A and B together can complete a piece of work in 10

Solution

ATQ, Let total work = 1 unit. Given: (A + B)'s 1 day work = 1/10 (B + C)'s 1 day work = 1/12 (A + C)'s 1 day work = 1/15 Add first two: (A + B) + (B + C) ⇒ A + 2B + C So, A + 2B + C = 1/10 + 1/12 = (6 + 5)/60 = 11/60 Subtract (A + C): (A + 2B + C) − (A + C) = 2B So, 2B = 11/60 − 1/15 1/15 = 4/60 2B = (11 − 4)/60 = 7/60 ⇒ B’s 1 day work = 7/120 Now, A + B = 1/10 ⇒ A = 1/10 − 7/120 1/10 = 12/120 A = (12 − 7)/120 = 5/120 = 1/24 Similarly, A + C = 1/15 ⇒ C = 1/15 − 1/24 1/15 = 8/120, 1/24 = 5/120 C = (8 − 5)/120 = 3/120 = 1/40 Now total 1 day work of A, B and C: A + B + C = 1/24 + 7/120 + 1/40 LCM of 24, 120, 40 = 120 = 5/120 + 7/120 + 3/120 = 15/120 = 1/8 So, all three together complete the work in 8 days.