A, B and C together can complete a piece of work in 10 days. All the three started working at it together and after 4 days A left. Then B and C together completed the work in 10 more days. A alone could complete the work in :

### Concept & Combined Work To find an individual's rate of work, subtract the rates of the other workers from the combined rate. $$ \text{Rate} = \frac{1}{\text{Time}} $$ ### Step-by-Step Solution * **Given:** A, B, and C can complete work in 10 days. * (A + B + C)'s 1 day work = $1/10$. * Work done by all three in 4 days = $4 \times (1/10) = 2/5$. * Remaining work = $1 - 2/5 = 3/5$. * B and C complete the remaining $3/5$ work in 10 days. * (B + C)'s 1 day work = $\frac{3/5}{10} = 3/50$. * A's 1 day work = (A + B + C)'s 1 day work - (B + C)'s 1 day work. * A's 1 day work = $1/10 - 3/50 = \frac{5 - 3}{50} = 2/50 = 1/25$. * Time taken by A alone = 25 days. ### Exam Strategy & Shortcut Total work = $10 \text{ units (days for all)}$. In 4 days, $40\%$ work is done. Remaining $60\%$ is done by B+C in 10 days, so their $100\%$ capacity is $10/0.6 = 50/3$ days. Use LCM method: Total work = $30$ units. $(A+B+C)$ efficiency = $3$. $(B+C)$ efficiency = $1.8$. $A$ efficiency = $3 - 1.8 = 1.2$. A's time = $30 / 1.2 = 25$. ### Common Pitfall Confusing the time taken to complete the *remaining* work with the time taken to complete the *whole* work. ### Final Answer Therefore, the correct answer is **25 days**.