Safety Math: Incident Rates, Noise Dose, Slopes, Lifting practice questions for the ASP exam
ASP11 Domain 1: Mathematical Calculations (1.4, 1.5, 1.8, 1.9 Calculations). 10 free questions below. Every answer quotes and links the formula source it rests on.
A manufacturing plant logged 18 OSHA-recordable injury and illness cases last year. Its 540 employees worked a total of 1,116,000 hours. What is the plant's total recordable incidence rate per 100 full-time workers?
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Answer: A. 3.23
Incidence rate = (N x 200,000) / hours worked = (18 x 200,000) / 1,116,000 = 3.23. The 200,000 base represents 100 employees working 40 hours a week for 50 weeks. Using 100,000 as the base would give half the correct value.
“(Number of injuries and illnesses X 200,000) / Employee hours worked = Incidence rate (The 200,000 hours in the formula represents the equivalent of 100 employees working 40 hours per week, 50 weeks per year, and provides the standard base for the incidence rates.)”
A manufacturing plant logged 17 OSHA-recordable injury and illness cases last year. Its 160 employees worked a total of 320,000 hours. What is the plant's total recordable incidence rate per 100 full-time workers?
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Answer: D. 10.63
Incidence rate = (N x 200,000) / hours worked = (17 x 200,000) / 320,000 = 10.63. The 200,000 base represents 100 employees working 40 hours a week for 50 weeks. Using 100,000 as the base would give half the correct value.
“(Number of injuries and illnesses X 200,000) / Employee hours worked = Incidence rate (The 200,000 hours in the formula represents the equivalent of 100 employees working 40 hours per week, 50 weeks per year, and provides the standard base for the incidence rates.)”
A warehouse recorded 10 OSHA-recordable cases in a year: 5 with days away from work, 3 with job transfer or restriction only, and 2 other recordable cases. Employees worked 330,000 hours. What is the DART rate (days away, restricted, or transferred) per 100 full-time workers?
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Answer: B. 4.85
DART uses the same formula as the total recordable rate but counts only cases with days away, restriction, or transfer: (5 + 3) x 200,000 / 330,000 = 4.85. Including the 2 other recordable cases gives the total recordable rate instead.
“The same formula can be used to compute the incidence rate for the most serious injury and illness cases, defined here as cases that result in workers taking time off from their jobs or being transferred to another job or doing lighter (restricted) duties.”
A warehouse recorded 21 OSHA-recordable cases in a year: 6 with days away from work, 3 with job transfer or restriction only, and 12 other recordable cases. Employees worked 500,000 hours. What is the DART rate (days away, restricted, or transferred) per 100 full-time workers?
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Answer: A. 3.60
DART uses the same formula as the total recordable rate but counts only cases with days away, restriction, or transfer: (6 + 3) x 200,000 / 500,000 = 3.60. Including the 12 other recordable cases gives the total recordable rate instead.
“The same formula can be used to compute the incidence rate for the most serious injury and illness cases, defined here as cases that result in workers taking time off from their jobs or being transferred to another job or doing lighter (restricted) duties.”
A maintenance mechanic's 8-hour shift includes 4 hours at 97 dBA, 2 hours at 88 dBA, 2 hours at 92 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: B. 185.0%
D = 100 x (C1/T1 + C2/T2 + ...), with T from Table G-16a: D = 100 x (4/3 + 2/10.6 + 2/6.1) = 185.0%. 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).
“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.”
A maintenance mechanic's 8-hour shift includes 4 hours at 85 dBA, 2 hours at 88 dBA, 2 hours at 102 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. 177.2%
D = 100 x (C1/T1 + C2/T2 + ...), with T from Table G-16a: D = 100 x (4/16 + 2/10.6 + 2/1.5) = 177.2%. 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).
“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.”
A noise dosimeter worn for a full 8-hour shift reads a dose of 120%. What is the employee's 8-hour time-weighted average (TWA) sound level?
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Answer: C. 91.3 dBA
TWA = 16.61 log10(D/100) + 90 = 16.61 x log10(120/100) + 90 = 91.3 dBA. A 100% dose corresponds to a TWA of 90 dBA.
“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.”
A noise dosimeter worn for a full 8-hour shift reads a dose of 180%. What is the employee's 8-hour time-weighted average (TWA) sound level?
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Answer: C. 94.2 dBA
TWA = 16.61 log10(D/100) + 90 = 16.61 x log10(180/100) + 90 = 94.2 dBA. A 100% dose corresponds to a TWA of 90 dBA.
“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.”
A crew will dig an 8-foot-deep trench in Type C soil with a 4-foot-wide bottom, using a simple slope on both sides at the maximum allowable slope (not short-term). What is the minimum width at the top of the excavation?
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Answer: A. 28 ft
Type C soil allows a maximum simple slope of 1 1/2:1 (horizontal:vertical). Each side extends 8 x 1.5 = 12 ft horizontally, so top width = 4 + 2 x 12 ft = 28 ft.
“All simple slope excavations 20 feet or less in depth shall have a maximum allowable slope of 1 1/2 :1.”
A crew will dig a 14-foot-deep trench in Type B soil with a 3-foot-wide bottom, using a simple slope on both sides at the maximum allowable slope (not short-term). What is the minimum width at the top of the excavation?
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Answer: C. 31 ft
Type B soil allows a maximum simple slope of 1:1 (horizontal:vertical). Each side extends 14 x 1 = 14 ft horizontally, so top width = 3 + 2 x 14 ft = 31 ft.
“All simple slope excavations 20 feet or less in depth shall have a maximum allowable slope of 1:1.”
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