Shortcut Trick
Total time = 11(1/3) days (B starts first). B works on days 1, 3, 5, 7, 9, 11 (6 days); A works on days 2, 4, 6, 8, 10 and 1/3 of day 12 (5 + 1/3 = 16/3 days).
Work done by A = (16/3) ÷ 48 = 1/9. Remaining work done by B = 1 − 1/9 = 8/9.
B finishes 8/9 work in 6 days ⇒ B finishes full work in (6 × 9)/8 = 27/4 days.
B finishes 4 times the work in (27/4) × 4 = 27 days.
∴ The correct answer is 27 days.
Alternate Method
Given: A alone = 48 days. Total alternate time = 11(1/3) days. B starts on day 1.
Distribution of work days:
⇒ B works on days: 1, 3, 5, 7, 9, 11 → Total 6 days.
⇒ A works on days: 2, 4, 6, 8, 10, and 1/3 of the 12th day → Total 5 + 1/3 = 16/3 days.
⇒ Work done by A in 16/3 days = (16/3) × (1/48) = 1/9 of total work.
⇒ Work done by B in 6 days = 1 − 1/9 = 8/9 of total work.
⇒ Time taken by B to finish 1 complete work = 6 ÷ (8/9) = (6 × 9) / 8 = 27/4 days.
⇒ Time taken by B to finish 4 times the same work = 4 × (27/4) = 27 days.
∴ The correct answer is 27 days.
Additional Information
Alternate Day Work
In alternate work, identify who starts first. If total time is T, and B starts, B works for ceil(T/2) days if T is not an even integer and cycle is 2 days.
Efficiency and Time
Time is inversely proportional to Efficiency. Work = Efficiency × Time. For multiple works, New Time = Original Time × Multiplier.