Multiple Choice Questions
1. One assumption inherent in the present worth method of analysis is that:
(a) The alternatives will be used only through the life of the shortest-lived alternative.
(b) The alternatives will be used only through the life of the longest-lived alternative.
(c) The cash flows of each alternative will change only by the inflation or deflation rate in succeeding life cycles.
(d) At least one of the alternatives will have a finite life.
2. When only one alternative can be selected from two or more, the alternatives are said to be:
(a) Mutually exclusive
(b) Independent alternatives
(c) Cost alternatives
(d) Revenue alternatives
3. For the mutually exclusive alternatives shown, the one(s) that should be selected are:
Alternative PW, $
A ......... €“ 25,000
B ......... €“12,000
C ......... 10,000
D ......... 15,000
(a) Only C
(b) Only A
(c) C and D
(d) Only D
4. The alternatives shown are to be compared on the basis of their present worth values. At an interest rate of 10% per year, the values of n that you should use in the uniform series factors to make a correct comparison by the present worth method are:
(a) n =3 years for A and n = 3 years for B
(b) n = 3 years for A and n = 6 years for B
(c) n =6 years for A and n =6 years for B
(d) None of the above
Problems 5 through 7 are based on the following information.

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 = €“250,000 + 40,000(P/A, 15%, 6) + 35,000(P/F, 15%, 6)
(b) PWJ = €“250,000 + 26,000(P/A, 15%, 6) + 35,000(P/F, 15%, 6)
(c) PWJ = €“250,000 €“ 26,000(P/A, 15%, 6) + 35,000(P/F, 15%, 6)
(d) PWJ = €“250,000 €“ 26,000(P/A, 15%, 6) €“ 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 = €“150,000 + 11,000(P/A, 15%, 3) + 25,000(P/F, 15%, 3)
(b) PWI = €“150,000 + 11,000(P/A, 15%, 6) + 25,000(P/F, 15%, 6)
(c) PWI = €“150,000 + 11,000(P/A, 15%, 6) + 175,000(P/F, 15%, 3) + 25,000(P/F, 15%, 6)
(d) PWI = €“150,000 + 11,000(P/A, 15%, 6) €“ 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 = €“80,000 €“ 15,000(P/A, 10%, 4) + 30,000(P/F, 10%, 4)
(b) PWX = €“80,000 €“ 20,000(P/A, 10%, 4) €“ 80,000(P/F, 10%, 2) + 10,000(P/F, 10%, 4)
(c) PWX = €“80,000 €“ 20,000(P/A, 10%, 2) + 10,000(P/F, 10%, 2)
(d) PWX = €“80,000 €“ 20,000(P/A, 10%, 4) €“ 70,000(P/F, 10%, 2) + 10,000(P/F, 10%, 4)

The interest rate is 10% peryear.

Alternative l Initial cost, S Annual Income, S per year Annual expenses, S per year Salvage value, S Life, years -150,000 20,000 9,000 25,000 Alternative J -250,000 40,000 -14 ,000 35,000 Initial cost,S Annual operating cost, S per year Salvage value, S Life, years Machine X 80,000 20,000 Machine Y -95,000 15,000 30,000 -20 10,000