This problem is based on the following cash flows for alternatives X and Y at an interest rate of 10% per year.
In comparing the alternatives on a present worth basis, the PW of machine X is closest to:
(a) $23,160
(b) $40,560
(c) $58,950
(d) $72,432
PWX = -146,000 + (80,000-15,000)(P/A,10%,6) - 136,000(P/F,10%,3) + 10,000(P/F,10%,6)
= -146,000 + 65,000(4.3553) - 136,000(0.7513) + 10,000(0.5645)
= $40,563
Answer is (b) $40,560
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