Question
A buffalo alone can plough field ‘A’ in 50 days. A
Bull alone can plough the field ‘A’ in 75 days. Find the number of days taken by 1 bulls and 3 buffalo to together plough field ‘B’ that is 12% larger than field ‘A’?Solution
Let the total work done to plough field ‘A’ = 150 units (LCM of 50 and 75) Then, efficiency of a buffalo = (150/50) = 3 units/day Efficiency of a bull = (150/75) = 2 units/day Total work required to plough field ‘B’ = 150 × 1.12 = 168 units Combined efficiency of 1 bulls and 2 buffalo = 1 × 2 + 2 × 3 = 8 units So, number of days required to plough field ‘B’ = 168/8 = 21 days
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