NEWEST CERTIFIED INDUSTRIAL HYGIENIST
(CIH) CERTIFICATION EXAM OFFERED BY
BOARD FOR GLOBAL EHS CREDENTIALING |
COMPLETE EXAM Q&A WITH RATIONALES
1. A lab technician wearing a 0.5 mm lead apron
receives an exposure of 200 mrem. The apron has a
half‑value layer (HVL) of 0.25 mm for the radiation in
use. What is the approximate dose without the apron
A) 100 mrem
B) 200 mrem
C) 400 mrem
D) 800 mrem
Correct answer: D
Rationale: Each HVL reduces dose by half. After 2
HVLs (0.5 mm total), the dose is ¼ of the unshielded
dose. Unshielded dose = 200 × 4 = 800 mrem.
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2. Noise dosimetry shows an 8‑hour TWA of 92 dBA.
Engineering controls are not feasible. Hearing
,protection with an NRR of 29 is available. Using the
OSHA method for NRR derating, what is the
estimated protected exposure level?
A) 89 dBA
B) 85 dBA
C) 88 dBA
D) 81 dBA
Correct answer: C
Rationale: OSHA requires derating NRR by 50% for
double protection, otherwise 50% generally. Derated
NRR = 29 × 0.5 = 14.5. Protected level = 92 – 14.5 =
77.5? That is too low. Actually OSHA: for muffs or
plugs, derate by 50%; field attenuation = (NRR – 7)/2,
then subtract from TWA. With NRR 29: derated NRR =
(29 – 7)/2 = 11. Protected exposure = 92 – 11 = 81
dBA.
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3. An area air sample for lead is 150 µg/m³ (8‑hour
TWA). The OSHA PEL for lead is 50 µg/m³. How many
,hours of respiratory protection would be required
daily if an employee works 8 hours in that area?
A) 0 hours
B) 2 hours
C) 4 hours
D) 8 hours
Correct answer: D
Rationale: PEL is 50 µg/m³; the measured 150 µg/m³
exceeds the PEL. Half‑mask respirators may be
insufficient; full facepiece or supplied air is required.
The employee must wear protection for all 8 hours
because the exposure exceeds the limit at all times.
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4. A worker’s daily noise exposure consists of 2
hours at 95 dBA and 6 hours at 85 dBA. What is the
8‑hour TWA?
A) 86 dBA
B) 88 dBA
, C) 89 dBA
D) 91 dBA
Correct answer: C
Rationale: Use formula: TWA = 10 log[(2×10^(95/10) +
6×10^(85/10))/8] = approx 89 dBA.
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5. The ACGIH TLV for carbon monoxide is 25 ppm as
an 8‑hour TWA. A welder is exposed to CO at 30 ppm
for 2 hours and at 10 ppm for the remaining 6 hours.
Is the TWA exceeded?
A) No, TWA is 20 ppm
B) No, TWA is 15 ppm
C) Yes, TWA is 33 ppm
D) Yes, TWA is 45 ppm
Correct answer: B
Rationale: TWA = [(30×2) + (10×6)] / 8 = (60 + 60)/8 =
120/8 = 15 ppm.
(CIH) CERTIFICATION EXAM OFFERED BY
BOARD FOR GLOBAL EHS CREDENTIALING |
COMPLETE EXAM Q&A WITH RATIONALES
1. A lab technician wearing a 0.5 mm lead apron
receives an exposure of 200 mrem. The apron has a
half‑value layer (HVL) of 0.25 mm for the radiation in
use. What is the approximate dose without the apron
A) 100 mrem
B) 200 mrem
C) 400 mrem
D) 800 mrem
Correct answer: D
Rationale: Each HVL reduces dose by half. After 2
HVLs (0.5 mm total), the dose is ¼ of the unshielded
dose. Unshielded dose = 200 × 4 = 800 mrem.
---
2. Noise dosimetry shows an 8‑hour TWA of 92 dBA.
Engineering controls are not feasible. Hearing
,protection with an NRR of 29 is available. Using the
OSHA method for NRR derating, what is the
estimated protected exposure level?
A) 89 dBA
B) 85 dBA
C) 88 dBA
D) 81 dBA
Correct answer: C
Rationale: OSHA requires derating NRR by 50% for
double protection, otherwise 50% generally. Derated
NRR = 29 × 0.5 = 14.5. Protected level = 92 – 14.5 =
77.5? That is too low. Actually OSHA: for muffs or
plugs, derate by 50%; field attenuation = (NRR – 7)/2,
then subtract from TWA. With NRR 29: derated NRR =
(29 – 7)/2 = 11. Protected exposure = 92 – 11 = 81
dBA.
---
3. An area air sample for lead is 150 µg/m³ (8‑hour
TWA). The OSHA PEL for lead is 50 µg/m³. How many
,hours of respiratory protection would be required
daily if an employee works 8 hours in that area?
A) 0 hours
B) 2 hours
C) 4 hours
D) 8 hours
Correct answer: D
Rationale: PEL is 50 µg/m³; the measured 150 µg/m³
exceeds the PEL. Half‑mask respirators may be
insufficient; full facepiece or supplied air is required.
The employee must wear protection for all 8 hours
because the exposure exceeds the limit at all times.
---
4. A worker’s daily noise exposure consists of 2
hours at 95 dBA and 6 hours at 85 dBA. What is the
8‑hour TWA?
A) 86 dBA
B) 88 dBA
, C) 89 dBA
D) 91 dBA
Correct answer: C
Rationale: Use formula: TWA = 10 log[(2×10^(95/10) +
6×10^(85/10))/8] = approx 89 dBA.
---
5. The ACGIH TLV for carbon monoxide is 25 ppm as
an 8‑hour TWA. A welder is exposed to CO at 30 ppm
for 2 hours and at 10 ppm for the remaining 6 hours.
Is the TWA exceeded?
A) No, TWA is 20 ppm
B) No, TWA is 15 ppm
C) Yes, TWA is 33 ppm
D) Yes, TWA is 45 ppm
Correct answer: B
Rationale: TWA = [(30×2) + (10×6)] / 8 = (60 + 60)/8 =
120/8 = 15 ppm.
