10 FE practice problems: time value of money, with solutions
Engineering economics appears in every FE specification, and time value of money is its core: moving amounts across time with the interest factors in the Engineering Economics chapter of the handbook. These ten problems use the factor formulas directly. Write the cash-flow diagram first, pick the factor by what you have and what you want, and check whether the answer should be larger or smaller than the amount you started with.
How to use these problems
Give each problem an honest attempt before you open the solution: write the given values with units, find the equation in the FE Reference Handbook PDF, and commit to an answer. Then compare line by line. If you picked a wrong option, read the note for that option; each one names the mistake that produces it.
The problems
Problem 1 · FE, Time value of money
A contractor plans to sell a crane for $153,000 at the end of year 15. At a MARR of 12% per year, the present worth of that sale is most nearly:
Handbook: Engineering Economics, factor formulas and Factor Tables, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B ($28,000)
- Given
- F = $153,000, n = 15 years, i = 12% per year
- Find
- present worth P of the resale ($)
- Handbook
- Engineering Economics, factor formulas (P/F) and Factor Table (i = 12%), pages 235 and 242
- Equation
P = F(P/F, i, n), with (P/F, i, n) = (1 + i)⁻ⁿ- Substitute
(P/F, 12%, 15) = 1/(1.12)^15 = 0.1827 (factor table, i = 12%)P = F(P/F, 12%, 15) = ($153,000)(0.1827) = $28,000
- Result
- $28,000, 3 significant figures
- Check
- P is less than F = $153,000, as a future amount is always worth less today when i > 0.
- Why the others are wrong
- A: used n + 1 = 16 years instead of 15
- C: used (F/P), which moves money forward in time, instead of (P/F)
- D: discounted with simple interest, F/(1 + ni), instead of compound (1 + i)⁻ⁿ
Problem 2 · FE, Time value of money
A county buys a dump truck for $274,000. The truck will be used for 18 years and will have no salvage value. At a MARR of 4% per year, the equivalent uniform annual cost of the purchase is most nearly:
Handbook: Engineering Economics, factor formulas and Factor Tables, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A ($21,600)
- Given
- P = $274,000, n = 18 years, i = 4% per year
- Find
- equivalent uniform annual cost A ($ per year)
- Handbook
- Engineering Economics, factor formulas (A/P) and Factor Table (i = 4%), pages 235 and 240
- Equation
A = P(A/P, i, n), with (A/P, i, n) = i(1 + i)ⁿ/[(1 + i)ⁿ - 1]- Substitute
(A/P, 4%, 18) = 0.04(1.04)^18/[(1.04)^18 - 1] = 0.0790 (factor table, i = 4%)A = P(A/P, 4%, 18) = ($274,000)(0.0790) = $21,600 per year
- Result
- $21,600, 3 significant figures
- Check
- A is more than P/n = $15,200, because the cost is recovered with a return at the MARR.
- Why the others are wrong
- B: used the sinking fund factor (A/F) instead of capital recovery (A/P)
- C: used n + 1 = 19 years instead of 18
- D: divided the first cost by n, which ignores the time value of money
Problem 3 · FE, Time value of money
A contractor plans to sell a crane for $122,000 at the end of year 16. At a MARR of 4% per year, the present worth of that sale is most nearly:
Handbook: Engineering Economics, factor formulas and Factor Tables, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B ($65,100)
- Given
- F = $122,000, n = 16 years, i = 4% per year
- Find
- present worth P of the resale ($)
- Handbook
- Engineering Economics, factor formulas (P/F) and Factor Table (i = 4%), pages 235 and 240
- Equation
P = F(P/F, i, n), with (P/F, i, n) = (1 + i)⁻ⁿ- Substitute
(P/F, 4%, 16) = 1/(1.04)^16 = 0.5339 (factor table, i = 4%)P = F(P/F, 4%, 16) = ($122,000)(0.5339) = $65,100
- Result
- $65,100, 3 significant figures
- Check
- P is less than F = $122,000, as a future amount is always worth less today when i > 0.
