Question

$A,B$ and $C$ can do a piece of work in $15,12$ and $20$ days respectively. They started the work together, but $C$ left after $2$ days. In how many days will the remaining work be competed by $A$ and $B$?

Answer 1

Work is done by $A$ in one day =$\frac{1}{15}$

By $B=\frac{1}{12}$

$C=\frac{1}{20}$

Together in 1 Day $=\frac{1}{15}+\frac{1}{12}+\frac{1}{20}=\frac{4+5+3}{60}=\frac{12}{60}=\frac{1}{5}$

$2$ days work = $\frac{2}{5}$

Remaining work = $1-\frac{2}{5}=\frac{3}{5}$

one day work of $A,B$ $\Rightarrow A+B=\frac{1}{15}+\frac{1}{12}=\frac{4+5}{60}=\frac{9}{60}=\frac{3}{20}$

$(A+B)$ work for $t$ days = $\frac{3}{20}t=\frac{3}{5}$

$t=4$ days

$A$ and $B$ will complete the work in $4$ days.