FE engineering economics practice problems with solutions
Below are six FE engineering economics practice problems: present worth, uniform series, annual cost, straight-line depreciation, breakeven and a benefit-cost ratio. Each has a step-by-step solution using the factor formulas or tables in the FE Reference Handbook 10.6, with the page number.
Last reviewed October 10, 2026 by Engineer Exam Lab. Exam facts link to NCEES. Problems are from our practice packs, written by a PhD environmental engineer, and every numeric answer was checked by a second, independent calculation.
Where engineering economics shows up
- FE Environmental: 5–8 questions. FE Civil: 5–8 questions (NCEES specifications).
- It is on every FE exam, so these points are the same whichever discipline you take (topics on every FE exam).
- Most questions use one factor from the handbook's Engineering Economics chapter: P/F, P/A, A/P, or a straight-line depreciation step.
How to work these
Read the timeline first: what happens now, what repeats every year, and what happens once at the end. Then pick the factor. The handbook gives both the factor formulas and factor tables by interest rate; the table is faster when the rate is listed, the formula when it isn't. Aim for about 3 minutes each.
Six FE engineering economics practice problems
Problem 1 · FE Environmental · Engineering Economics · Multiple choice
A contractor plans to sell a crane for $37,000 at the end of year 7. At a MARR of 10% per year, the present worth of that sale is most nearly:
Show the worked solution
Answer: B) $19,000
Equation: P = F(P/F, i, n), with (P/F, i, n) = (1 + i)⁻ⁿ
- (P/F, 10%, 7) = 1/(1.10)^7 = 0.5132 (factor table, i = 10%)
- P = F(P/F, 10%, 7) = ($37,000)(0.5132) = $19,000
Why the wrong choices are wrong:
- A used n + 1 = 8 years instead of 7.
- C took the future amount at face value, with no discounting.
- D discounted with simple interest, F/(1 + ni), instead of compound (1 + i)⁻ⁿ.
Handbook: FE Reference Handbook 10.6, Engineering Economics, factor formulas (P/F) and Factor Table (i = 10%), p. 235 and 241.
Problem 2 · FE Civil · Engineering Economics · Multiple choice
A water utility expects an energy upgrade at a pump station to save $42,000 per year for 16 years. Using a MARR of 12% per year, the present worth of these savings is most nearly:
Show the worked solution
Answer: B) $293,000
Equation: P = A(P/A, i, n), with (P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]
- (P/A, 12%, 16) = [(1.12)^16 - 1]/[0.12(1.12)^16] = 6.9740 (factor table, i = 12%)
- P = A(P/A, 12%, 16) = ($42,000)(6.9740) = $293,000
Why the wrong choices are wrong:
- A added the yearly savings without discounting them (A × n).
- C used (F/A), the future worth of the series, instead of (P/A).
- D discounted the total of all savings as one amount at year n with (P/F).
Handbook: FE Reference Handbook 10.6, Engineering Economics, factor formulas (P/A) and Factor Table (i = 12%), p. 235 and 242.
Problem 3 · FE Environmental · Engineering Economics · Multiple choice
A county buys a dump truck for $518,000. The truck will be used for 13 years and will have no salvage value. At a MARR of 6% per year, the equivalent uniform annual cost of the purchase is most nearly:
Show the worked solution
Answer: C) $58,500
Equation: A = P(A/P, i, n), with (A/P, i, n) = i(1 + i)ⁿ/[(1 + i)ⁿ - 1]
- (A/P, 6%, 13) = 0.06(1.06)^13/[(1.06)^13 - 1] = 0.1130 (factor table, i = 6%)
- A = P(A/P, 6%, 13) = ($518,000)(0.1130) = $58,500 per year
Why the wrong choices are wrong:
- A used n + 1 = 14 years instead of 13.
- B used the sinking fund factor (A/F) instead of capital recovery (A/P).
- D used n - 1 = 12 years instead of 13.
Handbook: FE Reference Handbook 10.6, Engineering Economics, factor formulas (A/P) and Factor Table (i = 6%), p. 235 and 240.
