A, B and C together can do a piece of work in 10 days. All three start the work together, but A leaves after 4 days. B and C together then complete the remaining work in 10 more days. In how many days can A alone complete the whole work?

Introduction / Context:
This problem uses combined work rates initially and then a reduced team to finish the remaining work. The information allows you to determine the shared rate of B and C and then extract A's individual rate. Finally, you compute the time A would take working alone.

Given Data / Assumptions:

• A, B and C together complete the work in 10 days.
• All three start together.
• A leaves after 4 days.
• B and C together finish the work in 10 more days.
• Work rate is constant for each worker.

Concept / Approach:
Let the total work be one unit. From the overall time for A, B and C, you get their combined daily work rate. Calculate how much work is done in the first 4 days. The remaining work is then handled by B and C alone over 10 days, which gives you the daily work rate of B + C. Subtracting this from the total combined rate gives A's individual rate and thus A's solo completion time.

Step-by-Step Solution:
Assume total work = 1 unit. A + B + C together finish in 10 days, so (a + b + c) = 1 / 10 per day. In the first 4 days, work done by A, B and C together = 4 * (1 / 10) = 4 / 10 = 2 / 5. Remaining work = 1 - 2 / 5 = 3 / 5. B and C together complete this 3 / 5 in 10 days. Thus (b + c) = (3 / 5) / 10 = 3 / 50 per day. Now (a + b + c) = 1 / 10 = 5 / 50. So A's rate a = (a + b + c) - (b + c) = 5 / 50 - 3 / 50 = 2 / 50 = 1 / 25 per day. Therefore A alone takes 25 days to finish the work.
Verification / Alternative check:
In 4 days, A does 4 * (1 / 25) = 4 / 25 of the work. In the same 4 days, B + C do 4 * (3 / 50) = 12 / 50 = 6 / 25. Total after 4 days = 4 / 25 + 6 / 25 = 10 / 25 = 2 / 5 of the work, which matches the earlier calculation. Then B and C do 10 * (3 / 50) = 30 / 50 = 3 / 5, completing the whole work.
Why Other Options Are Wrong:
24, 23 or 21 days would imply different daily rates for A, which would not match the observed combined and B + C work rates. Substituting these values back would either leave unfinished work or produce more than the total work.
Common Pitfalls:
Learners sometimes incorrectly treat "10 more days" as total days, not extra days after the first 4. Another mistake is miscomputing the remaining work fraction after the first phase. Always carefully separate phases of work and recompute remaining work step by step.
Final Answer:
A alone can finish the work in 25 days.