FE Industrial and Systems Exam Prep: 34 Free Worked Problems
Start your FE Industrial and Systems exam prep here: 34 original, unofficial problems that cover all 13 NCEES knowledge areas, each with a worked answer, followed by a topic priority map and an eight-week plan. It's a starter set for finding your gaps, not a full-length exam or a prediction of passing.
Practice problems
Question 1. Arrivals at a tool crib
5. Probability and Statistics · Single best answer
Requests arrive at a tool crib as a Poisson process with a mean of 3 per hour. What is the probability that exactly 2 requests arrive in a given hour?
- A. 0.050
- B. 0.149
- C. 0.224
- D. 0.667
Reveal answer and explanation
Answer: C — 0.224
The Poisson probability is p(x) = e^(−λ)·λ^x / x!. With λ = 3 and x = 2: e^(−3) × 9 ÷ 2 ≈ 0.0498 × 4.5 ≈ 0.224.
Why the others miss: A is P(0) = e^(−3). B is P(1) = 3e^(−3). D is 2/3, a ratio of counts rather than a probability from the distribution.
Source: NIST/SEMATECH e-Handbook, 1.3.6.6.19 Poisson Distribution
Original, unofficial practice. These are not NCEES questions. Work them with an approved calculator and your copy of the FE Reference Handbook.
Jump to: Topic map · Eight-week plan · Exam rules · Answer key
Work the rest untimed first, with your handbook open. Try each problem before you reveal the answer.
Question 2. Confidence interval with known σ
5. Probability and Statistics · Single best answer
A machined dimension is normally distributed with a known process standard deviation σ = 6.0 mm. A random sample of 36 parts has a mean of 50.0 mm. What is the 95% confidence interval for the process mean?
- A. 48.36 to 51.64 mm
- B. 38.24 to 61.76 mm
- C. 47.97 to 52.03 mm
- D. 48.04 to 51.96 mm
Reveal answer and explanation
Answer: D — 48.04 to 51.96 mm
With σ known, use z: x̄ ± z₀.₀₂₅ · σ/√n = 50.0 ± 1.96 × 6.0/√36 = 50.0 ± 1.96 × 1.0 = 50.0 ± 1.96 mm.
Why the others miss: A uses z = 1.645, which gives a 90% interval. B forgets to divide σ by √n. C uses a t value (about 2.03 for 35 degrees of freedom); that's the right tool only when σ has to be estimated from the sample, and here it's given. The interval estimates the process mean, not the range of individual parts.
Source: OpenStax Introductory Statistics 2e, 8.1 A Single Population Mean using the Normal Distribution
Question 3. Least-squares slope
5. Probability and Statistics · Numeric entry · answer to one decimal place
Five (x, y) pairs are (1, 2), (2, 4), (3, 5), (4, 4) and (5, 5). What is the slope b of the least-squares regression line ŷ = a + bx?
Reveal answer and explanation
Answer: 0.6
x̄ = 3 and ȳ = 4. The deviations are x: −2, −1, 0, 1, 2 and y: −2, 0, 1, 0, 1. So Σ(x − x̄)(y − ȳ) = 4 + 0 + 0 + 0 + 2 = 6 and Σ(x − x̄)² = 10. Then b = 6 ÷ 10 = 0.6, and the intercept is a = 4 − 0.6 × 3 = 2.2.
Common mistakes: 1.7 (about 10 ÷ 6) flips the ratio. 2.2 is the intercept, not the slope. Lay the deviations out in columns; most errors here are bookkeeping, not statistics.
Source: OpenStax Introductory Statistics 2e, 12.3 The Regression Equation
Question 4. Comparing machines by annual cost
4. Engineering Economics · Single best answer
Two machines do the same job for 5 years with no salvage value. Machine A costs $20,000 and $5,000 a year to operate. Machine B costs $30,000 and $3,000 a year to operate. Assume purchase costs are paid today and operating costs occur at each year-end. At an effective annual interest rate of 10%, which machine has the lower equivalent uniform annual cost (EUAC), and what is it?
- A. Machine A, about $10,276 per year
- B. Machine B, about $10,914 per year
- C. Neither; both cost $9,000 per year
- D. Machine B, because its operating cost is lower
Reveal answer and explanation
Answer: A — Machine A, about $10,276 per year
Turn each first cost into an annual amount with the capital-recovery factor: (A/P, 10%, 5) = 0.1(1.1)⁵ ÷ [(1.1)⁵ − 1] ≈ 0.263797. Machine A: 20,000 × 0.263797 + 5,000 ≈ $10,276. Machine B: 30,000 × 0.263797 + 3,000 ≈ $10,914. Machine A is cheaper by about $638 a year.
Why the others miss: B is Machine B's EUAC, the higher one. C spreads first cost evenly with no interest (20,000/5 + 5,000 = 30,000/5 + 3,000 = 9,000). The tie disappears once you include the time value of money. D looks only at operating cost.
Source: Penn State EME 460, Compound Interest Formulas III (P/A and A/P factors)
Question 5. Net present worth
4. Engineering Economics · Numeric entry · nearest dollar
A project costs $6,000 today and saves $2,500 at the end of each of the next three years. There is no salvage value or other cash flow. At an effective annual interest rate of 10%, what is the project's net present worth?
Reveal answer and explanation
Answer: $217 ($217.13 to the nearest cent)
Bring the three equal savings back to today with the uniform-series present-worth factor: (P/A, 10%, 3) = [(1.1)³ − 1] ÷ [0.1(1.1)³] ≈ 2.48685. Savings are worth 2,500 × 2.48685 ≈ $6,217.13 today. Subtract the $6,000 spent at time zero: NPW ≈ $217. It's positive, so the project clears a 10% hurdle.
Common mistakes: $1,500 adds the savings without discounting. $6,217 is the present worth of the savings before subtracting the first cost. Draw the cash-flow timeline first; most misses come from placing a cash flow in the wrong year.
Source: Penn State EME 460, Compound Interest Formulas III (P/A and A/P factors)
Question 6. Loan payment
4. Engineering Economics · Single best answer
A plant borrows $100,000 at 8% effective annual interest and repays it in 5 equal end-of-year payments. What is each payment, to the nearest dollar?
- A. $20,000
- B. $25,046
- C. $28,000
- D. $21,600
Reveal answer and explanation
Answer: B — $25,046
A = P(A/P, 8%, 5) = 100,000 × 0.08(1.08)⁵ ÷ [(1.08)⁵ − 1] ≈ 100,000 × 0.250456 ≈ $25,046.
