Question

A can do a piece of work in $14$ days and B in $21$ days. They begin together, but $3$ days before the completion of the work, A leaves off. In how many days is the work complete?

• A. $10days$
• B. $10\frac{1}{5}days$
• C. $15days$
• D. $7days$

Answer 1

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

$(A+B{)}^{\prime }s1$ day work
$=\frac{1}{14}+\frac{1}{21}=\frac{3+2}{42}=\frac{5}{42}$
At the end, B works for $3$ days so work done by B in $3days=3\times \frac{1}{21}=\frac{1}{7}$.
Remaining work $1-\frac{1}{7}=\frac{6}{7}$ which is to be done by A and B.
$\frac{5}{42}$ of work $(A+B)$ can do in $1$ days
$\therefore \frac{6}{7}$ of work $(A+B)$ can do in $\frac{1}{5/42}\times \frac{6}{7}=7\frac{1}{5}days$
$\therefore$ Time to complete the work
$=3+7\frac{1}{5}=10\frac{1}{5}days$.