Castleport Test Prep

Free FE Industrial and Systems Practice Test

Fifty original, unofficial FE Industrial and Systems practice questions, with a worked solution under every one and no signup.

Question 1 — Next month's forecast

FEIS-001 · Manufacturing, Service, and Other Production Systems · Choose one answer

A plant forecasts demand with single exponential smoothing, α = 0.2. The March forecast was 100 units, and actual March demand was 110 units. What is the April forecast?

  • A. 108 units
  • B. 102 units
  • C. 110 units
  • D. 105 units
Show answer and worked solution

Answer: B. 102 units

New forecast = α(actual) + (1 − α)(old forecast) = 0.2(110) + 0.8(100) = 102 units.

Another way to see it: the forecast was 10 units low, so it moves 20% of the way toward the actual value.

  • A (108) puts the 0.2 weight on the old forecast and 0.8 on the actual. That swaps the weights.
  • C (110) is a naïve forecast that just repeats last month's demand.
  • D (105) is a simple average of the two numbers, which ignores α.

Relation: F(t+1) = αA(t) + (1 − α)F(t). Handbook search words to try: exponential smoothing.

Source: NIST/SEMATECH e-Handbook, §6.4.3.1 Single Exponential Smoothing (smoothing equation; NIST indexes it as S_t = αy_{t−1} + (1−α)S_{t−1}).

Question 2 — A flagged part

FEIS-002 · Probability and Statistics · Choose one answer

2% of parts are defective. An automated test flags 95% of defective parts, and it also flags 4% of good parts by mistake. A part gets flagged. What is the probability it is actually defective?

  • A. 0.95
  • B. 0.058
  • C. 0.33
  • D. 0.019
Show answer and worked solution

Answer: C. 0.33

P(flagged) = (0.02)(0.95) + (0.98)(0.04) = 0.019 + 0.0392 = 0.0582.

P(defective | flagged) = 0.019 / 0.0582 = 0.326, or about 0.33.

Most flagged parts are good parts. Good parts are so common that their small false-alarm rate adds up.

  • A (0.95) is P(flagged | defective). The question asks the reverse.
  • B (0.058) is the probability that any part gets flagged.
  • D (0.019) is the joint probability that a part is defective and flagged, before dividing by P(flagged).

Relation: Bayes' theorem: P(D | F) = P(D)P(F | D) / P(F). Handbook search words to try: Bayes, conditional probability.

Source: Stanford CS 109, Conditional Probability and Bayes (April 8, 2024) (PDF p. 8: conditional probability and the product rule).

Question 3 — Choosing a machine by annual cost

FEIS-003 · Engineering Economics · Choose one answer

Two machines each last 5 years, and the effective annual interest rate is 10%. Machine A costs $40,000, has $8,000 per year in operating and maintenance (O&M) costs, and has a $5,000 salvage value. Machine B costs $60,000, has $4,000 per year in O&M, and has a $10,000 salvage value. O&M costs occur at each year-end, and salvage is received at the end of year 5. Which machine has the lower equivalent uniform annual cost (EUAC), and what is it?

  • A. Machine A, about $17,700 per year
  • B. Machine B, about $18,200 per year
  • C. Machine A, about $18,550 per year
  • D. They tie at $16,000 per year
Show answer and worked solution

Answer: A. Machine A, about $17,700 per year

Factors at 10%, 5 years: (A/P) = 0.26380 and (A/F) = 0.16380.

EUAC(A) = 40,000(0.26380) + 8,000 − 5,000(0.16380) = $17,733.

EUAC(B) = 60,000(0.26380) + 4,000 − 10,000(0.16380) = $18,190.

Machine A is cheaper, by about $460 per year.

  • B computes Machine B's EUAC correctly, but it is the higher of the two.
  • C is Machine A's EUAC with the salvage value left out.
  • D divides first cost evenly over 5 years and ignores both interest and salvage.

Relation: EUAC = P(A/P, i, n) + annual O&M − S(A/F, i, n). Handbook search words to try: capital recovery, sinking fund.

Source: Penn State EME 460, Compound Interest Formulas III (§6, Eq. 1-6: capital-recovery factor (A/P)); Penn State EME 460, Compound Interest Formulas II (§4, Eq. 1-4: sinking-fund factor (A/F)).

Question 4 — Two pumps and one controller

FEIS-004 · Systems Engineering, Analysis, and Design · Choose one answer

A system has two identical pumps in parallel, and either pump alone can do the job. Each pump has a mission reliability of 0.90. The pump pair is in series with a controller whose reliability over the same mission is 0.95. Failures are independent. What is the system reliability?

  • A. 0.7695
  • B. 0.855
  • C. 0.9405
  • D. 0.99
Show answer and worked solution

Answer: C. 0.9405

Pump pair: the pair fails only if both pumps fail, so R = 1 − (0.10)(0.10) = 0.99.

System: the pair and the controller must both work, so R = 0.99 × 0.95 = 0.9405.

  • A (0.7695) multiplies all three reliabilities, as if every component were in series.
  • B (0.855) counts only one pump.
  • D (0.99) is the pump pair alone. It leaves out the controller.

Relation: Parallel: R = 1 − Π(1 − Rᵢ). Series: R = ΠRᵢ. Both assume independent failures. Handbook search words to try: reliability, series, parallel.

Source: NIST/SEMATECH e-Handbook, §8.1.8.2 Series model (series system reliability); NIST/SEMATECH e-Handbook, §8.1.8.3 Parallel or redundant model (system CDF is the product of component CDFs; reliability = 1 − CDF).

Question 5 — How many stations?

FEIS-005 · Manufacturing, Service, and Other Production Systems · Choose one answer

An assembly line has 480 productive minutes per shift and must make 240 units per shift. The tasks for one unit add up to 9.0 minutes. What is the theoretical minimum number of workstations?

  • A. 4
  • B. 4.5
  • C. 5
  • D. 2
Show answer and worked solution

Answer: C. 5

Cycle time = 480 ÷ 240 = 2.0 minutes per unit.

Minimum stations = 9.0 ÷ 2.0 = 4.5, rounded up to 5.

That's a lower bound. Task times and precedence rules can force a real balance to use more stations.

  • A (4) rounds down. Four stations can't hold 9.0 minutes of work at a 2.0-minute cycle time.
  • B (4.5) isn't a whole number of stations.
  • D (2) is the cycle time in minutes, not a station count.

Relation: Cycle time = available time ÷ demand; N_min = ⌈total task time ÷ cycle time⌉. Handbook search words to try: line balancing, cycle time.

Source: King Saud University-hosted Line Balancing lecture (PDF p. 7: cycle time and theoretical minimum station count; pp. 9–10: examples).

Question 6 — Waiting at an inspection station

FEIS-006 · Modeling and Quantitative Analysis · Numerical entry

Parts arrive at a single inspection station at random (Poisson) at 12 per hour. Inspection times are exponential, with a service rate of 15 per hour. This is a steady-state M/M/1 queue with unlimited waiting space and independent arrival and service processes. On average, how many minutes does a part wait in line before inspection starts?

Your answer: ______ minutes
Enter a whole number of minutes.

Show answer and worked solution

Answer: 16 minutes

Wq = λ / [μ(μ − λ)] = 12 / (15 × 3) = 0.2667 hour = 16 minutes.

Check: total time in the system is W = 1/(μ − λ) = 1/3 hour = 20 minutes. Subtract the 4-minute average inspection (1/μ) and you get 16.

  • 20 is the time in the system, which includes the inspection itself.
  • 3.2 is Lq, the average number of parts waiting, not a time.
  • 4 is the average inspection time in minutes (1/μ), with no waiting included. It is also the numerical value of L, the average number of parts in the whole system.

Relation: M/M/1: Wq = λ/[μ(μ − λ)]; Little's law L = λW. Handbook search words to try: queueing, M/M/1.

Source: MIT 2.854, M/M/1 queue notes (PDF p. 1: steady-state M/M/1, λ < μ, infinite queue, Lq, Wq, and Little's law).

Question 7 — Critical path and float

FEIS-007 · Engineering Management · Choose one answer

A project has these activities (duration in days; immediate predecessors): A, 3 (none); B, 4 (A); C, 7 (A); D, 2 (B); E, 3 (C and D). All dependencies are finish-to-start with zero lag, and there are no resource constraints. What is the shortest project duration, and what is the total float of activity B?

  • A. 12 days; B has 0 days of float
  • B. 19 days; B has 0 days of float
  • C. 13 days; B has 0 days of float
  • D. 13 days; B has 1 day of float
Show answer and worked solution

Answer: D. 13 days; B has 1 day of float

Forward pass: A runs 0–3. B runs 3–7. C runs 3–10. D runs 7–9. E can't start until both C and D finish, so it starts at 10 and finishes at 13.

Backward pass: E's late start is 10. D's late finish is 10, so its late start is 8. B's late finish is 8, so its late start is 4.

