Question
A buffalo alone can plough field ‘A’ in 20 days. A
Bull alone can plough the field ‘A’ in 32 days. Find the number of days taken by 2 bulls and 2 buffalo to together plough field ‘B’ that is 30% larger than field ‘A’?Solution
Let the total work done to plough field ‘A’ = 160 units (LCM of 20 and 32) Then, efficiency of a buffalo = (160/20) = 8 units/day Efficiency of a bull = (160/32) = 5 units/day Total work required to plough field ‘B’ = 160 × 1.30 = 208 units Combined efficiency of 2 bulls and 2 buffalo = 2 × 5 + 2 × 8 = 26 units So, number of days required to plough field ‘B’ = 208/26 = 8 days
Digestibility of legume protein is
2*25 plough is move at speed of 5 Km/Hr how much time will it take to plough the field of size 300*300 meter if filed effiiciency of instrument is 60%
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