Free PE Industrial and Systems Practice Test

Answer: A. 6.0 days; 0.17%

Inventory in days of demand: 600 ÷ 150 = 4.0 days and 300 ÷ 150 = 2.0 days, so 6.0 working days of waiting.

One working day = 2 × 8 × 60 = 960 minutes. Lead time = 6.0 × 960 + 10 = 5,770 minutes (the 10 is the value-added time: 2 + 5 + 3).

Process cycle efficiency = 10 ÷ 5,770 = 0.17%.

Why the others miss:

• B: Uses 24-hour calendar days (6 × 1,440 + 10 = 8,650 min). The stem asks for working minutes.
• C: Counts only the 600-unit queue (4 × 960 + 10 = 3,850 min).
• D: Uses one 8-hour shift per day (6 × 480 + 10 = 2,890 min).

If you missed it: Write the time basis next to every number before you divide. Most lead-time errors are a days-versus-minutes slip.

Source: Derived calculation; method and arithmetic shown above

Answer: B. 3

Sort largest to smallest: scratches 120, dents 80, misaligned label 50, wrong color 30, loose cap 12, other 8.

Cumulative share: 120 ÷ 300 = 40.0%, then 200 ÷ 300 = 66.7%, then 250 ÷ 300 = 83.3% after three categories.

Why the others miss:

• A: Two categories reach only 66.7%.
• C: Adds wrong color after 80% is already reached (93.3%). The question asks for the fewest.
• D: Accumulates in the order the categories were listed, without sorting first. That takes five categories to pass 80%.

If you missed it: A Pareto chart only works after you sort. Rank first, then accumulate.

Source: Derived calculation; method and arithmetic shown above

Answer: A. Nominal group technique

Silent idea generation, round-robin sharing without discussion, clarification, and then voting to prioritize are the steps of the nominal group technique.

Why the others miss:

• B: An affinity diagram organizes a large number of ideas into their natural groupings. It sorts ideas; it doesn't rank them by vote.
• C: A Pareto analysis ranks categories by frequency or cost from data, not by a group vote.
• D: Value stream mapping traces material and information flow through a process.

If you missed it: The silent-writing step plus a ranking vote is the signature of nominal group technique.

Source: NIATx, Nominal Group Technique Instructions (steps 1–4); ASQ, What is an affinity diagram?

Answer: A. Quick-release fixtures, 51

Quick-release fixtures: 5(9) + 4(0) + 2(3) = 45 + 0 + 6 = 51.

Barcode verification: 5(1) + 4(9) + 2(3) = 5 + 36 + 6 = 47.

Visual work instructions: 5(3) + 4(3) + 2(9) = 15 + 12 + 18 = 45.

Why the others miss:

• B: Barcode verification's 47 is correct, but it is second.
• C: 15 is the unweighted sum of relationship scores (3 + 3 + 9). Unweighted sums ignore what customers care about most.
• D: Fixtures lead by 4 points. There is no tie.

If you missed it: Multiply each relationship by its row weight before adding down the column.

Source: Derived calculation; method and arithmetic shown above; ASQ, What is quality function deployment?

Answer: A, C, and E

A (space), C (budget), and E (an interface the design must use) limit the solution space. They are constraints.

B and D describe what the system must do. They are functional requirements.

Why the other choices are wrong:

• B: Scanning each pick is a function the system performs.
• D: Printing a packing list is a function the system performs.

If you missed it: Ask whether the statement describes an action the system takes (functional) or a boundary on any design you could choose (constraint). All-or-nothing scoring: you need A, C, and E, and nothing else.

Source: Definitions supplied in the stem; classification follows directly from them

Answer: A. 76.0%

Availability = run time ÷ planned time = (450 − 45) ÷ 450 = 405 ÷ 450 = 0.900.

Performance = (ideal cycle × total count) ÷ run time = (0.5 × 720) ÷ 405 = 0.889.

Quality = good ÷ total = 684 ÷ 720 = 0.950.

OEE = 0.900 × 0.889 × 0.950 = 0.760. Check: good units × ideal cycle ÷ planned time = 684 × 0.5 ÷ 450 = 0.760.

Why the others miss:

• B: Averages the three factors instead of multiplying them.
• C: Multiplies availability and performance but leaves out quality.
• D: Uses the full 480-minute shift as planned time. Scheduled breaks aren't planned production time.

If you missed it: The shortcut check (good units × ideal cycle ÷ planned time) catches most OEE slips in seconds.

Source: Derived calculation; method and arithmetic shown above

Answer: A. It rises by $120

30 extra hours is within the allowable increase of 50, so the $12 shadow price holds for every added hour.

Gain from the hours = 30 × $12 = $360. Extra overtime cost = 30 × $8 = $240.

Net change = $360 − $240 = +$120.

Why the others miss:

• B: Counts the value of the hours but not the extra $8 per hour.
• C: Counts the overtime cost but not the value the hours create.
• D: Re-solving is needed only when the change goes past the allowable range. 30 hours is inside it.

If you missed it: A shadow price is valid only inside the allowable range, and it already reflects costs in the model. Subtract only the costs the model doesn't include.

Source: Derived calculation; method and arithmetic shown above

Answer: 3.79 requests (accepted: 3.77 to 3.8)

Offered load a = λ/μ = 10/6 = 1.667. Utilization ρ = λ/(2μ) = 10/12 = 0.833.

P₀ = [1 + a + a²/(2(1 − ρ))]⁻¹ = [1 + 1.667 + 2.778/0.333]⁻¹ = [1 + 1.667 + 8.333]⁻¹ = 1/11 = 0.0909.

Lq = P₀ a² ρ / [2! (1 − ρ)²] = 0.0909 × 2.778 × 0.833 ÷ (2 × 0.0278) = 3.79.

Wq = Lq/λ = 0.379 h, about 22.7 minutes.

Common wrong answers:

• 4.17: Treats the crib as one fast server (M/M/1 with μ = 12/h). Two servers at 6/h aren't the same as one at 12/h, because one can sit idle while the other is busy.
• 0.83: Reports utilization, not the queue length.

If you missed it: For multi-server queues, compute P₀ first; everything else follows from it.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 9 minutes via S–A–C–T

Label nodes from S (Dijkstra): B = 2, D = 2 + 1 = 3, A = 4, C = min(4 + 3, 2 + 6, 3 + 6) = 7, T = min(7 + 2, 3 + 9) = 9.

Trace back: T from C, C from A, A from S, so the route is S–A–C–T.

Why the others miss:

• B: Follows the cheapest next link at every step (S→B, B→D, D→C, C→T). Greedy choices don't guarantee the shortest total.
• C: A feasible route, but 1 minute longer.
• D: A feasible route, but 3 minutes longer.

If you missed it: Keep a permanent label for each node and update neighbors before moving on. Don't commit to the cheapest single link.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 62

n = (z s / e)² = (1.96 × 12 / 3)² = (7.84)² = 61.5.

Round up to 62 so the half-width doesn't exceed ±3 minutes.

Why the others miss:

• B: Rounds 61.5 down, which falls just short of the precision you asked for.
• C: Forgets to square (7.84 rounded up).
• D: Uses (s/e)² without the z value.