- Why the others are wrong
- A: used n - 1 = 15 years instead of 16
- C: discounted with simple interest, F/(1 + ni), instead of compound (1 + i)⁻ⁿ
- D: used n + 1 = 17 years instead of 16
Problem 4 · FE, Time value of money
A county buys a dump truck for $435,000. The truck will be used for 12 years and will have no salvage value. At a MARR of 8% per year, the equivalent uniform annual cost of the purchase is most nearly:
Handbook: Engineering Economics, factor formulas and Factor Tables, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D ($57,700)
- Given
- P = $435,000, n = 12 years, i = 8% per year
- Find
- equivalent uniform annual cost A ($ per year)
- Handbook
- Engineering Economics, factor formulas (A/P) and Factor Table (i = 8%), pages 235 and 241
- Equation
A = P(A/P, i, n), with (A/P, i, n) = i(1 + i)ⁿ/[(1 + i)ⁿ - 1]- Substitute
(A/P, 8%, 12) = 0.08(1.08)^12/[(1.08)^12 - 1] = 0.1327 (factor table, i = 8%)A = P(A/P, 8%, 12) = ($435,000)(0.1327) = $57,700 per year
- Result
- $57,700, 3 significant figures
- Check
- A is more than P/n = $36,300, because the cost is recovered with a return at the MARR.
- Why the others are wrong
- A: multiplied P by i only, which leaves out recovering the first cost itself
- B: used n - 1 = 11 years instead of 12
- C: used n + 1 = 13 years instead of 12
Problem 5 · FE, Time value of money
A contractor plans to sell a crane for $49,000 at the end of year 19. At a MARR of 4% per year, the present worth of that sale is most nearly:
Handbook: Engineering Economics, factor formulas and Factor Tables, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A ($23,300)
- Given
- F = $49,000, n = 19 years, i = 4% per year
- Find
- present worth P of the resale ($)
- Handbook
- Engineering Economics, factor formulas (P/F) and Factor Table (i = 4%), pages 235 and 240
- Equation
P = F(P/F, i, n), with (P/F, i, n) = (1 + i)⁻ⁿ- Substitute
(P/F, 4%, 19) = 1/(1.04)^19 = 0.4746 (factor table, i = 4%)P = F(P/F, 4%, 19) = ($49,000)(0.4746) = $23,300
- Result
- $23,300, 3 significant figures
- Check
- P is less than F = $49,000, as a future amount is always worth less today when i > 0.
- Why the others are wrong
- B: used n - 1 = 18 years instead of 19
- C: took the future amount at face value, with no discounting
- D: used (F/P), which moves money forward in time, instead of (P/F)
Problem 6 · FE, Time value of money
A water utility expects an energy upgrade at a pump station to save $34,000 per year for 21 years. Using a MARR of 8% per year, the present worth of these savings is most nearly:
Handbook: Engineering Economics, factor formulas and Factor Tables, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A ($341,000)
- Given
- A = $34,000 per year, n = 21 years, i = 8% per year
- Find
- present worth P of the savings ($)
- Handbook
- Engineering Economics, factor formulas (P/A) and Factor Table (i = 8%), pages 235 and 241
- Equation
P = A(P/A, i, n), with (P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]- Substitute
(P/A, 8%, 21) = [(1.08)^21 - 1]/[0.08(1.08)^21] = 10.0168 (factor table, i = 8%)P = A(P/A, 8%, 21) = ($34,000)(10.0168) = $341,000
- Result
- $341,000, 3 significant figures
- Check
- P is less than the undiscounted total n × A = $714,000, as it must be when i > 0.
- Why the others are wrong
- B: used (F/A), the future worth of the series, instead of (P/A)
- C: added the yearly savings without discounting them (A × n)
- D: used n - 1 = 20 years instead of 21
Problem 7 · FE, Time value of money
A water utility expects an energy upgrade at a pump station to save $33,500 per year for 24 years. Using a MARR of 12% per year, the present worth of these savings is most nearly:
Handbook: Engineering Economics, factor formulas and Factor Tables, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C ($261,000)
- Given
- A = $33,500 per year, n = 24 years, i = 12% per year
- Find
- present worth P of the savings ($)
- Handbook
- Engineering Economics, factor formulas (P/A) and Factor Table (i = 12%), pages 235 and 242
- Equation
P = A(P/A, i, n), with (P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]- Substitute
(P/A, 12%, 24) = [(1.12)^24 - 1]/[0.12(1.12)^24] = 7.7843 (factor table, i = 12%)P = A(P/A, 12%, 24) = ($33,500)(7.7843) = $261,000
- Result
- $261,000, 3 significant figures
- Check
- P is less than the undiscounted total n × A = $804,000, as it must be when i > 0.
