Exposure, Noise, Containment and Lifting Calculations questions for the CSP
CSP11 Domain 6: Occupational Health and Applied Science (6.1 Exposure measurement; 6.4 Lifting; 6.5 Containment volumes). 14 free questions below. Every answer quotes and links the formula source it rests on.
Question 1Calculation
Air sampling shows a painter's exposure to acetone during an 8-hour shift was 4 hours at 1640 ppm, 1 hour at 793 ppm, 3 hours at 208 ppm. Using OSHA's cumulative exposure formula, what is the 8-hour equivalent exposure (E)?
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Answer: D. 997.1 ppm
E = (Ca Ta + Cb Tb + Cc Tc) ÷ 8 = (1640 x 4 + 793 x 1 + 208 x 3) ÷ 8 = 997.1 ppm. Averaging the three concentrations without weighting by time gives a different, incorrect value.
Answer D rests on
29 CFR 1910.1000(d)(1)(i)(opens ecfr.gov in a new tab)“The cumulative exposure for an 8-hour work shift shall be computed as follows: E = (Ca Ta + Cb Tb + . . .Cn Tn) ÷ 8 Where: E is the equivalent exposure for the working shift. C is the concentration during any period of time T where the concentration remains constant. T is the duration in hours of the exposure at the concentration C.”
Question 2Calculation
Air sampling shows a painter's exposure to xylene during an 8-hour shift was 3 hours at 154 ppm, 4 hours at 111 ppm, 1 hour at 19 ppm. Using OSHA's cumulative exposure formula, what is the 8-hour equivalent exposure (E)?
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Answer: C. 115.6 ppm
E = (Ca Ta + Cb Tb + Cc Tc) ÷ 8 = (154 x 3 + 111 x 4 + 19 x 1) ÷ 8 = 115.6 ppm. Averaging the three concentrations without weighting by time gives a different, incorrect value.
Answer C rests on
29 CFR 1910.1000(d)(1)(i)(opens ecfr.gov in a new tab)“The cumulative exposure for an 8-hour work shift shall be computed as follows: E = (Ca Ta + Cb Tb + . . .Cn Tn) ÷ 8 Where: E is the equivalent exposure for the working shift. C is the concentration during any period of time T where the concentration remains constant. T is the duration in hours of the exposure at the concentration C.”
Question 3Calculation
A degreasing area has an 8-hour mixture exposure of 177 ppm ethyl acetate (PEL 400 ppm), 72 ppm sec-butyl alcohol (PEL 150 ppm), 89 ppm n-heptane (PEL 500 ppm). Using OSHA's mixture formula, what is the equivalent exposure (Em), and is the combination within the limit?
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Answer: D. 1.10; over the limit
Em = 177/400 + 72/150 + 89/500 = 1.10. The value of Em shall not exceed unity (1), so the mixture is over the limit. Dividing total concentration by total PELs is not the OSHA method.
Answer D rests on
29 CFR 1910.1000(d)(2)(i)(opens ecfr.gov in a new tab)“In case of a mixture of air contaminants an employer shall compute the equivalent exposure as follows: Em = (C1 ÷ L1 + C2 ÷ L2) + . . .(Cn ÷ Ln) Where: Em is the equivalent exposure for the mixture. C is the concentration of a particular contaminant. L is the exposure limit for that substance specified in subpart Z of 29 CFR part 1910. The value of Em shall not exceed unity (1).”
Question 4Calculation
A degreasing area has an 8-hour mixture exposure of 309 ppm acetone (PEL 1000 ppm), 42 ppm methyl ethyl ketone (PEL 200 ppm), 187 ppm isopropyl alcohol (PEL 400 ppm). Using OSHA's mixture formula, what is the equivalent exposure (Em), and is the combination within the limit?
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Answer: C. 0.99; within the limit
Em = 309/1000 + 42/200 + 187/400 = 0.99. The value of Em shall not exceed unity (1), so the mixture is within the limit. Dividing total concentration by total PELs is not the OSHA method.
Answer C rests on
29 CFR 1910.1000(d)(2)(i)(opens ecfr.gov in a new tab)“In case of a mixture of air contaminants an employer shall compute the equivalent exposure as follows: Em = (C1 ÷ L1 + C2 ÷ L2) + . . .(Cn ÷ Ln) Where: Em is the equivalent exposure for the mixture. C is the concentration of a particular contaminant. L is the exposure limit for that substance specified in subpart Z of 29 CFR part 1910. The value of Em shall not exceed unity (1).”
Question 5Calculation
A hazardous waste container storage area holds 7 containers of 330 gallons and one 550-gallon container, all holding free liquids. Under the RCRA container storage rule, what is the minimum capacity the containment system must have?