If you missed it: Sample-size results always round up.

Source: Derived calculation; method and arithmetic shown above

Answer: A, C, and E

A, C, and E move material without interruption and measure output by volume or weight. They are continuous.

Why the other choices are wrong:

• B: Discrete: separate, countable vehicles.
• D: Batch: a fixed quantity is processed together, then the next load starts.
• F: A service process. Patients are discrete arrivals, not a continuous flow of material.

If you missed it: Ask how output is counted. Units counted one by one means discrete; separate loads means batch.

Source: Definition supplied in the stem

Answer: A. Demand volume

The decision flips only if the Make advantage can fall below $0.

Only demand volume's range crosses zero (−$20,000 at its low end). So demand volume can flip the decision.

Why the others miss:

• B: Material price has the widest bar, but its whole range stays positive. Width shows sensitivity, not whether the decision changes.
• C: Labor rate's range stays well above zero.
• D: Demand volume's range crosses zero.

If you missed it: On a tornado diagram, look for the bar that crosses the decision threshold, not just the longest bar.

Source: Derived calculation; method and arithmetic shown above

Answer: Q → T → S → P → R

Identify → Exploit → Subordinate → Elevate → Prevent inertia, then repeat.

Exploit (get the most from the constraint you already have) comes before Elevate (adding capacity), because the cheapest improvement is usually using the constraint better.

If you missed it: Exploit and subordinate cost little; elevate usually costs money. That's why elevate comes fourth. Drag-and-drop ordering is scored all-or-nothing.

Source: TOC Institute, The Five Focusing Steps

Answer: A. 48

Good units leaving Station 1 = 60 × 0.80 = 48 per hour.

Station 2 could handle 50 but only receives 48, and Station 3 can handle 55. Output = 48 good units per hour. Station 1 plus its scrap loss is the real constraint.

Why the others miss:

• B: Station 2's nominal capacity. It never gets 50 units an hour to work on.
• C: Applies the 20% scrap to Station 2's capacity instead of to Station 1's output.
• D: Station 3 is the second-fastest station, not the limit.

If you missed it: Convert every station to good-unit capacity at the same point in the flow before you pick the bottleneck.

Source: Derived calculation; method and arithmetic shown above

Answer: A. Gasket

Value indexes: motor 0.80, housing 0.72, gasket 0.33, fastener kit 0.83.

The gasket costs three times what its function is worth, so it ranks first.

Why the others miss:

• B: The motor has the largest dollar gap ($8) but a value index of 0.80. The team ranks by index, not by dollars.
• C: 0.72 is the second-lowest index.
• D: 0.83 is the highest index, closest to good value.

If you missed it: Value engineering targets cost that buys no essential function. A low worth-to-cost ratio flags that waste.

Source: Derived calculation; method and arithmetic shown above; FAR 48.101(a), value engineering

Answer: A. $800

Paths: A–C–E = 11 days (critical) and B–D–E = 10 days. Target 9 days.

Day 1: crash C, the cheapest critical activity ($200). Now both paths are 10 days.

Day 2: both paths must drop. E is on both, at $600; the cheapest pair is A + B at $800. Crash E.

Total = $200 + $600 = $800.

Why the others miss:

• B: Crashes C twice. C can only be crashed 1 day.
• C: Crashes C and then A. That shortens A–C–E to 9 days, but B–D–E is still 10, so the project only drops 1 day.
• D: Crashes E and A. That works, but costs $300 more than crashing C and E.

If you missed it: After every day of crashing, recheck which paths are critical. A second path often joins the critical set.

Source: Derived calculation; method and arithmetic shown above

Answer: A. CPI 0.80, SPI 0.875: over budget and behind schedule

CPI = EV ÷ AC = 140,000 ÷ 175,000 = 0.80. Below 1.0 means the work done cost more than planned.

SPI = EV ÷ PV = 140,000 ÷ 160,000 = 0.875. Below 1.0 means less work was done than planned.

Why the others miss:

• B: Inverts both ratios (AC ÷ EV and PV ÷ EV).
• C: Swaps the two indices.
• D: Computes 'CPI' as PV ÷ AC. Earned value, not planned value, goes in both numerators.

If you missed it: EV is in the numerator of both indices. Cost compares to actual cost; schedule compares to planned value.

Source: CDC Unified Process, Earned Value Practices Guide, p. 2 (CPI = EV/AC, SPI = EV/PV) (source link unavailable)

Answer: A. Layout 1, 4,250 load-ft per day

Layout 1: 30(50) + 10(100) + 25(50) + 5(100) = 1,500 + 1,000 + 1,250 + 500 = 4,250.

Layout 2: 30(100) + 10(50) + 25(50) + 5(50) = 3,000 + 500 + 1,250 + 250 = 5,000.

Why the others miss:

• B: Layout 2's total is correct, but it is higher.
• C: Adds distances without weighting them by flow.
• D: Leaves out the C→A backflow of 5 loads.

If you missed it: Every flow, in both directions, gets multiplied by its distance.

Source: Derived calculation; method and arithmetic shown above

The cited CDC earned-value guide is currently unavailable at its original address. The equations and calculations are shown in the worked solution above.

Answer: 1 → a, 2 → b, 3 → c, 4 → d

Buying the equipment is acquisition; getting it running and staff trained is implementation; keeping it running year after year is sustainment; taking it out of service is retirement.

If you missed it: Life-cycle cost analysis exists because the purchase price is only one of the four phases. Matching items are scored all-or-nothing.

Source: Phase names from the NCEES specification, item 2H

Answer: B. Layout 2, score 14

Layout 1 adjacencies: Receiving–Stores 16, Machining–Office −32, Receiving–Machining 2, Stores–Office 4 → −10.

Layout 2 adjacencies: Machining–Stores 8, Receiving–Office 0, Machining–Receiving 2, Stores–Office 4 → 14.

Layout 3 adjacencies: Stores–Machining 8, Receiving–Office 0, Stores–Receiving 16, Machining–Office −32 → −8.

In a 2 × 2 block, keeping Machining and Office apart forces them onto a diagonal, which puts Receiving and Stores on the other diagonal. Giving up the A pair beats paying the X penalty.

Why the others miss:

• A: Layout 1 gets the A pair but also puts the X pair side by side.
• C: Layout 3 also puts the X pair side by side.
• D: Layout 3 also puts the A pair side by side, so the premise is false. And an A adjacency earns 16 points while an X adjacency costs 32; the total score decides, not the single best pair.

If you missed it: Check the X pairs first. An undesirable adjacency usually costs more than any single good one earns.

Source: Derived calculation; method and arithmetic shown above

Answer: A. (4, 3)

For rectilinear distance, use the weighted median of each coordinate. Total weight = 14, so the median is where cumulative weight first reaches more than 7.

x sorted: 1 (4), 4 (cumulative 9) → x = 4.

y sorted: 1 (5), 3 (cumulative 8) → y = 3.

Optimal location: (4, 3).

Why the others miss:

• B: The weighted center of gravity. It is a reasonable start for straight-line travel, but it doesn't minimize rectilinear distance.
• C: The unweighted average of the coordinates.
• D: Swaps x and y.

