A and B together can complete a work in 12 days, B and C

Solution

Let daily work of A, B, C be a, b, c. Given: a + b = 1/12 …(i) b + c = 1/16 …(ii) a + c = 1/24 …(iii) Add (i) and (ii): a + 2b + c = 1/12 + 1/16 = 7/48 Subtract (iii): (a + 2b + c) − (a + c) = 7/48 − 1/24 2b = 7/48 − 2/48 = 5/48 ⇒ b = 5/96 From (i): a = 1/12 − 5/96 = (8 − 5)/96 = 3/96 = 1/32 From (iii): c = 1/24 − 1/32 = (4 − 3)/96 = 1/96 Rates: A: 1/32, B: 5/96, C: 1/96 All three together: a + b + c = 1/32 + 5/96 + 1/96 = (3 + 5 + 1)/96 = 9/96 = 3/32 of work/day Work done in first 3 days: = 3 × 3/32 = 9/32 Remaining work = 1 − 9/32 = 23/32 Let A and C work alone for t days. Rate(A + C) = 1/32 + 1/96 = 4/96 = 1/24 Work done by A and C alone = t × 1/24 = t/24 Remaining work before D joins = 23/32 − t/24 D’s efficiency = half of B ⇒ d = (5/96)/2 = 5/192 Rate(A + C + D) = 1/24 + 5/192 = 8/192 + 5/192 = 13/192 They work 2 days together: work = 2 × 13/192 = 26/192 = 13/96 So remaining work before D joined = 13/96. Hence, 23/32 − t/24 = 13/96 23/32 = 69/96 So, 69/96 − t/24 = 13/96 t/24 = (69 − 13)/96 = 56/96 = 14/24 ⇒ t = 14 days