Engineering Economics is 4-6 of the 110 questions on the FE Civil exam and is the most formulaic area on the test. Every problem reduces to placing cash flows on a timeline and applying a factor from the handbook interest tables. Candidates lose these points to sign errors and to picking the wrong factor, not to conceptual difficulty.
The problems below cover the patterns that actually recur: present worth comparison of two alternatives, equivalent uniform annual cost, a gradient series, straight-line and MACRS depreciation, and a benefit-cost ratio on a public project. Each solution shows the timeline first, then the factor notation, then the arithmetic.
Exam weight: NCEES lists Engineering Economics at 4-6 questions (4-6%) of the 110-question FE Civil exam. Work each problem below on paper first, then reveal the worked solution — reading a solution you have not attempted builds recognition, not recall.
What NCEES Tests in Engineering Economics
The specification covers time value of money, cost estimating, economic analyses, and depreciation. In practice that means single payment, uniform series, and gradient factors used to move money between present, future, and annual equivalents. Expect alternative comparison by present worth, future worth, or annual worth, plus rate of return and payback period. Public infrastructure framing is common, so benefit-cost ratio appears more often here than on other FE disciplines.
Depreciation questions center on straight-line and MACRS, with the recovery period tables printed in the handbook. Cost estimating covers escalation, inflation adjustment, and the difference between nominal and effective interest rates when compounding is not annual. Break-even analysis shows up as a crossover point between two alternatives. Numbers are usually clean because the factors do the work.
5 Free Engineering Economics Practice Problems
Each problem below comes from the PECivilClick FE Civil question bank and matches the style, difficulty and format of the real exam. Attempt each one under a three-minute limit — roughly the pace the exam demands.
Problem 1
A perpetual easement yields \(\\)9{,}500\( per year indefinitely. If it is purchased for \)\\(95{,}000\), what annual interest rate is earned on the investment?
A) 0.08
B) 0.09
C) 0.1
D) 0.12
Answer: C) 0.1
Step 1 – Identify given information: A perpetual easement yields \( A = \\)9{,}500 \( per year indefinitely and is purchased for \) P = \\(95{,}000 \). We need to find the annual interest rate \( i \).
Step 2 – Perpetuity formula: For a perpetuity (infinite series of equal payments), the FE Handbook gives:
\( P = \frac{A}{i} \quad \Rightarrow \quad i = \frac{A}{P} \)
Step 3 – Substitute and solve:
\( i = \frac{9{,}500}{95{,}000} = 0.10 = 10\% \)
Step 4 – Answer: The annual interest rate earned is \( 10\% \). Option (A) 8% and (B) 9% are too low, and (D) 12% is too high. The perpetuity formula is one of the simplest in engineering economics -- if the annual payment and present worth are known, the rate is simply their ratio.
Correct answer: (C)
Problem 2
A pump has an initial cost of \(\\)8{,}000\(, a salvage value of \)\\(1{,}000\) after 5 years, and annual maintenance of \(\\)600\(. If \) i = 8\% $, what is the EUAC?
A) \(\\)2{,}300$/yr
B) \(\\)2{,}430$/yr
C) \(\\)2{,}600$/yr
D) \(\\)2{,}800$/yr
Answer: B) \(\\)2{,}430$/yr
Step 1 – Identify the given information:
| Initial cost (P) | \( \\)8{,}000 \( |
| Salvage value (S) | \) \\(1{,}000 \) |
| Annual maintenance | \( \\)600\text{/yr} \( |
| Interest rate (i) | \) 8\% \( |
| Life (n) | \) 5 \text{ years} \( |
Step 2 – Write the EUAC formula. The Equivalent Uniform Annual Cost converts all costs to an equal annual series:
\) \text{EUAC} = P(A/P,\, i,\, n) + A_{\text{maint}} - S(A/F,\, i,\, n) \(
where:
\)(A/P, i, n)\( = capital recovery factor (converts present cost to annual)
\)(A/F, i, n)\( = sinking fund factor (converts future salvage to annual credit)
Step 3 – Look up factors from the FE Handbook at \) i = 8\%,\; n = 5 \(:
\) (A/P,\, 8\%,\, 5) = 0.2505 \(
\) (A/F,\, 8\%,\, 5) = 0.1705 \(
Step 4 – Substitute and compute:
\) \text{EUAC} = 8{,}000(0.2505) + 600 - 1{,}000(0.1705) \(
\) \text{EUAC} = 2{,}004 + 600 - 170.5 \(
\) \text{EUAC} = \\(2{,}433.50\text{/yr} \approx \\)2{,}430\text{/yr} $
Correct answer: (B)
Problem 3
Payments occur at the end of each year for 5 years: \(\\)1{,}000\( in year 1 increasing by \)\\(200\) each year. If \( i = 8\% \), what is the present worth?
