Question
A buffalo alone can plough field ‘A’ in 20 days. A
Bull alone can plough the field ‘A’ in 10 days. Find the number of days taken by 1 bulls and 4 buffalo to together plough field ‘B’ that is 35% larger than field ‘A’?Solution
Let the total work done to plough field ‘A’ = 20 units (LCM of 20 and 10) Then, efficiency of a buffalo = (20/20) = 1 units/day Efficiency of a bull = (20/10) = 2 units/day Total work required to plough field ‘B’ = 20 × 1.35 = 27 units Combined efficiency of 1 bulls and 4 buffalo = 2 × 1 + 4 × 1 = 6 units So, number of days required to plough field ‘B’ = 27/6 = 4.5 days
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