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    Question

    A buffalo alone can plough field ‘A’ in 10 days. A

    Bull alone can plough the field ‘A’ in 16 days. Find the number of days taken by 1 bulls and 3 buffalo to together plough field ‘B’ that is 45% larger than field ‘A’?
    A 2 days Correct Answer Incorrect Answer
    B 3 days Correct Answer Incorrect Answer
    C 4 days Correct Answer Incorrect Answer
    D 3.5 days Correct Answer Incorrect Answer
    E None of these Correct Answer Incorrect Answer

    Solution

    Let the total work done to plough field ‘A’ = 80 units (LCM of 10 and 16) Then, efficiency of a buffalo = (80/10) = 8 units/day Efficiency of a bull = (80/16) = 5 units/day Total work required to plough field ‘B’ = 80 × 1.45 = 116 units Combined efficiency of 1 bulls and 3 buffalo = 5 × 1 + 3 × 8 = 29 units So, number of days required to plough field ‘B’ = 116/29 = 4 days

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