Problem 4 · FE Environmental · Engineering Economics · Multiple choice
A treatment plant buys a centrifuge for $92,000. It has a 10-year life and a salvage value of $2,000. Using straight-line depreciation, the book value at the end of year 3 is most nearly:
Show the worked solution
Answer: B) $65,000
Equation: Dj = (P - S)/n; BV = P - Σ Dj
- Dj = (P - S)/n = ($92,000 - $2,000)/10 = $9,000 per year
- BV after 3 years = P - 3 × Dj = $92,000 - 3 × $9,000 = $65,000
Why the wrong choices are wrong:
- A subtracted only one year of depreciation.
- C is the yearly depreciation charge, not the book value.
- D is the accumulated depreciation, not the book value.
Handbook: FE Reference Handbook 10.6, Engineering Economics, Depreciation (Straight Line) and Book Value, p. 236.
Problem 5 · FE Environmental · Engineering Economics · Multiple choice
A precast plant has fixed costs of $527,000 per year. Each manhole section sells for $78 and costs $60 to make. The breakeven production is most nearly:
Show the worked solution
Answer: B) 29,300 units/yr
Equation: Q = F/(price - variable cost per unit)
- At breakeven, revenue = cost: price·Q = F + v·Q
- Q = F/(price - v) = $527,000/($78 - $60) = 29,300 units per year
Why the wrong choices are wrong:
- A divided the fixed cost by the price, ignoring the variable cost.
- C divided the fixed cost by the variable cost.
- D added the variable cost to the price.
Handbook: FE Reference Handbook 10.6, Engineering Economics, Breakeven Analysis, p. 236.
Problem 6 · FE Civil · Engineering Economics · Multiple choice
A county flood-control project costs $1,820,000 now and $51,500 per year to operate. It prevents damages worth $399,000 per year for 19 years. At a MARR of 6% per year, the conventional benefit-cost ratio is most nearly:
Show the worked solution
Answer: C) 1.86
Equation: B/C = PW(benefits)/PW(costs), with (P/A, i, n) = [(1 + i)ⁿ - 1]/[i(1 + i)ⁿ]
- (P/A, 6%, 19) = 11.1581 (factor table, i = 6%)
- PW of benefits = $399,000 × 11.1581 = $4,452,000
- PW of costs = $1,820,000 + $51,500 × 11.1581 = $2,395,000
- B/C = $4,452,000/$2,395,000 = 1.86
Why the wrong choices are wrong:
- A is the cost-to-benefit ratio (inverted).
- B used undiscounted totals.
- D is the modified ratio (benefits minus O&M over first cost), not the conventional one asked for.
Handbook: FE Reference Handbook 10.6, Engineering Economics, Benefit-Cost Analysis and factor formulas, p. 237 and 240.
The slips the wrong choices are built from
- Off-by-one years. An amount at the end of year 7 is discounted 7 years, not 8.
- Simple instead of compound interest. F/(1 + ni) is not F(1 + i)⁻ⁿ.
- Wrong direction. A/P spreads a present cost over the years; P/A brings a yearly amount back to today. Swapping them gives a far-off answer.
- Benefit-cost bookkeeping. In the conventional ratio, operating costs go in the denominator with the first cost, not subtracted from benefits.
Want 100 more like these, plus a timed 110-question mock? Each pack is $39, one time, with every solution tied to the handbook page.
What this page doesn't cover
Six problems can't cover every engineering economics subtopic. Rate of return, arithmetic gradients, capitalized cost and MACRS depreciation also appear on the exam. Each of our packs has 5 engineering economics problems in the practice set and 5 to 6 in the mock.
Questions
How many engineering economics questions are on the FE exam?
5 to 8 on both FE Civil and FE Environmental, according to the NCEES exam specifications. Engineering economics appears on every FE discipline exam.
Do I need to memorize interest factor formulas?
No. The FE Reference Handbook lists the factor formulas and factor tables for common interest rates. You need to know which factor fits the timeline and find it quickly.
Should I use the factor table or the formula?
Use the table when the interest rate is listed in it, since it is faster. Use the formula when the rate is not in the tables. Both give the same answer to the precision the choices need.
Are these real FE exam questions?
No. They are original practice problems from our packs, written in exam style. Each numeric answer was computed by code and recomputed independently.
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Engineer Exam Lab is not affiliated with or endorsed by NCEES. The problems here are original practice problems, not questions from a real exam.