A and B together can complete a piece of work in 10 days. B and C together can complete it in 15 days, and A and C together can complete it in 12 days. All three start working together. A leaves after x days. B leaves 2 days after A leaves, and C alone finishes the remaining work in 8 more days. In how many days (value of
x) did A work?

Let daily work of A, B, C be a, b, c. Total work = 1. Given: a + b = 1/10 b + c = 1/15 a + c = 1/12 Add all: 2(a + b + c) = 1/10 + 1/15 + 1/12 LCM(10,15,12) = 60 (1/10 + 1/15 + 1/12) = (6 + 4 + 5)/60 = 15/60 = 1/4 So a + b + c = 1/8 Now: a + b = 1/10 ⇒ (a + b + c) − (b + c) = a ⇒ a = 1/8 − 1/15 = (15 − 8)/120 = 7/120 Similarly, b = (a + b + c) − (a + c) = 1/8 − 1/12 = (3 − 2)/24 = 1/24 c = (a + b + c) − (a + b) = 1/8 − 1/10 = (5 − 4)/40 = 1/40 Timeline: Stage 1 (0 to x days): A + B + C → rate = 1/8 Work₁ = x × 1/8 = x/8 Stage 2 (next 2 days): B + C → rate = 1/15 Work₂ = 2 × 1/15 = 2/15 Stage 3 (next 8 days): only C → rate = 1/40 Work₃ = 8 × 1/40 = 8/40 = 1/5 Total work = 1: x/8 + 2/15 + 1/5 = 1 2/15 + 1/5 = 2/15 + 3/15 = 5/15 = 1/3 So: x/8 + 1/3 = 1 x/8 = 2/3 x = (2/3) × 8 = 16/3 days So A worked for 16/3 days.