Question : A can do a piece of work in 8 days which B can destroy in 3 days. A has worked for 6 days, during the last 2 days of which B has been destroying; how many days must A work alone to complete the remaining work?
Option 1: $7\ \text{days}$
Option 2: $7\frac{1}{3}\ \text{days}$
Option 3: $7\frac{2}{3}\ \text{days}$
Option 4: $8\ \text{days}$
Correct Answer: $7\frac{1}{3}\ \text{days}$
Solution :
Time taken by A alone to do the work = 8 days
Part of work done by A alone in a day = $\frac{1}{8}$
Time taken by B alone to destroy the work = 3 days
Part of work destroyed by B alone in a day = $\frac{1}{3}$
Part of work done by A in 6 days = $\frac{6}{8}$
Part of work destroyed by B in last 2 days = $\frac{2}{3}$
Part of work completed after 6 days = $\frac{6}{8}-\frac{2}{3}$ = $\frac{18-16}{24}$ = $\frac{2}{24}$
Remaining part of work to be completed by A alone = $1-\frac{2}{24}$ = $\frac{22}{24}$
So, the number of days needed to complete the remaining work by A alone = $\frac{\frac{22}{24}}{\frac{1}{8}}$ = $\frac{22}{3}$ = $7\frac{1}{3}\ \text{days}$
Hence, the correct answer is $7\frac{1}{3}\ \text{days}$.
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