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Answer: A. 550 gallons
Capacity must be the greater of 10% of total container volume (0.10 x 2,860 = 286 gal) or the largest container (550 gal), so 550 gallons.
Answer A rests on
40 CFR 264.175(b)(3)(opens ecfr.gov in a new tab)“The containment system must have sufficient capacity to contain 10% of the volume of containers or the volume of the largest container, whichever is greater. Containers that do not contain free liquids need not be considered in this determination;”
Question 6Calculation
A hazardous waste container storage area holds 3 containers of 110 gallons and one more 110-gallon container, all holding free liquids. Under the RCRA container storage rule, what is the minimum capacity the containment system must have?
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Answer: D. 110 gallons
Capacity must be the greater of 10% of total container volume (0.10 x 440 = 44 gal) or the largest container (110 gal), so 110 gallons.
Answer D rests on
40 CFR 264.175(b)(3)(opens ecfr.gov in a new tab)“The containment system must have sufficient capacity to contain 10% of the volume of containers or the volume of the largest container, whichever is greater. Containers that do not contain free liquids need not be considered in this determination;”
Question 7Calculation
A maintenance mechanic's 8-hour shift includes 2 hours at 105 dBA, 2 hours at 88 dBA, 4 hours at 100 dBA. Using OSHA's noise dose method (Table G-16a reference durations), what is the mechanic's daily noise dose?
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Answer: A. 418.9%
D = 100 x (C1/T1 + C2/T2 + ...), with T from Table G-16a: D = 100 x (2/1 + 2/10.6 + 4/2) = 418.9%. A dose of 50% or more is at or above the action level, so a hearing conservation program is required under 1910.95(c)(1).
Answer A rests on
29 CFR 1910.95, Appendix A(opens ecfr.gov in a new tab)“When the workshift noise exposure is composed of two or more periods of noise at different levels, the total noise dose over the work day is given by: D = 100(C1 / T1 + C2 / T2 + Cn / Tn), where Cn indicates the total time of exposure at a specific noise level, and Tn indicates the reference duration for that level as given by Table G-16a.”
Question 8Calculation
A maintenance mechanic's 8-hour shift includes 1 hour at 100 dBA, 6 hours at 105 dBA, 1 hour at 90 dBA. Using OSHA's noise dose method (Table G-16a reference durations), what is the mechanic's daily noise dose?
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Answer: C. 662.5%
D = 100 x (C1/T1 + C2/T2 + ...), with T from Table G-16a: D = 100 x (1/2 + 6/1 + 1/8) = 662.5%. A dose of 50% or more is at or above the action level, so a hearing conservation program is required under 1910.95(c)(1).
Answer C rests on
29 CFR 1910.95, Appendix A(opens ecfr.gov in a new tab)“When the workshift noise exposure is composed of two or more periods of noise at different levels, the total noise dose over the work day is given by: D = 100(C1 / T1 + C2 / T2 + Cn / Tn), where Cn indicates the total time of exposure at a specific noise level, and Tn indicates the reference duration for that level as given by Table G-16a.”
Question 9Calculation
A noise dosimeter worn for a full 8-hour shift reads a dose of 63%. What is the employee's 8-hour time-weighted average (TWA) sound level?
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Answer: C. 86.7 dBA
TWA = 16.61 log10(D/100) + 90 = 16.61 x log10(63/100) + 90 = 86.7 dBA. A 100% dose corresponds to a TWA of 90 dBA.
Answer C rests on
29 CFR 1910.95, Appendix A(opens ecfr.gov in a new tab)“The eight-hour time-weighted average sound level (TWA), in decibels, may be computed from the dose, in percent, by means of the formula: TWA = 16.61 log10 (D/100) + 90.”
Question 10Calculation
A noise dosimeter worn for a full 8-hour shift reads a dose of 75%. What is the employee's 8-hour time-weighted average (TWA) sound level?
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Answer: B. 87.9 dBA
TWA = 16.61 log10(D/100) + 90 = 16.61 x log10(75/100) + 90 = 87.9 dBA. A 100% dose corresponds to a TWA of 90 dBA.
Answer B rests on
29 CFR 1910.95, Appendix A(opens ecfr.gov in a new tab)“The eight-hour time-weighted average sound level (TWA), in decibels, may be computed from the dose, in percent, by means of the formula: TWA = 16.61 log10 (D/100) + 90.”
Question 11Calculation
A lifting task has H = 15 in, V = 30 in, D = 20 in, and an asymmetry angle of 0°. The frequency multiplier is 0.75 and the coupling multiplier is 0.9. Using the Revised NIOSH Lifting Equation (U.S. customary units), what is the Recommended Weight Limit (RWL)?