If you missed it: Rectilinear plus weights means weighted median, solved separately for x and y.

Source: Derived calculation; method and arithmetic shown above

Answer: 2,118 positions

Positions = 1,800 ÷ 0.85 = 2,117.6.

Round up: 2,118 positions.

Common wrong answers:

• 1530: Multiplies by 0.85 instead of dividing. Planning for partial fill needs more positions than pallets, not fewer.
• 2117: Rounds down and leaves the plan short by a fraction of a position.

If you missed it: A utilization factor below 1 always increases the space you need.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 90 ft²

Little's law: WIP = throughput × time in system = 90 parts/h × (40/60) h = 60 parts.

Racks = 60 ÷ 12 = 5. Floor space = 5 × 18 = 90 ft².

Why the others miss:

• B: Multiplies 90 parts per hour by 40 minutes without converting minutes to hours.
• C: Gives every part its own 18 ft² instead of 12 parts per rack.
• D: Reports the WIP count (60 parts) as square feet.

If you missed it: Match the time units in Little's law before multiplying.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 14

Units to start = 4,200 ÷ 0.95 = 4,421.1. Hours needed = 4,421.1 × 6 ÷ 60 = 442.1 h.

Effective hours per operator present = 40 × 0.90 = 36 h. Operators present = 442.1 ÷ 36 = 12.28.

Staff for absenteeism: 12.28 ÷ (1 − 0.08) = 13.35. Round up to 14.

Why the others miss:

• B: Rounds up 12.28 but forgets absenteeism.
• C: Rounds 12.28 down and forgets absenteeism.
• D: Multiplies by the 90% efficiency instead of dividing by it.

If you missed it: Scrap, efficiency, and absenteeism all push the requirement up. If an adjustment lowers your answer, its direction is wrong.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 5

Load = 50 × (45 + 120 × 2.5) = 50 × 345 = 17,250 min = 287.5 h.

Capacity per machine = 80 × 0.85 = 68 h. Machines = 287.5 ÷ 68 = 4.23. Round up: 5.

Why the others miss:

• B: Either ignores setups (250 h ÷ 68 = 3.68 → 4) or rounds 4.23 down.
• C: Ignores setups and availability (250 ÷ 80 = 3.1) and rounds down.
• D: Uses 40 available hours instead of 80 (287.5 ÷ 34 = 8.46 → 9).

If you missed it: Setup time is real machine time. Add it per batch, not per unit.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 1.50; justified

Conventional B/C = (benefits − disbenefits) ÷ costs = (180,000 − 30,000) ÷ 100,000 = 1.50.

B/C > 1.0, so the project is justified.

Why the others miss:

• B: Adds disbenefits to costs (180 ÷ 130). That's a different ratio from the one asked for, even though the decision happens to match here.
• C: Ignores the disbenefits.
• D: Inverts the ratio (cost ÷ net benefit).

If you missed it: Read which convention the question names. The decision often matches across conventions, but the number doesn't.

Source: Derived calculation; method and arithmetic shown above

Answer: B. Site P, 78

Screen first: R fails the rail must-have and is out.

P: 0.5(80) + 0.3(70) + 0.2(85) = 40 + 21 + 17 = 78. Q: 35 + 24 + 15 = 74.

Choose Site P.

Why the others miss:

• A: R scores highest (85.5) but fails a must-have, so its score doesn't matter.
• C: Q's 74 is correct but lower than P's 78.
• D: Adds R's unweighted scores, and R is still disqualified.

If you missed it: Apply knockout criteria before you weight and score anything.

Source: Derived calculation; method and arithmetic shown above

Answer: A. $425

Optimal plan: DC1 → C2 35 and C3 25; DC2 → C1 30 and C2 10.

Cost = 35(6) + 25(3) + 30(3) + 10(5) = 210 + 75 + 90 + 50 = $425.

Why it's optimal: each customer's cheaper source is used up to that DC's capacity. DC2 is $1 cheaper than DC1 for both C1 and C2, but $4 more expensive for C3, so DC2's 40 units go to C1 and C2.

Why the others miss:

• B: The northwest-corner starting plan (DC1 → C1 30, C2 30; DC2 → C2 15, C3 25). It's feasible but not optimized.
• C: Sends each customer's full demand from its cheapest DC, which needs 75 units from DC2. DC2 can only ship 40.
• D: A feasible plan (DC1 → C2 45, C3 15; DC2 → C1 30, C3 10) that sends 10 units on the expensive DC2→C3 lane.

If you missed it: A starting solution isn't an optimal one. Check whether shifting units between lanes lowers cost.

Source: Derived calculation; method and arithmetic shown above

Answer: 25 pallets

Floor space: turn the 48 in side across the width (48 + 48 = 96 in ≤ 100 in), so 2 pallets per row. Rows = 624 in ÷ 40 in = 15.6 → 15 rows → 30 pallets.

Weight: 44,000 ÷ 1,750 = 25.1 → 25 pallets.

Weight governs: 25 pallets.

Common wrong answers:

• 30: Checks floor space only. Weight runs out first.
• 26: Uses the other orientation for floor space and ignores weight.

If you missed it: A unit-load plan checks both cube and weight. The smaller limit wins.

Source: Derived calculation; method and arithmetic shown above

Answer: A. Y, about $356,700

X: 250,000 + 18,000(5.3349) + 40,000(0.4665) = 250,000 + 96,028 + 18,660 = $364,700.

Y: 205,000 + 28,000(5.3349) + 5,000(0.4665) = 205,000 + 149,377 + 2,333 = $356,700.

Y costs about $8,000 less in present worth.

Why the others miss:

• B: Leaves out X's retirement cost. Dropping that one phase reverses the decision.
• C: X's present worth is correct, but it is the higher of the two.
• D: Adds the cash flows without discounting. They happen to tie undiscounted, which shows why timing matters.

If you missed it: Include all four life-cycle phases, and discount each to the same point in time.

Source: Derived calculation; method and arithmetic shown above

Answer: B and D

Each truck must pass three checks: capacity ≥ 3,000 lb, lift ≥ 20 ft, aisle ≤ 10 ft.

B passes all three. D passes all three (3,000 lb meets 3,000 lb exactly).

Why the other choices are wrong:

• A: Lifts only 18 ft and needs a 12 ft aisle.
• C: Doesn't lift to rack height.
• E: Capacity 2,500 lb and lift 16 ft are both short.

If you missed it: Screen every option against every requirement. One failure is enough to drop it.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 10

Time per delivery = 300/200 + 300/200 + 1.0 = 1.5 + 1.5 + 1.0 = 4.0 min.

Vehicle-minutes needed per hour = 120 × 4.0 = 480. Each AGV supplies 60 × 0.85 = 51.

480 ÷ 51 = 9.41. Round up: 10 AGVs.

Why the others miss:

• B: Leaves out the empty return trip (2.5 min per delivery).
• C: Ignores the 85% factor (480 ÷ 60).
• D: Rounds 9.41 down.

If you missed it: Every loaded trip has an empty trip back unless the stem says vehicles chain jobs.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 3.75 min

Expected inspections per good unit = 1 ÷ (1 − 0.20) = 1.25 (a geometric series: 1 + 0.2 + 0.2² + …).

