CQE Practice Questions (2026) — Free Sample Questions with Answers

Answer: A

1. 1. Total cost per casting = unit price + (quality-related cost + delivery-related cost)/annual volume.
2. 2. Annual volume = 10,000 castings for each supplier.
3. 3. Northfield 43.00 + 2,000/10,000; Ridgeway 38.50 + 52,000/10,000; Lakeshore 41.00 + 8,000/10,000; Pinecrest 39.50 + 40,000/10,000.
4. 4. Totals: Northfield $43.20, Ridgeway $43.70, Lakeshore $41.80, Pinecrest $43.50 per casting.
5. 5. Lakeshore Metals ranks first; the lowest price (Ridgeway) and the lowest nonperformance cost (Northfield) both mislead.

Lakeshore Metals: Correct. Total cost per casting = $41.00 + ($5,000 + $3,000)/10,000 = $41.80, the lowest of the four.

Ridgeway Castings: Has the lowest unit price ($38.50), but its $52,000 of quality and delivery costs add $5.20 per casting, for $43.70; ranking on price alone ignores nonperformance costs.

Pinecrest Alloys: Looks best ($40.10) if delivery costs are left out, but adding its $34,000 of delivery-related cost gives $43.50.

Northfield Foundry: Has the lowest nonperformance costs, but its high price gives $43.00 + $0.20 = $43.20, above Lakeshore.

A cost-based supplier rating adds the costs of poor quality and late delivery to the purchase price; the lowest total cost, not the lowest price, ranks first.

Answer: B

1. 1. Simple payback = initial investment / net annual savings.
2. 2. Net annual savings = scrap savings + labor savings − recurring costs = 22,000 + 14,000 − 6,000 = $30,000 per year (no table constants needed).
3. 3. Payback = $84,000 / $30,000 per year.
4. 4. Payback = 2.80 years.
5. 5. The investment is recovered in under three years; recurring costs reduce the savings, not the one-time investment.

2.33 years: Ignores the annual maintenance cost: 84,000 / 36,000 = 2.33 years overstates the net benefit.

2.80 years: Correct. Net annual savings = 22,000 + 14,000 − 6,000 = $30,000; payback = 84,000 / 30,000 = 2.80 years.

3.00 years: Double-counts maintenance: it is added to the investment (90,000) and also subtracted from the savings (90,000 / 30,000).

3.82 years: Uses only the scrap savings (84,000 / 22,000) and omits the labor reduction and maintenance.

Simple payback divides the one-time investment by net annual savings, so recurring costs are subtracted from savings exactly once.

Answer: D

configuration identification: Identification defines the configuration items and their documentation; the new board revision was identified and released.

configuration change control: Change control evaluates and approves changes; the board approved this change properly.

configuration audit: A configuration audit verifies that a product conforms to its documented configuration; it might detect the gap, but the failure is in recording status.

configuration status accounting: Correct. Status accounting records and reports the approved configuration and the implementation status of each change for every unit, which is exactly the missing serial-number information.

Configuration management has four functions: identification, change control, status accounting (who has what), and audit (verifying conformance).

Answer: C

1. Formula: series Rs = ΠRi; active parallel Rp = 1 − Π(1 − Ri).
2. No table constants are needed. R_A = 0.95, R_B = R_C = 0.90, R_D = 0.98.
3. R_BC = 1 − (0.10)(0.10) = 0.99; R_sys = 0.95 × 0.99 × 0.98.
4. R_sys = 0.92169 ≈ 0.9217.
5. Interpretation: about 92.2% of missions succeed. The single-point blocks A and D now limit system reliability.

0.7541: This treats all four blocks as series (0.95 × 0.90 × 0.90 × 0.98). With redundancy, B and C should be combined in parallel.

0.8379: This counts only one branch of the redundant pair (0.95 × 0.90 × 0.98) and ignores the backup that C provides.

0.9217: Correct. Parallel pair: 1 − (1 − 0.90)² = 0.99. System: 0.95 × 0.99 × 0.98 = 0.9217.