Why the others miss: A ignores interest. C adds a full year's interest on the original $100,000 to every payment, even though the balance shrinks. D charges one year of interest and then splits the total over five years.
Question 7. M/M/1 time in system
6. Modeling and Quantitative Analysis · Single best answer
A single inspection station has Poisson arrivals at λ = 8 jobs per hour and exponential service at μ = 10 jobs per hour (an M/M/1 queue). In steady state, what is the average time a job spends in the system, waiting plus service?
- A. 6 minutes
- B. 24 minutes
- C. 30 minutes
- D. 4 jobs
Reveal answer and explanation
Answer: C — 30 minutes
Since λ < μ, the steady-state M/M/1 formula applies: W = 1/(μ − λ) = 1/(10 − 8) = 0.5 h = 30 min. Check it with Little's law: utilization ρ = 0.8, the average number in the system is L = ρ/(1 − ρ) = 4 jobs, and W = L/λ = 4/8 = 0.5 h.
Why the others miss: A is the mean service time, 1/μ. B is the time waiting in line only, Wq = λ/[μ(μ − λ)] = 0.4 h. D is L, the average number in the system, which isn't a time.
Question 8. Linear program optimum
6. Modeling and Quantitative Analysis · Single best answer
Maximize Z = 40x + 30y subject to x + y ≤ 10, 2x + y ≤ 16, and x, y ≥ 0. What is the maximum value of Z?
- A. 300
- B. 320
- C. 360
- D. 400
Reveal answer and explanation
Answer: C — 360
For this bounded feasible region, at least one optimal solution occurs at a corner. The corners are (0, 0), (8, 0), (0, 10), and the point where x + y = 10 meets 2x + y = 16, which is (6, 4). Their Z values are 0, 320, 300 and 240 + 120 = 360. The maximum is 360 at (6, 4).
Why the others miss: A and B are other corners that are feasible but worse. D uses (10, 0), which breaks 2x + y ≤ 16. Check feasibility before you compare objective values.
Source: LibreTexts, Applied Finite Mathematics (Sekhon & Bloom), 3.1 Maximization Applications
Question 9. Long-run machine availability (Markov)
6. Modeling and Quantitative Analysis · Single best answer
A machine is checked once per shift. If it's up, it is up next shift with probability 0.9 and down with probability 0.1. If it's down, it is repaired and up next shift with probability 0.6 and still down with probability 0.4. In the long run, at what fraction of shift checks is it up?
- A. 0.900
- B. 0.857
- C. 0.600
- D. 0.143
Reveal answer and explanation
Answer: B — 0.857
Solve for the steady-state vector π with πT = π and π_up + π_down = 1. In steady state the flow out of "up" equals the flow in: π_up × 0.1 = π_down × 0.6. So π_up = 0.6 ÷ (0.1 + 0.6) ≈ 0.857.
Why the others miss: A and C are one-step transition probabilities, not long-run results. D is the long-run fraction of shift checks at which the machine is down.
Source: LibreTexts, Applied Finite Mathematics (Sekhon & Bloom), 10.3 Regular Markov Chains
Question 10. Process capability index
12. Quality · Single best answer
A shaft diameter has specification limits of 9.4 to 10.6 mm. The process is stable and normally distributed, with mean 10.1 mm and standard deviation 0.15 mm. What is Cpk?
- A. 1.33
- B. 1.11
- C. 1.56
- D. 0.75
Reveal answer and explanation
Answer: B — 1.11
Cpk = min[(USL − μ)/3σ, (μ − LSL)/3σ] = min[(10.6 − 10.1)/0.45, (10.1 − 9.4)/0.45] ≈ min[1.11, 1.56] = 1.11. The mean sits closer to the upper limit, and that's the side that controls.
Why the others miss: A is Cp = (USL − LSL)/6σ = 1.2/0.9, which ignores centering. C is the lower-side value, the larger of the two. D inverts Cp.
Source: NIST/SEMATECH e-Handbook, 6.1.6 What is Process Capability?
Question 11. X-bar chart limits
12. Quality · Single best answer
An X̄–R chart uses subgroups of n = 5. The grand mean is 50.00 and the average range R̄ is 4.0. For n = 5, A₂ = 0.577. What are the X̄-chart control limits?
- A. 47.69 and 52.31
- B. 38.00 and 62.00
- C. 48.85 and 51.15
- D. 0 and 8.46
Reveal answer and explanation
Answer: A — 47.69 and 52.31
UCL and LCL = x̿ ± A₂R̄ = 50.00 ± 0.577 × 4.0 ≈ 50.00 ± 2.31.
Why the others miss: B uses ±3R̄, treating the range as if it were σ. C uses half of A₂R̄. D gives R-chart limits (D₃R̄ = 0 and D₄R̄ = 2.115 × 4.0 = 8.46), not X̄-chart limits. Control limits come from process variation, not from specifications.
Source: NIST/SEMATECH e-Handbook, 6.3.2.1 Shewhart X-bar and R and S Control Charts (factor table)
Question 12. Probability of accepting a lot
12. Quality · Single best answer
A single sampling plan draws n = 20 items from a large lot and accepts the lot if it finds at most 1 defective item. If the lot is 5% defective, what is the probability of acceptance? Use the binomial model.
- A. 0.358
- B. 0.377
- C. 0.950
- D. 0.736
Reveal answer and explanation
Answer: D — 0.736
Pa = P(0) + P(1) = (0.95)²⁰ + 20(0.05)(0.95)¹⁹ ≈ 0.358486 + 0.377354 ≈ 0.736. That's one point on the plan's operating characteristic (OC) curve.
Why the others miss: A is P(0) alone, which would be the answer for a c = 0 plan. B is P(1) alone. C is the fraction of good items in the lot, not the plan's acceptance probability.
Source: NIST/SEMATECH e-Handbook, 6.2.3.2 Choosing a Sampling Plan with a given OC Curve
Question 13. Exponential smoothing forecast
8. Manufacturing, Service, and Other Production Systems · Single best answer
Last period's forecast was 100 units and actual demand was 120 units. Using simple exponential smoothing with α = 0.2, what is the forecast for next period?
- A. 104
- B. 116
- C. 110
- D. 120
Reveal answer and explanation
Answer: A — 104
New forecast = α × actual + (1 − α) × previous forecast = 0.2(120) + 0.8(100) = 24 + 80 = 104. Put another way, the forecast moves 20% of the way from 100 toward 120.
Why the others miss: B puts the 0.8 weight on actual demand, so α is applied backwards. C is a simple average of the two numbers. D is a naive forecast that just repeats the last actual value.
