1 man $=$ work of $\frac{7}{3}$ women
Work of 7 men $=$ work of $\frac{7\times 7}{3}$
$=\frac{49}{3}$ women
$\therefore$ Work of 7 men + 5 women
$=$ work of $\frac{49}{3}+5$
$=\frac{64}{3}$ women
$\begin{array}{cc}Women& Days\\ 7& 32\\ \frac{64}{3}& x\end{array}$
$\therefore \frac{64}{3}:7::32:x$
$\therefore \frac{64}{3}\times x=7\times 32$
$\therefore x=\frac{7\times 32\times 3}{64}=10.5days$
$\therefore$ So, 7 men and 5 women can do piece of work twice as large in $21$ days.