Oklahoma Wastewater Operator
Class A Exam Practice Questions And
Correct Answers (Verified Answers)
Plus Rationale 2027 Q&A| Instant
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1. A Class A wastewater operator is evaluating the overall
performance of an activated-sludge treatment plant. The influent
flow is 2.0 MGD, the aeration basin contains 2.5 million gallons, and
the mixed liquor suspended solids concentration is 2,500 mg/L.
Which operational calculation is most directly used to evaluate the
relationship between the mass of microorganisms in the aeration
basin and the organic loading entering the process?
A. Hydraulic loading rate
B. Food-to-microorganism ratio
C. Surface overflow rate
D. Solids capture rate
B. Food-to-microorganism ratio
,The food-to-microorganism ratio compares the biodegradable organic
loading entering the biological process with the quantity of
microorganisms available for treatment and is a fundamental activated-
sludge process-control parameter.
2. A wastewater treatment facility has an aeration basin receiving an
influent BOD₅ load of 2,000 lb/day. The aeration basin contains
3,000 lb of mixed-liquor volatile suspended solids. What is the
approximate F/M ratio?
A. 0.33 day⁻¹
B. 0.67 day⁻¹
C. 1.50 day⁻¹
D. 6.00 day⁻¹
B. 0.67 day⁻¹
The F/M ratio is calculated by dividing the pounds of BOD₅ applied per
day by the pounds of microorganisms in the aeration basin, giving 2,000
÷ 3,000 = approximately 0.67 day⁻¹.
3. An operator notices that the secondary clarifier effluent has
become increasingly turbid while the return activated sludge
concentration has declined. Microscopic examination indicates a
predominance of dispersed, poorly settling organisms rather than
well-formed floc. Which condition is most consistent with this
observation?
,A. Excessive nitrification
B. Pin floc or poor floc formation
C. Excessive grit removal
D. Excessive chlorine residual
B. Pin floc or poor floc formation
Poorly formed biological floc can remain suspended and escape with the
clarifier effluent, producing elevated turbidity and suspended-solids
concentrations even when the biological process is receiving adequate
aeration.
4. A wastewater plant is experiencing filamentous bulking, and the
sludge volume index has increased substantially. Which operating
measurement should the Class A operator examine most closely
when evaluating whether the aeration process is receiving sufficient
oxygen?
A. Dissolved oxygen concentration
B. Chlorine contact time
C. Primary clarifier surface area
D. Sludge-hauling frequency
A. Dissolved oxygen concentration
Dissolved oxygen is a critical process-control measurement because
inadequate oxygen can favor filamentous organisms and impair floc
formation and settling.
, 5. A treatment plant has an average flow of 4.0 MGD and a primary
clarifier with a liquid volume of 500,000 gallons. What is the
approximate hydraulic detention time?
A. 0.5 hour
B. 1.5 hours
C. 3.0 hours
D. 6.0 hours
B. 3.0 hours
Detention time is calculated as volume divided by flow; 500,000 gallons
÷ 4,000,000 gallons/day equals 0.125 day, or approximately 3 hours.
6. A Class A operator is reviewing the performance of a primary
clarifier. The clarifier is receiving a substantially higher flow than
normal, and suspended solids are beginning to appear in the
primary effluent. Which hydraulic parameter is most likely to have
increased?
A. Solids retention time
B. Surface overflow rate
C. Sludge age
D. Volatile solids destruction
B. Surface overflow rate
Class A Exam Practice Questions And
Correct Answers (Verified Answers)
Plus Rationale 2027 Q&A| Instant
Download Pdf.
1. A Class A wastewater operator is evaluating the overall
performance of an activated-sludge treatment plant. The influent
flow is 2.0 MGD, the aeration basin contains 2.5 million gallons, and
the mixed liquor suspended solids concentration is 2,500 mg/L.
Which operational calculation is most directly used to evaluate the
relationship between the mass of microorganisms in the aeration
basin and the organic loading entering the process?
A. Hydraulic loading rate
B. Food-to-microorganism ratio
C. Surface overflow rate
D. Solids capture rate
B. Food-to-microorganism ratio
,The food-to-microorganism ratio compares the biodegradable organic
loading entering the biological process with the quantity of
microorganisms available for treatment and is a fundamental activated-
sludge process-control parameter.
2. A wastewater treatment facility has an aeration basin receiving an
influent BOD₅ load of 2,000 lb/day. The aeration basin contains
3,000 lb of mixed-liquor volatile suspended solids. What is the
approximate F/M ratio?
A. 0.33 day⁻¹
B. 0.67 day⁻¹
C. 1.50 day⁻¹
D. 6.00 day⁻¹
B. 0.67 day⁻¹
The F/M ratio is calculated by dividing the pounds of BOD₅ applied per
day by the pounds of microorganisms in the aeration basin, giving 2,000
÷ 3,000 = approximately 0.67 day⁻¹.
3. An operator notices that the secondary clarifier effluent has
become increasingly turbid while the return activated sludge
concentration has declined. Microscopic examination indicates a
predominance of dispersed, poorly settling organisms rather than
well-formed floc. Which condition is most consistent with this
observation?
,A. Excessive nitrification
B. Pin floc or poor floc formation
C. Excessive grit removal
D. Excessive chlorine residual
B. Pin floc or poor floc formation
Poorly formed biological floc can remain suspended and escape with the
clarifier effluent, producing elevated turbidity and suspended-solids
concentrations even when the biological process is receiving adequate
aeration.
4. A wastewater plant is experiencing filamentous bulking, and the
sludge volume index has increased substantially. Which operating
measurement should the Class A operator examine most closely
when evaluating whether the aeration process is receiving sufficient
oxygen?
A. Dissolved oxygen concentration
B. Chlorine contact time
C. Primary clarifier surface area
D. Sludge-hauling frequency
A. Dissolved oxygen concentration
Dissolved oxygen is a critical process-control measurement because
inadequate oxygen can favor filamentous organisms and impair floc
formation and settling.
, 5. A treatment plant has an average flow of 4.0 MGD and a primary
clarifier with a liquid volume of 500,000 gallons. What is the
approximate hydraulic detention time?
A. 0.5 hour
B. 1.5 hours
C. 3.0 hours
D. 6.0 hours
B. 3.0 hours
Detention time is calculated as volume divided by flow; 500,000 gallons
÷ 4,000,000 gallons/day equals 0.125 day, or approximately 3 hours.
6. A Class A operator is reviewing the performance of a primary
clarifier. The clarifier is receiving a substantially higher flow than
normal, and suspended solids are beginning to appear in the
primary effluent. Which hydraulic parameter is most likely to have
increased?
A. Solids retention time
B. Surface overflow rate
C. Sludge age
D. Volatile solids destruction
B. Surface overflow rate