Source: NIST/SEMATECH e-Handbook, 6.4.3.1 Single Exponential Smoothing
Question 14. Minimum number of workstations
8. Manufacturing, Service, and Other Production Systems · Single best answer
An assembly line must produce 300 units in a 7.5-hour shift of available production time. The tasks for one unit total 6.0 minutes. What is the theoretical minimum number of workstations?
- A. 3
- B. 4
- C. 5
- D. 1.5
Reveal answer and explanation
Answer: B — 4
Cycle time = available time ÷ required output = 450 min ÷ 300 = 1.5 min per unit. Minimum stations = total task time ÷ cycle time = 6.0 ÷ 1.5 = 4. When the division doesn't come out even, round up to a whole number of workstations. A real balance may need more than the minimum when tasks don't split neatly.
Why the others miss: A is too few: three stations at 1.5 minutes each give only 4.5 minutes of work time per unit. C adds a station that the minimum doesn't need. D is the cycle time, not a station count.
Source: King Saud University IE 314, Assembly-Line Balancing lecture notes
Question 15. Economic order quantity
8. Manufacturing, Service, and Other Production Systems · Numeric entry · nearest whole unit
Annual demand is 12,000 units, the cost to place an order is $50, and holding cost is $3 per unit per year. Assume the basic EOQ model: constant known demand, instant replenishment, no shortages and no quantity discounts. What is the economic order quantity?
Reveal answer and explanation
Answer: 632 units (632.46 to two decimal places)
EOQ = √(2DS/H) = √(2 × 12,000 × 50 ÷ 3) = √400,000 ≈ 632 units, which works out to about 19 orders a year.
Common mistakes: 400,000 forgets the square root. 447 drops the factor of 2. Make sure H is per unit per year so it matches annual demand.
Question 16. Bottleneck throughput
8. Manufacturing, Service, and Other Production Systems · Single best answer
Every unit passes through stations A, B and C in that order. Each station has one machine, and the stations work at the same time on different units. Processing times are 2.0, 3.0 and 2.5 minutes per unit. Assume sufficient input and demand, enough buffer space, and no downtime or scrap. After startup, what is the maximum throughput?
- A. 8 units/hour
- B. 24 units/hour
- C. 30 units/hour
- D. 20 units/hour
Reveal answer and explanation
Answer: D — 20 units/hour
Station capacities are A: 60/2 = 30, B: 60/3 = 20 and C: 60/2.5 = 24 units per hour. The slowest station, B, is the bottleneck, so the line can't produce more than 20 units an hour.
Why the others miss: A assumes one unit must clear all three stations before the next starts (60 ÷ 7.5). B and C are capacities of stations that aren't the bottleneck. Speeding up A or C wouldn't raise output at all.
Source: MIT OpenCourseWare 15.772J, Capacity Planning and Production Flow Control (lecture 6), pp. 15–21
Question 17. How many machines?
9. Facilities and Supply Chain · Single best answer
Weekly demand is 1,200 parts, and each part needs 6 minutes of machine time. Each machine is scheduled 40 hours a week and provides productive capacity equal to 80% of those hours. How many whole machines are needed?
- A. 3
- B. 3.75
- C. 4
- D. 5
Reveal answer and explanation
Answer: C — 4
Required machine time = 1,200 × 6 ÷ 60 = 120 hours a week. Each machine delivers 40 × 0.80 = 32 effective hours. 120 ÷ 32 = 3.75, which rounds up to 4 machines.
Why the others miss: A ignores the efficiency loss (120 ÷ 40 = 3). B is the fractional machine-equivalent requirement; four whole machines are needed. D rounds up more than the math requires.
Source: Derived from the definitions in the problem: required hours ÷ effective hours per machine
Question 18. Rectilinear distance
9. Facilities and Supply Chain · Single best answer
In a plant laid out on a grid of aisles, a forklift travels from a dock at (2, 3) to a cell at (7, 9). Coordinates are in tens of meters, and the forklift can only move parallel to the aisles. What is the shortest travel distance, in the same units?
- A. 7.81
- B. 11
- C. 21
- D. 61
Reveal answer and explanation
Answer: B — 11 (110 m)
Aisle travel uses rectilinear (Manhattan) distance: |x₁ − x₂| + |y₁ − y₂| = |2 − 7| + |3 − 9| = 5 + 6 = 11.
Why the others miss: A is the straight-line Euclidean distance, √(5² + 6²), which the forklift can't drive. C adds the coordinates instead of their differences. D is the sum of the squares, 25 + 36, without the square root.
Source: NIST Dictionary of Algorithms and Data Structures, Manhattan distance
Question 19. Comparing two layouts by load-distance
9. Facilities and Supply Chain · Single best answer
Only two material flows matter for this comparison, and every load costs the same to move per meter. A to B: 30 loads a day, 10 m apart in Layout X and 20 m in Layout Y. B to C: 20 loads a day, 30 m apart in Layout X and 10 m in Layout Y. Which layout has the smaller total load-distance, and what is it?
- A. Layout Y, 800 load-meters/day
- B. Layout X, 900 load-meters/day
- C. Layout X, 800 load-meters/day
- D. Layout Y, 900 load-meters/day
Reveal answer and explanation
Answer: A — Layout Y, 800 load-meters/day
Weight each distance by its flow. Layout X: 30(10) + 20(30) = 900. Layout Y: 30(20) + 20(10) = 800. Since the cost per load-meter is the same for both flows, Y has the lower handling cost.
Why the others miss: B is Layout X's correct total, but it's the higher one. C and D pair each layout with the other layout's total. A real layout decision would also weigh space, safety and other constraints; this question asks only about handling.
Source: University of Texas at Austin (P. A. Jensen), Facility Layout: Evaluation
Question 20. Centralizing safety stock
9. Facilities and Supply Chain · Single best answer
Four regional warehouses each hold 100 units of safety stock. Their demands are independent and identically distributed. The firm will serve all four regions from one central warehouse at the same service level. Assume replenishment lead times are equal and unchanged and the same safety factor is used before and after pooling. Using the square-root law, how much safety stock does the central warehouse need?
- A. 200
- B. 400
- C. 100
- D. 25
Reveal answer and explanation
Answer: A — 200
When independent demands are pooled, combined standard deviation grows with √n, not n. Central safety stock = √4 × 100 = 200 units, half the original 400. That reduction is the risk-pooling benefit.
Why the others miss: B adds the four stocks with no pooling benefit. C assumes pooling removes all the extra variability. D divides instead of multiplying. The law depends on independence; when regional demands rise and fall together, the benefit shrinks.
