Question
A buffalo alone can plough field ‘A’ in 30 days. A
Bull alone can plough the field ‘A’ in 36 days. Find the number of days taken by 2 bulls and 1 buffalo to together plough field ‘B’ that is 20% larger than field ‘A’?Solution
Let the total work done to plough field ‘A’ = 180 units (LCM of 30 and 36) Then, efficiency of a buffalo = (180/30) = 6 units/day Efficiency of a bull = (180/36) = 5 units/day Total work required to plough field ‘B’ = 180 × 1.20 = 216 units Combined efficiency of 2 bulls and 1 buffalo = 5 × 2 + 6 × 1 = 16 units So, number of days required to plough field ‘B’ = 216/16 = 13.5 days
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Conclusion
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II: R ≥ E
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...
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