Subtopic: Cost analysis of implements and tractors
Question Statement:
A farmer has a choice of buying a 4 bottom × 41 cm mould board plough for Rs 8570 or a 5 bottom × 45 cm mould board plough for Rs 12000. Each plough has a life of 15 years. Neglect salvage value, interest charges and other costs on the ploughs. With either plough the operating speed is 6.5 km h⁻¹ and field efficiency is 82%. Assume that the cost per hectare for tractor energy to be same for both ploughs. If the labor cost is Rs 10 per hour, the minimum number of hectares that would justify the purchase of the larger plough (i.e. break even point) is
Options:
(A) 73.7
(B) 89.9
(C) 737.3
(D) 899.4
Official Answer Key:
A
Step-by-Step Solution & Explanation
Effective field capacity, FC = (width×speed×field eff.)/10 (ha/h). FC_A = (4×0.41×6.5×0.82)/10 = 0.8741 ha/h; FC_B = (5×0.45×6.5×0.82)/10 = 1.1993 ha/h. Annual fixed cost (straight-line, no interest/salvage) = purchase price/life: f_A = 8570/15 = Rs 571.3; f_B = 12000/15 = Rs 800.0 (extra fixed cost of B = Rs 228.7/yr). Labor cost per ha = (Rs 10/h)/FC: labor_A = 10/0.8741 = Rs 11.44/ha; labor_B = 10/1.1993 = Rs 8.34/ha (saving of Rs 3.10/ha with plough B, tractor-energy cost being equal cancels out). Break-even area = extra fixed cost / labor saving per ha = 228.7/3.10 = 73.7 ha.
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