B's total float = late start − early start = 4 − 3 = 1 day. The critical path is A–C–E.

  • A uses path A–B–D–E (12 days), which isn't the longest path.
  • B adds every duration, as if nothing could run at the same time.
  • C has the right duration but assumes every activity is critical.

Relation: Total float = LS − ES (or LF − EF). Handbook search words to try: critical path, CPM.

Source: NASA, Schedule Management Handbook, NASA/SP-2010-3403 (January 2010) (§5.4 p. 34: finish-to-start; §5.8.3 p. 47: total float; §7.4 pp. 59–60: critical path).

Question 8 — An off-center process

FEIS-008 · Quality · Choose one answer

A diameter has specification limits of 10.00 ± 0.05 mm. The process is stable and approximately normal, with a mean of 10.01 mm and a within-process standard deviation of 0.01 mm. What is Cpk?

  • A. 1.33
  • B. 1.67
  • C. 2.00
  • D. 0.67
Show answer and worked solution

Answer: A. 1.33

Cpk uses the closer specification limit:

Upper side: (10.05 − 10.01) / (3 × 0.01) = 1.33. Lower side: (10.01 − 9.95) / (3 × 0.01) = 2.00.

Cpk = the smaller value = 1.33. (Cp = 0.10 / 0.06 = 1.67.)

  • B (1.67) is Cp. It ignores the fact that the mean is off-center toward the upper limit.
  • C (2.00) uses the farther (lower) limit.
  • D (0.67) divides the upper-side distance by 6σ instead of 3σ.

Relation: Cp = (USL − LSL)/6σ; Cpk = min[(USL − μ)/3σ, (μ − LSL)/3σ]. Handbook search words to try: process capability.

Source: NIST/SEMATECH e-Handbook, §6.1.6 What is Process Capability? (stable/normal process conditions and Cp/Cpk definitions).

Question 9 — Lifting index for a box transfer

FEIS-009 · Human Factors, Ergonomics, and Safety · Choose one answer

Use the Revised NIOSH Lifting Equation in metric units (load constant 23 kg). For a single-task, two-handed lift that meets the equation's applicability conditions: horizontal distance at the origin H = 40 cm, vertical origin V = 75 cm, travel distance D = 50 cm, and no twisting (asymmetry multiplier 1.00). Use a frequency multiplier of 0.85 and a coupling multiplier of 1.00. The box weighs 15 kg, and no significant control is required at the destination. What is the lifting index at the origin?

  • A. 0.84
  • B. 0.74
  • C. 0.65
  • D. 1.35
Show answer and worked solution

Answer: D. 1.35

HM = 25/H = 25/40 = 0.625. VM = 1 − 0.003|75 − 75| = 1.00. DM = 0.82 + 4.5/D = 0.82 + 0.09 = 0.91.

RWL = 23 × 0.625 × 1.00 × 0.91 × 1.00 × 0.85 × 1.00 = 11.1190625 kg (about 11.1 kg).

Lifting index = load ÷ RWL = 15 ÷ 11.1190625 ≈ 1.35. The load is about a third over the recommended weight limit for this lift at the origin.

  • A (0.84) leaves out the horizontal multiplier. Holding the load 40 cm out is what drives this RWL down.
  • B (0.74) divides RWL by the load. That's upside down.
  • C (0.65) divides the load by the 23 kg constant and skips every multiplier.

Relation: RWL = LC × HM × VM × DM × AM × FM × CM; LI = load ÷ RWL. Handbook search words to try: lifting, NIOSH.

Source: NIOSH, Applications Manual for the Revised NIOSH Lifting Equation, Pub. No. 94-110 (revised 2021) (§1.1.2 printed p. 1 (PDF p. 17): lifting index; §1.2 pp. 5–6: applicability; §1.3 p. 7 (PDF p. 23): metric multipliers; p. 8: destination control).

Question 10 — Standard time from a time study

FEIS-010 · Work Design · Choose one answer

A time study gives an average observed time of 2.40 minutes. The worker was rated at 110%. Allowances are 15%, expressed as a percentage of normal time. What is the standard time?

  • A. 3.04 minutes
  • B. 3.11 minutes
  • C. 2.64 minutes
  • D. 2.76 minutes
Show answer and worked solution

Answer: A. 3.04 minutes

Normal time = observed time × rating = 2.40 × 1.10 = 2.64 minutes.

Standard time = normal time × (1 + allowance) = 2.64 × 1.15 = 3.04 minutes.

Read the allowance basis carefully. If the 15% were a share of the total workday, you would divide by (1 − 0.15) instead.

  • B (3.11) uses NT ÷ (1 − 0.15). That formula is for allowances stated as a percentage of total time, which isn't what this stem says.
  • C (2.64) stops at normal time and adds no allowance.
  • D (2.76) applies the allowance to the observed time and skips the rating.

Relation: NT = observed time × rating; ST = NT(1 + A) when A is a fraction of normal time. Handbook search words to try: time study, standard time, allowance.

Source: T. Armstrong, University of Michigan, Time Study notes (Normal Time & Performance Ratings; Standard Time & Allowances (both allowance bases)).

Question 11 — Five years of growing output

FEIS-011 · Mathematics · Choose one answer

A line produces 1,000 units in year 1, and output grows 5% each year after that. Using unrounded model values for each year, what is total output over the 5 years, rounded to the nearest whole unit?

  • A. 5,000 units
  • B. 5,526 units
  • C. 1,216 units
  • D. 5,500 units
Show answer and worked solution

Answer: B. 5,526 units

Each year is 1.05 times the year before, so this is a geometric series with a₁ = 1,000 and r = 1.05.

S₅ = 1,000(1.05⁵ − 1) / (1.05 − 1) = 5,525.63125 ≈ 5,526 units.

  • A (5,000) ignores the growth.
  • C (1,216) is year 5's output alone (1,000 × 1.05⁴).
  • D (5,500) adds 50 units per year, which is linear growth, not 5% compounding.

Relation: Sₙ = a₁(1 − rⁿ)/(1 − r). Handbook search words to try: geometric series, progression.

Source: OpenStax College Algebra 2e, Chapter 9 Key Equations (sum of a finite geometric series).

Question 12 — Asked to design outside your field

FEIS-012 · Ethics and Professional Practice · Choose one answer

A licensed industrial engineer has no education or experience in structural design. The plant manager asks her to design and seal a structural steel mezzanine to save time. Under the NCEES Model Rules, what should she do?

  • A. Do it, because a license lets an engineer seal any engineering work
  • B. Do it, but ask a coworker for an informal look
  • C. Decline the structural design and have it done by a licensee qualified in that field
  • D. Do it if the schedule would otherwise slip
Show answer and worked solution

Answer: C. Decline the structural design and have it done by a licensee qualified in that field

The Model Rules say licensees undertake assignments only when qualified by education or experience in the specific technical field (§240.15 B.1). Protecting public health, safety, and welfare comes first (§240.15 A.1).

She can still coordinate the project. The structural work belongs with someone qualified to do it.

The Model Rules are a model that NCEES publishes for licensing boards. Your own board's rules govern your practice.

  • A treats a license as unlimited competence. It isn't.
  • B: an informal look doesn't make her qualified in structural design.
  • D: a deadline doesn't change the competence rule.

Relation: NCEES Model Rules §240.15 A.1 and B.1. Handbook search words to try: ethics, licensure.

Source: NCEES Model Rules (August 2026) (§240.15 A.1 and B.1–3, printed pp. 11–12 (PDF pp. 15–16)).

Question 13 — Placing a new warehouse

FEIS-013 · Facilities and Supply Chain · Choose one answer

A new warehouse should minimize total weighted rectilinear distance to four customers. Locations (x, y) and weekly trips: C1 at (2, 8), 30 trips; C2 at (6, 2), 15 trips; C3 at (10, 6), 40 trips; C4 at (14, 10), 15 trips. Where should the warehouse go?

  • A. (7.6, 6.6)
  • B. (10, 6)
  • C. (6, 8)
  • D. (8, 6)
Show answer and worked solution

Answer: B. (10, 6)

With rectilinear distance, solve x and y separately. Each optimum is the weighted median: the coordinate where the running total of weights first reaches half the total (100 ÷ 2 = 50).

x, in order: 2 (30), 6 (running total 45), 10 (running total 85). The total passes 50 at x = 10.

y, in order: 2 (15), 6 (running total 55). The total passes 50 at y = 6.

Best location: (10, 6).

  • A (7.6, 6.6) is the weighted average (center of gravity). That's the answer for squared straight-line distance, not rectilinear distance.
  • C and D aren't weighted medians in both coordinates. Their weighted-distance totals are 600 and 560 trip-distance units, respectively, versus 540 at (10, 6).

Relation: Rectilinear minisum location: weighted median in x and in y. Handbook search words to try: facility location, rectilinear.

Source: Georgia Tech ISyE, Continuous Point Location (PDF pp. 21–22: rectilinear minisum decomposition and median conditions (March 16, 2003)).

