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

Answer: D

Shift 2 has the highest defect rate because it has the most defects.: Compared raw counts; Shift 2's rate is 36 ÷ 3,000 = 1.2%, the lowest of the three.

Shift 1 and Shift 2 have the same defect rate.: Shift 1 is 30 ÷ 2,000 = 1.5% and Shift 2 is 1.2%; the rates differ.

Shift 3 has the lowest defect rate because it has the fewest defects.: Ignored Shift 3's smaller output; fewer defects on fewer pouches is a higher rate.

Shift 3 has the highest defect rate even though it has the fewest defects.: Correct. Rates: Shift 1 30 ÷ 2,000 = 1.5%; Shift 2 36 ÷ 3,000 = 1.2%; Shift 3 24 ÷ 1,200 = 2.0%. Shift 3 is highest.

A bar chart of counts compares totals only; divide by the amount produced before comparing groups of different size.

Answer: B

1. 1. Formula: chord between adjacent holes = D × sin(180° ÷ N).
2. 2. Given: bolt-circle diameter D = 5.000 in; N = 8 holes.
3. 3. Substitute: 5.000 × sin(22.5°) = 5.000 × 0.38268.
4. 4. Result: x = 1.913 in.
5. 5. Adjacent hole centers should measure 1.913 in apart.

0.957 in: Used the radius (2.500 in) in place of the bolt-circle diameter in D × sin(180°/N).

1.913 in: Correct. Chord = D × sin(180°/N) = 5.000 × sin(180°/8) = 5.000 × sin 22.5° = 5.000 × 0.38268 = 1.913 in.

3.536 in: Used sin(360°/N) = sin 45°; the half-angle 180°/N belongs in the chord formula.

4.619 in: Used cosine instead of sine; cosine gives the distance along the bisector, not the chord.

The adjacent-hole chord on a bolt circle is D × sin(180°/N): half the angle between holes, times the full diameter.

Answer: C

−0.0480 mm: Rounded to three decimal places (−0.048), treating the leading zeros as significant digits.

−0.0477 mm: This is the value rounded to three significant digits; truncation drops the extra digits without rounding.

−0.0476 mm: Correct. Leading zeros are not significant, so the significant digits are 4, 7, 6, 8. Truncating keeps 4, 7, 6 and drops the 8, giving −0.0476 mm; the sign is unchanged.

−0.0470 mm: Truncated to three decimal places (−0.047), counting the leading zero as a significant digit.

Leading zeros are never significant. Truncating drops the digits beyond the last one kept without rounding, and the sign of a negative number stays the same.

Answer: C

location A: At A the shadow edge runs between the two tolerance lines, so the profile conforms there.

location B: At B, on the angled segment, the edge stays between the two lines; the sloped band only looks narrower.

location C: Correct. At C the edge of the shadow rises above the upper tolerance line, so material extends beyond the tolerance zone at that location.

location D: At D the edge sits back between the lines after the bump, so the profile conforms there.

On an optical comparator the part shadow must lie entirely between the overlay chart's tolerance lines, which are drawn at the screen magnification.

Answer: A

1. 1. H = L × sin θ.
2. 2. Given: L = 5.000 in between roll centers; θ = 20°.
3. 3. H = 5.000 × sin 20° = 5.000 × 0.34202.
4. 4. H = 1.7101 in (to 0.0001 in).
5. 5. A 1.7101 in stack under the upper roll sets the bar to 20°.

1.7101 in: Correct. H = L · sin θ = 5.000 × sin 20° = 5.000 × 0.34202 = 1.7101 in.

1.8199 in: Uses tan 20° instead of sin 20°.

4.5647 in: Calculator left in radian mode: 5.000 × sin(20 rad).

4.6985 in: Uses cos 20° instead of sin 20°.

For a sine bar, the stack height is the bar length times the sine of the angle; make sure the calculator is in degree mode.

Answer: B

the thread is oversize because the NOT GO member entered: Applies plain plug logic; a thread NOT GO member is allowed partial engagement up to three turns.

the thread is acceptable: Correct. Under the unified screw-thread gauging rules (ASME B1.2) the NOT GO thread plug may enter, but not more than three complete turns; two turns with a full-length GO is acceptable.

the thread must be measured with a pitch micrometer before a decision: Gauging already gives an accept decision; extra variables measurement is not required for this result.

the thread is undersize because the NOT GO member stopped: The NOT GO stopping is the expected result; an undersize thread would show up as the GO member failing to enter.

A tapped hole passes when the GO thread plug enters the full length and the NOT GO plug enters no more than three complete turns.

Answer: C

1. 1. Reading = numbered sleeve value + (exposed 0.025 in divisions × 0.025) + thimble line on the index.
2. 2. Given: last numbered sleeve line "3" = 0.300 in; 2 small divisions visible past it; thimble line 12 aligned with the index line.
3. 3. 0.300 + 2 × 0.025 + 12 × 0.001 = 0.300 + 0.050 + 0.012.
4. 4. Reading = 0.362 in.
5. 5. The spacer measures 0.362 in thick.