Source: Georgia Tech (Spyros Reveliotis), The square-root law, pp. 1–3, equations 8–12
Question 21. Critical path float
7. Engineering Management · Single best answer
A project has these activities, with durations in days: A (3, no predecessor); B (5, after A); C (2, after A); D (4, after C); E (2, after both B and D). What is the total float of activity B?
- A. 0 days
- B. 1 day
- C. 2 days
- D. 11 days
Reveal answer and explanation
Answer: B — 1 day
There are two paths. A–B–E takes 3 + 5 + 2 = 10 days, and A–C–D–E takes 3 + 2 + 4 + 2 = 11 days. The longer path is critical, so the project takes 11 days. E can't start until D finishes on day 9, but B finishes on day 8, so B can slip 1 day without moving the finish.
Why the others miss: A treats B as if it were on the critical path. C is activity C's duration, not B's float. D is the project duration.
Source: GAO Schedule Assessment Guide (GAO-16-89G), Best Practices 6–7: critical path and total float
Question 22. Earned value status
7. Engineering Management · Select all that apply
At a status date, a project's planned value (PV) is $50,000, its earned value (EV) is $40,000, and its actual cost (AC) is $45,000. Which statements are correct? Select all that apply.
- A. CPI = 0.89, so the project is over budget for the work done
- B. SPI = 0.80, so the project is behind schedule
- C. The project is under budget because AC is less than PV
- D. Cost variance is −$5,000
Reveal answer and explanation
Answer: A, B and D (all three required)
CPI = EV/AC = 40/45 ≈ 0.89 and SPI = EV/PV = 40/50 = 0.80. Values below 1.0 are unfavorable for both. Cost variance = EV − AC = −$5,000.
Why C is wrong: C compares AC with the wrong number. Cost performance compares AC with EV, the budgeted value of the work actually done. Spending less than the plan while finishing even less work isn't being under budget.
Source: GAO Cost Estimating and Assessment Guide (GAO-20-195G), Ch. 19, EVM performance indexes
Question 23. NIOSH recommended weight limit
10. Human Factors, Ergonomics, and Safety · Single best answer
A two-handed lift (U.S. units) has the hands 20 in. out from the midpoint between the ankles horizontally (H), starts 30 in. above the floor (V), and travels 10 in. vertically (D), with an asymmetry angle A = 0°. The frequency multiplier is 0.85 and the coupling is good (multiplier 1.0). Using the revised NIOSH lifting equation with its 51-lb load constant, what is the recommended weight limit (RWL) at the lift origin?
- A. 43.4 lb
- B. 25.5 lb
- C. 21.7 lb
- D. 30.0 lb
Reveal answer and explanation
Answer: C — 21.7 lb
RWL = LC × HM × VM × DM × AM × FM × CM. HM = 10/H = 0.50; VM = 1 − 0.0075|30 − 30| = 1.00; DM = 0.82 + 1.8/10 = 1.00; AM = 1.00 with A = 0°; CM = 1.00. So RWL = 51 × 0.50 × 0.85 = 21.675 lb ≈ 21.7 lb. Using this origin RWL, a hypothetical 30-lb box gives L/RWL = 30 ÷ 21.675 ≈ 1.38; NIOSH's design goal is a lifting index of 1.0 or less. A full task assessment must also check the destination when significant control is required there and apply the equation's task limitations.
Why the others miss: A leaves out the horizontal multiplier. B leaves out the frequency multiplier. D is a load weight, not a limit. The biggest penalty in this calculation is the horizontal reach; reducing H would increase the calculated RWL if the other multipliers stay the same.
Source: NIOSH, Applications Manual for the Revised NIOSH Lifting Equation, Pub. 94-110 (revised Sept. 2021), §1.3–1.4, §2.1.1 and §2.3; NIOSH, Revised NIOSH Lifting Equation: interpreting results
Question 24. Noise dose
10. Human Factors, Ergonomics, and Safety · Single best answer
Under OSHA's general-industry noise standard, a worker is exposed to 95 dBA for 4 hours and 90 dBA for 4 hours. What is the daily noise dose?
- A. 92.5%
- B. 100%
- C. 200%
- D. 150%
Reveal answer and explanation
Answer: D — 150%
OSHA's reference duration is T = 8 ÷ 2^((L − 90)/5): 8 hours at 90 dBA and 4 hours at 95 dBA. Dose = 100(C₁/T₁ + C₂/T₂) = 100(4/4 + 4/8) = 150%. Anything over 100% exceeds the permissible exposure. That dose equals an 8-hour time-weighted average of about 92.9 dBA. OSHA's hearing-conservation action level is lower: an 8-hour TWA of 85 dBA, equivalent to a 50% dose.
Why the others miss: A averages decibels as if they were linear, but decibels are logarithmic. B counts only the 95-dBA period. C treats both periods as 95 dBA.
Source: OSHA, 29 CFR 1910.95 Appendix A, Noise Exposure Computation (Table G-16a); OSHA, 29 CFR 1910.95(c)(1)–(2), hearing conservation
Question 25. Standard time
11. Work Design · Single best answer
A time study gives an average observed time of 2.00 minutes. The operator was rated at 110% performance. The allowance is 15%, applied as a percentage of normal time. What is the standard time?
- A. 2.20 min
- B. 2.53 min
- C. 2.59 min
- D. 2.30 min
Reveal answer and explanation
Answer: B — 2.53 min
Normal time = observed time × rating = 2.00 × 1.10 = 2.20 min. Standard time = normal time × (1 + allowance) = 2.20 × 1.15 = 2.53 min.
Why the others miss: A stops at normal time. D adds the allowance to observed time and skips the rating. C (2.20 ÷ 0.85) is the other convention, where the allowance is a percentage of total time rather than of normal time. Textbooks use both, so read which one the problem states.
Source: University of Michigan (T. Armstrong), Time Study: Normal Time & Performance Ratings; Standard Time & Allowances; Eastern Mediterranean University IENG 301, Ch. 22 notes (the other allowance convention)
Question 26. Learning curve: unit 4
11. Work Design · Single best answer
Under an 80% unit learning curve (unit theory), the first unit takes 100 labor-hours. How many hours does the fourth unit take?
- A. 80 h
- B. 64 h
- C. 51.2 h
- D. 60 h
Reveal answer and explanation
Answer: B — 64 h
Under unit theory, each doubling of quantity multiplies the unit time by the learning rate. Unit 2 takes 100 × 0.8 = 80 h, and unit 4 takes 80 × 0.8 = 64 h. Using the formula Y = A·X^b with b = ln 0.8 ÷ ln 2 ≈ −0.322 gives the same 100 × 4^(−0.322) ≈ 64 h.