Question 14 — Best product mix

FEIS-014 · Modeling and Quantitative Analysis · Choose one answer

For continuous variables x and y, maximize Z = 3x + 5y subject to x + 2y ≤ 40, 3x + 2y ≤ 60, and x, y ≥ 0. What are the optimal point and the maximum Z?

  • A. (0, 20), Z = 100
  • B. (20, 0), Z = 60
  • C. (10, 15), Z = 105
  • D. (20, 20), Z = 160
Show answer and worked solution

Answer: C. (10, 15), Z = 105

The optimum is at a corner of the feasible region. The corners are (0, 0), (20, 0), (0, 20), and the point where the two constraints cross.

Subtract the first constraint from the second: 2x = 20, so x = 10 and y = 15.

Z at each corner: 0, 60, 100, and 105. The maximum is at (10, 15).

  • A and B are real corners, but neither gives the largest Z.
  • D breaks both constraints: 20 + 2(20) = 60 > 40, and 3(20) + 2(20) = 100 > 60.

Relation: For a linear program with a nonempty bounded feasible region, at least one optimum occurs at a vertex. Handbook search words to try: linear programming.

Source: LibreTexts, Applied Finite Mathematics (Sekhon and Bloom), §3.1 Maximization Applications (fundamental theorem of linear programming (optimum at a vertex)).

Question 15 — Confidence interval for cycle time

FEIS-015 · Probability and Statistics · Choose one answer

An independent random sample of n = 16 cycle times from a normal population has a mean of 50.2 s and a standard deviation of 2.0 s. The population standard deviation is unknown. Using t(0.025, 15) = 2.131, what is the two-sided 95% confidence interval for the mean?

  • A. 49.13 to 51.27 s
  • B. 49.22 to 51.18 s
  • C. 45.94 to 54.46 s
  • D. 49.93 to 50.47 s
Show answer and worked solution

Answer: A. 49.13 to 51.27 s

Standard error = s/√n = 2.0/4 = 0.5 s. Margin = 2.131 × 0.5 = 1.0655 s.

Interval = 50.2 ± 1.0655 = 49.1345 to 51.2655 s, or 49.13 to 51.27 s.

This interval estimates the mean. It does not say 95% of individual cycles fall inside it.

  • B uses z = 1.96. With σ unknown and a small sample, use t.
  • C forgets to divide s by √n.
  • D divides s by n instead of √n.

Relation: x̄ ± t(α/2, n−1) · s/√n. Handbook search words to try: confidence interval, t-distribution.

Source: NIST/SEMATECH e-Handbook, §1.3.5.2 Confidence Limits for the Mean (t-based interval); OpenStax Introductory Statistics 2e, §8.2 A Single Population Mean Using the Student t Distribution (opening conditions and Properties of the Student t-Distribution).

Question 16 — Funding maintenance forever

FEIS-016 · Engineering Economics · Choose one answer

A city wants to fund $12,000 of maintenance at the end of every year, forever. At 8% interest per year, how much must it invest now?

  • A. $96,000
  • B. $960
  • C. $162,000
  • D. $150,000
Show answer and worked solution

Answer: D. $150,000

For a perpetual annual amount, capitalized cost P = A ÷ i = 12,000 ÷ 0.08 = $150,000.

Check: $150,000 earning 8% throws off exactly $12,000 a year without touching the principal.

  • A ($96,000) multiplies $12,000 by 8.
  • B ($960) is $12,000 × 0.08. That's one year of interest on the payment, not the fund.
  • C ($162,000) adds one extra payment, as if the first withdrawal happened today.

Relation: Capitalized cost: P = A/i. Handbook search words to try: capitalized cost.

Source: OpenStax Principles of Finance, §8.1 Perpetuities (Eq. 8.1: present value of a perpetuity; College Endowment example).

Question 17 — Reading earned value

FEIS-017 · Engineering Management · Choose one answer

At a status date, planned value (PV) is $50,000, earned value (EV) is $45,000, and actual cost (AC) is $48,000. Which statement is correct?

  • A. CPI = 0.94, over budget; SPI = 0.90, behind schedule
  • B. CPI = 1.07, under budget; SPI = 1.11, ahead of schedule
  • C. CPI = 1.04, under budget; SPI = 0.90, behind schedule
  • D. CPI = 0.90, over budget; SPI = 0.94, behind schedule
Show answer and worked solution

Answer: A. CPI = 0.94, over budget; SPI = 0.90, behind schedule

Cost performance index: CPI = EV/AC = 45,000/48,000 = 0.94. Below 1 means the work done cost more than budgeted.

Schedule performance index: SPI = EV/PV = 45,000/50,000 = 0.90. Below 1 means less work is done than planned.

SPI alone doesn't tell you how many days late the project will finish.

  • B flips both ratios.
  • C uses PV/AC for CPI.
  • D swaps the two indexes.

Relation: CPI = EV/AC; SPI = EV/PV. Handbook search words to try: earned value.

Source: U.S. DOE, EVMS Gold Card (Performance Indices: CPI and SPI).

Question 18 — Limits for a defectives chart

FEIS-018 · Quality · Choose one answer

A p chart tracks the fraction defective. The established baseline fraction defective is 0.04, each sample has n = 200 units, and unit classifications are independent. What is the 3-sigma upper control limit?

  • A. 0.082
  • B. 0.068
  • C. 0.628
  • D. 0.054
Show answer and worked solution

Answer: A. 0.082

σ_p = √[0.04 × 0.96 / 200] = 0.01386.

UCL = 0.04 + 3(0.01386) = 0.082.

The lower limit works out negative, so it's set to 0.

  • B (0.068) uses 2 sigma.
  • C (0.628) forgets to divide by n inside the square root.
  • D (0.054) uses 1 sigma.

Relation: p chart: p̄ ± 3√[p̄(1 − p̄)/n]. Handbook search words to try: control chart, p chart.

Source: NIST/SEMATECH e-Handbook, §6.3.3.2 Proportions Control Charts (p-chart limits).

Question 19 — A day's noise dose

FEIS-019 · Human Factors, Ergonomics, and Safety · Choose one answer

Under federal OSHA's general-industry noise standard, evaluate the permissible exposure limit (PEL) using a 90 dBA sampling threshold and a 5 dB exchange rate. An 8-hour shift includes 4 hours at 90 dBA, 2 hours at 95 dBA, and 2 hours at 85 dBA. Based on these exact levels and durations, what is the PEL noise dose, and how does it compare with the 100% limit?

  • A. 100%, exactly at the limit
  • B. 112.5%, over the limit
  • C. 125%, over the limit
  • D. 90%, below the limit
Show answer and worked solution

Answer: A. 100%, exactly at the limit

For this PEL evaluation, OSHA's 90 dBA sampling threshold excludes the 2 hours at 85 dBA. The reference duration is T = 8 ÷ 2^((L − 90)/5) hours: 8 hours at 90 dBA and 4 hours at 95 dBA.

PEL dose = 100 × (4/8 + 2/4) = 100 × (0.5 + 0.5) = 100%, exactly at the stated limit.

Hearing conservation uses an 80 dBA sampling threshold and a 50% dose action level. Meeting the PEL does not remove those separate requirements.

  • B (112.5%) includes 2/16 for the 85 dBA period. That gives the dose using the hearing-conservation sampling threshold, not the PEL dose asked for here.
  • C (125%) both includes the below-threshold 85 dBA period and incorrectly uses 8 hours as its reference duration.
  • D (90%) confuses an arithmetic average of decibel readings with a percentage dose. OSHA's dose method uses exposure-duration fractions.

Relation: PEL dose D = 100 × Σ(Cᵢ/Tᵢ), counting only levels at or above the 90 dBA threshold; T = 8/2^((L − 90)/5). Handbook search words to try: noise, OSHA.

Source: OSHA, 29 CFR 1910.95 Occupational Noise Exposure ((b)(1) and Table G-16: permissible exposure; (c)(1) and (d)(2)(i): hearing-conservation action level and measurement range); OSHA, 29 CFR 1910.95 Appendix A (noise-dose formula, reference duration, and Table G-16a); OSHA, Differentiation between the 80 dBA threshold for hearing conservation and the 90 dBA PEL (September 26, 2001) (reply distinguishing the 90 dBA PEL sampling threshold from the 80 dBA hearing-conservation threshold).

Question 20 — Moment from an angled pull

FEIS-020 · Engineering Sciences · Choose one answer

A mechanic pulls with 40 lbf on the end of a wrench handle 12 in long. The force acts at 60° to the handle. What is the magnitude of the moment about the bolt?

  • A. 40.0 lbf·ft
  • B. 3.33 lbf·ft
  • C. 20.0 lbf·ft
  • D. 34.6 lbf·ft
Show answer and worked solution

Answer: D. 34.6 lbf·ft

Convert the handle to feet: 12 in = 1.0 ft.