0.9900: This is the reliability of the redundant pair alone. A and D are in series with it and must be multiplied in.

Reduce the diagram step by step: combine parallel blocks with 1 − Π(1 − R), then multiply the series blocks.

Answer: D

1. Formula: total position tolerance = stated tolerance + bonus, where bonus = |actual size − MMC size|.
2. MMC of an internal feature (hole) is its smallest size: 10.00 − 0.10 = 9.90 mm.
3. Bonus = 10.06 − 9.90 = 0.16 mm.
4. Total = 0.20 + 0.16 = ⌀0.36 mm.
5. Interpretation: the hole axis may lie anywhere inside a ⌀0.36 mm zone centered on its true position.

⌀0.16 mm: This is only the bonus (10.06 − 9.90). The stated ⌀0.20 tolerance from the frame was left out.

⌀0.20 mm: This forgets the MMC bonus. The Ⓜ modifier lets the tolerance grow as the hole departs from its MMC size.

⌀0.26 mm: Here the bonus was measured from the nominal size (10.06 − 10.00 = 0.06). Bonus is measured from MMC, which for a hole is the smallest size, 9.90.

⌀0.36 mm: Correct. The hole is at MMC at its smallest size, 10.00 − 0.10 = 9.90. Bonus = 10.06 − 9.90 = 0.16, so the total is 0.20 + 0.16 = ⌀0.36 mm.

At MMC (Ⓜ), bonus tolerance equals how far the feature's actual size has moved from MMC toward LMC. For a hole, MMC is the smallest size.

Answer: B

1. 1. GRR = √(EV² + AV²); TV = √(GRR² + PV²); %GRR = 100 × GRR/TV.
2. 2. EV = 0.012, AV = 0.009, PV = 0.036 mm (no table constants needed: values are already standard deviations).
3. 3. GRR = √(0.000144 + 0.000081) = √0.000225 = 0.015; TV = √(0.000225 + 0.001296) = √0.001521 = 0.039.
4. 4. %GRR = 0.015/0.039 × 100 = 38.5%.
5. 5. Over 30% (and ndc = 3 < 5), so the measurement system is not acceptable; repeatability is the larger component, pointing to the gage.

14.8%: This is the variance ratio GRR²/TV² (% contribution), not %GRR of total variation.

38.5%: Correct. GRR = √(0.012² + 0.009²) = 0.015; TV = √(0.015² + 0.036²) = 0.039; %GRR = 0.015/0.039 = 38.5%, above 30%, so the system is unacceptable (ndc = 1.41 × 0.036/0.015 = 3).

41.7%: Divides GRR by part variation (0.015/0.036) instead of total variation.

53.8%: Adds EV and AV as standard deviations (0.021/0.039); components combine as the root sum of squares.

Gage R&R components combine as variances; under AIAG guidance %GRR below 10% is acceptable, 10–30% may be acceptable, and above 30% is unacceptable.

Answer: D

1. 1. α = 1 − Pa(AQL); β = Pa(LTPD).
2. 2. Read the curve: at p = 1%, Pa ≈ 0.95; at p = 6%, Pa ≈ 0.14 (Poisson check: np = 0.8 gives Pa = e^−0.8(1 + 0.8 + 0.32) = 0.953; np = 4.8 gives Pa = e^−4.8(1 + 4.8 + 11.52) = 0.143).
3. 3. α = 1 − 0.95 = 0.05.
4. 4. β = 0.14.
5. 5. The plan rejects about 5% of lots at the AQL and accepts about 14% of lots at the LTPD.

α ≈ 0.95; β ≈ 0.14: Reports Pa at the AQL as α; the producer's risk is the probability of rejecting a good lot, 1 − Pa.

α ≈ 0.05; β ≈ 0.86: Takes the complement at the LTPD; β is the probability of accepting a bad lot, which is Pa itself (≈ 0.14), not 1 − Pa.

α ≈ 0.14; β ≈ 0.05: Mirror swap: assigns the LTPD value to the producer and the AQL value to the consumer.