0.312 in: Omits the two exposed 0.025 in sleeve divisions past the "3" line.

0.337 in: Counts only one 0.025 in sleeve division past the "3" line.

0.362 in: Correct. The sleeve shows 0.300 plus 2 small divisions (2 × 0.025 = 0.050) for 0.350, and thimble line 12 is on the index line: 0.350 + 0.012 = 0.362 in.

0.387 in: Counts a third 0.025 in division that is still covered by the thimble edge.

Read a micrometer in three parts: the last numbered sleeve line (0.100 in each), the exposed 0.025 in divisions, then the thimble line on the index (0.001 in each).

Answer: A

flinching: Correct. Recording borderline parts as exactly on the limit, and so passing them, is flinching; the pile-up of readings at 0.7480 and 0.7500 with none beyond is the classic sign.

gauge bias: Both inspectors used the same micrometer, so a gauge offset would affect them equally; the difference is in the person, not the instrument.

cosine error: Cosine error comes from a tilted indicator and makes readings consistently high; it would not cluster readings exactly at both tolerance limits.

fatigue: Fatigue causes random misses and errors late in a shift, not a consistent pattern of rounding marginal readings into tolerance.

Flinching is the tendency to give marginal parts the benefit of the doubt, recording readings just outside a limit as on the limit; a pile-up of readings exactly at the limits is its signature.

Answer: B

1. (1) Formula: for a plunger indicator tilted θ from the measuring direction, true displacement = indicated reading × cos θ.
2. (2) Given: indicated reading = 0.0120 in; θ = 25°; cos 25° = 0.9063.
3. (3) Substitute: 0.0120 in × 0.9063 = 0.010876 in.
4. (4) Result: 0.0109 in (rounded to 0.0001 in).
5. (5) Interpretation: the tilted setup over-reads by about 0.0011 in, so the true rise is smaller than the dial shows; square the spindle to the surface before measuring again.

0.0051 in: Used sin 25° (0.4226) instead of cos 25°; the correction for spindle tilt is the cosine of the angle.

0.0109 in: Correct. Cosine error: true rise = 0.0120 in × cos 25° = 0.0120 × 0.9063 = 0.0109 in. The tilted spindle over-reads the real movement.

0.0120 in: The uncorrected dial reading; it ignores the cosine error caused by the 25° tilt.

0.0132 in: Divided by cos 25° instead of multiplying; a tilted plunger over-reads, so the true value is smaller than the reading, not larger.

A tilted plunger indicator travels farther than the surface moves, so it over-reads; multiply the reading by the cosine of the tilt angle to get the true value, or better, set the spindle perpendicular to the surface.

Answer: D

reduce inspection cost for the customer: That is the purpose of reduced inspection, used when quality is consistently good.

replace corrective action by the supplier: Tightened inspection does not substitute for the supplier’s corrective action.

give the supplier time to recover without penalty: Tightened inspection raises, not lowers, the chance of rejection.

protect the customer by making acceptance harder: Correct — tightened inspection (a smaller acceptance number or larger sample) protects the consumer when the supplier’s quality appears to have deteriorated.

Normal, tightened and reduced inspection let a scheme respond to quality history: tighten to protect the consumer when quality slips, reduce to save cost when it is consistently good.

Answer: C

1. Formula: total position tolerance = stated tolerance + bonus; for an external feature, bonus = MMC size − actual size.
2. Given: pin ⌀12.00 ±0.05 → MMC = 12.05; measured ⌀11.96; position ⌀0.08 Ⓜ.
3. Bonus = 12.05 − 11.96 = 0.09.
4. Total = 0.08 + 0.09 = ⌀0.17 mm.
5. Interpretation: the pin axis may lie anywhere in a ⌀0.17 zone centered on true position.

⌀0.08: Forgot the bonus that the Ⓜ modifier allows as the pin departs from MMC.

⌀0.09: Used the hole rule (MMC = smallest size, 11.95), giving a bonus of only 0.01; a pin’s MMC is its largest size.

⌀0.17: Correct — a pin’s MMC is its largest size, ⌀12.05; bonus = 12.05 − 11.96 = 0.09, so the total is 0.08 + 0.09 = ⌀0.17.

⌀0.18: Added the full size tolerance (0.10) as bonus, which applies only at LMC (⌀11.95), not at ⌀11.96.

A pin (external feature) is at MMC at its largest size, a hole at its smallest; bonus is the departure of the actual size from MMC.

Answer: A

1. Formulas: actual position (diametral) = 2√(Δx² + Δy²); allowed = stated tolerance + bonus, bonus = actual size − MMC size for a hole.
2. Given: basic X 30.00, Y 20.00; measured X 30.04, Y 19.97; hole ⌀8.10 ±0.05 (MMC = 8.05), measured ⌀8.12; position ⌀0.10 Ⓜ.
3. Δx = 0.04, Δy = −0.03 → 2√(0.0016 + 0.0009) = 2 × 0.05 = ⌀0.10. Bonus = 8.12 − 8.05 = 0.07.
4. Allowed = 0.10 + 0.07 = ⌀0.17 mm.
5. Interpretation: ⌀0.10 ≤ ⌀0.17, so the hole location is acceptable.