Expected inspection time = 1.25 × 3 = 3.75 min.

Why the others miss:

• B: Adds one round of rework (1.2 inspections) but ignores units that fail a second time.
• C: Ignores rework entirely.
• D: Divides by the failure rate instead of the pass rate.

If you missed it: With repeated independent trials until success, expected trials = 1 ÷ (probability of success).

Source: Derived calculation; method and arithmetic shown above

Answer: A. Site 1; 50,000 units

At 40,000 units: Site 1 = 300,000 + 12(40,000) = $780,000; Site 2 = 500,000 + 8(40,000) = $820,000. Site 1 is cheaper.

Break-even: 300,000 + 12Q = 500,000 + 8Q → 4Q = 200,000 → Q = 50,000 units. Above that, Site 2's lower variable cost wins.

Why the others miss:

• B: Correct crossover, but Site 2 is cheaper only above 50,000 units.
• C: Divides the fixed-cost difference by Site 2's variable cost ($8) instead of the variable-cost difference ($4).
• D: Divides the fixed-cost difference by Site 1's variable cost ($12).

If you missed it: Crossover volume = difference in fixed cost ÷ difference in variable cost.

Source: Derived calculation; method and arithmetic shown above

Answer: A. About 5.0; outside the limits, and the forecast runs too low

Running sum of errors = 12 + 8 − 4 + 10 + 6 + 9 = 41.

MAD = (12 + 8 + 4 + 10 + 6 + 9) ÷ 6 = 49 ÷ 6 = 8.17.

Tracking signal = 41 ÷ 8.17 = 5.0, outside ±4.

Errors are actual − forecast and mostly positive, so actual demand keeps beating the forecast: the forecast runs too low.

Why the others miss:

• B: Right value, wrong direction. Positive actual − forecast errors mean the forecast is too low.
• C: Divides by the sum of absolute errors (49) instead of their average.
• D: Reports the running sum as the tracking signal.

If you missed it: Check how the error is defined before reading the sign.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 40.3

t̄ = 3, ȳ = 28. Deviations of t: −2, −1, 0, 1, 2. Deviations of y: −8, −4, −1, 5, 8.

Slope b = Σ(t − t̄)(y − ȳ) ÷ Σ(t − t̄)² = (16 + 4 + 0 + 5 + 16) ÷ 10 = 4.1.

Intercept a = 28 − 4.1(3) = 15.7. Forecast for period 6 = 15.7 + 4.1(6) = 40.3.

Why the others miss:

• B: Extends the average change from first to last period ((36 − 20) ÷ 4 = 4 per period). That isn't least squares.
• C: The mean of the data, with no trend.
• D: Forecasts period 7 instead of period 6.

If you missed it: Center the time index first; the arithmetic gets much shorter.

Source: Derived calculation; method and arithmetic shown above

Answer: 1,450 units

Seasonalized forecast = deseasonalized forecast × seasonal index = 1,250 × 1.16 = 1,450 units.

Common wrong answers:

• 1078: Divides by the index. You divide to remove seasonality from data, and multiply to put it back into a forecast.

If you missed it: An index above 1 means a stronger-than-average season, so the forecast should rise.

Source: Derived calculation; method and arithmetic shown above

Answer: A. Level; $4,000 vs. chase $100,000

Level: 50 workers make 1,000 units each month. Ending inventory: 400, 400, 0, 0. Holding = (400 + 400) × $5 = $4,000. No hires or layoffs.

Chase: workers needed 30, 50, 70, 50. Changes from 50: lay off 20, hire 20, hire 20, lay off 20. Cost = 40 × $1,000 + 40 × $1,500 = $100,000.

The level plan is cheaper by $96,000.

Why the others miss:

• B: Counts the chase plan's zero inventory but leaves out its hiring and layoff costs.
• C: Counts the chase plan's hiring but not its layoffs.
• D: Counts the chase plan's layoffs but not its hiring.

If you missed it: List every cost the plan triggers each month before comparing totals.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 170 units in week 3

Net requirement in week 5 = 300 − 80 on hand − 50 scheduled receipt = 170.

Lot-for-lot orders exactly 170. Offset by the 2-week lead time: release 170 units in week 3.

Why the others miss:

• B: Places the order in the week it's needed, with no lead-time offset.
• C: Ignores the scheduled receipt.
• D: Ignores the on-hand inventory.

If you missed it: Net first (subtract on hand and scheduled receipts), then size the lot, then offset by lead time.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 2,309 units

EPQ = √[2DS ÷ (H(1 − D/P))] = √[2(24,000)(400) ÷ (6 × (1 − 0.4))] = √(19,200,000 ÷ 3.6) = √5,333,333 = 2,309 units.

Because inventory builds gradually while production runs, average inventory is lower than with instant delivery, so the lot is larger than the EOQ.

Why the others miss:

• B: The basic EOQ, which assumes the whole lot arrives at once.
• C: Multiplies by (1 − D/P) instead of dividing.
• D: Drops the 2 under the square root.

If you missed it: When production feeds inventory at a finite rate, maximum inventory is Q(1 − D/P), not Q.

Source: Derived calculation; method and arithmetic shown above

Answer: A. Y, because the incremental investment earns about 11.6%, above the MARR

Both projects beat 10% alone: IRR(X) ≈ 15.2%, IRR(Y) ≈ 13.9%.

Increment Y − X: $60,000 now for $16,500 per year. (P/A, i, 5) = 60,000 ÷ 16,500 = 3.636, so i ≈ 11.6%. That beats 10%, so the extra $60,000 is worth spending. Choose Y.

Check with present worth at 10%: X = 30,000(3.7908) − 100,000 = $13,724; Y = 46,500(3.7908) − 160,000 = $16,272. Y is higher.

Why the others miss:

• B: Picking the higher IRR ignores scale. The question is whether the extra investment earns the MARR.
• C: The incremental return (11.6%) is above the MARR, not below it.
• D: There's no 20% hurdle; the MARR is 10%.

If you missed it: For mutually exclusive choices, compare by incremental ROR or by present worth, never by ranking IRRs.

Source: Derived calculation; method and arithmetic shown above

Answer: A. B, about $18,180 per year

A: 40,000(0.30192) + 8,000 = $20,077 per year.

B: 70,000(0.17401) + 6,000 = $18,181 per year. B is cheaper by about $1,900 per year.

Why the others miss:

• B: Compares present worths over different lives. A's present worth covers only 4 years of service; B's covers 8.
• C: First cost alone ignores operating cost and how long each machine lasts.
• D: Spreads B's first cost with no interest (70,000 ÷ 8 + 6,000).

If you missed it: With repeatable unequal lives, compare annual worth, or use a common multiple of the lives.

Source: Derived calculation; method and arithmetic shown above

Answer: 64.0 $ per unit

Activity rates: $400 per setup, $60 per inspection, $20 per machine hour.

L's overhead = 30(400) + 200(60) + 400(20) = 12,000 + 12,000 + 8,000 = $32,000.

Per unit = 32,000 ÷ 500 = $64.00.

