How to take an FE practice exam that tells you something
Take a full FE practice exam under real conditions: 110 questions in 5 hours 20 minutes with a 25-minute break, the searchable handbook only, and score it by knowledge area rather than just a total. Below is how to set it up, plus a 5-question timed set to start with.
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.
Set it up like exam day
- 110 questions, computer-based, at Pearson VUE test centers year-round (NCEES).
- 6-hour appointment: 2-minute nondisclosure agreement, 8-minute tutorial, 5 hours 20 minutes of exam time, and a 25-minute scheduled break (NCEES).
- That is about 2.9 minutes per question (320 minutes ÷ 110).
- You get a searchable electronic FE Reference Handbook on screen; you can't bring your own (our handbook guide).
- $225 exam fee paid to NCEES; one attempt per quarterly testing window and no more than three in 12 months (NCEES).
- Book 6 hours. Quiet room, phone away, an NCEES-approved calculator, and the FE Reference Handbook 10.6 PDF open on screen with no other notes.
- Split it like the real exam. Work about half, take the 25-minute break, then finish. Total exam time 5 hours 20 minutes.
- Pace check. At 2.9 minutes a question you should pass question 55 at about 2 hours 40 minutes. Flag and skip anything over 5 minutes.
- Score by knowledge area. Not just a total. The NCEES diagnostic report after a failed attempt is also by area, so this is how you find your weak spots.
- Review every miss. Was it a concept, a unit slip, or time spent finding the equation? Each needs a different fix.
Try a 5-question timed set (15 minutes)
Five problems from our practice sets, across environmental and civil areas. Set a 15-minute timer.
Problem 1 · FE Environmental · Water and Wastewater · Multiple choice
An activated sludge aeration basin has a volume of 3,100 m³. It receives 10,000 m³/d of primary effluent with a BOD5 of 195 mg/L, and the mixed liquor suspended solids (MLSS) concentration is 2,450 mg/L. The food-to-microorganism (F:M) ratio, based on MLSS, is most nearly:
Show the worked solution
Answer: A) 0.257 kg BOD5/(kg MLSS·d)
Equation: F:M = Q0·S0/(Vol·XA)
- F:M = Q0·S0/(V·XA) = (10,000 m³/d)(195 mg/L)/[(3,100 m³)(2,450 mg/L)] = 0.257 per day
- (mg/L cancels, so the units are kg BOD5 per kg MLSS per day)
Why the wrong choices are wrong:
- B is the volumetric organic loading (kg BOD5/m³·d), not F:M.
- C treated the daily flow as hourly.
- D converted MLSS to an assumed MLVSS (0.8) that the problem did not give.
Handbook: FE Reference Handbook 10.6, Environmental Engineering, Activated Sludge, Organic loading rate (F:M), p. 340.
Problem 2 · FE Environmental · Health Hazards and Risk Assessment · Multiple choice
A resident drinks 2.0 L/day of water containing 0.116 mg/L of a noncarcinogenic chemical, 350 days/year for 9 years. Body weight = 70 kg; for a noncarcinogen the averaging time equals the exposure duration (365 days/year). The reference dose is 0.0003 mg/kg·day. The hazard quotient is most nearly:
Show the worked solution
Answer: A) 10.6
Equation: CDI = (CW)(IR)(EF)(ED)/[(BW)(AT)]; HQ = CDI/RfD (the hazard index sums HQs)
- AT = ED × 365 = 9 × 365 = 3,285 days
- CDI = (0.116)(2.0)(350)(9)/[(70)(3,285)] = 3.178 × 10⁻³ mg/(kg·day)
- HQ = CDI/RfD = 3.178 × 10⁻³/0.0003 = 10.6
Why the wrong choices are wrong:
- B divided the water concentration by the RfD (no intake calculation).
- C assumed exposure every day (EF left out).
- D averaged over 70 years (the cancer convention) instead of the exposure duration.
Handbook: FE Reference Handbook 10.6, Safety, Exposure (ingestion in drinking water) and Risk Assessment/Toxicology (noncarcinogens, RfD), p. 26 and 29.