A) \(\\)5{,}200$
B) \(\\)5{,}470$
C) \(\\)5{,}800$
D) \(\\)6{,}100$
Answer: B) \(\\)5{,}470$
Step 1 – Identify given information:
| Base annual payment (A) | \( \\)1{,}000 \( |
| Annual gradient (G) | \) \\(200\text{/yr} \) |
| Interest rate (i) | \( 8\% \) |
| Number of years (n) | \( 5 \) |
Payments: \(\\)1{,}000\(; \)\\(1{,}200\); \(\\)1{,}400\(; \)\\(1{,}600\); \(\\)1{,}800\(.
Step 2 – Arithmetic gradient present worth formula: From the FE Handbook, a gradient series is split into a uniform base plus an increasing gradient:
\) P = A(P/A,\, i,\, n) + G(P/G,\, i,\, n) \(
Step 3 – Substitute and solve:
Look up at \) i = 8\%,\; n = 5 \(: \) (P/A) = 3.9927 \(, \) (P/G) = 7.3724 \(
\) P = 1{,}000(3.9927) + 200(7.3724) \(
\) P = 3{,}993 + 1{,}474 = \\(5{,}467 \)
Step 4 – Answer: The present worth is approximately \( \\)5{,}470 \(, matching option (B). The gradient factor \) (P/G) \( accounts for the increasing portion only; the base amount is handled separately by \) (P/A) \(. The gradient starts at zero in year 1 and increases by \) G $ each year.
Correct answer: (B)
Problem 4
A replacement will cost \(\\)4{,}000$ today and is expected to inflate at 5% annually. If the replacement occurs in 3 years and the discount rate is 8%, what is the present worth?
A) \(\\)3{,}400$
B) \(\\)3{,}680$
C) \(\\)4{,}000$
D) \(\\)4{,}320$
Answer: B) \(\\)3{,}680$
Step 1 – Identify given information:
| Current cost (\(P_0\)) | \( \\)4{,}000 \( |
| Inflation rate (f) | \) 5\%\text{/yr} \( |
| Discount rate (d) | \) 8\% \( |
| Replacement year (n) | \) 3 \( |
Step 2 – Present worth with inflation: The future cost inflates at rate \) f \(, then is discounted at the market rate \) d \(:
\) P = P_0 \left(\frac{1+f}{1+d}\right)^n \(
This combines two steps: inflating the cost to year \) n \(, then discounting it back to year 0.
Step 3 – Substitute and solve:
\) P = 4{,}000 \left(\frac{1.05}{1.08}\right)^3 = 4{,}000(0.9722)^3 \(
\) (0.9722)^3 = 0.9190 \(
\) P = 4{,}000 \times 0.9190 = \\(3{,}676 \)
Step 4 – Answer: The present worth is approximately \( \\)3{,}680 \(, matching option (B). Since the discount rate exceeds the inflation rate, the present worth is less than today's cost. If inflation equaled the discount rate, the present worth would be exactly \) \\(4{,}000 \).
Correct answer: (B)
Problem 5
\(\\)2{,}000$ is deposited today in an account earning 5% annual interest. No withdrawals are made. What is the account balance after 4 years?