Common wrong answers:

• 32.80: Spreads all $410,000 of overhead on machine hours alone ($41 per hour × 400 ÷ 500). That's the traditional allocation, not ABC.

If you missed it: ABC charges each product for the activities it actually uses. Low-volume products often use lots of setups per unit.

Source: Derived calculation; method and arithmetic shown above

Answer: A. It undercosts L by $31.20 per unit

Traditional: 400 MH × $41 ÷ 500 units = $32.80 per unit.

ABC: [30(400) + 200(60) + 400(20)] ÷ 500 = 32,000 ÷ 500 = $64.00 per unit.

Traditional is lower by 64.00 − 32.80 = $31.20 per unit, so it undercosts L.

Why the others miss:

• B: Reverses the direction. The traditional figure is the lower one.
• C: The total is the same, but the split between products changes. That shift is the point of ABC.
• D: Reports the full ABC overhead as the difference.

If you missed it: Volume-based allocation shifts overhead from low-volume, setup-heavy products onto high-volume ones.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 2

Moore–Hodgson: start in due-date order and remove the longest job whenever one goes late.

J1 done 4 (on time), J2 done 7 (on time), J3 done 12 (late, due 9). Remove the longest so far, J3 (5 h); completion falls to 7.

J4 done 9 (on time), J5 done 15 (late, due 12). Remove the longest of J1, J2, J4, J5: J5 (6 h).

On-time sequence J1–J2–J4, then J3 and J5 late: 2 late jobs. Checking all 120 sequences confirms none does better.

Why the others miss:

• B: Earliest-due-date order alone (J3, J4, and J5 late). EDD minimizes maximum lateness, not the count of late jobs.
• C: Not achievable here: J1, J2, J3, and J4 can't all be on time, and adding J5 doesn't help.
• D: Total work is 20 hours against a latest due date of 12.

If you missed it: Match the rule to the goal: SPT for mean flow time, EDD for maximum lateness, Moore–Hodgson for number of late jobs.

Source: Derived calculation; method and arithmetic shown above

Answer: A. B–D–C–E–A; 30 hours

Pick the smallest remaining time. If it's on M1, schedule that job as early as possible; if on M2, as late as possible.

B (1, M1) first; A (2, M2) last; D (3, M1) second; E (4, M2) second-to-last; C in the middle: B–D–C–E–A.

M1 finishes: 1, 4, 13, 23, 28. M2 finishes: 7, 15, 22, 27, 30. Makespan = 30 hours, the minimum.

Why the others miss:

• B: Alphabetical order, with no rule applied.
• C: Sorts by shortest M1 time. That ignores M2.
• D: Applies the rule backward (short M1 times last).

If you missed it: Short on M1 goes early so M2 starts fast; short on M2 goes late so M2 finishes fast.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 1,000 units; about $97,460 per year

EOQ at $9.60 (H = $1.92) = √(2 × 10,000 × 50 ÷ 1.92) = 722, below the 1,000 break, so it isn't valid at that price.

EOQ at $10.00 (H = $2.00) = √(1,000,000 ÷ 2) = 707. Total = 100,000 + 707 + 707 = $101,414.

At the break, 1,000 units: 96,000 + (10 × 50) + (500 × 1.92) = 96,000 + 500 + 960 = $97,460. Order 1,000.

Why the others miss:

• B: The best plan without the discount, about $3,950 per year more expensive.
• C: 722 is below 1,000, so it can't get the $9.60 price.
• D: Ordering plus holding is lower at 707, but purchase cost dominates. Total cost decides.

If you missed it: With all-units discounts, compare total cost at each valid EOQ and at each price break.

Source: Derived calculation; method and arithmetic shown above

Answer: A. About 875

Underage cost Cu = 75 − 30 = $45 (profit lost per unit short). Overage cost Co = 30 − 10 = $20 (loss per unit left over).

Critical ratio = Cu ÷ (Cu + Co) = 45 ÷ 65 = 0.692, so z ≈ 0.50.

Q = 800 + 0.50 × 150 = about 875.

Why the others miss:

• B: Swaps Cu and Co (ratio 0.308, z ≈ −0.50).
• C: Orders the mean, which is right only when Cu = Co.
• D: Uses Cu ÷ price (45 ÷ 75 = 0.60) as the ratio.

If you missed it: Shortages here cost more than leftovers, so you should order above the mean.

Source: Derived calculation; method and arithmetic shown above

Answer: A, B, and C

Direct: 3 × 4 = 12 shipments × $520 = $6,240 per week.

Cross-dock: 3 × 900 + 4,800 × 0.25 + 4 × 350 = 2,700 + 1,200 + 1,400 = $5,300. Shipments: 3 inbound + 4 outbound = 7. Savings = $940.

At $0.40 handling: 2,700 + 1,920 + 1,400 = $6,020, still below $6,240.

At $0.50 handling: 2,700 + 2,400 + 1,400 = $6,500, now above $6,240.

Why the other choices are wrong:

• D: At $0.50 per carton, cross-docking costs $6,500, more than direct.
• E: 12 shipments × $520 = $6,240, not $5,040.

If you missed it: Cross-docking trades many small shipments for fewer full ones plus handling. The handling rate decides the break-even.

Source: Derived calculation; method and arithmetic shown above

Answer: A. Dedicated 1,400; randomized 900

Dedicated storage reserves positions for each SKU's own peak: 400 + 300 + 500 + 200 = 1,400.

Randomized storage lets any pallet use any open position, so it needs only the largest total on hand at once: 900.

Why the others miss:

• B: Dedicated slots can't be shared, so they must cover each SKU's separate peak.
• C: Randomized storage shares slots, so it never needs the sum of separate peaks.
• D: 350 is the average of the peaks, not the largest total on hand.

If you missed it: Dedicated: sum of maximums. Randomized: maximum of the sum.

Source: Derived calculation; method and arithmetic shown above

Answer: Q → P → R

In-transit cost per day = 20,000 × 500 × 0.25 ÷ 365 = $6,849.

Truck: 100,000 + 4 × 6,849 = $127,397. Rail: 60,000 + 12 × 6,849 = $142,192. Air: 240,000 + 6,849 = $246,849.

Lowest to highest: truck, rail, air.

If you missed it: Ranking by freight alone puts rail first. Slow transit ties up high-value inventory, and that cost can outweigh cheaper freight.

Source: Derived calculation; method and arithmetic shown above

Answer: A, B, and C

Cycle = 2.0 + 6.0 + 1.0 = 9.0 min.

Operator works 2.0 + 1.5 + 1.0 = 4.5 min, so idle 9.0 − 4.5 = 4.5 min (50% utilization).

Machine runs 6.0 of 9.0 min = 66.7%; it is stopped 3.0 min per cycle.

Why the other choices are wrong:

• D: Operator utilization is 4.5 ÷ 9.0 = 50%.
• E: The machine is stopped during loading and unloading, 3.0 min per cycle.

