Question

$A$ and $B$ can do a piece of work in $12$ days, $B$ and $C$ in $15$ days and $C$ and $A$ in $20$ days. How much time will $A$ alone take to finish the job?

Answer 1

Time taken by $A$ and $B$ to complete a unit of work $=12$ days

Therefore, work done by $A$ and $B$ in $1$ day $=\frac{1}{12}$ units of work

Similarly,

Work done by $B$ and $C$ is $1$ day $=\frac{1}{15}$ units of work

Work done by $A$ and $C$ in $1$ day $=\frac{1}{20}$ units of work

Therefore, total work done by $A,B$ and $C$ $=\frac{1}{2}\times (\frac{1}{12}+\frac{1}{15}+\frac{1}{20})$

LCM of $12,15$ and $20$ is $60$

$=\frac{1}{2}\times \left(\frac{1\times 5+1\times 4+1\times 3}{60}\right)$

$=\frac{1}{2}\times \left(\frac{12}{60}\right)$

$=\frac{1}{10}$ units of work

Therefore, work done by $A$ alone in $1$ day $=$ Total work done $-$ Work done by $B$ and $C$

$=\frac{1}{10}-\frac{1}{15}$

$=\frac{1\times 3-1\times 2}{30}$

$=\frac{1}{30}$ units of work

Hence, time taken by $A$ to complete a unit of work $=30$ days