## Answers

From the given information, it can be noticed that the life of both the machines is not equal. The Machine A has a life of 4 years and the life of Machine B is 6 years. In case of unequal lives, we have to use the common multiple method and convert the unequal life into equal life and then can evaluate. The LCM of 4 years and 6 years is 12 years. Therefore, the Machine A is to be repeated 3 times and Machine B is repeated 2 times.

As per the question we have to calculate the annual worth of machine B.

First cost = -77,000

Annual Operating Cost = -21,000

Salvage Value = 10,000

MARR = 15%

Step – 1

Calculate the PW

PW = -77,000 – 77,000 (P/F, 15%, 6) – 21,000 (P/A, 15%, 12) + 10,000 (P/F, 15%, 6) + 10,000 (P/F, 15%, 12)

PW = -77,000 – 77,000 (0.43233) – 21,000 (5.42062) + 10,000 (0.43233) + 10,000 (0.18691)

PW = -217,930

Step – 2

Calculate the annual worth

AW = PW (A/P, 15%, 12)

AW = -217,930 (0.18448)

AW = -40,203.7

**Answer - C) $-40,000**