If you missed it: Draw the operator and machine columns side by side; idle time shows up as gaps.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 5 machines, about $12.65 per unit

Ideal number n′ = (load + run) ÷ (load + walk) = (1.5 + 10) ÷ (1.5 + 0.5) = 5.75. Compare 5 and 6.

n = 5: machines set the pace, so cycle = 11.5 min for 5 units. Cost = ($30 + 5 × $60) ÷ 60 × 11.5 ÷ 5 = $5.50/min × 2.3 = $12.65.

n = 6: the operator sets the pace, so cycle = 6 × 2.0 = 12 min for 6 units. Cost = $6.50/min × 2.0 = $13.00.

Why the others miss:

• B: The cost for 6 machines is right, but it's higher than 5.
• C: Keeps the 11.5-minute cycle with 6 machines. With 6, the operator can't get around in 11.5 minutes, so the cycle stretches to 12.
• D: Four machines leave the operator idle and cost more per unit.

If you missed it: Above n′ the operator is the bottleneck; below it the machines are. Check both integers next to n′.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 4 stations; 75%

Total work = 150 s. Theoretical minimum = ⌈150 ÷ 50⌉ = 3.

Station 1: A (20), then C (15) = 35; B (35) and E (30) won't fit. Station 2: B (35); D (25) and E (30) won't fit. Station 3: E (30); D (25) won't fit. Station 4: D, F, G = 50.

Efficiency = 150 ÷ (4 × 50) = 75%.

Three stations is impossible here: any station holding A can't also hold B or E, which leaves at least 130 s for two stations of 50 s.

Why the others miss:

• B: The theoretical minimum, not what precedence and task sizes allow.
• C: Divides 150 by 4 stations × 47 s (an average), not by the 50 s cycle.
• D: Over-counts stations; the rule produces four.

If you missed it: Efficiency always uses the cycle time, even if no station is fully loaded.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 562 units

The slowest station sets the pace: 48 s per unit.

Output = 7.5 × 3,600 ÷ 48 = 562.5 → 562 complete units.

Why the others miss:

• B: Uses the average station time (43.5 s).
• C: Uses the fastest station (39 s).
• D: Treats the line as one station doing all 174 s of work.

If you missed it: A line can't move faster than its slowest station.

Source: Derived calculation; method and arithmetic shown above

Answer: 50.4 seconds (accepted: 50.3 to 50.5)

1 TMU = 0.00001 h = 0.036 s. Normal time = 1,250 × 0.036 = 45.0 s.

Standard time = 45.0 × 1.12 = 50.4 s.

Common wrong answers:

• 45.0: Leaves out the allowance.
• 0.014: Reports hours (1,250 × 0.00001 × 1.12) when the question asks for seconds.

If you missed it: Convert TMU once, carefully: 100,000 TMU per hour, 0.036 s per TMU.

Source: MTM Association, MTM research publication (1 TMU = 0.036 seconds). The no-performance-rating condition used here is stated directly in the question.

Answer: A. 9

Allowed error = 5% of the mean = 0.05 × 0.80 = 0.040 min.

n = (z s ÷ e)² = (1.96 × 0.06 ÷ 0.040)² = (2.94)² = 8.64. Round up: 9.

Why the others miss:

• B: Rounds 8.64 down.
• C: Forgets to square 2.94.
• D: Treats ±5% as ±0.05 min instead of 5% of the mean.

If you missed it: '±5%' is relative to the planning mean. Convert it to minutes before using the sample-size formula.

Source: Derived calculation; method and arithmetic shown above

Answer: A. About 669 hours

b = log(0.85) ÷ log(2) = −0.2345.

T₁ = 200, T₂ = 170, T₃ = 200 × 3^−0.2345 = 154.6, T₄ = 144.5.

Total = 200 + 170 + 154.6 + 144.5 = 669 hours.

Why the others miss:

• B: Uses the cumulative-average model (4 × 200 × 0.85²). The stem specifies the unit model.
• C: The time for the fourth unit alone.
• D: Ignores learning.

If you missed it: Unit model: sum each unit's time. Cumulative-average model: average × count. Check which one the problem names.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 323

Infinite-population size: n₀ = z² p(1 − p) ÷ e² = 3.8416 × 0.09 ÷ 0.0009 = 384.2.

Correct for the finite lot: n = n₀ ÷ [1 + (n₀ − 1) ÷ N] = 384.2 ÷ 1.1916 = 322.4. Round up: 323.

Why the others miss:

• B: Skips the finite population correction (384.2 rounded up).
• C: Rounds 322.4 down.
• D: Halves the uncorrected size, which has no basis.

If you missed it: When the sample is a noticeable share of the lot (here about 19%), apply the correction.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 1.72 min

Working time = 480 × 0.85 = 408 min. Normal time per unit = 408 × 1.10 ÷ 300 = 1.496 min.

Standard time = 1.496 × 1.15 = 1.72 min.

Why the others miss:

• B: Leaves out the 110% rating.
• C: Normal time, with no allowance.
• D: Rates the whole 480-minute shift instead of only the working time.

If you missed it: Work sampling first finds the share of time spent working, then rates only that time.

Source: Derived calculation; method and arithmetic shown above

Answer: A, B, C, and E

87 dBA is at or above the 85 dBA 8-hour TWA that triggers a hearing conservation program, (c)(1). That program includes monitoring (d), audiometric testing (g), hearing protectors made available at no cost (i)(1), and training (k).

87 dBA is below the 90 dBA, 8-hour permissible exposure in Table G-16, so the controls in (b)(1) aren't triggered.

Why the other choices are wrong:

• D: Table G-16 allows 90 dBA for 8 hours. At 87 dBA the PEL isn't exceeded.
• F: Wearing protectors is mandatory at 85 dBA or more only for employees without a baseline audiogram or with a standard threshold shift, (i)(2). Neither applies here.

If you missed it: Keep the two numbers apart: 85 dBA triggers the hearing conservation program; 90 dBA is the 8-hour permissible exposure.

Source: 29 CFR 1910.95(b)(1), Table G-16, (c)(1), (d)(1), (g)(1), (i)(1)–(2), (k)(1)

Answer: A. 2 hours

Table G-16 permits 3 h at 97 dBA and 6 h at 92 dBA.

Used so far: 2 ÷ 3 = 0.667. Remaining fraction: 1 − 0.667 = 0.333.

Allowed time at 92 dBA = 0.333 × 6 = 2 hours.

Why the others miss:

• B: Subtracts 2 hours from the 6-hour limit, mixing exposures at different levels as if they were the same.
• C: Ignores the exposure already received.
• D: Assumes 2 hours at 97 dBA used up the whole daily allowance. It used 2/3 of it.

If you missed it: Mixed exposures add as fractions of each level's permitted time.

Source: 29 CFR 1910.95(b)(1), Table G-16 and footnote

Answer: 1 → a, 2 → b, 3 → c, 4 → d

RULA grand scores run from 1 to 7: 1–2 is action level 1, 3–4 is level 2, 5–6 is level 3, and 7 is level 4.

If you missed it: Two scores per level, except 7, which stands alone at the top.

Source: Gómez-Galán et al., Musculoskeletal Risks: RULA Bibliometric Review (2020), action levels; McAtamney, Notes for using RULA (grand score 1–7)

Answer: A. 57.4 N·m

Moment = Σ (mass × g × horizontal distance) = (8.0 × 0.60 + 3.5 × 0.30) × 9.81 = (4.80 + 1.05) × 9.81 = 57.4 N·m.

