Problems 6.52–6.54 are based on the following cash flows and an interest rate of 10% per year. Alternative X Y First cost, $ −200,000 −800,000 Annual cost, $/year −60,000 −10,000 Salvage value, $ 20,000 150,000 Life, years 5 [infinity] 6.52 In comparing the alternatives by the annual worth method, the AW of X is determined by the following equation: (a) −200,000(0.10) − 60,000 + 20,000(0.10) (b) −200,000(A∕P,10%,5) − 60,000 + 20,000(A∕F,10%,5) (c) −200,000(A∕P,10%,5) − 60,000 − 20,000(A∕F,10%,5) (d) −200,000(0.10) − 60,000 + 20,000(A∕F,10%,5) 6.53 The annual worth of perpetual service for alternative X is represented by the following equation: (a) −200,000(0.10) − 60,000 + 20,000(0.10) (b) −200,000(A∕P,10%,5) − 60,000 + 20,000(A∕F,10%,5) (c) −200,000(A∕P,10%,5) − 60,000 − 20,000(A∕F,10%,5) (d) −200,000(0.10) − 60,000 + 20,000(A∕F,10%,5) 6.54 The annual worth of alternative Y is closest to: (a) $−50,000 (b) $−76,625 (c) $−90,000 (d) $−92,000