- Why the others are wrong
- A: used (F/A), the future worth of the series, instead of (P/A)
- B: discounted the total of all savings as one amount at year n with (P/F)
- D: added the yearly savings without discounting them (A × n)
Problem 8 · FE, Time value of money
A contractor plans to sell a crane for $75,000 at the end of year 15. At a MARR of 4% per year, the present worth of that sale is most nearly:
Handbook: Engineering Economics, factor formulas and Factor Tables, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A ($41,600)
- Given
- F = $75,000, n = 15 years, i = 4% per year
- Find
- present worth P of the resale ($)
- Handbook
- Engineering Economics, factor formulas (P/F) and Factor Table (i = 4%), pages 235 and 240
- Equation
P = F(P/F, i, n), with (P/F, i, n) = (1 + i)⁻ⁿ- Substitute
(P/F, 4%, 15) = 1/(1.04)^15 = 0.5553 (factor table, i = 4%)P = F(P/F, 4%, 15) = ($75,000)(0.5553) = $41,600
- Result
- $41,600, 3 significant figures
- Check
- P is less than F = $75,000, as a future amount is always worth less today when i > 0.
- Why the others are wrong
- B: discounted with simple interest, F/(1 + ni), instead of compound (1 + i)⁻ⁿ
- C: used n - 1 = 14 years instead of 15
- D: took the future amount at face value, with no discounting
Problem 9 · FE, Time value of money
A water utility expects an energy upgrade at a pump station to save $59,500 per year for 13 years. Using a MARR of 10% per year, the present worth of these savings is most nearly:
Handbook: Engineering Economics, factor formulas and Factor Tables, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C ($423,000)
- Given
- A = $59,500 per year, n = 13 years, i = 10% per year
- Find
- present worth P of the savings ($)
- Handbook
- Engineering Economics, factor formulas (P/A) and Factor Table (i = 10%), pages 235 and 241
- Equation
P = A(P/A, i, n), with (P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]- Substitute
(P/A, 10%, 13) = [(1.10)^13 - 1]/[0.10(1.10)^13] = 7.1034 (factor table, i = 10%)P = A(P/A, 10%, 13) = ($59,500)(7.1034) = $423,000
- Result
- $423,000, 3 significant figures
- Check
- P is less than the undiscounted total n × A = $773,500, as it must be when i > 0.
- Why the others are wrong
- A: used n - 1 = 12 years instead of 13
- B: discounted the total of all savings as one amount at year n with (P/F)
- D: used n + 1 = 14 years instead of 13
Problem 10 · FE, Time value of money
A contractor plans to sell a crane for $21,000 at the end of year 5. At a MARR of 4% per year, the present worth of that sale is most nearly:
Handbook: Engineering Economics, factor formulas and Factor Tables, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A ($17,300)
- Given
- F = $21,000, n = 5 years, i = 4% per year
- Find
- present worth P of the resale ($)
- Handbook
- Engineering Economics, factor formulas (P/F) and Factor Table (i = 4%), pages 235 and 240
- Equation
P = F(P/F, i, n), with (P/F, i, n) = (1 + i)⁻ⁿ- Substitute
(P/F, 4%, 5) = 1/(1.04)^5 = 0.8219 (factor table, i = 4%)P = F(P/F, 4%, 5) = ($21,000)(0.8219) = $17,300
- Result
- $17,300, 3 significant figures
- Check
- P is less than F = $21,000, as a future amount is always worth less today when i > 0.
- Why the others are wrong
- B: used n - 1 = 4 years instead of 5
- C: used n + 1 = 6 years instead of 5
- D: took the future amount at face value, with no discounting
A new problem posts every day inside the lab, with the full worked solution the same evening.
Frequently asked questions
Should I use the factor tables or the formulas?
The handbook prints both. Table values are rounded, so a factor computed from the formula can differ slightly in the last digit; either way, carry enough digits through the calculation.
Is engineering economics on every FE exam?
Yes. Every FE specification includes engineering economics (FE Chemical names it Economics).
What is the most common time value of money mistake?
Picking the inverse factor, for example using the present worth factor when the question asks for a future amount. A quick size check catches it.
Sources
- NCEES FE Civil CBT exam specifications (PDF). Retrieved October 3, 2026.
- NCEES FE Mechanical CBT exam specifications (PDF). Retrieved October 3, 2026.
- NCEES FE Industrial and Systems CBT exam specifications (PDF). Retrieved October 3, 2026.
- NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.