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

Answer: A

Likely indoor growth, as the taxa differ and Stachybotrys is only indoors: Correct. Penicillium/Aspergillus dominates indoors but not outdoors, and Stachybotrys, a water-damage indicator, appears only indoors. Both point to indoor amplification, so look for moisture sources.

Acceptable air, since no spore count exceeds a numerical OEL: There are no consensus numerical OELs for airborne fungal spores; the indoor/outdoor comparison is the usual basis.

Outdoor infiltration as the main indoor spore source: With infiltration, indoor taxa would mirror outdoor taxa (Cladosporium dominant), not Penicillium/Aspergillus.

No amplification, since the indoor and outdoor totals match: Similar totals hide the composition: indoor taxa differ in rank order from outdoor taxa.

Interpret bioaerosol data by comparing indoor and outdoor taxa and rank order, not just totals; water-damage indicator fungi found only indoors suggest amplification.

Answer: D

1. Permitted time: T = 8 / 2^((L − 90)/5): 90 dBA → 8 h; 95 dBA → 4 h; 100 dBA → 2 h; 72 dBA is below the 80-dBA threshold and adds no dose.
2. D = C1/T1 + C2/T2 + C3/T3 = 4/8 + 2/4 + 1/2 = 1.50 = 150%.
3. TWA = 16.61 log(D/100) + 90 = 16.61 log(150/100) + 90 = 16.61 × 0.176 + 90.
4. TWA = 92.9 dBA.

87.9 dBA: Used 85 dBA as the constant in the OSHA TWA equation instead of 90 dBA.

90.0 dBA: Time-averaged the decibel values arithmetically; decibels must be converted to dose.

91.8 dBA: Applied the 10-log (3-dB exchange) coefficient to an OSHA 5-dB dose.

92.9 dBA: Correct. D = 4/8 + 2/4 + 1/2 = 1.50 (150%); TWA = 16.61 log(150/100) + 90 = 92.9 dBA.

Convert each level to a fraction of its OSHA permitted time, sum to a dose, then convert with TWA = 16.61 log(D/100) + 90.

Answer: B

1. Fiber density: E = (F/nf − B/nb) ÷ Af = (62/100 − 5/100) ÷ 0.00785 mm² = 0.57 ÷ 0.00785 = 72.6 fibers/mm².
2. Average flow: (2.1 + 1.9) ÷ 2 = 2.0 L/min; V = 2.0 L/min × 240 min = 480 L.
3. Concentration: C = E × Ac ÷ (V × 10³) = 72.6 f/mm² × 385 mm² ÷ (480 L × 1,000 cc/L).
4. C = 27,950 fibers ÷ 480,000 cc = 0.058 f/cc.

0.055 f/cc: Used the pre-calibration flow (2.1 L/min, V = 504 L) instead of the average flow.

0.058 f/cc: Correct. E = (62/100 − 5/100) ÷ 0.00785 mm² = 72.6 f/mm²; V = 2.0 L/min × 240 min = 480 L; C = 72.6 × 385 ÷ (480 × 10³) = 0.058 f/cc.

0.063 f/cc: Did not subtract the field-blank count.

0.068 f/cc: Added the field-blank count instead of subtracting it.

Blank-correct the per-field count, divide by the field area to get fibers/mm², then scale by the effective filter area and divide by air volume in cc.

Answer: D

The PID must be recalibrated with isobutylene: Recalibrating with isobutylene does not change the lamp energy; the solvent still will not ionize.

The vapor concentration is below the detection limit: A strong odor and a near-zero PID reading point to an instrument limitation, not a low concentration.

High humidity is quenching the detector signal: Humidity can reduce a PID response, but it does not explain no response to a compound the lamp cannot ionize.

The lamp energy is too low; use 11.7 eV or FID: Correct. A PID responds only to compounds whose ionization potential is below the lamp energy; 11.2 eV exceeds 10.6 eV, so the reading means nothing. An 11.7-eV lamp or a flame ionization detector will respond.

A PID detects only compounds with an ionization potential below the lamp energy (10.6 eV is typical; 11.7 eV is wider-range). Use an FID or a higher-energy lamp otherwise.

Answer: C

Pump flow drifted high, raising the cut size: A higher cyclone flow lowers the cut size and would collect less mass; the post-calibration also showed no drift.

The PVC filter gained moisture during the shift: The field blank barely changed, so a moisture or weighing artifact of this size is unlikely.

The cyclone was inverted during sampling: Correct. NMAM 0600 warns the sampler must never be inverted or tilted past horizontal; grit in the cyclone grit pot can then fall onto the filter and bias the respirable result high.

His breathing zone had more respirable dust: The co-workers did the same tasks on the same vehicles with breathing zones within 1 m of his, so a real threefold difference in respirable dust is not plausible.

Keep a cyclone sampler upright (never past horizontal); inversion dumps non-respirable material onto the filter and biases results high.

