Question

A can finish a piece of work in 14 days. The same work can be finished by B in 21 days. They start the work together, but 3 days before the completion of the work, A leaves.
What is the total number of days required to complete the work, had both of them worked together?

• A.
• B.
• C.
• D.
• E.

Answer 1

The correct option is C

B’s 3 days’ work $=(\frac{1}{21}\times 3)=\frac{1}{7}$

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

(A+B)’s 1 day’s work $=(\frac{1}{14}+\frac{1}{21})=\frac{5}{42}$

Now, $\frac{5}{42}$ work is done by A and B in 1 day.

∴ $\frac{6}{7}$ work is done by A and B in $(\frac{42}{5}\times \frac{6}{7})=\frac{36}{5}days$

Hence, total time taken $=(3+\frac{36}{5})$ days = $10\frac{1}{5}$ days