α ≈ 0.05; β ≈ 0.14: Correct. α = 1 − Pa at the AQL = 1 − 0.95 = 0.05; β = Pa at the LTPD ≈ 0.14.

Producer's risk is the rejection probability at the AQL; consumer's risk is the acceptance probability at the LTPD.

Answer: A

1. OEE = A × P × Q, with A = run time ÷ planned time, P = (ideal cycle time × total count) ÷ run time, Q = good count ÷ total count.
2. No table constants are needed; run time = 480 − 60 = 420 min.
3. A = 420 ÷ 480 = 0.875; P = (0.5 × 700) ÷ 420 = 350 ÷ 420 = 0.833; Q = 665 ÷ 700 = 0.950.
4. OEE = 0.875 × 0.833 × 0.950 = 0.693 = 69.3% (check: 0.5 × 665 ÷ 480 = 0.693).
5. Availability is the largest loss (12.5 points), so downtime reduction gives the biggest OEE gain.

69.3%: Correct. OEE = availability × performance × quality = 0.875 × 0.833 × 0.950 = 0.693, or 69.3%.

72.9%: This is availability × performance only (0.875 × 0.833); the 35 defective parts (quality = 0.950) were left out.

79.2%: This is performance × quality only (0.833 × 0.950); the 60 minutes of downtime (availability = 0.875) were left out.

83.1%: This is availability × quality only (0.875 × 0.950); the speed loss (performance = 0.833) was left out.

OEE multiplies availability, performance and quality; a quick check is ideal cycle time × good count ÷ planned production time.

Answer: C

Customer complaints: With 5 in and 0 out, complaints are the key outcome (the result of the other issues), the mirror image of a driver.

Overtime: Overtime has the most total arrows (6), but they are split 3 in and 3 out; total count marks a hub, not the driver.

Operator turnover: Correct. The issue with the most outgoing arrows (5 out, 0 in) influences every other issue and is influenced by none, so it is the key driver (root cause).

Unclear work instructions: It is a strong secondary driver (4 out), but operator turnover has more outgoing arrows and no incoming ones.

In an interrelationship digraph, the most arrows out marks the key driver to work on first; the most arrows in marks the key outcome, a symptom.

Answer: C

1. Cumulative % for the top k categories = (sum of the k largest counts) ÷ total count × 100.
2. No table constants are needed; total = 64 + 40 + 24 + 16 + 10 + 6 = 160.
3. Top two = 64 + 40 = 104; 104 ÷ 160 = 0.650.
4. Cumulative percentage = 65.0%.
5. Two of six categories carry about two-thirds of the defects, so they are the "vital few" to attack first.

25.0%: This is insufficient solder alone (40 ÷ 160); it leaves out the largest bar.

40.0%: This is solder bridges alone (64 ÷ 160), the first bar instead of the first two.

65.0%: Correct. The two tallest bars are solder bridge (64) and insufficient solder (40): (64 + 40) ÷ 160 = 65.0% cumulative.

80.0%: This is the cumulative share of the top three categories ((64 + 40 + 24) ÷ 160); the question asks for two.

On a Pareto chart, read the cumulative percentage by adding the bars from the left in descending order and dividing by the total; the vital few usually sit to the left of the 70–80% point.

Answer: A

1. 1. Formula: σ̂ = R̄/d2; Cpk = min(USL − µ, µ − LSL)/(3σ̂).
2. 2. Constant for n = 5: d2 = 2.326.
3. 3. σ̂ = 0.186/2.326 = 0.080 mm; USL − µ = 0.22 mm; µ − LSL = 0.38 mm.
4. 4. Cpk = 0.22/0.24 = 0.92.
5. 5. Interpretation: Cpk < 1.00, so the process makes oversize parts; centering it at 10.00 mm would raise Cpk to Cp = 1.25.

0.92: Correct. σ = R̄/d2 = 0.186/2.326 = 0.080 mm; Cpk = min(10.30 − 10.08, 10.08 − 9.70)/(3 × 0.080) = 0.22/0.24 = 0.92, so the process is not capable even though Cp = 1.25.