Why the others miss: A is unit 2. C is unit 8. D subtracts 20 hours per doubling instead of 20 percent. Watch for cumulative-average (Wright) wording, which gives different unit times.
Source: GAO Cost Estimating and Assessment Guide (GAO-20-195G), Appendix VII, Unit Formulation, pp. 378–379
Question 27. Series–parallel reliability
13. Systems Engineering, Analysis, and Design · Single best answer
Two identical pumps, each with mission reliability 0.90, run in parallel, and the pair works if either pump works. That pair is in series with a controller of reliability 0.95. Failures are independent. What is the system reliability?
- A. 0.770
- B. 0.855
- C. 0.990
- D. 0.941
Reveal answer and explanation
Answer: D — 0.941
Parallel pair: R = 1 − (1 − 0.90)² = 1 − 0.01 = 0.99. In series with the controller: 0.99 × 0.95 = 0.9405.
Why the others miss: A treats all three components as series (0.9 × 0.9 × 0.95). B uses one pump with the controller. C is the pump pair alone; it leaves out the controller.
Source: NIST/SEMATECH e-Handbook, 8.1.8.2 Series model; NIST/SEMATECH e-Handbook, 8.1.8.3 Parallel or redundant model
Question 28. Inherent availability
13. Systems Engineering, Analysis, and Design · Numeric entry · three decimal places
A conveyor has a mean time between failures (MTBF) of 400 hours and a mean time to repair (MTTR) of 10 hours. What is its steady-state inherent availability?
Reveal answer and explanation
Answer: 0.976 (40/41 ≈ 0.97561)
Inherent availability = MTBF ÷ (MTBF + MTTR) = 400 ÷ 410 ≈ 0.976.
Common mistakes: 0.975 computes 1 − 10/400, which uses the wrong denominator. Inherent availability counts only corrective repair time; waiting for parts or crews falls under other availability measures.
Source: NASA Kennedy Space Center, Formulas: Inherent Availability and Reliability with Constant Failure and Repair Rates; NASA Kennedy Space Center, Availability—Types and Comparison, pp. 2–3
Question 29. Solving a 2 × 2 system
1. Mathematics · Single best answer
Solve 2x + y = 10 and x + 3y = 15.
- A. x = 3, y = 4
- B. x = 4, y = 3
- C. x = 5, y = 0
- D. x = 2, y = 6
Reveal answer and explanation
Answer: A — x = 3, y = 4
From the first equation, y = 10 − 2x. Substituting: x + 3(10 − 2x) = 15, so −5x = −15 and x = 3, which gives y = 4. Check both: 2(3) + 4 = 10 and 3 + 3(4) = 15.
Why the others miss: C and D satisfy only the first equation. B swaps the two values. Always check your answer in both equations.
Source: OpenStax College Algebra 2e, 7.1 Systems of Linear Equations: Two Variables
Question 30. Infinite geometric series
1. Mathematics · Single best answer
What is the sum of 1,000 + 800 + 640 + 512 + … continued without end?
- A. 1,800
- B. 4,000
- C. 5,000
- D. The series diverges
Reveal answer and explanation
Answer: C — 5,000
This is a geometric series with first term a₁ = 1,000 and ratio r = 0.8. Since |r| < 1, it converges to S = a₁ ÷ (1 − r) = 1,000 ÷ 0.2 = 5,000.
Why the others miss: A adds only the first two terms. B uses a₁r ÷ (1 − r), which leaves out the first term. D would be true only if |r| were 1 or more.
Source: OpenStax College Algebra 2e, Chapter 9 Key Equations (infinite geometric series)
Question 31. Balancing a lever
2. Engineering Sciences · Single best answer
A rigid, weightless bar pivots at a fulcrum. A 400 N load hangs 0.5 m to the left of the fulcrum. What downward force applied 2.0 m to the right of the fulcrum holds the bar in equilibrium?
- A. 100 N
- B. 200 N
- C. 400 N
- D. 1,600 N
Reveal answer and explanation
Answer: A — 100 N
For the bar not to rotate, the net torque about the pivot must be zero: 400 × 0.5 = F × 2.0, so F = 100 N.
Why the others miss: D multiplies by the distance ratio instead of dividing. C ignores the lever arms. B uses a 1.0 m arm instead of 2.0 m.
Source: OpenStax University Physics Vol. 1, 12.1 Conditions for Static Equilibrium
Question 32. Annual energy cost of a motor load
2. Engineering Sciences · Single best answer
A 7.5 kW electrical load runs 16 hours a day, 250 days a year. Electricity costs $0.12 per kWh. What is the annual energy cost?
- A. $14.40
- B. $3,600
- C. $5,256
- D. $3,600,000
Reveal answer and explanation
Answer: B — $3,600
Energy = power × time = 7.5 kW × 16 h/day × 250 days = 30,000 kWh. Cost = 30,000 × $0.12 = $3,600.
Why the others miss: A is one day's cost. C runs the load 365 days. D enters 7,500 W as if it were 7,500 kW.
Source: OpenStax University Physics Vol. 2, 9.5 Electrical Energy and Power
Question 33. A family stake in a bid evaluation
3. Ethics and Professional Practice · Single best answer
A licensed engineer is hired by a client to evaluate three bids for a warehouse automation system. The engineer's spouse owns a significant share of one bidder. Under the NCEES Model Rules, what must the engineer do?
- A. Nothing, as long as the evaluation is technically sound
- B. Disclose the conflict of interest to the client
- C. Score that bidder lower to offset any bias
- D. Disclose it only if that bidder wins
Reveal answer and explanation
Answer: B — Disclose the conflict of interest to the client
§240.15 B.6 requires licensees to disclose to employers or clients all known or potential conflicts of interest, or other circumstances that could influence or appear to influence their judgment. Disclosure does not waive the engineer's other professional duties.
Why the others miss: A ignores the "appear to influence" standard. C swaps one bias for another. D discloses too late, after the client has already relied on the evaluation.
Source: NCEES Model Rules (revised Aug. 2026), §240.15 B.6 and A.10
Question 34. Speaking for a paying interest
3. Ethics and Professional Practice · Select all that apply
A licensed engineer is paid by a forklift manufacturer to speak at a public hearing on warehouse safety rules. Under the NCEES Model Rules, which statements are correct? Select all that apply.
