Question

$A$ can do a piece of work in $25$ days, which $B$ alone can do in $20$ days. A started the work and was joined by $B$ after $10$ days. The work lasted for
(a) $12\frac{1}{2}$ days
(b) $15$ days
(c) $16\frac{2}{3}$ days
(d) $14$ days

Answer 1

$A$ can do the piece of work in $25$ days.

The part of the work done by $A$ in $1$ day=$\frac{1}{25}$

$B$ can do the same work in $20$ days.

The part of the work done by $B$ in $1$ day $=\frac{1}{20}$

Since, $A$ worked as alone $10$ days.

So, the part of the work done by $A$ in $10$ days $=\frac{10}{25}=\frac{2}{5}$

$\Rightarrow$ The remaining work $=1-\frac{2}{5}=\frac{3}{5}$---(1)

The part of the work done by $A+B$ $=\frac{1}{25}+\frac{1}{20}$

$=\frac{4+5}{100}$

$=\frac{9}{100}$

Hence, The part of the work done by $A+B=\frac{9}{100}$

Since, the remaining part of the work is $\frac{3}{5}$

$\Rightarrow$ Time taken to finish the remaining work if they work together

$=\frac{3}{5}÷\frac{9}{100}$

$=\frac{3}{5}\times \frac{100}{9}$

$=\frac{100}{15}$

$=6\frac{10}{15}$

$=6\frac{2}{3}$

So they will finish the remaining work in $6\frac{2}{3}$ days.

Here $A$ already worked in $10$ days.

So, Total number of days to finish the work $=(10+6\frac{2}{3})$ days

$=(10+6+\frac{2}{3})$ days

$=16\frac{2}{3}$ days

Hence, Option C is correct.