Question
A buffalo alone can plough field ‘A’ in 40 days. A
Bull alone can plough the field ‘A’ in 80 days. Find the number of days taken by 3 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’ = 80 units (LCM of 40 and 80) Then, efficiency of a buffalo = (80/40) = 2 units/day Efficiency of a bull = (80/80) = 1 units/day Total work required to plough field ‘B’ = 80 × 1.25 = 100 units Combined efficiency of 3 bulls and 1 buffalo = 1 × 3 + 2 × 1 = 5 units So, number of days required to plough field ‘B’ = 100/5 = 20 days
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