- A. The engineer must identify the manufacturer as the party they're speaking for
- B. The engineer must reveal any interest they have in the matter
- C. Paid statements on engineering matters are prohibited in all cases
- D. The engineer's public opinion must rest on adequate knowledge of the facts and a competent evaluation
Reveal answer and explanation
Answer: A, B and D (all three required)
§240.15 A.6 bars statements inspired or paid for by interested parties unless the licensee explicitly identifies those parties and reveals any interest they have. §240.15 A.5 requires public opinions to rest on adequate knowledge of the facts and a competent evaluation.
Why C is wrong: C overstates the rule. Paid statements aren't banned; they require disclosure.
Source: NCEES Model Rules (revised Aug. 2026), §240.15 A.5–A.6
Answer key
Show all answers
| # | Knowledge area | Answer |
|---|---|---|
| 1 | 5. Probability and Statistics | C — 0.224 |
| 2 | 5. Probability and Statistics | D — 48.04 to 51.96 mm |
| 3 | 5. Probability and Statistics | 0.6 |
| 4 | 4. Engineering Economics | A — Machine A, about $10,276 per year |
| 5 | 4. Engineering Economics | $217 ($217.13 to the nearest cent) |
| 6 | 4. Engineering Economics | B — $25,046 |
| 7 | 6. Modeling and Quantitative Analysis | C — 30 minutes |
| 8 | 6. Modeling and Quantitative Analysis | C — 360 |
| 9 | 6. Modeling and Quantitative Analysis | B — 0.857 |
| 10 | 12. Quality | B — 1.11 |
| 11 | 12. Quality | A — 47.69 and 52.31 |
| 12 | 12. Quality | D — 0.736 |
| 13 | 8. Manufacturing, Service, and Other Production Systems | A — 104 |
| 14 | 8. Manufacturing, Service, and Other Production Systems | B — 4 |
| 15 | 8. Manufacturing, Service, and Other Production Systems | 632 units (632.46 to two decimal places) |
| 16 | 8. Manufacturing, Service, and Other Production Systems | D — 20 units/hour |
| 17 | 9. Facilities and Supply Chain | C — 4 |
| 18 | 9. Facilities and Supply Chain | B — 11 (110 m) |
| 19 | 9. Facilities and Supply Chain | A — Layout Y, 800 load-meters/day |
| 20 | 9. Facilities and Supply Chain | A — 200 |
| 21 | 7. Engineering Management | B — 1 day |
| 22 | 7. Engineering Management | A, B and D (all three required) |
| 23 | 10. Human Factors, Ergonomics, and Safety | C — 21.7 lb |
| 24 | 10. Human Factors, Ergonomics, and Safety | D — 150% |
| 25 | 11. Work Design | B — 2.53 min |
| 26 | 11. Work Design | B — 64 h |
| 27 | 13. Systems Engineering, Analysis, and Design | D — 0.941 |
| 28 | 13. Systems Engineering, Analysis, and Design | 0.976 (40/41 ≈ 0.97561) |
| 29 | 1. Mathematics | A — x = 3, y = 4 |
| 30 | 1. Mathematics | C — 5,000 |
| 31 | 2. Engineering Sciences | A — 100 N |
| 32 | 2. Engineering Sciences | B — $3,600 |
| 33 | 3. Ethics and Professional Practice | B — Disclose the conflict of interest to the client |
| 34 | 3. Ethics and Professional Practice | A, B and D (all three required) |
Scoring: give yourself one point per problem. Select-all problems (22 and 34) count only if you choose every correct option and nothing else, the same no-partial-credit rule NCEES uses. Numeric answers (3, 5, 15 and 28) count when they match the stated rounding. For this practice set, round half-up: an exact halfway value rounds away from zero. Keep extra precision until the final step.
How to read your result: each knowledge area has only two to four problems here. That's enough to point you toward a weak spot, not enough to measure one. NCEES doesn't publish a passing score, so no number of correct answers on this page translates into "pass" or "fail." Use your misses to decide where your first weeks go. Count a problem you got right but solved slowly, or guessed, as a miss for planning.
What's on the FE Industrial and Systems exam: the 13-area priority map
NCEES publishes a question range for each of the 13 knowledge areas in the FE Industrial and Systems exam specifications, effective beginning with the July 2020 exams. The ranges below come straight from that document. The tiers and the "practice first" column are our study recommendations, not NCEES weights or difficulty ratings.
| # | NCEES knowledge area | Questions | Our tier | Practice first | Problems above |
|---|---|---|---|---|---|
| 5 | Probability and Statistics | 10–15 | Core | Binomial, Poisson and normal distributions; central limit theorem; confidence intervals; hypothesis tests and error types; regression; ANOVA and factorial designs | 1–3 |
| 4 | Engineering Economics | 9–14 | Core | Factor formulas; PW, EAC, FW, IRR and benefit-cost; break-even, sunk cost and replacement; depreciation (straight line and MACRS) and after-tax cash flow | 4–6 |
| 6 | Modeling and Quantitative Analysis | 9–14 | Core | LP formulation and corner-point solutions; M/M/1 queues and Little's law; Markov chains; inverse-transform simulation; flowcharts and algorithms | 7–9 |
| 12 | Quality | 9–14 | Core | Control charts; Cp and Cpk; sampling plans and OC curves; Six Sigma, QFD and house of quality; Taguchi loss | 10–12 |
| 8 | Manufacturing, Service, and Other Production Systems | 9–14 | Core | Forecasting; line balancing; throughput and bottlenecks; EOQ, MRP and aggregate planning; theory of constraints; lean; processes | 13–16 |
| 9 | Facilities and Supply Chain | 9–14 | Core | From/to charts and layout types; rectilinear vs. Euclidean distance; machine and labor capacity; risk pooling; distribution networks | 17–20 |
| 7 | Engineering Management | 8–12 | Next | CPM and PERT; earned value; WBS and agile; KPIs and balanced scorecard; decision trees and utility | 21–22 |
| 10 | Human Factors, Ergonomics, and Safety | 8–12 | Next | NIOSH lifting equation; noise dose; displays and controls; anthropometry; cumulative trauma; safety programs | 23–24 |
| 11 | Work Design | 7–11 | Next | Time study and allowances; predetermined time systems; work sampling; learning curves; motion economy | 25–26 |
| 13 | Systems Engineering, Analysis, and Design | 8–12 | Next | Series and parallel reliability; MTTF and availability; FMEA and fault trees; requirements and life-cycle | 27–28 |
| 1 | Mathematics | 6–9 | Keep warm | Calculus, series, matrices and vectors, analytic geometry | 29–30 |
| 2 | Engineering Sciences | 4–6 | Repair early if rusty | Thermodynamics and fluids; statics, dynamics and materials; circuits | 31–32 |
| 3 | Ethics and Professional Practice | 4–6 | Keep warm | Model Rules duties, contracts, public protection | 33–34 |
Why these six come first: their published ranges add up to 55 to 85 of the 110 questions (our sum of the low ends and of the high ends). Three of them, Probability and Statistics, Engineering Economics, and Modeling, cover 28 to 43 questions on their own, making them useful areas for practicing calculations with the handbook open.
