Question

$A$ and $B$ can do a piece of work in $6$ days and $4$ days respectively. ${}^{\prime }{A}^{\prime }$ started the work; worked at it for $2$ days and then was joined by $B$. Find the total time taken to complete the work.

Answer 1

$A^{\prime }s$ $1$ day work

$B^{\prime }s$ $1$ day work $=\frac{1}{4}$

$A$ and day work $=\frac{1}{6}+\frac{1}{4}$=16

$=\frac{2+3}{12}=\frac{5}{12}$

$A^{\prime }s2$ day work $=\frac{1}{6}\times 2=\frac{1}{3}$

Remaining work $=1-\frac{1}{3}=\frac{2}{3}$ [$\because$ whole work $=1$]

Now $A$ and $B$ can do $\frac{5}{12}$th work in $1$ day

They will do $1$ work (Whole work) in $1\times \frac{12}{5}$ daysB′s

and they will do $\frac{2}{3}$th1 work $=\frac{12}{5}\times \frac{2}{3}$

$=\frac{8}{5}=1\frac{3}{5}$ days

$\therefore$ Total work was finished in $=1\frac{3}{5}+2$

$=3\frac{3}{5}$ days [$\because a\frac{b}{c}+d=(a+d)\frac{b}{c}$]