Engineering Economics practice questions

Answer: C) $134

• AIncorrect. That's one year.
• BIncorrect. That's simple growth.
• CCorrect. \(100(1.03)^{10}\).
• DIncorrect. Recheck the exponent.

Answer: A) $3,000 a year

• ACorrect. ($20,000 − $2,000) ÷ 6.
• BIncorrect. Salvage reduces depreciation.
• CIncorrect. That isn't the calculation.
• DIncorrect. That's the total depreciable amount.

Answer: A) 4 years

• ACorrect. $1,000 ÷ $250 per year = 4 years.
• BIncorrect. That's inverted.
• CIncorrect. Recheck the division.
• DIncorrect. Recheck the division.

Answer: C) $3,355

• AIncorrect. That's a future worth.
• BIncorrect. That ignores interest.
• CCorrect. \(500(P/A, 8\%, 10) = 500(6.7101)\).
• DIncorrect. Recheck the series factor.

Answer: A) 10.38%

• ACorrect. \((1.025)^4 - 1\).
• BIncorrect. Quarterly compounding raises it.
• CIncorrect. That's the quarterly rate.
• DIncorrect. That's semiannual compounding.

Answer: D) $8,758

• AIncorrect. Recheck the discount factor.
• BIncorrect. Discount the salvage value.
• CIncorrect. Subtract the salvage value.
• DCorrect. 10,000 − 2,000(0.6209).

Answer: C) 4 years

• AIncorrect. That multiplies instead.
• BIncorrect. That reverses it.
• CCorrect. $12,000 ÷ $3,000.
• DIncorrect. That subtracts instead.

Answer: C) $7,451

Equal payments that repay a present amount come from the capital recovery factor: A = P(A/P, 8%, 10) = $50,000 × 0.14903 ≈ $7,451.

• AIncorrect. This adds 10 years of simple interest ($40,000) to the principal and divides by 10. It ignores that the balance, and so the interest owed, falls as the loan is paid down.
• BIncorrect. $50,000/10 repays only the principal and leaves out interest entirely.
• CCorrect. A = P(A/P, 8%, 10) = $50,000 × [0.08(1.08)^10/((1.08)^10 − 1)] = $50,000 × 0.14903 ≈ $7,451.
• DIncorrect. This uses the sinking-fund factor (A/F, 8%, 10) = 0.06903, which treats $50,000 as a future amount rather than a present loan.

Answer: A) It ignores the time value of money and cash flows after payback

Simple payback ignores the time value of money and any cash flows after the payback point, so it can favor quick-payback projects even when they are less profitable.

• ACorrect. Simple payback just adds up undiscounted cash flows until the investment is recovered. Anything earned after that point is ignored, so it can favor short-lived projects over more profitable long-lived ones.
• BIncorrect. Simple payback uses no interest rate. Discounted payback is the variant that needs one.
• CIncorrect. It does the opposite: it ranks projects that recover their cost quickly higher.
• DIncorrect. Depreciation handling depends on how the cash flows are prepared, not on the payback method. Its real weakness is ignoring timing and later cash flows.

Answer: C) $18,216

Use the uniform-series present worth factor: P = A(P/A,i,n) = 2,000 × 9.1079 ≈ $18,216. The other values come from not discounting, discounting as a single lump sum, or computing future worth.

• AIncorrect. This treats the $30,000 total as one lump sum received at year 15 and discounts it with (P/F,7%,15) = 0.3624. The payments actually arrive every year, so the uniform-series factor is needed.
• BIncorrect. This is the undiscounted sum, 15 × $2,000. It ignores the time value of money.
• CCorrect. P = A(P/A,7%,15) = 2,000 × [1 − 1.07^−15]/0.07 = 2,000 × 9.1079 ≈ $18,216.
• DIncorrect. This is the future worth at year 15: F = 2,000 × (F/A,7%,15) = 2,000 × 25.129. The question asks for the present worth.