The trap for IE grads: Mathematics, Engineering Sciences and Ethics total 14 to 21 questions. If you haven't touched Engineering Sciences (thermo, fluids, statics, circuits) since your early coursework, refresh it early. That's why the plan below gives it its own block in week 4.
These range sums are outer bounds, not a real 110-question form. Counting how many subtopics an area lists won't tell you how heavily it's tested; use the published ranges.
Before you start the plan, mark each of the 13 areas as can explain, needs review or not yet tried, based on your results above and your coursework. That's a self-rating for planning, not a score. Revisit it at the end of each week.
Your eight-week FE Industrial and Systems study plan
Assumptions: about 8 hours a week (for example, an hour on four weeknights and 2 hours each weekend day), 64 hours in all, for a senior or a graduate a few years out. This is our editorial plan, not an NCEES requirement, and no number of hours guarantees a result.
| Week | Where the 8 hours go | What to do |
|---|---|---|
| 1 | Starter set 2 h · Probability and Statistics 5 h · error log 1 h | Download the current FE Reference Handbook from your MyNCEES account. Work all 34 problems above untimed, and log every miss, guess and slow solve. Then drill distributions, confidence intervals, hypothesis tests, regression and ANOVA. |
| 2 | Engineering Economics 5 h · Mathematics 2 h · review 1 h | Practice factor formulas until they're automatic. Then PW, EAC and IRR comparisons, benefit-cost, replacement and sunk cost, depreciation and after-tax cash flow. Then calculus, series and matrices. |
| 3 | Modeling and Quantitative Analysis 5 h · Ethics 1.5 h · review 1.5 h | LP formulation and corner-point solutions, M/M/1 and Little's law, Markov steady state, inverse-transform simulation. Then read NCEES Model Rules §240.15 end to end. |
| 4 | Quality 5 h · Engineering Sciences 2 h · review 1 h | X̄–R and attribute charts, Cp and Cpk, sampling plans and OC curves, Taguchi loss, QFD. Then a focused refresh of statics, basic thermo and fluids, and circuits. |
| 5 | Manufacturing/Service/Production 5 h · Work Design 2 h · review 1 h | Forecasting and tracking signals, line balancing and bottlenecks, EOQ and MRP, aggregate planning, theory of constraints, lean. Then time study, allowances, work sampling and learning curves. |
| 6 | Facilities and Supply Chain 4 h · Engineering Management 3 h · review 1 h | From/to charts and load-distance, layout types, distance metrics, capacity math, risk pooling, network design. Then CPM and PERT, earned value, decision trees. |
| 7 | Human Factors/Ergonomics/Safety 3 h · Systems Engineering 3 h · mixed review 2 h | NIOSH lifting equation, noise dose, displays and controls, anthropometry. Then reliability networks, availability, FMEA and fault trees. Finish with mixed problems across all earlier weeks. |
| 8 | Timed rerun 1.75 h · targeted drills 3.75 h · review misses 2 h · exam-day check 0.5 h | Rework the 34 problems in about 99 minutes. Spend the drill block on the two weakest areas in your log, and rework every miss from a blank page. Confirm your ID, calculator model and appointment. |
That's 64 planned hours, and every one of the 13 areas gets its own block. The 99-minute target is scaled from the exam's average pace: 320 minutes ÷ 110 questions × 34 ≈ 99 minutes. Rerunning problems you've already seen checks your method and speed on them. It doesn't show how you'd do on new problems, so keep fresh practice and review of familiar problems in separate columns of your log.
Where fresh practice comes from: the 34 problems here are checkpoints, not a 64-hour question bank. Fill the practice blocks with problems from your own coursework and textbooks and the free teaching sources linked under each answer.
What one study hour looks like: about 10 minutes reviewing one concept, 10 minutes finding it in the handbook, 25 minutes solving problems, 10 minutes checking solutions, and 5 minutes logging what went wrong. Shift the minutes when your log tells you to.
Adjust the plan to your situation
- Recent graduate: try each week's checkpoint problems before spending the full review block. If you can solve them, explain them and find the formulas quickly, move some time to a weaker area. One correct problem doesn't clear a whole area, though.
- Out of school three or more years: add two weeks at the start for math, statistics and Engineering Sciences, for a 10-week plan. A longer calendar isn't falling behind.
- Only 4 hours a week: stretch each week over two calendar weeks, about 16 weeks in all.
- Exam in four weeks: combine the weeks in pairs (1+2, 3+4, 5+6, 7+8), which means about 16 hours a week. With less time than that, cut by your error log, not evenly.
- Retaking after a fail: start from your NCEES diagnostic report. It shows each knowledge area on a 0–15 scale next to the average of passing examinees (Examinee Guide, pp. 20–21). Move your weakest areas into weeks 1–3 and keep a lighter pass over the rest. Our NCEES retake policy guide covers attempt limits and testing windows.
Does timing matter? NCEES's 2014–25 data for first-time takers from ABET-accredited programs under NCEES member-board jurisdiction, across all FE disciplines, shows 71.34% passing before graduation, 70.62% within 12 months of graduation, and 63.43% at 12 months or later (NCEES Squared 2025, p. 15). Those are aggregate figures, not Industrial and Systems results, and they don't show that timing causes a pass.
Turn misses into your next session
Write down why an answer went wrong, not just that it did. A units slip and a wrong model need different fixes.
| Problem | Fresh or familiar? | What went wrong | Correct setup | Error type | Next action |
|---|---|---|---|---|---|
| Example: Problem 2 | Fresh | Used t instead of z | σ is known, so use z = 1.96 with σ/√n | Model choice | Write a one-line "z or t?" check and work three more interval problems |
Useful error types: concept, model choice, setup, units, handbook lookup, calculator, reading, time.
Using the FE Reference Handbook on exam day
NCEES supplies the current FE Reference Handbook on screen as a searchable PDF, and it's your only reference. You search it with the search box on the left of the reference window; Ctrl+F doesn't work (Examinee Guide, May 2026, p. 10). Download the version NCEES shows for your exam date from your MyNCEES account. You may print it for personal study but not share or post it without NCEES's written permission (NCEES help: exam reference handbooks).