Problem 3 · FE Environmental · Engineering Economics · Multiple choice
A county buys a dump truck for $829,000. The truck will be used for 24 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:
Show the worked solution
Answer: D) $78,800
Equation: A = P(A/P, i, n), with (A/P, i, n) = i(1 + i)ⁿ/[(1 + i)ⁿ - 1]
- (A/P, 8%, 24) = 0.08(1.08)^24/[(1.08)^24 - 1] = 0.0950 (factor table, i = 8%)
- A = P(A/P, 8%, 24) = ($829,000)(0.0950) = $78,800 per year
Why the wrong choices are wrong:
- A used the sinking fund factor (A/F) instead of capital recovery (A/P).
- B divided the first cost by n, which ignores the time value of money.
- C multiplied P by i only, which leaves out recovering the first cost itself.
Handbook: FE Reference Handbook 10.6, Engineering Economics, factor formulas (A/P) and Factor Table (i = 8%), p. 235 and 241.
Problem 4 · FE Civil · Transportation Engineering · Multiple choice
A circular horizontal curve has a radius of 1,750 ft and an intersection angle (deflection between tangents) of 44°. The length of the curve is most nearly:
Show the worked solution
Answer: B) 1,340 ft
Equation: L = RIπ/180 (I in degrees)
- L = RIπ/180 = (1,750 ft)(44°)(π/180) = 1,340 ft
Why the wrong choices are wrong:
- A added the two tangent distances.
- C is the tangent distance T, not the curve length.
- D used π/360 instead of π/180.
Handbook: FE Reference Handbook 10.6, Civil Engineering, Transportation, Horizontal Curves, p. 308.
Problem 5 · FE Civil · Construction Engineering · Multiple choice
A small project has these activities (duration, predecessors): A (6 days, no predecessor); B (12 days, after A); C (4 days, after A); D (11 days, after B); E (11 days, after C); F (8 days, after D and E). Using the critical path method, the total float of activity C is most nearly:
Show the worked solution
Answer: B) 8 days
Equation: ES = latest EF of predecessors; EF = ES + d; LF = earliest LS of successors; LS = LF - d; Float = LS - ES
- Forward pass (ES = latest EF of predecessors; EF = ES + duration): A: ES 0, EF 6, B: ES 6, EF 18, C: ES 6, EF 10, D: ES 18, EF 29, E: ES 10, EF 21, F: ES 29, EF 37
- Project duration = 37 days
- Backward pass (LF = earliest LS of successors; LS = LF - duration): A: LS 0, LF 6, B: LS 6, LF 18, C: LS 14, LF 18, D: LS 18, LF 29, E: LS 18, LF 29, F: LS 29, LF 37
- Float of C = LS - ES = 14 - 6 = 8 days
Why the wrong choices are wrong:
- A is the late start time, not the float.
- C is the length of the middle of the critical path, not a float.
- D took LF - ES, which leaves the duration in.
Handbook: FE Reference Handbook 10.6, Civil Engineering, Construction, CPM Precedence Relationships (activity on node), p. 316.
Where full practice exams come from
- NCEES interactive practice exams: $50 per volume, 50 questions, written from retired exam items, with a timed option (MyNCEES). The closest look at real questions.
- Our packs: a 110-question mock in the NCEES proportions plus 100 problems by area, for FE Environmental or FE Civil, $39 each.
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
How many weeks to study and in what order is in our 30-day and 60-day study plans.
Questions
How long is an FE practice exam?
A realistic one matches the real exam: 110 questions and 5 hours 20 minutes of exam time, with a 25-minute break (NCEES).
What score should I aim for on a practice exam?
NCEES does not publish a passing percentage; results are pass or fail. Use the practice exam to find weak knowledge areas, not to predict a score.
Is the NCEES practice exam worth it?
Yes. It is written from retired exam items, so it is the closest look at real questions. Use it alongside other practice.
How many practice exams should I take?
At least one full timed run before exam day, and a second after you work your weak areas, to check the gap closed.
Related guides
- FE exam format and timing
- Time per question on the FE
- How the FE is scored
- FE exam sample questions by type
Engineer Exam Lab is not affiliated with or endorsed by NCEES. The problems here are original practice problems, not questions from a real exam.