A can do a piece of work in 32 days and B can do the same work in 48 days. They work together for 8 days and then A leaves. In how much time (in days) will B complete 60% of the remaining work?

A’s one day work = $\frac{1}{32}$

B’s one day work = $\frac{1}{48}$

Together one day work:

$= \frac{1}{32}+\frac{1}{48}$

LCM of $32$ and $48 = 96$

$= \frac{3+2}{96}$

$= \frac{5}{96}$

Work done together in $8$ days:

$= 8 \times \frac{5}{96}$

$= \frac{5}{12}$

Remaining work:

$= 1-\frac{5}{12}$

$= \frac{7}{12}$

B has to complete $60%$ of remaining work:

$60%=\frac{3}{5}$

Required work:

$= \frac{3}{5}\times \frac{7}{12}$

$= \frac{7}{20}$

Time taken by B:

$= \frac{\frac{7}{20}}{\frac{1}{48}}$

$= \frac{7}{20}\times 48$

$= \frac{84}{5}$

$= 16\frac{4}{5}$ days

Correct Answer: $16\frac{4}{5}$