For every practice problem:
- Name the model first: queue, LP, control chart, capital recovery. Then search for it.
- Search a short concept term, not numbers or wording from the problem.
- Read the variable definitions and conditions before plugging in. Is it the sample or the population formula? Is the allowance a percentage of normal time or of total time?
- Check that your answer is the quantity asked for, in the units asked for.
15-minute lookup drill: find what you'd need for a capital-recovery factor, a Poisson probability, X̄-chart limits, Cpk, M/M/1 time in system, a learning-curve unit time and a noise dose. For each one, note the search term that found it fastest in your copy, the units of every input, and one assumption that limits it. Record search terms, not page numbers, since pages shift between handbook versions. Repeat the drill in week 8 and compare your times.
Some questions won't be solved by a formula lookup at all. Ethics, contracts, layout choices and reading a control chart depend on understanding.
FE exam rules that affect how you prepare
| Rule | What NCEES says | What it means for you |
|---|---|---|
| Length | 110 questions in 5 hours 20 minutes of exam time (FE exam page) | About 2.9 minutes (roughly 2 min 55 s) per question on average. That's our arithmetic for planning, not a per-question limit. |
| Appointment | The FE page calls the appointment 6 hours. The Examinee Guide's table, p. 16, lists 5 h 55 min: a 2-minute nondisclosure agreement, an 8-minute tutorial, the exam, and a 25-minute scheduled break. | Plan for about 6 hours and follow your appointment confirmation. |
| Two sections | After about half the questions, you review and submit them, and they lock. The scheduled break comes after that. Unscheduled breaks come out of your exam time. (Guide, pp. 11–12) | Review your flagged questions before you submit the first half. |
| Question types | Multiple choice plus multiple-correct, point-and-click, drag-and-drop and fill-in-the-blank. No partial credit. Some unscored pretest items are mixed in. (Guide, p. 11) | Practice numeric-entry and select-all problems like 3, 5, 15, 22, 28 and 34. |
| Scoring | Based on the number of correct answers, with no deduction for wrong ones, converted to a scaled score. NCEES doesn't publish the passing score. (NCEES exam scoring) | Answer every question. |
| Calculator | One approved model. For 2026 exams the approved models are Casio fx-115 and fx-991 models, the HP 33s and HP 35s, and TI-30X and TI-36X models. An on-screen TI-30XS is also available. (NCEES 2026 calculator list, p. 1; Guide, p. 8) | Practice on the model you'll bring, and recheck the list if you test in a later year. |
| Scratch work | Two reusable booklets and three markers, supplied at the test center. (Guide, p. 9) | Practice without pencil and paper. |
| Fee | $225, paid to NCEES. Your board may charge its own application fee. (FE exam page) | Check your board's process before you register. |
| Results | Typically 7–10 days, reported as pass or fail. If you don't pass, you get a diagnostic report by knowledge area. (NCEES exam scoring) | Use the report to reorder your plan. |
| Retakes | One attempt per testing window (Jan–Mar, Apr–Jun, Jul–Sep, Oct–Dec) and no more than three in any 12 months. Some boards are stricter. (Guide, p. 5) | A fail costs you at least the rest of that window. |
| Eligibility | Set by your licensing board, not NCEES. (Guide, p. 2) | Check your board through the state selector on the FE exam page. |
Industrial and Systems pass rates: in NCEES's fiscal year from October 1, 2024, to September 30, 2025, 62% of 547 first-time takers passed, and 25% of 118 repeat takers passed. Among first-time takers with an EAC/ABET-accredited bachelor's degree, 64% of 456 passed (NCEES Squared 2025, pp. 10–11). These are one year's group results, not anyone's individual odds.
Common questions
Is FE Industrial and Systems harder than the other FE exams? By pass rate, it's in the same range. Those figures compare different candidate groups; they don't establish which exam would be harder for you. In FY2024–25, first-time pass rates ran from 61% (Civil and Other Disciplines) to 69% (Mechanical), with Industrial and Systems at 62% (Squared 2025, p. 10). Statistics, economics and modeling have prominent question ranges in the Industrial and Systems specification.
Is this the PE Industrial and Systems exam? No. The PE is a separate 85-question exam for engineers with several years of experience (NCEES PE Industrial and Systems). Everything on this page is for the FE.
Do I need to pass the FE to become a PE? The FE is generally the first exam step toward a P.E. license, but engineer intern (EI or EIT) certification and licensure rules belong to your state board. Passing the FE doesn't by itself let you practice as a professional engineer (NCEES FE exam page).
How do I register or reschedule? Our NCEES exam registration guide walks through board approval, fees and scheduling with Pearson. If you need testing accommodations, see the NCEES exam accommodations guide before you schedule.
Taking a different FE discipline, or still choosing? Our FE exam prep hub has subject maps for all seven FE exams.
Sources and independence
By Castleport Test Prep Editorial Team · Last verified: October 7, 2026
On that date we checked the FE Industrial and Systems specification, the exam format and timing, question types, scoring, results timing, handbook access, the 2026 calculator list, the fee, retake rules and the pass rates against the NCEES sources below. We also checked each practice problem's underlying principle against the source linked with it and recalculated every numerical answer. AI tools helped with drafting, source checking and calculation checks. That's source and arithmetic checking, not a professional engineering review. See our editorial standards and how we verify exam facts.
- NCEES, FE Industrial and Systems CBT Exam Specifications, effective beginning with the July 2020 exams, pp. 1–3
- NCEES, FE exam page
- NCEES, Examinee Guide, May 2026 (current guide: ncees.org/examinee-guide)
- NCEES, Exam scoring
- NCEES, 2026 approved calculators, member-board memo dated October 20, 2025, p. 1
- NCEES Help, Exam reference handbooks
- NCEES, Squared 2025, fiscal year October 1, 2024–September 30, 2025
- NCEES, Model Rules, revised August 2026
- Teaching sources for each problem are linked with its answer.
Found an error? Our corrections page explains how to report it.
Castleport Test Prep is an independent exam prep publisher, not affiliated with, endorsed by, or approved by the National Council of Examiners for Engineering and Surveying (NCEES), Pearson, or any licensing board. The practice problems on this page are original and unofficial, not NCEES exam questions. Exam and credential names are used only to identify the exam; trademarks belong to their respective owners. This page doesn't guarantee an exam result, determine eligibility, or lead to licensure on its own.