M = rF sin θ = (1.0 ft)(40 lbf)(sin 60°) = 34.6 lbf·ft.

Only the part of the force perpendicular to the handle turns the bolt.

  • A (40.0) assumes the force is perpendicular (sin 90° = 1).
  • B (3.33) divides the force by the length in inches.
  • C (20.0) uses cos 60° instead of sin 60°.

Relation: M = rF sin θ. Handbook search words to try: moment, torque.

Source: OpenStax University Physics Vol. 1, §10.6 Torque (Eq. 10.22).

Question 21 — Order quantity for a purchased part

FEIS-021 · Manufacturing, Service, and Other Production Systems · Numerical entry

Annual demand for a part is 12,000 units. Each order costs $100 to place, and holding one unit for a year costs $2.40. Assume steady demand, instant replenishment, no shortages, and no quantity discounts. What is the economic order quantity (EOQ)?

Your answer: ______ units
Enter a whole number of units.

Show answer and worked solution

Answer: 1,000 units

EOQ = √(2DS/H) = √(2 × 12,000 × 100 / 2.40) = √1,000,000 = 1,000 units.

At 1,000 units, yearly ordering cost (12 orders × $100) equals yearly holding cost (500 average units × $2.40). Both are $1,200.

  • 707 leaves the 2 out of the formula.
  • Mixing monthly demand with a yearly holding cost gives a wildly wrong answer. Put D and H on the same time basis.

Relation: EOQ = √(2DS/H). Handbook search words to try: economic order quantity.

Source: MIT 2.854, Inventory (Fall 2016) (slides 44–50: economic order quantity assumptions and formula).

Question 22 — Pooling safety stock

FEIS-022 · Facilities and Supply Chain · Choose one answer

Four regional warehouses each face independent weekly demand with a standard deviation of 100 units. Lead time is fixed at 1 week before and after pooling, and the service target uses z = 1.65. If the four are combined into one central warehouse, how much safety stock does it need?

  • A. 330 units
  • B. 660 units
  • C. 165 units
  • D. 83 units
Show answer and worked solution

Answer: A. 330 units

Independent variances add: σ_pooled = √(4 × 100²) = 200 units of one-week demand.

Safety stock = z × σ × √(lead time) = 1.65 × 200 × 1 = 330 units.

Four separate warehouses would hold 4 × 165 = 660 units. Pooling cuts that in half here.

  • B (660) is the total for four separate warehouses.
  • C (165) is one warehouse's safety stock.
  • D (83) divides σ by √4 instead of multiplying.

Relation: σ_pooled = √(Σσᵢ²) for independent demand; SS = zσ√L. Handbook search words to try: safety stock, risk pooling.

Source: FIU-hosted Management of Uncertainty lecture (slides 10–12: pooled demand variance and safety stock).

Question 23 — Learning on a new assembly

FEIS-023 · Work Design · Choose one answer

Unit 1 of a new assembly takes 100 labor-hours. Production follows an 80% unit learning curve. How long does unit 8 take?

  • A. 64.0 hours
  • B. 51.2 hours
  • C. 41.0 hours
  • D. 80.0 hours
Show answer and worked solution

Answer: B. 51.2 hours

On a unit curve, the time for one unit drops to 80% each time the unit number doubles. Unit 8 is three doublings from unit 1.

T₈ = 100 × 0.8³ = 51.2 hours. (Same result from Tₙ = T₁nᵇ with b = log 0.8 / log 2 = −0.322.)

  • A (64.0) is unit 4.
  • C (41.0) is unit 16.
  • D (80.0) is unit 2.

Relation: Tₙ = T₁nᵇ, b = log(rate)/log 2. Handbook search words to try: learning curve.

Source: NASA, Guidelines for Application of Learning/Cost Improvement Curves, NASA TM X-64968 (October 31, 1975) (§IV.B, printed pp. 6–7 (PDF pp. 10–11): Crawford unit learning-curve model).

Question 24 — Which requirements can be verified?

FEIS-024 · Systems Engineering, Analysis, and Design · Select all that apply

Select all that apply. Which of these requirements give a measurable way to verify that they've been met?

  • A. After a 30-second warm-up, the scale shall read a 1.000 kg reference mass within ±0.005 kg at 20 ± 1 °C.
  • B. The interface shall be user-friendly.
  • C. The emergency stop shall respond as quickly as possible.
  • D. In an acceptance test of 1,000 labeled parcels, the sorter shall route at least 995 to the correct outlet.
  • E. The design shall use modern technology.

Counts as correct only if you choose every correct option and no others.

Show answer and worked solution

Answer: A, D

A states the test condition, the reference value, and the tolerance. D states the test and the pass threshold. Either one can be checked with a clear yes or no.

NASA's systems engineering guidance asks whether each requirement is stated with enough information to verify it once the product is built.

  • B, C, and E use words with no measurable target (user-friendly, as quickly as possible, modern). Nobody can test them as written.
  • Being verifiable doesn't make a requirement the right requirement. That's a separate validation question.

Relation: Requirements should be verifiable: stated so the finished product can be checked against them. Handbook search words to try: requirements, verification.

Source: NASA Systems Engineering Handbook, §4.2 Technical Requirements Definition (§4.2.1.2.4, step 5: Are the requirements verifiable?).

Question 25 — What counts in a replacement decision

FEIS-025 · Engineering Economics · Select all that apply

Select all that apply. A plant is deciding whether to keep a machine it bought 3 years ago or replace it. Ignoring taxes, which items are relevant to the decision?

  • A. The original purchase price, $80,000
  • B. The existing machine's current market value, $30,000
  • C. The existing machine's current book value, $44,000
  • D. Future operating costs of both machines
  • E. The new machine's purchase price

Counts as correct only if you choose every correct option and no others.

Show answer and worked solution

Answer: B, D, E

Only future cash flows that differ between the choices matter.

B counts because keeping the old machine means giving up the $30,000 you could sell it for. D and E are future costs that depend on the choice.

  • A is a sunk cost. That money is spent either way.
  • C is an accounting number. In an after-tax study it matters through depreciation and taxes, but this question ignores taxes.

Relation: Ignore sunk costs; compare future, differing cash flows. Handbook search words to try: replacement analysis, sunk cost.

Source: OpenStax Principles of Managerial Accounting, §10.1 Identify Relevant Information for Decision-Making (sunk costs are not relevant).

End of Section 1. Good place for a break. On the real exam, after about half the questions you review and submit them before the optional scheduled break, and you can't go back to that first half afterward (NCEES Examinee Guide, pp. 11–12). Nothing is locked on this page.

Section 2: Questions 26–50

Question 26 — Area under a curve

FEIS-026 · Mathematics · Choose one answer

What is the area between y = 4 − x² and the x-axis from x = 0 to x = 2?

  • A. 8
  • B. 16/3 ≈ 5.33
  • C. 8/3 ≈ 2.67
  • D. 4
Show answer and worked solution

Answer: B. 16/3 ≈ 5.33

∫₀² (4 − x²) dx = [4x − x³/3] from 0 to 2 = 8 − 8/3 = 16/3 ≈ 5.33.

The curve is on or above the axis on this whole interval, so the integral equals the area.

  • A (8) integrates only the 4 and drops the x² term.
  • C (2.67) is the x² part alone.
  • D (4) is the curve's height at x = 0.

Relation: Area = definite integral when f(x) ≥ 0. Handbook search words to try: integral, area.

Source: OpenStax Calculus Vol. 1, §5.2 The Definite Integral (definite integral and area).

Question 27 — Picking an audit team

FEIS-027 · Probability and Statistics · Choose one answer

Three of 8 inspectors will form an audit team. Order doesn't matter. How many different teams are possible?

  • A. 24
  • B. 56
  • C. 336
  • D. 512
Show answer and worked solution

Answer: B. 56

C(8, 3) = 8! / (3! × 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 56.

  • A (24) just multiplies 8 × 3.
  • C (336) is the number of ordered arrangements (permutations). A team has no order.
  • D (512) is 8³, which allows the same inspector to be picked more than once.

Relation: C(n, r) = n!/[r!(n − r)!]. Handbook search words to try: combinations, permutations.

Source: OpenStax College Algebra 2e, §9.5 Counting Principles (Find the Number of Combinations Using the Formula: combinations when order does not matter; C(n,r) = n!/[r!(n−r)!]).

Question 28 — Long-run machine uptime

FEIS-028 · Modeling and Quantitative Analysis · Choose one answer

A machine follows a time-homogeneous Markov model with two states, Up and Down, each day. If it's Up today, it's Up tomorrow with probability 0.9. If it's Down today, it's Up tomorrow with probability 0.3. In the long run, what fraction of days is it Up?

  • A. 0.90
  • B. 0.60
  • C. 0.25
  • D. 0.75
Show answer and worked solution

Answer: D. 0.75

In steady state, the probability flowing out of Up equals the probability flowing in: 0.1·π_Up = 0.3·π_Down.

