Question

$A$ can do a piece of work in $20$ days which $B$ alone can do in $12$ days. Both started working together, $B$ worked at it for $9$ days. $A$ can finish the remaining work in

(a) $3$ days

(b) $5$ days

(c) $7$ days

(d) $11$ days

Answer 1

The correct option is (b) $5$ days

$A^{\prime }s$ $1$ day's work $=\frac{1}{20}$

$B^{\prime }s$ $1$ day's work $=\frac{1}{12}$

Both's $1$ day's work $=\frac{1}{20}+\frac{1}{12}=\frac{1\times 3}{20\times 3}+\frac{1\times 5}{12\times 5}$ [$\because$ L.C.M of $20$ and $12$ is $60$]

$=\frac{3+5}{60}=\frac{8}{60}=\frac{2}{15}$

$B^{\prime }s$ $9$ day's work $=\frac{1}{12}\times 9=\frac{3}{4}$

Remaining work (to do by $A$) $=1-\frac{3}{4}=\frac{1}{4}$

$A$ can finish $\frac{1}{20}$ work in $1$ day.

$\therefore$ $A$ will finish $\frac{1}{4}$ work in $\frac{1}{\frac{1}{20}}\times \frac{1}{4}=20\times \frac{1}{4}=5$ days