Why the others miss:

• B: Leaves out the arm's own weight.
• C: Forgets g, leaving kg·m instead of N·m.
• D: Places the arm's mass at the tool's distance (0.60 m) instead of at its own center of mass.

If you missed it: Every mass gets its own moment arm, and the moment arm is the horizontal distance to the joint.

Source: Derived calculation; method and arithmetic shown above

Answer: A. From about 1.40 to about 0.84

Before: RWL = 51 × (10/20) × 0.70 = 17.9 lb. LI = 25 ÷ 17.9 = 1.40.

After: RWL = 51 × (10/12) × 0.70 = 29.8 lb. LI = 25 ÷ 29.8 = 0.84.

Bringing the load closer raises the horizontal multiplier, which raises the recommended weight limit and drops the index below 1.0.

Why the others miss:

• B: Divides RWL by the load. The lifting index is load ÷ RWL.
• C: Assumes only vertical height matters. Horizontal distance has its own multiplier.
• D: Leaves out the other multipliers (the 0.70 product), which overstates the recommended weight limit both times.

If you missed it: A lifting index above 1.0 means the load is more than the recommended weight limit for those conditions.

Source: Cole & McGlothlin, Ergonomics Case Study: Revised NIOSH Lifting Equation (2009), p. D74, citing NIOSH Pub. 94-110

Answer: A. Feed rate

Link values: alarm acknowledge 10, feed rate 24, spindle speed 20, coolant 18. Feed rate ranks first.

Why the others miss:

• B: Highest importance alone, but used only twice an hour (link value 10).
• C: Highest frequency alone, but low importance (link value 18).
• D: Second-highest link value (20).

If you missed it: Combine the criteria the way the problem defines them before ranking.

Source: Ranking criterion supplied in the stem; arithmetic shown

Answer: A. 77.6 in

95th percentile = 70.0 + 1.645 × 2.8 = 74.6 in.

Clearance = 74.6 + 3.0 = 77.6 in.

Why the others miss:

• B: Designs for the average worker plus allowance. Half the workforce would be taller.
• C: Uses the 90th percentile (z ≈ 1.28).
• D: Leaves out the allowance.

If you missed it: Clearances are set by the tall end of the distribution; reaches are set by the short end.

Source: Derived calculation; method and arithmetic shown above

Answer: A. Fixed-value (grouping and counting) method

The fixed-value method counts repetitions or parts, and it can weigh an item to confirm it's complete. Weighing the bag confirms the 12-screw count: fixed-value.

Why the others miss:

• B: The contact method checks a physical characteristic of the part itself, such as diameter or temperature. Here the weight is only a stand-in for the screw count.
• C: The motion-step method checks that process steps happen in the right order.
• D: Successive inspection is a type of inspection, not one of the setting functions that detects the error.

If you missed it: Ask what the device is really verifying: a feature (contact), a count (fixed-value), or a sequence (motion-step).

Source: ASQ, What is mistake proofing?

Answer: A. UCL 0.138, LCL 0.022

σp = √[p̄(1 − p̄) ÷ n] = √(0.08 × 0.92 ÷ 200) = 0.0192.

UCL = 0.08 + 3(0.0192) = 0.138. LCL = 0.08 − 3(0.0192) = 0.022.

Why the others miss:

• B: Uses 2-sigma limits.
• C: Uses √(p̄ ÷ n) and drops the (1 − p̄) term.
• D: Uses 1-sigma limits.

If you missed it: The binomial standard error is √[p(1 − p)/n]. Don't drop the (1 − p).

Source: NIST/SEMATECH e-Handbook §6.3.3.2, p control charts

Answer: Point 7

Point 5 (52.4) is the first point above 52, but it's alone.

At point 7 (52.6), two of the last three points (5 and 7) are above 52: rule 2 fires. Points 3–7 also have four of five above 51 (51.4, 52.4, 51.2, 52.6): rule 3 fires at the same point.

The first signal is point 7. Point 15 (53.4) later breaks rule 1, but by then the process had already signaled.

Common wrong answers:

• 5: The first point beyond 2σ, but one point in that zone isn't a signal by itself.
• 15: The most extreme point, but not the first signal.

If you missed it: Scan in order and test every rule at each new point. The most dramatic point is rarely the first signal.

Source: NIST/SEMATECH e-Handbook §6.3.2, Western Electric control chart rules

Answer: A. UCL 10, LCL 0; 9 is in control

UCL = c̄ + 3√c̄ = 4 + 3(2) = 10. The raw lower limit is 4 − 6 = −2, so for a count chart it is set to 0.

9 ≤ 10, so the panel is in control.

Why the others miss:

• B: Uses 1-sigma limits (4 ± 2).
• C: Uses 2-sigma limits (4 + 4 = 8).
• D: Uses 3c̄ instead of 3√c̄. Poisson standard deviation is √c̄.

If you missed it: For Poisson counts, the variance equals the mean, so σ = √c̄.

Source: NIST/SEMATECH e-Handbook §6.3.3.1, c control charts

Answer: 0.0117 in (accepted: 0.0116 to 0.0117)

σ̂ = R̄ ÷ d₂ = 0.024 ÷ 2.059 = 0.0117 in.

Common wrong answers:

• 0.024: Reports R̄ itself as σ.
• 0.0494: Multiplies by d₂ instead of dividing.

If you missed it: The range of a small subgroup is larger than σ, so σ̂ must come out smaller than R̄.

Source: NIST/SEMATECH e-Handbook §6.3.2.1, Shewhart X̄ and R and S charts (σ̂ = R̄/d₂)

Answer: A. Cpu = 1.33

With one upper limit, use Cpu = (USL − μ) ÷ 3σ = (5.0 − 3.8) ÷ 0.9 = 1.33.

Why the others miss:

• B: Divides by 6σ. One-sided indices use 3σ.
• C: Divides the USL itself by 3σ, ignoring where the mean sits.
• D: One-sided specifications use Cpu or Cpl. Cp is the index that needs both limits.

If you missed it: Each one-sided index measures the distance from the mean to that limit, in units of 3σ.

Source: NIST/SEMATECH e-Handbook §6.1.6, process capability (Cpu, Cpl, one-sided specifications)

Answer: A. 7,560 ppm

z(USL) = (25.10 − 25.00) ÷ 0.04 = 2.5 → 0.00621 above.

z(LSL) = (24.88 − 25.00) ÷ 0.04 = −3.0 → 0.00135 below.

Total = 0.00756 = 7,560 ppm.

Why the others miss:

• B: Counts only the upper tail.
• C: Counts only the lower tail.
• D: Doubles the upper tail, as if the process were centered between the limits. It isn't.

If you missed it: An off-center process has two different tails. Compute each one.

Source: Derived calculation; method and arithmetic shown above

Answer: A. 0.364

Accept on the first sample only with 0 defectives: P(0) = 0.98⁵⁰ = 0.364.