With π_Up + π_Down = 1: π_Up = 0.3/(0.1 + 0.3) = 0.75.

  • A (0.90) is a one-day transition probability, not the long-run share.
  • B (0.60) averages 0.9 and 0.3.
  • C (0.25) is the long-run fraction of Down days.

Relation: Steady state: πP = π, Σπ = 1. Handbook search words to try: Markov chain, steady state.

Source: Susan Holmes, Stanford Stats 366 / Stats 166 Course Notes, Markov Chains (2012) (two-state weather example and final stationary-distribution paragraphs, αP = α).

Question 29 — Three-point estimate

FEIS-029 · Engineering Management · Choose one answer

An activity's optimistic time is 4 days, most likely time is 6 days, and pessimistic time is 14 days. What is its PERT expected duration?

  • A. 6.0 days
  • B. 7.0 days
  • C. 8.0 days
  • D. 1.67 days
Show answer and worked solution

Answer: B. 7.0 days

tₑ = (a + 4m + b)/6 = (4 + 24 + 14)/6 = 7.0 days.

The long pessimistic tail pulls the expected time above the most likely time.

  • A (6.0) is the most likely time.
  • C (8.0) is a plain average of the three estimates.
  • D (1.67) is the classic PERT estimate of the standard deviation, (b − a)/6.

Relation: tₑ = (a + 4m + b)/6; σ = (b − a)/6. Handbook search words to try: PERT.

Source: D. L. Bricker, University of Iowa, PERT (July 23, 1998) (PDF p. 1, The BETA distribution: boxed mean and standard-deviation formulas).

Question 30 — Net requirement for a component

FEIS-030 · Manufacturing, Service, and Other Production Systems · Numerical entry

For one MRP period, a component has a gross requirement of 500 units. On-hand inventory is 120 units, a scheduled receipt of 80 units is available in time for that period's requirements, and 50 units of safety stock must stay on hand. What is the net requirement?

Your answer: ______ units
Enter a whole number of units.

Show answer and worked solution

Answer: 350 units

Net requirement = gross requirement + safety stock − (on hand + scheduled receipts) = 500 + 50 − (120 + 80) = 350 units.

  • 300 forgets the safety stock.
  • 430 forgets the scheduled receipt that's already coming.

Relation: Net = gross + safety stock − (on hand + scheduled receipts), floored at 0. Handbook search words to try: MRP, net requirements.

Source: University of Houston Bauer College, Operations Management text, MRP 1 (PDF p. 2, §4.1 MRP inventory plan (MRP1): projected beginning inventory and net requirements).

Question 31 — The cost of being slightly off target

FEIS-031 · Quality · Choose one answer

A quadratic (Taguchi) loss function applies. A part 0.5 mm off target causes a $50 loss. What is the loss for a part 0.2 mm off target?

  • A. $20
  • B. $0
  • C. $4
  • D. $8
Show answer and worked solution

Answer: D. $8

L = k(y − m)². From the first part: k = 50 / 0.5² = 200 dollars per mm².

For 0.2 mm: L = 200 × 0.2² = $8.

  • A ($20) scales the loss linearly (0.2/0.5 × $50).
  • B ($0) assigns no loss despite the nonzero deviation. The stated quadratic model gives positive loss away from target.
  • C ($4) gets k from a linear ratio (50/0.5 = 100) and then squares the deviation.

Relation: L(y) = k(y − m)². Handbook search words to try: Taguchi, loss function.

Source: George Mason University OM 456, Taguchi Loss Function (L(x) = k(x − T)²).

Question 32 — Reaching a touchscreen button

FEIS-032 · Human Factors, Ergonomics, and Safety · Choose one answer

Movement time follows Fitts' law, MT = a + b·log₂(2D/W), with a = 50 ms and b = 150 ms per bit. The distance to the button center is D = 16 cm, and its width along the direction of movement is W = 2 cm. What is the predicted movement time?

  • A. 650 ms
  • B. 500 ms
  • C. 200 ms
  • D. 2,450 ms
Show answer and worked solution

Answer: A. 650 ms

Index of difficulty = log₂(2 × 16 / 2) = log₂(16) = 4 bits.

MT = 50 + 150 × 4 = 650 ms.

In this model, a wider or closer target has a lower index of difficulty.

  • B (500) uses log₂(D/W) = 3. The formula in the stem uses 2D.
  • C (200) adds a and b and skips the index of difficulty.
  • D (2,450) uses 2D/W = 16 directly, without the logarithm.

Relation: MT = a + b·log₂(2D/W). Handbook search words to try: Fitts, human factors.

Source: I. Scott MacKenzie, Movement Time Prediction in Human-Computer Interfaces (Graphics Interface 1992) (A Brief Tour of Fitts' Law, Eqs. 1 and 3; Extension to Two Dimensions).

Question 33 — A promise with nothing in return

FEIS-033 · Ethics and Professional Practice · Choose one answer

An engineer offers in writing to do a plant-layout study for a former professor's business at no charge, and the business accepts. The engineer seeks no payment, promise, performance, or forbearance in exchange, and the business supplies none. Which contract concept means the bargained-for exchange missing here?

  • A. An offer
  • B. Acceptance
  • C. Consideration
  • D. A notarized signature
Show answer and worked solution

Answer: C. Consideration

Consideration is a bargained-for promise, performance, or forbearance. Here, the business supplies none in exchange for the engineer's promise.

Acceptance alone does not create that exchange. Whether another legal doctrine could make the promise enforceable is a separate question.

  • A and B: the stem expressly states an offer and an acceptance.
  • D: notarizing a signature does not supply the missing bargained-for exchange.

Relation: Consideration: a bargained-for promise, performance, or forbearance. Handbook search words to try: contracts, agreements.

Source: Hamer v. Sidway, 124 N.Y. 538 (1891), New York Court of Appeals (reported pp. 545–546: consideration and bargained-for forbearance).

Question 34 — How many machines?

FEIS-034 · Facilities and Supply Chain · Choose one answer

Weekly demand is 2,000 parts. Standard time is 3.0 minutes per part. Identical machines work in parallel. Each is scheduled 40 hours per week and produces at the stated standard rate for 80% of that time, with no additional setup or scrap losses. What is the minimum number of machines?

  • A. 3
  • B. 4
  • C. 5
  • D. 2
Show answer and worked solution

Answer: B. 4

Hours needed = 2,000 × 3.0 ÷ 60 = 100 hours per week.

Effective hours per machine = 40 × 0.80 = 32 hours.

Machines = 100 ÷ 32 = 3.125, rounded up to 4.

  • A (3) comes from ignoring efficiency (100 ÷ 40 = 2.5, rounded up) or from rounding 3.125 down. Either way, three machines fall short.
  • C (5) is more than the minimum.
  • D (2) truncates 2.5 and ignores efficiency.

Relation: Machines = hours required ÷ effective hours per machine, rounded up. Handbook search words to try: capacity, number of machines.

Source: MIT 15.772J, Lecture 6: Capacity Planning and Production Flow Control (Fall 2014) (slides 17–22: capacity, productive time, and resources; numerical inputs and the 80% time factor are supplied in this question).

Question 35 — Work sampling sample size

FEIS-035 · Work Design · Numerical entry

A work-sampling study estimates that a machine is idle about 20% of the time. You want the estimate within ±0.04 (an absolute margin) at 95% confidence, using z = 1.96. Assume independent random observations and use the normal-approximation planning formula with p = 0.20. What is the minimum number of observations?

Your answer: ______ observations
Enter a whole number.

Show answer and worked solution

Answer: 385 observations

n = z² p(1 − p) / E² = 1.96² × 0.20 × 0.80 / 0.04² = 384.16.

Round this minimum sample size up: 385 observations.

  • 384 rounds down, which falls just short of the target precision.
  • 196 forgets to square z.

Relation: n = z²p(1 − p)/E². Handbook search words to try: work sampling, sample size.

Source: OpenStax Introductory Statistics 2e, §8.3 A Population Proportion (Calculating the Sample Size n; normal-approximation and random-sampling conditions).

Question 36 — Ranking failure modes

FEIS-036 · Systems Engineering, Analysis, and Design · Choose one answer

An FMEA rates two failure modes on 1–10 scales. Mode X: severity 8, occurrence 3, detection 6. Mode Y: severity 5, occurrence 6, detection 5. Which has the higher risk priority number (RPN)?

  • A. Mode X, RPN 144
  • B. Mode Y, RPN 150
  • C. Mode X, RPN 17
  • D. They tie
Show answer and worked solution

Answer: B. Mode Y, RPN 150

RPN = severity × occurrence × detection. X: 8 × 3 × 6 = 144. Y: 5 × 6 × 5 = 150.

Y ranks higher by RPN. Still, RPN is a screening number. A severity-8 mode like X deserves its own look even when its RPN is lower.

  • A is X's correct RPN, but it's the smaller one.
  • C (17) adds the ratings instead of multiplying.
  • D: 144 and 150 aren't equal.

