Question

$A$ and $B$ can finish a piece of work in $16$ days respectively. $A$ started the work and worked at it for $2$ days. He was then joined by $B$. Find the total time taken to finish the work.

Answer 1

We know that,
Number of days $A$ required to do piece of work $=16$ days

Number of days $B$ required to do piece of work $=12$ days

$\therefore$ We can calculate, work done by $A$ in $1$ day $=1/16$

Work done by $B$ in $1$ day $=1/12$

$A$ alone works for $2$ days so, the amount of work completed by $A$ in $2$ days $=2\times 1/16=1/8$

$\therefore$ work left $=1-1/8=7/8$

Work done by $A,B$ together in a one day $=1/16+1/12=7/48$
$\therefore$ they can complete the work together in $48/7$ days
But only $7/8^{th}$ of work is to be completed by both $A$ and $B$

$\therefore$ time required to complete $7/8^{th}$ work together by $A$ and $B=7/8\times 48/7=6$ days

Time taken to finish the work $=6+2=8$ days ( $2$ is added because $1/8$ work is done by $A$ alone)
$\therefore$ Total time taken to complete the work $=8$ days.