Question

$A$ alone can do a piece of work in $12$ days and $B$ alone can do the same work in $16$ days. They worked together on it for $3$ days and then $A$ left. How long did $B$ take to finish the remaining work?

Answer 1

A's 1 day work $=\frac{1}{12}$
B's 1 day work $=\frac{1}{16}$
(A+B)'s 1 day work$=\frac{1}{12}+\frac{1}{16}$
$=\frac{4+3}{48}$
$=\frac{7}{48}$

(A+B)'s 3 days work = $3\times \frac{7}{48}$
=$\frac{7}{16}$
Remaining work left=$1-\frac{7}{16}$
$=\frac{16-7}{16}$
$=\frac{9}{16}$
Since, B can do one work in 16 days
Therefore, B will finish remaining work=$\frac{9}{16}\times 16$
$=9$ days