Question

$A$ can do a piece of work in $14$ days which $B$ can do in $21$ days. They begin together, but $3$ days before the completion of the work, $A$ leaves off. The total number of days to complete the work is

• A. $6\frac{3}{5}$
• B. $8\frac{1}{2}$
• C. $10\frac{1}{5}$
• D. $13\frac{1}{2}$

Answer 1

The correct option is C $10\frac{1}{5}$

$B^{\prime }s3$ days work $=(\frac{1}{21}\times 3)=\frac{1}{7}$

Remaining work $=(1-\frac{1}{7})=\frac{6}{7}$

$(A+B{)}^{\prime }s1$ day work $=(\frac{1}{14}+\frac{1}{21})$
$=\frac{5}{42}$

Now, $\frac{5}{42}$ work is done by $A$ and $B$ in $1$ day.
$\therefore$ $\frac{6}{7}$ work is done by $A$ and $B$ in
$(\frac{42}{5}\times \frac{6}{7})=\frac{36}{5}$ days
Hence, total time taken $=(3+\frac{36}{5})=10\frac{1}{5}$ days