Engineering Economics practice questions

Answer: B) B, because its net present worth exceeds that of A

For mutually exclusive alternatives, choose the one with the highest positive NPW: B ($7,000) beats A ($5,000). The incremental B/C of 1.2 confirms that B's extra cost is justified.

• AIncorrect. The B/C ratios are right, but ranking mutually exclusive alternatives by their overall B/C ratios can give the wrong choice. NPW (or incremental analysis) is the correct criterion, and it favors B.
• BCorrect. NPW(A) = $15,000 − $10,000 = $5,000 and NPW(B) = $27,000 − $20,000 = $7,000. With mutually exclusive alternatives, select the one with the larger positive NPW.
• CIncorrect. Benefit per dollar is just the B/C ratio again. Maximizing NPW shows that B's extra $10,000 of cost buys $12,000 of extra benefit, so B is worth the added investment.
• DIncorrect. The incremental B/C is ($27,000 − $15,000)/($20,000 − $10,000) = 1.2, which is above 1.0. The incremental analysis also favors B, and both alternatives have positive NPW.

Answer: C) $8,954

Use the single-payment compound amount factor: F = P(1 + i)^n = 5,000 × 1.06^10 ≈ $8,954.

• AIncorrect. This is simple interest: 5,000 × (1 + 0.06 × 10) = $8,000. It leaves out interest earned on interest.
• BIncorrect. This compounds for only 9 periods: 5,000 × 1.06^9 ≈ $8,447. The money is invested for 10 years.
• CCorrect. F = P(F/P, 6%, 10) = 5,000 × 1.06^10 = 5,000 × 1.7908 ≈ $8,954.
• DIncorrect. This compounds for 11 periods: 5,000 × 1.06^11 ≈ $9,492. That is one year too many.

Answer: B) $30,000

• AIncorrect. Salvage affects annual depreciation.
• BCorrect. Annual depreciation is $10,000.
• CIncorrect. That isn't the calculation.
• DIncorrect. That's after four years.

Answer: A) $115.93

• ACorrect. $100 × 1.03⁵.
• BIncorrect. Inflation compounds.
• CIncorrect. It's five years.
• DIncorrect. That overstates it.

Answer: B) The lowest acceptable return on an investment

• AIncorrect. That isn't MARR.
• BCorrect. Projects below it are usually rejected.
• CIncorrect. MARR includes more than inflation.
• DIncorrect. MARR is set by the organization.

Answer: C) $7,514

Use the capital recovery factor to convert a present cost into equivalent annual payments: A = 30,000 × (A/P, 8%, 5) ≈ $7,514.

• AIncorrect. This uses the sinking fund factor (A/F, 8%, 5) = 0.17046. That factor turns a future amount into annual payments, not a present cost.
• BIncorrect. This is $30,000 / 5, which leaves out the time value of money.
• CCorrect. A = P(A/P, 8%, 5) = 30,000 × [0.08(1.08)^5 / ((1.08)^5 − 1)] = 30,000 × 0.25046 ≈ $7,514.
• DIncorrect. This adds a full year of interest on the original cost ($2,400) to straight-line ($6,000). It overstates the cost because the unrecovered balance shrinks each year.

Answer: C) $6,139

Present worth of a single future amount is P = F/(1 + i)^n = 10,000/(1.05)^10 ≈ $6,139.

• AIncorrect. This is 10,000 × (1.05)^10, the future worth of $10,000 today. The problem needs the reverse: discount the future amount back to the present.
• BIncorrect. This uses simple interest: 10,000/(1 + 0.05×10). The problem says the interest is compounded annually.
• CCorrect. P = F(P/F, 5%, 10) = 10,000/(1.05)^10 = 10,000/1.6289 ≈ $6,139.
• DIncorrect. This is 10,000/1.05, which discounts the amount by only one year instead of ten.

Answer: B) At least the MARR

• AIncorrect. Then the extra money isn't justified.
• BCorrect. The extra investment then earns enough.
• CIncorrect. That means no return on the extra money.
• DIncorrect. It must meet the MARR.

Answer: B) $7,451

• AIncorrect. That ignores interest (straight division).
• BCorrect. \(A = P(A/P, 8\%, 10) = 50{,}000 \times 0.14903\).
• CIncorrect. That's interest only, with no capital recovery.
• DIncorrect. Recheck the capital recovery factor.

Answer: B) 12.68%

• AIncorrect. Monthly compounding raises the effective rate.
• BCorrect. \((1.01)^{12} - 1\).
• CIncorrect. Recheck the calculation.
• DIncorrect. That's the monthly rate.