Question

$A$ and $B$ can do a piece of work in $12$ days, $B$ and $C$ in $15$ days, $C$ and $A$ in $20$ days. In how many days will they finish it together and separately?

Answer 1

Work done by $A$ and $B$ in $1$ day $=\frac{1}{12}$
Work done by $B$ and $C$ in $1$ day $=\frac{1}{15}$
Work done by $C$ and $A$ in $1$ day $=\frac{1}{20}$
Work done by $(A+B)+(B+C)+(C+A)$ in $1$ day $=\frac{1}{12}+\frac{1}{15}+\frac{1}{20}$
Work done by $(2A+2B+2C)$ in $1$ day $=\frac{5+4+3}{60}$
Work done by $2(A+B+C)$ in $1$ day $=\frac{12}{60}$
Work done by $A$, $B$ and $C$ together in $1$ day $=\frac{1}{2}\times \frac{12}{60}=\frac{1}{10}$
$\therefore$ $A$, $B$ and $C$ will finish the work in $10$ days.
Work done by $A$ in $1$ day
(i.e.) $[{(A+B+C)}^{\prime }swork-{(B+C)}^{\prime }swork]=\frac{1}{10}-\frac{1}{15}=\frac{3-2}{30}=\frac{1}{30}$
$\therefore$ $A$ will finish the work in $30$ days.
Work done by $B$ in $1$ day
(i.e.) $[{(A+B+C)}^{\prime }swork-{(C+A)}^{\prime }swork]=\frac{1}{10}-\frac{1}{20}=\frac{2-1}{20}=\frac{1}{20}$
$\therefore$ $B$ will finish the work in $20$ days.
Work done by $C$ in $1$ day
(i.e.) $[{(A+B+C)}^{\prime }swork-{(A+B)}^{\prime }swork]=\frac{1}{10}-\frac{1}{12}=\frac{6-5}{60}=\frac{1}{60}$
$\therefore$ $C$ will finish the work in $60$ days.