Question
A buffalo alone can plough field ‘A’ in 80 days. A
Bull alone can plough the field ‘A’ in 160 days. Find the number of days taken by 2 bulls and 1 buffalo to together plough field ‘B’ that is 25% larger than field ‘A’?Solution
Let the total work done to plough field ‘A’ = 160 units (LCM of 80 and 160) Then, efficiency of a buffalo = (160/80) = 2 units/day Efficiency of a bull = (160/160) = 1 units/day Total work required to plough field ‘B’ = 160 × 1.25 = 200 units Combined efficiency of 2 bulls and 1 buffalo = 2 × 1 + 2 × 1 = 4 units So, number of days required to plough field ‘B’ = 200/4 = 50 days
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