Relation: RPN = S × O × D. Handbook search words to try: FMEA, risk priority number.

Source: NPTEL Module 5, Lecture 1: FMEA (RPN = Severity × Occurrence × Detection).

Question 37 — Predicting from a fitted line

FEIS-037 · Probability and Statistics · Choose one answer

For a set of (x, y) data: x̄ = 10, ȳ = 50, Sxy = 120, and Sxx = 40. Here Sxy = Σ(x − x̄)(y − ȳ) and Sxx = Σ(x − x̄)². Using the least-squares line, what is the predicted y at x = 12?

  • A. 56
  • B. 36
  • C. 50
  • D. 116
Show answer and worked solution

Answer: A. 56

Slope b = Sxy/Sxx = 120/40 = 3. Intercept a = ȳ − b·x̄ = 50 − 30 = 20.

ŷ = 20 + 3(12) = 56.

  • B (36) leaves out the intercept.
  • C (50) is just ȳ.
  • D (116) computes the intercept as ȳ + b·x̄.

Relation: b = Sxy/Sxx; a = ȳ − b·x̄. Handbook search words to try: regression, least squares.

Source: OpenStax Introductory Statistics 2e, §12.3 The Regression Equation (least-squares slope and intercept formulas immediately before Residuals Plots).

Question 38 — Two processes, one crossover

FEIS-038 · Engineering Economics · Choose one answer

Process A costs $20,000 per year plus $8 per unit. Process B costs $50,000 per year plus $5 per unit. At what annual volume do they cost the same, and which is cheaper above that volume?

  • A. 6,000 units; B is cheaper above it
  • B. 10,000 units; A is cheaper above it
  • C. 3,750 units; B is cheaper above it
  • D. 10,000 units; B is cheaper above it
Show answer and worked solution

Answer: D. 10,000 units; B is cheaper above it

Set the costs equal: 20,000 + 8Q = 50,000 + 5Q, so 3Q = 30,000 and Q = 10,000 units.

Above that volume, B's lower per-unit cost outweighs its higher fixed cost, so B is cheaper.

  • A (6,000) divides the $30,000 fixed-cost gap by B's $5 variable cost.
  • B has the right volume but the wrong side.
  • C (3,750) divides the gap by A's $8 variable cost.

Relation: Crossover: F_A + v_AQ = F_B + v_BQ. Handbook search words to try: break-even, fixed cost, variable cost.

Source: OpenStax Principles of Managerial Accounting, §3.2 Calculate a Break-Even Point (fixed and variable cost behavior).

Question 39 — Generating a service time

FEIS-039 · Modeling and Quantitative Analysis · Choose one answer

Service times are exponential with a mean of 5 minutes, so F(x) = 1 − e^(−x/5). A simulation sets F(x) = U and solves for x (the inverse transform). With U = 0.30, what service time is generated?

  • A. 1.78 minutes
  • B. 6.02 minutes
  • C. 1.50 minutes
  • D. 0.07 minutes
Show answer and worked solution

Answer: A. 1.78 minutes

Solve 1 − e^(−x/5) = 0.30: x = −5 ln(1 − 0.30) = −5 ln(0.70) = 1.78 minutes.

  • B (6.02) is −5 ln(U). That shortcut also produces exponential values in a simulation, but it isn't F⁻¹(U), which is what the stem asks for.
  • C (1.50) is just 5 × 0.30.
  • D (0.07) uses the rate (0.2 per minute) where the mean belongs.

Relation: Inverse transform: x = F⁻¹(U); for the exponential, x = −mean × ln(1 − U). Handbook search words to try: simulation, inverse transform, exponential.

Source: Brown University CS 1951k, Inverse Transform Sampling notes (p. 2, Example 3.2 (exponential)).

Question 40 — What a perfect forecast is worth

FEIS-040 · Engineering Management · Choose one answer

Payoffs in $1,000s: Expand earns 100 if demand is high (probability 0.6) and −20 if demand is low (probability 0.4). Hold earns 40 either way. What is the expected value of perfect information (EVPI)?

  • A. $24,000
  • B. $76,000
  • C. $52,000
  • D. $36,000
Show answer and worked solution

Answer: A. $24,000

Without more information: EV(Expand) = 0.6(100) + 0.4(−20) = 52. EV(Hold) = 40. Best choice: Expand, worth 52.

With perfect information, you'd Expand when demand is high and Hold when it's low: 0.6(100) + 0.4(40) = 76.

EVPI = 76 − 52 = 24, or $24,000. Under the expected-monetary-value criterion, that's the most a perfect forecast is worth paying for.

  • B ($76,000) is the expected value with perfect information, before subtracting.
  • C ($52,000) is the best expected value without it.
  • D ($36,000) subtracts Hold's value (40) instead of the best choice's value (52).

Relation: EVPI = EV with perfect information − best EV without it. Handbook search words to try: decision analysis, expected value.

Source: George E. Apostolakis, MIT ESD.72, DA 2. The Value of Perfect Information (Spring 2007) (slides 9–11: expected value with perfect information, best value without it, and their difference).

Question 41 — Sequencing jobs on one machine

FEIS-041 · Manufacturing, Service, and Other Production Systems · Choose one answer

Five jobs are all ready at time 0 on one continuously available machine. Run each job to completion, with no idle time or separate setup time. Processing times, in arrival order, are 6, 2, 8, 3, and 5 hours. Using the shortest processing time (SPT) rule, what is the mean flow time?

  • A. 14.6 hours
  • B. 4.8 hours
  • C. 11.4 hours
  • D. 17.4 hours
Show answer and worked solution

Answer: C. 11.4 hours

SPT order: 2, 3, 5, 6, 8. Completion times: 2, 5, 10, 16, 24.

Mean flow time = (2 + 5 + 10 + 16 + 24) / 5 = 57 / 5 = 11.4 hours.

Under these single-machine, all-jobs-ready assumptions, SPT gives the lowest possible mean flow time.

  • A (14.6) keeps the arrival order (first come, first served).
  • B (4.8) is the average processing time, not the average time to completion.
  • D (17.4) runs the longest job first.

Relation: Flow time = completion time − ready time; SPT minimizes the mean under the stated single-machine assumptions. Handbook search words to try: scheduling, SPT.

Source: Nesim K. Erkip, Bilkent University IE375, Scheduling I (Fall 2020) (PDF p. 12: SPT; p. 15: single-machine assumptions; p. 16: minimum mean-flow theorem).

Question 42 — Will the lot pass?

FEIS-042 · Quality · Choose one answer

A single sampling plan inspects n = 20 parts and accepts the lot if 1 or fewer are defective. The lot is 5% defective. Assume random sampling from a lot much larger than the sample. Using the binomial distribution, what is the probability the lot is accepted?

  • A. 0.36
  • B. 0.74
  • C. 0.95
  • D. 0.26
Show answer and worked solution

Answer: B. 0.74

P(0 defective) = 0.95²⁰ ≈ 0.358486.

P(1 defective) = 20 × 0.05 × 0.95¹⁹ ≈ 0.377354.

P(accept) ≈ 0.735840 ≈ 0.74.

  • A (0.36) counts only zero defectives. The plan also accepts one.
  • C (0.95) is the chance a single part is good.
  • D (0.26) is the probability of rejecting the lot.

Relation: Pa = Σ (d = 0 to c) C(n, d)pᵈ(1 − p)ⁿ⁻ᵈ. Handbook search words to try: acceptance sampling, OC curve.

Source: NIST/SEMATECH e-Handbook, §6.2.3.2 Choosing a Sampling Plan with a given OC Curve (Number of defectives is approximately binomial; The binomial distribution: acceptance probability).

Question 43 — Sizing a doorway

FEIS-043 · Human Factors, Ergonomics, and Safety · Choose one answer

User stature is normally distributed with a mean of 176 cm and a standard deviation of 7 cm. In a simplified stature-only estimate, what clearance height accommodates 95% of users, one-sided, before adding allowances for shoes and movement? Use z₀.₉₅ = 1.645.

  • A. 164.5 cm
  • B. 189.7 cm
  • C. 187.5 cm
  • D. 183.0 cm
Show answer and worked solution

Answer: C. 187.5 cm

Clearance should fit the tall end of the population, so use the 95th percentile.

176 + 1.645 × 7 = 187.5 cm.

  • A (164.5) is the 5th percentile of stature, so approximately 95% of this population is taller than that height. It is not a reach measurement.
  • B (189.7) uses z = 1.96. That value goes with a two-sided 95% range, not a one-sided 95th percentile.
  • D (183.0) is the mean plus one standard deviation, roughly the 84th percentile.

Relation: Percentile value = mean + z·σ. Handbook search words to try: anthropometry, percentile.

Source: City University of Hong Kong, Online Anthropometry, §5.5 Computation of Percentiles (percentile = mean + k·SD); NASA, Anthropometrics and Crew Physical Characteristics, OCHMO-TB-049 (July 21, 2025) (p. 2 final bullet: percentiles differ across body dimensions; pp. 3 and 5: clearance and reach context).

