Question
A and B together can complete a work in 6 days, B and C
together in 8 days, and C and A together in 12 days. In how many days can each of them complete the work alone?Solution
Let daily work rates be a, b, c. a + b = 1/6 b + c = 1/8 c + a = 1/12 Add all three: 2(a + b + c) = 1/6 + 1/8 + 1/12 LCM(6, 8, 12) = 24 1/6 = 4/24, 1/8 = 3/24, 1/12 = 2/24 So 2(a + b + c) = (4 + 3 + 2)/24 = 9/24 = 3/8 ⇒ a + b + c = 3/16 Now, a = (a + b + c) − (b + c) = 3/16 − 1/8 = 3/16 − 2/16 = 1/16 So A alone takes 16 days. b = (a + b + c) − (c + a) = 3/16 − 1/12 LCM(16,12) = 48 3/16 = 9/48, 1/12 = 4/48 ⇒ b = (9 − 4)/48 = 5/48 So B alone takes 48/5 days. c = (a + b + c) − (a + b) = 3/16 − 1/6 3/16 = 9/48, 1/6 = 8/48 ⇒ c = (9 − 8)/48 = 1/48 So C alone takes 48 days.
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