⌀0.10 actual; ⌀0.17 allowed: Correct — actual position = 2√(0.04² + 0.03²) = ⌀0.10; MMC of the hole is ⌀8.05, so bonus = 8.12 − 8.05 = 0.07 and allowed = 0.10 + 0.07 = ⌀0.17.

⌀0.10 actual; ⌀0.10 allowed: Forgot the bonus tolerance that Ⓜ allows when the hole is larger than MMC.

⌀0.05 actual; ⌀0.17 allowed: Used the radial deviation √(Δx² + Δy²) = 0.05 and forgot the ×2 for a diametral zone.

⌀0.14 actual; ⌀0.17 allowed: Added the deviations instead of using the Pythagorean formula: 2 × (0.04 + 0.03) = 0.14.

Diametral position deviation is twice the radial distance from true position; with Ⓜ on a hole, add the difference between the actual size and the smallest (MMC) size as bonus.

Answer: A

magnetic particle: Correct. Carbon steel is ferromagnetic, and magnetic particle testing detects surface and near-surface cracks that distort the magnetic field.

dye penetrant: Penetrant can only enter flaws that are open to the surface, so cracks just below the surface are missed.

visual inspection with a magnifier: Visual inspection sees only what reaches the surface.

profile inspection on an optical comparator: Profile inspection checks the part's contour, not cracks in the material.

For ferromagnetic parts, magnetic particle finds surface and near-surface flaws; dye penetrant finds only flaws open to the surface.

Answer: B

1. 1. Formula: Cp = (USL − LSL)/(6σ), with σ the within-subgroup standard deviation.
2. 2. Given: USL = 380 g, LSL = 368 g, σ = 1.6 g.
3. 3. Substitute: (380 − 368)/(6 × 1.6) = 12/9.6.
4. 4. Result: Cp = 1.25.
5. 5. Interpretation: the process spread uses 80% of the tolerance; Cp ignores centering, so Cpk is needed to see whether the mean is off target.

0.63: Used half the tolerance (6 g) in the numerator, which is the form used for Cpk, not Cp.

1.25: Correct. Cp = (USL − LSL)/6σ = (380 − 368)/(6 × 1.6) = 12/9.6 = 1.25.

2.50: Divided by 3σ instead of 6σ.

7.50: Divided the tolerance by σ alone and omitted the 6.

Cp compares the tolerance width with the 6σ process spread and ignores centering; Cpk also accounts for where the mean sits. Cp and Cpk use within-subgroup σ, while Pp and Ppk use overall σ.

Answer: C

1. 1. Rule: a signal occurs at the 8th consecutive point on one side of the center line (CL = 2.40 mm).
2. 2. Subgroup 9 = 2.39 mm (below CL), so any run above the CL starts at subgroup 10.
3. 3. Subgroups 10–17 = 2.42, 2.44, 2.41, 2.43, 2.42, 2.44, 2.41, 2.43, all above 2.40.
4. 4. The 8th point of the run is subgroup 17.
5. 5. Interpretation: the process mean has shifted upward from subgroup 10 on, though no point is beyond the limits; look for a special cause around subgroup 10.

10: Subgroup 10 is where the run starts; one point above the center line is not yet a signal.

16: Applied a seven-point rule; with the stated eight-point rule, subgroup 16 is only the seventh point in the run.

17: Correct. Subgroup 9 is below the center line; subgroups 10 through 17 are the first eight in a row above it, so the signal occurs at subgroup 17.

18: Counted nine points; the eighth consecutive point (subgroup 17) already meets the rule.

A run of points on one side of the center line signals a shift in the process mean even when every point is inside the control limits; count from the first point of the run.

Answer: C

1. 1. Formula: cumulative % = (sum of the counts in the leading categories ÷ total count) × 100.
2. 2. Given: short shot 64, flash 48, sink mark 30; total of all bars = 64 + 48 + 30 + 22 + 10 + 6 = 180.
3. 3. Substitute: (64 + 48 + 30)/180 = 142/180.
4. 4. Result: 0.7889 × 100 = 78.9%.
5. 5. Interpretation: three of six categories cause almost 80% of the defects, so improvement effort should start with them (the vital few).

16.7%: This is sink mark alone (30/180), the percentage of the third bar rather than the cumulative percentage.

62.2%: Stopped after two categories: (64 + 48)/180 = 62.2%.

78.9%: Correct. Short shot, flash and sink mark total 64 + 48 + 30 = 142 of 180 defects, and 142/180 = 78.9%.

91.1%: Added a fourth category (splay): 164/180 = 91.1%.

On a Pareto chart, the share removed by fixing the leading bars is their combined count divided by the grand total (the cumulative percentage); the "vital few" bars are where improvement starts.