Question

$A$ can finish a work in 24 days, $B$ in 9 days and $C$ in 12 days. $B$ and $C$ start the work but are forced to leave after 3 days. In how many days was the remaining work done by $A$ ?

Answer 1

$(B+C)$'s 1 days' work $=\frac{1}{9}+\frac{1}{12}=\frac{4+3}{36}=\frac{7}{36}$

$\therefore (B+C)$'s 3 days' work $=3\times \frac{7}{36}=\frac{7}{12}$

$\therefore$ Remaining part of the work $=1-\frac{7}{12}=\frac{5}{12}$
$A$ will finish the whole work (1) in $24$ days.
$\therefore A$ will do $\frac{5}{12}$ part of the whole work in $\frac{5}{12}\times 24=10$ days.