Question 44 — Inverting a 2×2 matrix

FEIS-044 · Mathematics · Choose one answer

A = [[2, 1], [5, 3]] (rows listed in order). What is A⁻¹?

  • A. [[3, 1], [5, 2]]
  • B. [[2, −1], [−5, 3]]
  • C. [[3, −1], [−5, 2]]
  • D. [[−3, 1], [5, −2]]
Show answer and worked solution

Answer: C. [[3, −1], [−5, 2]]

The determinant is (2)(3) − (1)(5) = 1.

For [[a, b], [c, d]], swap a and d, change the signs of b and c, then divide by the determinant: A⁻¹ = [[3, −1], [−5, 2]].

Check: A × A⁻¹ = [[6 − 5, −2 + 2], [15 − 15, −5 + 6]] = the identity matrix.

  • A swaps the diagonal but forgets the sign changes.
  • B changes the signs but doesn't swap the diagonal.
  • D has every sign backward. That's −A⁻¹.

Relation: A⁻¹ = (1/det A)[[d, −b], [−c, a]], provided det A ≠ 0. Handbook search words to try: matrix inverse, determinant.

Source: OpenStax College Algebra 2e, Chapter 7 Key Equations (Finding the Inverse of a 2×2 Matrix; nonzero determinant condition).

Question 45 — Flow through a reducer

FEIS-045 · Engineering Sciences · Choose one answer

Water flows steadily through full pipes with no branches or leaks. Its cross-section-average speed is 6 ft/s in a pipe with a 4 in inside diameter, then it passes into a pipe with a 2 in inside diameter. Treating the water as incompressible, what is the average velocity in the smaller pipe?

  • A. 3 ft/s
  • B. 12 ft/s
  • C. 24 ft/s
  • D. 1.5 ft/s
Show answer and worked solution

Answer: C. 24 ft/s

Continuity: A₁v₁ = A₂v₂. Area goes with diameter squared, so v₂ = v₁(D₁/D₂)² = 6 × (4/2)² = 24 ft/s.

The diameters are both in inches, so the units cancel in the ratio.

  • B (12) uses the diameter ratio instead of the area ratio.
  • A (3) and D (1.5) apply the ratio backward. The fluid speeds up in a smaller pipe.

Relation: A₁v₁ = A₂v₂. Handbook search words to try: continuity, flow rate.

Source: OpenStax University Physics Vol. 1, §14.5 Fluid Dynamics (Q = A v̄ explanation and Eq. 14.14, incompressible continuity).

Question 46 — Comparing two layouts

FEIS-046 · Facilities and Supply Chain · Choose one answer

Daily material trips: A→B 30, A→C 10, B→C 20. Layout 1 distances: A–B 20 m, A–C 40 m, B–C 20 m. Layout 2 distances: A–B 20 m, A–C 20 m, B–C 40 m. Which layout has the lower total load-distance, and what is it?

  • A. Layout 2, 1,600 trip·m per day
  • B. Layout 1, 1,600 trip·m per day
  • C. They're equal, because the distances add to the same total
  • D. Layout 1, 1,400 trip·m per day
Show answer and worked solution

Answer: D. Layout 1, 1,400 trip·m per day

Layout 1: 30(20) + 10(40) + 20(20) = 600 + 400 + 400 = 1,400.

Layout 2: 30(20) + 10(20) + 20(40) = 600 + 200 + 800 = 1,600.

Layout 1 wins because it puts the heavier B–C flow on the shorter distance.

  • A picks the layout with the higher total.
  • B pairs Layout 1 with Layout 2's total.
  • C ignores the number of trips. Matching total distances doesn't mean matching load-distance.

Relation: Load-distance = Σ(trips × distance). Handbook search words to try: layout, from-to chart.

Source: J. E. Beasley, OR-Notes: Facility Location (Brunel University) (load-distance score).

Question 47 — Top event in a fault tree

FEIS-047 · Systems Engineering, Analysis, and Design · Choose one answer

In a fault tree, the top event occurs if A occurs OR if both B AND C occur. The basic events are independent: P(A) = 0.01, P(B) = 0.05, and P(C) = 0.10. What is the probability of the top event?

  • A. 0.06
  • B. 0.16
  • C. 0.00005
  • D. about 0.0150
Show answer and worked solution

Answer: D. about 0.0150

AND gate (independent events): P(B and C) = 0.05 × 0.10 = 0.005.

OR gate: P(top) = 1 − (1 − 0.01)(1 − 0.005) = 0.01495, about 0.0150.

For small probabilities, simply adding them (0.01 + 0.005 = 0.015) is a close upper-bound approximation.

  • A (0.06) adds P(A) and P(B) and ignores C.
  • B (0.16) adds all three event probabilities, ignoring both the B-AND-C requirement and overlap. Even an all-OR calculation would be 1 − (0.99)(0.95)(0.90) = 0.15355.
  • C (0.00005) treats every event as feeding an AND gate.

Relation: For independent events, AND: P = ΠPᵢ; OR: P = 1 − Π(1 − Pᵢ). Handbook search words to try: fault tree, probability.

Source: NASA, Fault Tree Handbook with Aerospace Applications, Version 1.1 (August 2002) (§6.1 printed pp. 72–76; Appendix B Eq. B.18, printed p. 175); OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability (multiplication rule for independent events; addition rule).

Question 48 — F statistic for three methods

FEIS-048 · Probability and Statistics · Choose one answer

A one-way ANOVA compares 3 assembly methods with 5 observations each. The treatment sum of squares is 40, and the error sum of squares is 60. What is the F statistic?

  • A. 0.67
  • B. 3.33
  • C. 4.00
  • D. 4.67
Show answer and worked solution

Answer: C. 4.00

Treatment df = k − 1 = 2. Error df = N − k = 15 − 3 = 12.

MS_treatment = 40/2 = 20. MS_error = 60/12 = 5.

F = 20/5 = 4.00.

  • A (0.67) divides the sums of squares without converting to mean squares.
  • B (3.33) uses k and N as the degrees of freedom instead of k − 1 and N − k.
  • D (4.67) uses N − 1 = 14 for the error degrees of freedom.

Relation: F = MS_treatment/MS_error; df = (k − 1, N − k). Handbook search words to try: ANOVA, F test.

Source: NIST/SEMATECH e-Handbook, §7.4.3.3 The ANOVA table (F = MST/MSE and degrees of freedom).

Question 49 — Monthly deposits

FEIS-049 · Engineering Economics · Choose one answer

$500 is deposited at the end of every month for 3 years. Interest is a nominal annual rate of 6%, compounded monthly. What is the account worth at the end of 3 years, to the nearest dollar?

  • A. $18,000
  • B. $19,102
  • C. $19,668
  • D. $19,080
Show answer and worked solution

Answer: C. $19,668

Monthly rate i = 6%/12 = 0.5%. Number of deposits n = 36.

F = 500 × [(1.005³⁶ − 1)/0.005] = 500 × 39.336 = $19,668.

Match the interest period to the payment period before using a factor.

  • A ($18,000) ignores interest.
  • B ($19,102) treats the deposits as $6,000 per year at 6% annual interest. That ignores interest earned within each year.
  • D ($19,080) adds 6% simple interest to the total once.

Relation: F = A[(1 + i)ⁿ − 1]/i with i and n per month. Handbook search words to try: future worth, uniform series.

Source: Penn State EME 460, Compound Interest Formulas II (Eq. 1-3: uniform-series compound-amount factor (F/A) and end-of-period payments).

Question 50 — Matching quality tools

FEIS-050 · Quality · Matching

Match each quality tool to its main use.

Question 50 — Matching quality tools
ToolMain use
House of quality (QFD)?
Cause-and-effect (fishbone) diagram?
Pareto chart?
Control chart?

Uses to place: Rank problem categories from most to least frequent to find the vital few · Relate customer requirements to technical design characteristics · Plot process data over time against a center line and control limits · Sort possible causes of a problem into categories

Counts as correct only if all four matches are right.

Show answer and worked solution

Answer:

House of quality (QFD) links what customers want to how the design will deliver it.

Fishbone diagram organizes brainstormed possible causes into categories.

Pareto chart sorts counted problem categories from largest to smallest.

Control chart tracks a process over time against a center line and upper and lower control limits.

  • A common slip is swapping fishbone and Pareto. The fishbone generates possible causes. The Pareto chart ranks problems you've already counted.

Relation: Basic quality tools and QFD. Handbook search words to try: quality tools, QFD.

Source: ASQ, House of Quality (definition); ASQ, Fishbone Diagram (definition); ASQ, Pareto Chart (definition); ASQ, Control Chart (definition).

Score yourself

Give yourself one point for each question you got completely right. Question 24 and Question 25 count only if you picked every correct option and nothing extra, and Question 50 counts only if all four matches are right. That's how NCEES scores its own multiple-correct and drag-and-drop items: right or wrong, no partial credit (Examinee Guide, p. 11).

