Two pumps can be used for pumping a corrosive liquid. A pump with a brass impell

Question

Two pumps can be used for pumping a corrosive liquid. A pump with a brass impeller costs $800 and lasts and lasts 3 years. A pump with a stainless steel impeller costs $1,900 and lasts 5 years. A rebuild costing 300 is required after 2,000 operating hours for the brass impeller pump, while the stainless steel impeller pump requires a $700 overhaul after 8,000 hours. If the operating cost of each pump is $1 per hour, how many hours per year must the pump operate to justify the more expensive type? use i=10%

Explanation / Answer

Solution: Let x = hours per year -800(A/P,10%,3) - (300/2000)x -1.0x = -1,900(A/P,10%,5) - (700/8000)x - 1.0 -800(0.40211) - 0.15x - 1.0x = -1,900(0.2638) - 0.0875x - 1.0x 0.0625x = 179.532, so x = 2873 hours per year

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