Why the others miss:

• B: P(2 or fewer). Finding 1 or 2 defectives sends you to a second sample; it isn't an acceptance.
• C: P(1 or 2), the chance of needing a second sample.
• D: P(3 or more), the chance of rejecting on the first sample.

If you missed it: Map each first-sample outcome to accept, reject, or continue before adding probabilities.

Source: NIST/SEMATECH e-Handbook §6.2.4, double sampling

Answer: A. 2.45%

AOQ = Pa × p × (N − n) ÷ N = 0.85 × 0.03 × 1,920 ÷ 2,000 = 0.0245 (2.45%).

Defectives remain only in the uninspected part of accepted lots.

Why the others miss:

• B: Drops the (N − n)/N factor; the 80 inspected units are also cleaned up.
• C: The incoming quality. Rectifying inspection improves it.
• D: p × (1 − Pa), the share of defectives removed by rejecting lots.

If you missed it: Outgoing defectives come only from accepted lots, and only from the units nobody inspected.

Source: NIST/SEMATECH e-Handbook §6.2.2, LASP definitions (AOQ, AOQL)

Answer: A, B, D, and F

A, B, and D describe the system as NIST summarizes it: AQL is the foundation, the standard issues a code letter, and plans switch among normal, tightened, and reduced inspection. NIST notes the basic tables are the same in 105D and 105E.

F is also true: the current DoD cancellation notice says MIL-STD-105E is canceled, while the current NCEES PE Industrial and Systems specification still lists MIL STD 105E as an acceptance-sampling example.

Why the other choices are wrong:

• C: Inspection begins at normal and moves to tightened when quality falls short.
• E: Acceptance is probabilistic. A lot at the AQL can still be rejected (producer's risk), and worse lots are sometimes accepted.

If you missed it: An AQL plan protects the producer at the AQL; it doesn't certify any single accepted lot.

Source: NIST/SEMATECH e-Handbook §6.2.3.1, MIL Standard 105D; Defense Logistics Agency, MIL-STD-105E Notice 3 (cancellation); NCEES PE Industrial and Systems Exam Specifications, Quality Engineering 5C; NIST/SEMATECH e-Handbook §6.2.2, producer's risk

Answer: 1 → a, 2 → b, 3 → c, 4 → d

PDSA is Plan-Do-Study-Act. ISO 9001 is titled Quality management systems — Requirements. Benchmarking compares against recognized leaders. A fishbone (cause-and-effect) diagram sorts possible causes into categories.

Description e is a Pareto analysis.

If you missed it: ISO 9001 is the requirements standard; ISO 9000 covers fundamentals and vocabulary.

Source: The W. Edwards Deming Institute, PDSA Cycle; ISO, ISO 9001:2026 Quality management systems — Requirements; ASQ, What is benchmarking?; ASQ, What is a fishbone diagram?

Answer: A. 8 runs; effect of C = 3

A 2³ design has 2 × 2 × 2 = 8 runs.

C high: (62 + 74 + 58 + 84) ÷ 4 = 69.5. C low: (60 + 72 + 54 + 80) ÷ 4 = 66.5.

Main effect of C = 69.5 − 66.5 = 3.

Why the others miss:

• B: 12 is the contrast (sum at high minus sum at low) before dividing by 4.
• C: Counts 2 × 3 instead of 2³.
• D: 19 is the main effect of A.

If you missed it: Main effect = average at the high level minus average at the low level.

Source: NIST/SEMATECH e-Handbook §5.3.3.3.2, full factorial example (2³ = 8 runs); NIST/SEMATECH e-Handbook §5.3.3.4.1, estimating effects

Answer: A. $1.28

Find k: 8 = k(0.10)², so k = 800 $/mm².

Loss at 25.04: L = 800 × (0.04)² = 800 × 0.0016 = $1.28.

Why the others miss:

• B: Scales the loss linearly (8 × 0.04 ÷ 0.10). Taguchi loss grows with the square of the deviation.
• C: The goalpost view: no loss inside tolerance. The quadratic loss model charges for any deviation from target.
• D: Computes k with an unsquared deviation (8 ÷ 0.10 = 80).

If you missed it: Halving the deviation cuts the loss to a quarter, not a half.

Source: Dassault Systèmes Isight documentation (hosted by MIT), Taguchi Robust Design: quality loss function

Answer: A, B, C, and E

ASQ's procedure builds a cross-functional team, sets the scope, identifies functions and failure modes, identifies effects, rates severity (1–10, recording the highest severity per mode), and identifies root causes. Failures are then prioritized and acted on, highest priority first.

Why the other choices are wrong:

• D: Scope comes before failure modes. Brainstorming without a defined scope wanders.
• F: ASQ calls for a cross-functional team with knowledge of the process, product, and customer needs.

If you missed it: FMEA runs in order: team, scope, functions, failure modes, effects, severity, causes, then prioritized action.

Source: ASQ, What is FMEA?

Answer: A. 0.779

R(600) = exp[−(600/1,200)²] = exp(−0.25) = 0.779.

Why the others miss:

• B: Treats life as exponential (γ = 1): exp(−0.5).
• C: Reports (t/α)^γ, the exponent, instead of the reliability.
• D: 1 − R, the probability of failure.

If you missed it: A shape parameter above 1 means wear-out: early survival is higher than the exponential model predicts.

Source: NIST/SEMATECH e-Handbook §8.1.6.2, Weibull

Answer: A. 0.972

R = Σ (i = 2 to 3) C(3, i) R^i (1 − R)^(3−i) = 3(0.90)²(0.10) + (0.90)³ = 0.243 + 0.729 = 0.972.

Why the others miss:

• B: The 1-of-3 parallel result, 1 − 0.10³.
• C: The series result: all three must work.
• D: Only the case of exactly two working.

If you missed it: For r-out-of-n, add the binomial terms from r up to n.

Source: NIST/SEMATECH e-Handbook §8.1.8.4, r out of n model

Answer: A. Replace every 2,000 h, about $0.59 per hour

1,000 h: (0.90 × 400 + 0.10 × 2,400) ÷ 980 = 600 ÷ 980 = $0.612/h.

2,000 h: (0.65 × 400 + 0.35 × 2,400) ÷ 1,850 = 1,100 ÷ 1,850 = $0.595/h.

Run to failure: 2,400 ÷ 2,500 = $0.960/h.

Why the others miss:

• B: Correct for 1,000 h, but slightly higher than 2,000 h.
• C: Run to failure is the most expensive here because each failure costs six times a planned replacement.
• D: The fewest failures isn't the goal; the lowest cost per hour is.

If you missed it: Compare policies on cost per hour of operation, not cost per cycle.

Source: Derived calculation; method and arithmetic shown above

Answer: A. The missing setpoint-verification step in the maintenance checklist

The five whys drill past symptoms to the cause that, once fixed, stops the problem from coming back. Resetting the heater fixes today's jam; adding a setpoint check to the checklist stops the next maintenance job from causing it again.

Why the others miss:

• B: A symptom on the way down the chain.
• C: The problem statement itself.
• D: An intermediate cause. Fixing it once doesn't prevent a repeat.

If you missed it: Stop asking why when you reach a cause you can control that would prevent recurrence, usually a process or system gap.

Source: ASQ, Five whys