Answer: C

1. Basis 100 g: toluene 50 g ÷ 92.14 = 0.5427 mol; o-xylene 50 g ÷ 106.17 = 0.4709 mol.
2. Mole fraction of toluene X = 0.5427 / 1.0136 = 0.5354.
3. Raoult's law (BGC sheet): P(toluene) = X × VP = 0.5354 × 28 mmHg = 14.99 mmHg.
4. ppm = (P / Patm) × 10⁶ = (14.99 / 760) × 10⁶ = 19,725 ≈ 19,700 ppm.

17,100 ppm: Applied the xylene mole fraction (0.465) to toluene's vapor pressure.

18,400 ppm: Used the mass fraction (0.50) instead of the mole fraction.

19,700 ppm: Correct. Mole fraction of toluene = 0.5427/(0.5427 + 0.4709) = 0.535; partial pressure = 0.535 × 28 = 15.0 mmHg; (15.0/760) × 10⁶ = 19,700 ppm.

36,800 ppm: Forgot the mole fraction and used pure-toluene vapor pressure.

For liquid mixtures, convert mass to mole fraction, apply Raoult's law (Pi = Xi·VPi), then ppm = (Pi/760) × 10⁶.

Answer: C

1. Stokes settling velocity (BGC sheet): V = g·d²·(ρp − ρa) / (18η).
2. Convert diameter: 5.0 µm = 5.0 × 10⁻⁴ cm; d² = 2.5 × 10⁻⁷ cm².
3. Numerator: 981 cm/s² × 2.5 × 10⁻⁷ cm² × 2.65 g/cm³ = 6.50 × 10⁻⁴ g/s².
4. Denominator: 18 × 1.81 × 10⁻⁴ g/(cm·s) = 3.258 × 10⁻³ g/(cm·s).
5. V = 6.50 × 10⁻⁴ ÷ 3.258 × 10⁻³ = 0.199 cm/s.

0.0499 cm/s: Used the radius (2.5 µm) instead of the diameter, which divides the result by 4.

0.0753 cm/s: Used unit density (1 g/cm³) instead of 2.65 g/cm³ for quartz.

0.199 cm/s: Correct. V = g·d²·(ρp − ρa)/(18η) = 981 × (5.0 × 10⁻⁴ cm)² × 2.65 ÷ (18 × 1.81 × 10⁻⁴) = 0.199 cm/s.

3.59 cm/s: Omitted the 18 in the denominator.

Stokes velocity grows with particle density and the square of the diameter: V = g·d²·(ρp − ρa)/(18η). Work in consistent CGS units.

Answer: B

Metal fume fever from pigment fume: Metal fume fever is a flu-like illness from freshly formed metal oxide fume (e.g. welding galvanized steel), not sprayed paint.

Diisocyanate sensitization asthma: Correct. Polyurethane hardeners contain diisocyanates (e.g. HDI), a leading cause of occupational asthma; once sensitized, very low exposures trigger symptoms that improve away from work.

Silicosis from sanding body filler: Silicosis is a chronic fibrotic disease with no work-week/vacation pattern.

Acute solvent narcosis from thinners: Solvent narcosis causes dizziness and CNS depression, not delayed wheezing that tracks the work week.

Work-related asthma that improves on vacation in a polyurethane spray painter points to diisocyanate sensitization.

Answer: D

Lack of absorptive lining inside the enclosure: A lining limits sound build-up inside the enclosure and is worth a few decibels, but cannot explain a 17-dB shortfall with 15% open area.

Wall panels with too little surface mass: Higher mass raises the transmission loss of the panels, but the openings, not the panels, control the result here.

No vibration isolation under the press: Structure-borne flanking can reduce performance, but the large openings are the dominant and predictable path.

Sound leaking through the 15% open area: Correct. Openings have essentially zero transmission loss; with 15% open area the composite TL cannot exceed about 10 log(1/0.15) ≈ 8 dB, whatever the wall rating.

Openings dominate enclosure performance: about 1% open area limits TL to roughly 20 dB, and 10% to about 10 dB. Seal penetrations and use lined, baffled openings.

Answer: B

1. Higher NRR = 33 (plugs). Derated attenuation for A-weighted data = NRR − 7 = 33 − 7 = 26 dB.
2. Dual-protection credit: 26 dB + 5 dB = 31 dB.
3. Estimated exposure = TWA − attenuation = 108 dBA − 31 dB = 77 dBA.

64 dBA: Derated and subtracted both NRRs (26 + 18 dB); the attenuation of combined protectors is not additive.

77 dBA: Correct. Use the higher NRR: (33 − 7) + 5 = 31 dB of attenuation, so 108 − 31 = 77 dBA.

82 dBA: Applied (NRR − 7) for the plug alone and omitted the 5-dB credit for the second protector.

87 dBA: Added the 5 dB to the exposure instead of to the attenuation.

OSHA OTM dual protection: estimated exposure = TWA − [(higher NRR − 7) + 5]; never add the two NRRs.

