Question |
The interest rate is 15% per year. 5. In comparing alternatives I and J by the present worth method, the value of n that must be used in 11,000(P/A, i, n) for alternative I is: (a) 3 (b) 6 (c) 18 (d) 36 6. In comparing alternatives I and J by the present worth method, the equation that yields the present worth of alternative J is: (a) PWJ = â??250,000 + 40,000(P/A, 15%, 6) + 35,000(P/F, 15%, 6) (b) PWJ = â??250,000 + 26,000(P/A, 15%, 6) + 35,000(P/F, 15%, 6) (c) PWJ = â??250,000 â?? 26,000(P/A, 15%, 6) + 35,000(P/F, 15%, 6) (d) PWJ = â??250,000 â?? 26,000(P/A, 15%, 6) â?? 35,000(P/F, 15%, 6) 7. In comparing alternatives I and J by the present worth method, the equation that yields the present worth of alternative I is: (a) PWI = â??150,000 + 11,000(P/A, 15%, 3) + 25,000(P/F, 15%, 3) (b) PWI = â??150,000 + 11,000(P/A, 15%, 6) + 25,000(P/F, 15%, 6) (c) PWI = â??150,000 + 11,000(P/A, 15%, 6) + 175,000(P/F, 15%, 3) + 25,000(P/F, 15%, 6) (d) PWI = â??150,000 + 11,000(P/A, 15%, 6) â?? 125,000(P/F, 15%, 3) + 25,000(P/F, 15%, 6) 8. The equation that will calculate the present worth of machine X is: (a) PWX = â??80,000 â?? 15,000(P/A, 10%, 4) + 30,000(P/F, 10%, 4) (b) PWX = â??80,000 â?? 20,000(P/A, 10%, 4) â?? 80,000(P/F, 10%, 2) + 10,000(P/F, 10%, 4) (c) PWX = â??80,000 â?? 20,000(P/A, 10%, 2) + 10,000(P/F, 10%, 2) (d) PWX = â??80,000 â?? 20,000(P/A, 10%, 4) â?? 70,000(P/F, 10%, 2) + 10,000(P/F, 10%, 4) The interest rate is 10% peryear. |