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    Question

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

    days. B and C together can complete it in 16 days. C and A together can complete it in 24 days. They all start working together. After 4 days A leaves, and 6 days after that B also leaves. In how many total days (from the start) is the work completed?
    A 42 days Correct Answer Incorrect Answer
    B 28 dys Correct Answer Incorrect Answer
    C 34 days Correct Answer Incorrect Answer
    D 54 days Correct Answer Incorrect Answer
    E None of these Correct Answer Incorrect Answer

    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) βˆ’ (1/A + 1/B) = 3/32 βˆ’ 1/12 = (9 βˆ’ 8)/96 = 1/96 β‡’ C = 96 days. From (iii): 1/A + 1/C = 1/24 β‡’ 1/A = 1/24 βˆ’ 1/96 = (4 βˆ’ 1)/96 = 3/96 = 1/32 β‡’ A = 32 days. From (ii): 1/B + 1/C = 1/16 β‡’ 1/B = 1/16 βˆ’ 1/96 = (6 βˆ’ 1)/96 = 5/96 β‡’ B = 96/5 days. Take total work = LCM of 32 and 96 = 96 units. Then rates: A: 96/32 = 3 units/day B: 96 Γ· (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 Γ— 9 = 36 units. Remaining = 96 βˆ’ 36 = 60 units. Next 6 days (A left; B + C work): B + C = 5 + 1 = 6 units/day Work in 6 days = 6 Γ— 6 = 36 units. Total done till now = 36 + 36 = 72 units. Remaining = 96 βˆ’ 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.

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