Keep three counts: correct, incorrect, and unanswered. They should add up to 50. Separately mark questions as reviewed if you opened the solution before answering. A score out of 50 means the most when you answered every question before opening any solution.

Score yourself
NCEES knowledge areaQuestions on the real examQuestions hereYour score
1. Mathematics6–9Q11, Q26, Q44___ / 3
2. Engineering Sciences4–6Q20, Q45___ / 2
3. Ethics and Professional Practice4–6Q12, Q33___ / 2
4. Engineering Economics9–14Q3, Q16, Q25, Q38, Q49___ / 5
5. Probability and Statistics10–15Q2, Q15, Q27, Q37, Q48___ / 5
6. Modeling and Quantitative Analysis9–14Q6, Q14, Q28, Q39___ / 4
7. Engineering Management8–12Q7, Q17, Q29, Q40___ / 4
8. Manufacturing, Service, and Other Production Systems9–14Q1, Q5, Q21, Q30, Q41___ / 5
9. Facilities and Supply Chain9–14Q13, Q22, Q34, Q46___ / 4
10. Human Factors, Ergonomics, and Safety8–12Q9, Q19, Q32, Q43___ / 4
11. Work Design7–11Q10, Q23, Q35___ / 3
12. Quality9–14Q8, Q18, Q31, Q42, Q50___ / 5
13. Systems Engineering, Analysis, and Design8–12Q4, Q24, Q36, Q47___ / 4
Total11050___ / 50

Official ranges come from the FE Industrial and Systems specification, pp. 1–3. They're ranges for each area, not percentages, and their endpoints don't add up to 110. We gave each area a share of the 50 questions roughly in proportion to the middle of its range, with at least two per area.

What your score means

It's your result on these 50 original questions, nothing more. NCEES counts correct answers on scored items, converts that count to a scaled score, and reports pass or fail. It doesn't publish a passing score (Examinee Guide, p. 14), so no percentage here can be turned into "ready" or "not ready." With two to five questions per area, your area scores point you somewhere to look. They don't measure your strengths.

Answer key

Show answer key
ASWB exam resource table
QuestionAnswer
1B
2C
3A
4C
5C
616 minutes
7D
8A
9D
10A
11B
12C
13B
14C
15A
16D
17A
18A
19A
20D
211,000 units
22A
23B
24A, D
25B, D, E
26B
27B
28D
29B
30350 units
31D
32A
33C
34B
35385 observations
36B
37A
38D
39A
40A
41C
42B
43C
44C
45C
46D
47D
48C
49C
50QFD → customer requirements to design characteristics; fishbone → sort possible causes; Pareto → rank by frequency; control chart → data over time vs. control limits

Turn misses into a review list

For every question you missed, guessed, or opened early, fill in one row. Name the real cause, then fix that cause, not just the one question.

Turn misses into a review list
QuestionWhat went wrongCorrected setup (equation and units)Next action
e.g., Q10Definition: used the wrong allowance basisST = NT(1 + A) when A is a share of normal timeRework two time-study problems, one with each allowance basis
Concept · Setup · Units · Arithmetic · Definition · Handbook lookup
Turn misses into a review list
CauseWhat fixes it
ConceptExplain the governing idea in your own words and when it applies, then solve a fresh example.
SetupWrite the unknown, the givens, and the model's assumptions before touching a number.
UnitsPut every rate and time on the same basis (per hour vs. per minute, per year vs. per month) before calculating.
ArithmeticRecompute at full precision and round only the final answer.
DefinitionUnderline the words that change the model: unit vs. cumulative curve, allowance basis, time in queue vs. in system, independent, select all.
Handbook lookupPractice finding the topic in the handbook and reading its variable definitions, without solving anything.

Then work in this order:

  1. Start with the bigger areas. Probability and Statistics carries 10–15 questions on the real exam; Engineering Sciences carries 4–6. When two gaps look about the same, work on the one from the bigger area first. A two-question sample can't tell you which area will cost you more points.
  2. Rebuild each missed setup with the handbook. On exam day you search the handbook with a search box, and Ctrl+F doesn't work (Examinee Guide, p. 10). Use the search words under each question. If one finds nothing, try another.
  3. Watch for definition traps. This set has several on purpose: Q6 (time in queue vs. in system), Q10 (allowance basis), Q19 (PEL vs. hearing-conservation sampling threshold), Q23 (unit learning curve), Q39 (F⁻¹(U) vs. a shortcut generator), and Q43 (one-sided vs. two-sided z).
  4. Retry without the solution a few days later. If you can set it up cold, try a fresh problem on the same concept.

How close is this to the real FE Industrial and Systems exam?

The set is unofficial and shorter than the real exam, and your score here doesn't predict a pass.

  • Topics. The real exam has 110 questions across the 13 knowledge areas in the NCEES specification, in effect since the July 2020 exams. This set touches every area but can't cover every subtopic, and NCEES's difficulty mix is its own.
  • Question types. Most real questions are single-answer multiple choice. NCEES also uses multiple-correct, point-and-click, drag-and-drop, and fill-in-the-blank items (Examinee Guide, p. 11). This set has 43 single-answer questions, 4 numerical-entry questions (Q6, Q21, Q30, Q35), 2 select-all questions (Q24, Q25), and 1 matching question (Q50). It has no point-and-click item.
  • Units. The exam uses both SI and U.S. customary units. Questions 20 and 45 use U.S. customary units; the rest use SI or plain counts and dollars.
  • Unscored items. NCEES includes a limited number of unscored pretest items that you can't identify (Examinee Guide, p. 11). Every question here counts toward your practice tally.

Exam-day facts that change how you practice

Exam-day facts that change how you practice
WhatWhat NCEES says
Length110 questions in 5 hours 20 minutes of exam time. The appointment adds a 2-minute nondisclosure agreement, an 8-minute tutorial, and an optional 25-minute scheduled break, for 5 hours 55 minutes (Examinee Guide, p. 16). The FE exam page describes the appointment as 6 hours.
Two sectionsYou get all the exam time at the start. After about half the questions you review and submit them, and you can't go back to them (p. 11).
ReferenceThe current FE Reference Handbook appears on screen as a searchable PDF. Search with the box on the left; Ctrl+F isn't available (p. 10). You can download the same handbook free through MyNCEES to practice with.
CalculatorBring one NCEES-approved model. An on-screen TI-30XS is also available (Examinee Guide, p. 8).
Scratch workThe test center gives you 2 reusable booklets and 3 markers (p. 9).
ScoringYour score is based on correct answers, with no deduction for wrong ones (p. 14). Never leave a question blank.

For booking, fees, results, and retakes, see our guides to NCEES exam registration, NCEES exam results, and the NCEES retake policy. Your licensing board decides whether you can sit for the exam, and passing the FE isn't a PE license (Examinee Guide, pp. 2 and 24; FE exam overview).

Is a new FE Industrial and Systems specification coming?

The July 2020 specification is the current one. In 2026, NCEES opened a content review of the FE Industrial and Systems exam and surveyed engineers and faculty through April 27, 2026. Check the NCEES FE exam page again before your test date.

How to get the most from this set

Have a calculator, scratch paper, and the free FE Reference Handbook open. On exam day that handbook is your only reference, so look things up there instead of in your notes.

  • First pass: untimed. Work each question, then read the whole solution, including on questions you got right. The wrong options show the setup, unit, and definition mistakes the solutions are built around.
  • Second pass: timed. The real exam averages about 2.9 minutes per question (320 minutes ÷ 110). At that pace, 50 questions take about 2 hours 25 minutes. That's our pacing suggestion, not an NCEES rule, and a second attempt measures how well you reviewed, not a fresh start.
  • Need the bigger picture? Our FE exam prep guide has the Industrial and Systems subject map and a 12-week study plan.

Use original practice questions, not shared or recalled exam content. NCEES treats disclosing nonpublic exam questions or answers, or NCEES practice exam questions and answers, as an exam irregularity that can lead to invalidated results (Examinee Guide, p. 13).

Sources

Exam facts on this page were checked against NCEES sources on October 7, 2026. Each solution links the source for its underlying principle; the scenario numbers are invented for practice.

All 50 questions are original practice items written for this page. We checked them with an AI-assisted source and calculation review: every numerical answer and distractor was recalculated, and each principle was checked against the linked source. That isn't a review by a licensed engineer. See our methodology and editorial standards. Found an error? Report a correction.

Castleport Test Prep is an independent exam prep publisher and is not affiliated with, endorsed by, or approved by NCEES (National Council of Examiners for Engineering and Surveying). These practice questions are original and unofficial; they are not real or recalled exam items. Exam and credential names identify the subjects discussed, and trademarks belong to their respective owners. Practice results don't guarantee passing the exam or licensure.

By the Castleport Test Prep Editorial Team · Last verified October 7, 2026

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