A and B together can complete a piece of work in 12

Solution

ATQ, Let individual times be A, B, C days. Then: 1/A + 1/B = 1/12 …(i) 1/B + 1/C = 1/16 …(ii) 1/C + 1/A = 1/24 …(iii) Add all three: 2(1/A + 1/B + 1/C) = 1/12 + 1/16 + 1/24. Right-hand side: LCM of 12,16,24 is 48. 1/12 = 4/48 1/16 = 3/48 1/24 = 2/48 Sum = (4 + 3 + 2)/48 = 9/48 = 3/16. Thus: 1/A + 1/B + 1/C = 3/32. Now find each: From (i): 1/A + 1/B = 1/12 So 1/C = (1/A + 1/B + 1/C) − (1/A + 1/B) = 3/32 − 1/12 = (9 − 8)/96 = 1/96 ⇒ C = 96 days. From (iii): 1/A + 1/C = 1/24 ⇒ 1/A = 1/24 − 1/96 = (4 − 1)/96 = 3/96 = 1/32 ⇒ A = 32 days. From (ii): 1/B + 1/C = 1/16 ⇒ 1/B = 1/16 − 1/96 = (6 − 1)/96 = 5/96 ⇒ B = 96/5 days. Take total work = LCM of 32 and 96 = 96 units. Then rates: A: 96/32 = 3 units/day B: 96 ÷ (96/5) = 5 units/day C: 96/96 = 1 unit/day. First 4 days (A + B + C): Daily work = 3 + 5 + 1 = 9 units Work in 4 days = 4 × 9 = 36 units. Remaining = 96 − 36 = 60 units. Next 6 days (A left; B + C work): B + C = 5 + 1 = 6 units/day Work in 6 days = 6 × 6 = 36 units. Total done till now = 36 + 36 = 72 units. Remaining = 96 − 72 = 24 units. Now only C works: C’s rate = 1 unit/day Time to finish 24 units = 24 days. Total time = 4 + 6 + 24 = 34 days.