Given those two annual costs of 9255 and 9196, the correct conclusion is that
Answer: C
A difference of well under one per cent cannot be resolved by data of ordinary accuracy, so the honest report says the alternatives are equivalent and lets other considerations decide. Both lives are ten years, so no equalising is needed.
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Two alternatives have lives of 4 and 6 years. Under the least common multiple approach the study period is
Answer: D
Correct answer: 12 years
The least common multiple of 4 and 6 is 12, so the first alternative is repeated three times and the second twice. Choosing 24 would work arithmetically but is unnecessarily long, and using either life alone gives the unequal horizons the method exists to avoid.
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A project's present worth is 25 000 over a 20-year life at 8 per cent. Its annual worth is about
Answer: A
Correct answer: 2546
\[ AW = 25{,}000 \times (A/P, 8\%, 20) = 25{,}000 \times 0.10185 = 2{,}546. \] Dividing 25,000 by 20 would give 1,250, which is the distractor: it ignores the interest that must be earned on the capital as well as its return.
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A machine bought for 80 000 three years ago now has a market value of 30 000 and a book value of 44 000. In a replacement study the figure to use is
Answer: B
Correct answer: 30 000
The market value is the amount forgone by keeping the machine, and forgoing money one could have is economically the same as spending it. The 80,000 paid three years ago is sunk, the 44,000 book value is an accounting figure, and the 14,000 difference between them matters only through its effect on tax.
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A project has a present worth of 5000 at 8 per cent and -2000 at 14 per cent. Its crossover with do-nothing lies
Answer: C
Correct answer: between 8 and 14 per cent
Present worth falls as the discount rate rises, so a value of +5,000 at 8 per cent and −2,000 at 14 per cent means the curve crosses zero somewhere between. Linear interpolation puts it near 12.3 per cent, though the curve is convex so the true figure is a little different.
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Two alternatives cost 40 000 and 60 000, with annual savings of 9000 and 12 500 over 8 years. At 10 per cent the increment of 20 000 buying 3500 a year has a present worth of about
Answer: D
Correct answer: -1330
\[ PW = -20{,}000 + 3{,}500 \times (P/A, 10\%, 8) = -20{,}000 + 3{,}500 \times 5.3349 = -1{,}328. \] The extra 20,000 does not earn its keep at 10 per cent, so the cheaper alternative is preferred - even though the dearer one saves more in absolute terms every year.
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Match the method with what it computes.
1. Present worth 2. Annual worth 3. Capitalised cost
P. Equivalent uniform amount per period Q. Equivalent value at time zero R. Endowment funding a perpetual cost
Answer: A
Correct answer: 1-Q, 2-P, 3-R
Three expressions of the same equivalence. Present worth places the value at time zero; annual worth expresses it as a uniform amount per period; and capitalised cost is the endowment that would fund a cost for ever, which is the present worth of a perpetuity. All three rank alternatives identically.
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Match the situation with the correct treatment of unequal lives.
1. Identical replacement is realistic 2. Technology will have changed 3. Each alternative repeats indefinitely
P. Set a study period and estimate market values Q. Compare over the least common multiple R. Compare on annual worth
Answer: B
Correct answer: 1-Q, 2-P, 3-R
The right treatment of unequal lives depends on what will actually happen. Where identical replacement is realistic, the least common multiple is legitimate. Where technology will have moved on, a defined study period with estimated market values is honest. And where each alternative genuinely repeats indefinitely, annual worth gives the answer with no repetition arithmetic at all.
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Put the steps of a present worth comparison in order.
1. Discount each alternative's cash flows at the chosen rate 2. Establish a common study period 3. Select the alternative with the greatest present worth 4. Set the minimum attractive rate of return
Answer: C
Correct answer: 4, 2, 1, 3
The rate comes first because nothing can be discounted without it; the study period comes next because it decides which cash flows are included; only then can the discounting be done and the alternatives ranked. Discounting before the horizon is settled is the common way of arriving at an incomparable pair of figures.
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Consider the following statements about worth analysis.
1. Annual worth avoids the need for a common study period 2. Present and annual worth can give opposite decisions 3. Sunk costs are irrelevant to a current decision
Answer: D
Correct answer: 1 and 3 only are correct
Statements 1 and 3 are correct: annual worth is already per period, and sunk costs cannot bear on a present decision. Statement 2 is false, since \( AW = PW \times (A/P, i, n) \) and the factor is always positive - the two measures cannot disagree about a sign.
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