A) \(\\)2{,}400$
B) \(\\)2{,}430$
C) \(\\)2{,}550$
D) \(\\)2{,}700$
Answer: B) \(\\)2{,}430$
Step 1 – Identify given information:
| Present deposit (P) | \( \\)2{,}000 \( |
| Interest rate (i) | \) 5\% \( |
| Number of years (n) | \) 4 \( |
Find the future value \) F \( (single payment compound amount).
Step 2 – Single payment compound amount formula (F/P): From the FE Handbook:
\) F = P(1 + i)^n \(
Step 3 – Substitute and solve:
\) F = 2{,}000(1.05)^4 \(
\) (1.05)^2 = 1.1025 \(
\) (1.05)^4 = (1.1025)^2 = 1.21551 \(
\) F = 2{,}000 \times 1.21551 = \\(2{,}431 \)
Step 4 – Answer: The account balance after 4 years is approximately \( \\)2{,}431 \(, which matches (B). Option (A) \)\\(2{,}400\) would result from simple interest, not compound interest. With compound interest, you earn interest on previously earned interest, producing a slightly higher result.
Correct answer: (B)
Using the FE Reference Handbook for Engineering Economics
The Engineering Economics section ends with several pages of interest factor tables at standard rates. Learn the factor notation before exam day, because searching for a plain phrase will not find them: the table headers read (P/F, i%, n), (A/P, i%, n), and so on. Read the notation as what you want over what you have. If the problem uses a rate not tabulated, the closed-form equations sit just above the tables.
Four Mistakes That Cost Points
- Mixing period length and interest period When a problem gives an annual rate but monthly payments, the interest rate and the number of periods must both convert. Using 6 percent with 60 monthly periods instead of 0.5 percent per month produces a confidently wrong answer that often appears as a choice. Convert the rate and n together, every time.
- Confusing nominal and effective interest rates Twelve percent compounded monthly is not twelve percent effective. Problems state the compounding frequency for a reason, and the effective annual rate formula is in the handbook. Whenever a problem mentions compounding more often than annually, compute the effective rate before comparing alternatives or the comparison itself is invalid.
- Getting the gradient series timing wrong The arithmetic gradient factor assumes the first gradient increment occurs at the end of period two, not period one. Candidates routinely apply the factor to a series that starts increasing immediately. Draw the cash flow diagram and separate the uniform base amount from the gradient before selecting any factor.
- Comparing alternatives with unequal lives Present worth comparison requires equal study periods. When one alternative lasts 5 years and another 10, you must use the least common multiple with repeated cycles, or switch to equivalent uniform annual cost. Comparing raw present worths across unequal lives is the single most common setup error in alternative selection problems.
Frequently Asked Questions
How many engineering economics questions are on the FE Civil exam?
NCEES specifies 4-6 questions from Engineering Economics out of 110, about 4-6 percent of the exam. Because the problems follow a small number of repeating patterns, this is one of the highest return areas per hour of study. Candidates who drill twenty or thirty problems typically move from unreliable to consistently correct on this material.
Do I need to memorize the interest factor formulas?
No. Both the closed-form equations and the tabulated factors are in the handbook. What you must know cold is the notation, so that you can translate a problem into (P/A, i%, n) without hesitating. The lookup takes seconds once you know which factor you need. Deciding which factor you need is the actual skill being tested.
Which depreciation methods appear on the FE Civil exam?
Straight-line and MACRS are the two that matter. Straight-line is a one-line calculation from cost, salvage value, and useful life. MACRS requires the recovery period and the percentage table, both of which are in the handbook. Know the half-year convention in the first year, since that is where MACRS problems most often catch candidates.
How is benefit-cost ratio tested on the FE Civil exam?
Usually on a public works scenario: a bridge, a road improvement, or a drainage project with annual benefits, annual costs, and an initial capital cost. Convert everything to the same basis, typically annual, then take benefits over costs. A ratio above one justifies the project. Watch for disbenefits, which belong in the numerator as a reduction, not added to costs.
Keep Going
Work through the other FE Civil knowledge areas with more free practice problems, review the full FE Civil exam topic breakdown, or plan your preparation with our FE exam study guide and study timeline.