1.25: This is Cp = (USL − LSL)/6σ = 0.60/0.48, which ignores the 0.08-mm offset from center.

1.58: Uses the distance to the farther limit (10.08 − 9.70 = 0.38); Cpk uses the nearer limit.

2.75: Omits the factor of 3: 0.22/0.080 is the z distance to the USL, not Cpk.

Cpk uses the nearer specification limit and within-subgroup σ = R̄/d2; the gap between Cp and Cpk shows how much is lost to off-centering.

Answer: A

The process is out of control: subgroups 11–18 form a run of eight above the center line: Correct. Subgroups 11 through 18 are all above 25.00 mm, meeting rule 2; this signals an upward shift in the mean that should be investigated.

The process is in control because every point lies within the control limits: Rule 1 is not the only test; a run of eight on one side is a nonrandom pattern even when all points are within limits.

The process is out of control because subgroup 17 is close to the upper control limit: Subgroup 17 (25.36 mm) is well inside the UCL of 25.60 mm; nearness to a limit is not one of the stated rules.

The limits should first be recalculated with subgroups 11–18 included before any conclusion is drawn: Recalculating limits to absorb a signal hides the shift; the run should be investigated against the existing limits.

Apply every stated rule: a sustained run on one side of the center line signals a mean shift even without any point beyond the limits.

Answer: C

1. 1. Formula: z = (x − μ)/σ for each specification limit; p = P(Z < z_L) + P(Z > z_U).
2. 2. Table values given: P(Z > 2.50) = 0.0062; P(Z < −2.00) = 0.0228.
3. 3. z_U = (25.05 − 25.00)/0.02 = 2.50; z_L = (24.96 − 25.00)/0.02 = −2.00.
4. 4. p = 0.0062 + 0.0228 = 0.0290 (about 29,000 PPM).
5. 5. Interpretation: about 2.9% of shafts are out of specification, mostly below the LSL, which is only 2.00σ from the mean versus 2.50σ for the USL.

0.0062: Counts only the upper tail (above 25.05 mm) and ignores shafts below 24.96 mm.

0.0228: Counts only the lower tail (below 24.96 mm) and ignores shafts above 25.05 mm.

0.0290: Correct. z_U = (25.05 − 25.00)/0.02 = 2.50 and z_L = (24.96 − 25.00)/0.02 = −2.00; the two tails add: 0.0062 + 0.0228 = 0.0290.

0.9710: This is the fraction conforming, the complement of the answer.

For two-sided specifications, compute a z for each limit and add both tail areas; an off-center mean makes the tails unequal.

Answer: C

the risk file lacks an initial list of hazards: Nothing in the evidence shows a missing hazard list; the file was completed before transfer.

the complaint-handling process is not documented: Complaints were logged and closed, so a complaint process exists; the gap is what happens to the data.

post-launch field information is not fed back into the risk assessment: Correct. Risk management is a life-cycle process: recurring complaints should trigger review of the occurrence estimate and controls, and the static risk file shows the feedback loop is broken.

the dFMEA should have been replaced by a fault tree analysis at design transfer: Choice of analysis tool is not a finding; either tool must still be updated with field data.

Auditing a risk management system checks that field and production data flow back into the risk assessment, not just that the file exists.

Answer: B

a design FMEA of each component: A dFMEA works bottom-up from single failure modes; it does not model the AND/OR combinations that lead to one top event.

a fault tree analysis: Correct. Fault tree analysis works top-down from an undesired event, uses AND/OR gates to model combinations of causes, and combines basic-event probabilities into a top-event probability.

a hazard and operability study: HAZOP applies guide words to deviations of process parameters; it does not quantify one top event from combined failures.

a preliminary hazard analysis: A PHA is an early, broad screen of hazards; it does not model detailed failure combinations or calculate their probability.

Use fault tree analysis (top-down, AND/OR logic, quantifiable) for combinations of failures leading to one hazard; FMEA works bottom-up from single failure modes.