Engineering Board Exam Reviewer · Present Worth Analysis · medium
Machine A costs P250,000, has annual costs of P30,000, and a salvage value of P50,000. Machine B costs P300,000, has annual costs of P22,000, and a salvage value of P70,000. Both last 8 years. At 12% interest, find the present worth of cost of the better machine.
- P381,016.25
- P378,835.03
- P399,029.19
- P440,000
Answer
B. P378,835.03
Solution
PWA=250,000+30,000(4.9676)-50,000(0.403883)=P378,835.03. PWB=300,000+22,000(4.9676)-70,000(0.403883)=P381,016.25. Machine A has the lower cost of the two, so the answer is P378,835.03.
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