A buffalo alone can plough field ‘A’ in 15 days. A Bull alone can plough the field ‘A’ in 25 days. Find the number of days taken by 2 bulls and 3 buffalo to together plough field ‘B’ that is 26% larger than field ‘A’?
Let the total work done to plough field ‘A’ = 75 units (LCM of 15 and 25) Then, efficiency of a buffalo = (75/15) = 5 units/day Efficiency of a bull = (75/25) = 3 units/day Total work required to plough field ‘B’ = 75 × 1.26 = 94.5 units Combined efficiency of 2 bulls and 3 buffalo = 3 × 2 + 3 × 5 = 21 units So, number of days required to plough field ‘B’ = 94.5/21 = 4.5 days
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MICR stands for
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