Answer: D

1. Hazard ratio = exposure ÷ OEL = 180 ppm ÷ 25 ppm = 7.2.
2. The required APF must be ≥ 7.2. OSHA Table 1: quarter-mask APR 5, half-mask APR 10, full-facepiece APR 50.
3. Check the half mask: MUC = APF × OEL = 10 × 25 ppm = 250 ppm ≥ 180 ppm, and 250 ppm < IDLH of 500 ppm.
4. Lowest adequate class: half-mask air-purifying respirator.

Quarter-mask air-purifying respirator: A quarter mask has an APF of 5, which gives a maximum use concentration of only 125 ppm, below the 180 ppm exposure.

Pressure-demand SCBA with a full facepiece: An SCBA is required for IDLH or unknown atmospheres; 180 ppm is well below the 500 ppm IDLH, so it is not the lowest adequate class.

Full-facepiece air-purifying respirator: A full facepiece (APF 50) is adequate but is more protection than the minimum class required.

Half-mask air-purifying respirator: Correct. The hazard ratio is 180 ÷ 25 = 7.2, which is within the half-mask APR APF of 10 (MUC 250 ppm), and the atmosphere is not IDLH.

Select the respirator whose APF is at least the hazard ratio (exposure ÷ OEL), and confirm that MUC = APF × OEL is not above the IDLH.

Answer: B

1. Convert each VP to velocity, V = 4005√VP: 2,194, 2,775, 3,154, 3,375, 3,491, 3,468, 3,327, 3,050, 2,657, 2,081 fpm.
2. Average velocity = 2,957 fpm (average the velocities, not the VPs).
3. A = π(1.167 ft)²/4 = 1.069 ft².
4. Q = VA = 2,957 fpm × 1.069 ft² = 3,161 cfm ≈ 3,160 cfm.

2,400 cfm: Multiplied 4005 by the average VP without taking the square root.

3,160 cfm: Correct. Mean of the point velocities = 2,957 fpm; × 1.069 ft² = 3,160 cfm.

3,200 cfm: Averaged the VPs first and converted once; the square-root relationship makes this overstate the velocity.

4,020 cfm: Used the diameter squared as the duct area instead of πd²/4.

In a Pitot traverse, convert every point to velocity first and then average; averaging VPs (a square-law quantity) overstates the mean velocity.

Answer: D

Hole in the duct near the fan: A leak near the fan would make the fan inlet SP less negative, not more negative.

Slipping fan belt: A slower fan lowers SP magnitude everywhere, including the fan inlet; fan speed was also verified.

Damaged hood entry before the tap: Extra entry loss at the hood would make SPh more negative relative to the duct VP, and would not raise the fan inlet SP this much.

Blockage between the elbow and the fan: Correct. Upstream readings dropped (less airflow) while the fan inlet became more negative, which places added resistance between the mid-duct tap and the fan.

An obstruction lowers SP magnitude upstream of it (less flow) and raises SP magnitude between it and the fan (the fan pulls harder against the blockage).

Answer: D

1. Plain opening (IVM Table 6-2): Q = V(10X² + A).
2. A = 0.5 ft × 1.0 ft = 0.50 ft²; X = 18 in. = 1.5 ft.
3. Q = 100 fpm × [10 × (1.5 ft)² + 0.50 ft²] = 100 × (22.5 + 0.50) = 2,300 cfm.

1,550 cfm: Did not square the capture distance (used 10X instead of 10X²).

1,730 cfm: Used the flanged-opening equation 0.75V(10X² + A); this opening has no flange.

2,250 cfm: Dropped the face-area term A from the plain-opening equation.

2,300 cfm: Correct. Plain opening, IVM Table 6-2: Q = V(10X² + A) = 100(10 × 1.5² + 0.50) = 2,300 cfm.

For a plain opening, Q = V(10X² + A): the distance term is squared and dominates, so capture airflow rises steeply as the source moves away.

Answer: B

Keep the findings confidential, because the client owns the data: Confidentiality under the BGC Code is not absolute; this is one of its stated exceptions.

Disclose what is needed to prevent serious harm: Correct. The BGC Code keeps sensitive information confidential unless an exception applies, including when nondisclosure would likely result in death or serious physical harm.

Wait for the client's written authorization to release it: Client authorization is one exception, but the imminent-harm exception does not depend on it.

Release the full report to the news media at once: Disclosure should be limited to what is necessary to prevent the harm, made to those who can act.

BGC Code exceptions to confidentiality: apparent unlawful activity, a lawful court or agency order, express client/employer authorization, or likely death or serious physical harm.

Answer: A

Listen, acknowledge concerns and share the data: Correct. When hazard is low and outrage high, the priority is to reduce outrage: listen, show respect, be open with data and make accountable commitments.

Present toxicology data proving the odors are harmless: Leads with technical proof and dismisses the concern; it does not address outrage or the odor nuisance.

Refer all questions to the regulatory agency: Deflecting responsibility erodes trust; the plant is the source and should engage.

Show that the risk is lower than that of driving or smoking: Comparing an involuntary risk with voluntary ones usually increases outrage and distrust.

Perceived risk = hazard + outrage; when outrage drives concern, use two-way dialogue, transparency and commitments rather than technical rebuttal.