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    Question

    Quantity I: A and B can complete a work together in 12

    days. After working together for 6 days, A leaves, and B finishes the remaining work in 10 more days. How many days would it take for A alone to complete the entire work? Quantity II: C and D together can complete a job in 15 days, but if C worked alone, he would take 24 days. D worked alone for 6 days before C joined to complete the remaining job together. How many days in total did it take to finish the work?
    A Quantity I < Quantity II Correct Answer Incorrect Answer
    B Quantity I ≥ Quantity II Correct Answer Incorrect Answer
    C Quantity I ≤ Quantity II Correct Answer Incorrect Answer
    D Quantity I = Quantity II Correct Answer Incorrect Answer
    E Quantity I > Quantity II Correct Answer Incorrect Answer

    Solution

    Quantity I: Let the total work be 1 unit. Combined rate of A and B = 1/12 units per day. In 6 days, they complete 6 * (1/12) = 1/2 of the work. Remaining work = 1 - 1/2 = 1/2 unit, which B completes in 10 days. B’s rate = (1/2) / 10 = 1/20 units per day. Rate of A alone = (1/12) - (1/20) = (5 - 3)/60 = 1/30 units per day. Time taken by A alone to complete the work = 1 / (1/30) = 30 days. Quantity II: Let the total work be 1 unit. Combined rate of C and D = 1/15 units per day. C’s rate alone = 1/24 units per day, so D’s rate = (1/15) - (1/24) = 1/40 units per day. D worked alone for 6 days, completing 6 * (1/40) = 3/20 of the work. Remaining work = 1 - 3/20 = 17/20. Time for C and D together to complete 17/20 work = (17/20) / (1/15) = (17/20) * 15 = 12.75 days. Total time = 6 + 12.75 = 18.75 days. Answer: A (Quantity I < Quantity II)

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