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Answer: C. 20.9 lb
RWL = 51 x HM x VM x DM x AM x FM x CM. HM = 10/15 = 0.667, VM = 1 - 0.0075|30 - 30| = 1.000, DM = 0.82 + 1.8/20 = 0.910, AM = 1 - 0.0032 x 0 = 1.000. RWL = 20.9 lb. Leaving out FM and CM is a common mistake.
Answer C rests on
NIOSH, Applications Manual for the Revised NIOSH Lifting Equation (DHHS Pub. No. 94-110)(opens ecfr.gov in a new tab)“The RWL is defined by the following equation: RWL = LC X HM X VM X DM X AM X FM X CM ... Load Constant LC 23kg 51lb; Horizontal Multiplier HM (25/H) (10/H); Vertical Multiplier VM 1-(.003|V-75|) 1-(.0075|V-30|); Distance Multiplier DM .82 + (4.5/D) .82 + (1.8/D); Asymmetric Multiplier AM 1-(.0032A) 1-(.0032A)”
Question 12Calculation
A worker lifts a 45-lb carton. The task's multipliers are HM = 0.50, VM = 0.925, DM = 0.880, AM = 0.856, FM = 0.94, and CM = 0.95. What is the Lifting Index (LI)?
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Answer: C. 2.84
RWL = 51 x 0.50 x 0.925 x 0.880 x 0.856 x 0.94 x 0.95 = 15.9 lb. LI = load / RWL = 45 / 15.9 = 2.84. Inverting the ratio (RWL / load) is the classic error.
Answer C rests on
NIOSH, Applications Manual for the Revised NIOSH Lifting Equation (DHHS Pub. No. 94-110)(opens ecfr.gov in a new tab)“The RWL is defined by the following equation: RWL = LC X HM X VM X DM X AM X FM X CM ... Load Constant LC 23kg 51lb; Horizontal Multiplier HM (25/H) (10/H); Vertical Multiplier VM 1-(.003|V-75|) 1-(.0075|V-30|); Distance Multiplier DM .82 + (4.5/D) .82 + (1.8/D); Asymmetric Multiplier AM 1-(.0032A) 1-(.0032A)”
Question 13Calculation
A worker lifts a 30-lb carton. The task's multipliers are HM = 0.67, VM = 0.925, DM = 0.910, AM = 1.000, FM = 0.85, and CM = 1. What is the Lifting Index (LI)?
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Answer: A. 1.23
RWL = 51 x 0.67 x 0.925 x 0.910 x 1.000 x 0.85 x 1 = 24.3 lb. LI = load / RWL = 30 / 24.3 = 1.23. Inverting the ratio (RWL / load) is the classic error.
Answer A rests on
NIOSH, Applications Manual for the Revised NIOSH Lifting Equation (DHHS Pub. No. 94-110)(opens ecfr.gov in a new tab)“The RWL is defined by the following equation: RWL = LC X HM X VM X DM X AM X FM X CM ... Load Constant LC 23kg 51lb; Horizontal Multiplier HM (25/H) (10/H); Vertical Multiplier VM 1-(.003|V-75|) 1-(.0075|V-30|); Distance Multiplier DM .82 + (4.5/D) .82 + (1.8/D); Asymmetric Multiplier AM 1-(.0032A) 1-(.0032A)”
Question 14Calculation
A lifting task has H = 12 in, V = 20 in, D = 25 in, and an asymmetry angle of 0°. The frequency multiplier is 0.85 and the coupling multiplier is 0.9. Using the Revised NIOSH Lifting Equation (U.S. customary units), what is the Recommended Weight Limit (RWL)?
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Answer: A. 26.8 lb
RWL = 51 x HM x VM x DM x AM x FM x CM. HM = 10/12 = 0.833, VM = 1 - 0.0075|20 - 30| = 0.925, DM = 0.82 + 1.8/25 = 0.892, AM = 1 - 0.0032 x 0 = 1.000. RWL = 26.8 lb. Leaving out FM and CM is a common mistake.
Answer A rests on
NIOSH, Applications Manual for the Revised NIOSH Lifting Equation (DHHS Pub. No. 94-110)(opens ecfr.gov in a new tab)“The RWL is defined by the following equation: RWL = LC X HM X VM X DM X AM X FM X CM ... Load Constant LC 23kg 51lb; Horizontal Multiplier HM (25/H) (10/H); Vertical Multiplier VM 1-(.003|V-75|) 1-(.0075|V-30|); Distance Multiplier DM .82 + (4.5/D) .82 + (1.8/D); Asymmetric Multiplier AM 1-(.0032A) 1-(.0032A)”
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