Engineering Economics practice questions

Answer: B) 12.68%

• AIncorrect. That's the nominal rate.
• BCorrect. \((1.01)^{12} - 1\).
• CIncorrect. That's the monthly rate.
• DIncorrect. Recheck: \((1.01)^{12}\).

Answer: D) 5 years

• AIncorrect. Divide, don't subtract.
• BIncorrect. That's inverted.
• CIncorrect. Recheck the division.
• DCorrect. 30,000/6,000.

Answer: C) 1.2

• AIncorrect. That's a difference.
• BIncorrect. That reverses it.
• CCorrect. Benefits ÷ costs.
• DIncorrect. That isn't the calculation.

Answer: C) 5.77%

Solving i = i' + f + i'f for i' gives i' = (i − f)/(1 + f) = 0.06/1.04 ≈ 5.77%. The 6% simple difference is only an approximation.

• AIncorrect. Adding the inflation rate to the market rate goes the wrong way. The market rate already includes inflation, so you remove inflation rather than add it.
• BIncorrect. 10% − 4% = 6% is the linear approximation. It ignores the cross term i'f, which the exact relationship includes.
• CCorrect. i' = (i − f)/(1 + f) = (0.10 − 0.04)/1.04 ≈ 0.0577, which is the same as (1.10/1.04) − 1 ≈ 5.77%.
• DIncorrect. This divides by (1 + i) instead of (1 + f): 0.06/1.10 ≈ 5.45%. Solving i = i' + f + i'f for i' gives a denominator of (1 + f).

Answer: C) $6,793

Use the capital recovery factor to turn a present amount into an equivalent annual amount: A = P(A/P, 6%, 10) = 50,000 × 0.13587 ≈ $6,793.

• AIncorrect. This uses the sinking fund factor (A/F, 6%, 10) = 0.07587, which converts a future amount into annual payments. The $50,000 is a present amount, so the capital recovery factor (A/P) is needed.
• BIncorrect. This is $50,000/10. It spreads the cost evenly but ignores the time value of money at 6%.
• CCorrect. A = P(A/P, 6%, 10) = $50,000 × [0.06(1.06)^10]/[(1.06)^10 − 1] = $50,000 × 0.13587 ≈ $6,793.
• DIncorrect. This adds $5,000 of straight-line recovery to $3,000 of interest on the full $50,000 every year. That overcharges interest, because the unrecovered balance shrinks each year.

Answer: B) $15,000

For SOYD, D_j = (n − j + 1)/[n(n + 1)/2] × (cost − salvage). For year 1 with n = 3: (3/6)($30,000) = $15,000.

• AIncorrect. This is $30,000/3, the straight-line charge. SOYD front-loads depreciation, so the year-1 charge is larger.
• BCorrect. The sum of the years' digits is 1 + 2 + 3 = 6. The year-1 fraction is 3/6, so D1 = (3/6)($30,000) = $15,000.
• CIncorrect. This uses 1/6, the fraction for year 3. In year 1 the numerator is the full remaining life (3).
• DIncorrect. This is (2/3)($30,000), the year-1 charge under double-declining-balance depreciation, not SOYD.

Answer: A) $4,500

• ACorrect. (50,000 − 5,000)/10.
• BIncorrect. Subtract the salvage value first.
• CIncorrect. Subtract, don't add, the salvage value.
• DIncorrect. Recheck the division by 10.

Answer: D) $25,000

• AIncorrect. That uses 2.5%.
• BIncorrect. That multiplies instead.
• CIncorrect. That adds one year of interest.
• DCorrect. For a perpetuity, P = A/i = 1,000/0.04.

Answer: C) $200,000

• AIncorrect. Recheck: 5,000/0.05 = 100,000.
• BIncorrect. Capitalize the perpetual maintenance.
• CCorrect. 100,000 + 5,000/0.05.
• DIncorrect. Recheck: 5,000/0.05 = 100,000.

Answer: B) $11,000

• AIncorrect. Depreciate to salvage, not to zero.
• BCorrect. 20,000 − 3(3,000).
• CIncorrect. That's the accumulated depreciation.
• DIncorrect. Annual depreciation is $3,000.