Florida Wastewater Treatment Plant
Operator Class C Certification
Examination Practice Questions &
[Verified Answers], Plus Explained
Rationales|2026 Latest Update| Instant
Download PDF
1. A wastewater treatment plant receives an average daily flow of 2.0
MGD. Approximately how many gallons of wastewater does the plant
receive in one hour?
A. 20,833 gallons
B. 41,667 gallons
C. 83,333 gallons
D. 120,000 gallons
Answer: B. 41,667 gallons
Rationale: To calculate hourly flow, divide the daily flow by 24 hours.
2,000,000 ÷ 24 = approximately 83,333 gallons per hour. Therefore,
the correct answer is actually C, not B.
Correct Answer: C. 83,333 gallons
2. What is the primary purpose of a bar screen at the headworks of a
wastewater treatment plant?
1|Page
,A. Remove dissolved organic matter
B. Remove large debris and screenings
C. Reduce ammonia concentration
D. Disinfect the wastewater
Answer: B. Remove large debris and screenings
Rationale: Bar screens provide preliminary treatment by removing
large objects such as rags, sticks, plastics, and other debris. Removing
these materials protects pumps and downstream equipment from
clogging or mechanical damage.
3. Which treatment process primarily removes settleable solids from
wastewater?
A. Primary sedimentation
B. Chlorination
C. Aeration
D. Ultraviolet disinfection
Answer: A. Primary sedimentation
Rationale: Primary sedimentation allows heavier settleable solids to
settle by gravity while floatable materials are removed by skimming.
This reduces the solids and organic loading placed on downstream
biological treatment.
4. A plant has a flow of 1.5 MGD and a primary clarifier with a surface
area of 10,000 square feet. What is the approximate surface overflow
rate?
A. 100 gpd/ft²
B. 150 gpd/ft²
2|Page
,C. 250 gpd/ft²
D. 1,500 gpd/ft²
Answer: B. 150 gpd/ft²
Rationale: Surface overflow rate equals flow in gallons per day
divided by clarifier surface area. 1.5 MGD = 1,500,000 gallons/day.
1,500,000 ÷ 10,000 = 150 gpd/ft².
5. What is the primary purpose of an aeration basin in an activated-
sludge process?
A. Remove grit
B. Provide oxygen and mixing for biological treatment
C. Disinfect final effluent
D. Remove large floating debris
Answer: B. Provide oxygen and mixing for biological treatment
Rationale: Aeration basins provide an environment where
microorganisms can consume biodegradable organic matter. Aeration
supplies oxygen for aerobic biological activity while mixing keeps
microorganisms and wastewater in contact.
6. Which parameter is commonly used to indicate the amount of
oxygen required to biologically stabilize biodegradable organic
matter?
A. BOD
B. pH
C. TSS
D. Turbidity
3|Page
, Answer: A. BOD
Rationale: Biochemical oxygen demand, or BOD, represents the
amount of dissolved oxygen microorganisms require to biologically
oxidize biodegradable organic material under specified test
conditions.
7. A wastewater sample has a BOD of 240 mg/L and a TSS
concentration of 300 mg/L. What is the BOD-to-TSS ratio?
A. 0.40
B. 0.60
C. 0.80
D. 1.25
Answer: C. 0.80
Rationale: Divide BOD by TSS: 240 ÷ 300 = 0.80. Ratios such as this can
help operators evaluate wastewater characteristics and treatment
performance.
8. What is the main purpose of returning activated sludge from the
secondary clarifier to the aeration basin?
A. Increase chlorine residual
B. Maintain an appropriate concentration of microorganisms in the
biological process
C. Remove grit from the influent
D. Increase primary sludge production
Answer: B. Maintain an appropriate concentration of microorganisms
in the biological process
4|Page
Operator Class C Certification
Examination Practice Questions &
[Verified Answers], Plus Explained
Rationales|2026 Latest Update| Instant
Download PDF
1. A wastewater treatment plant receives an average daily flow of 2.0
MGD. Approximately how many gallons of wastewater does the plant
receive in one hour?
A. 20,833 gallons
B. 41,667 gallons
C. 83,333 gallons
D. 120,000 gallons
Answer: B. 41,667 gallons
Rationale: To calculate hourly flow, divide the daily flow by 24 hours.
2,000,000 ÷ 24 = approximately 83,333 gallons per hour. Therefore,
the correct answer is actually C, not B.
Correct Answer: C. 83,333 gallons
2. What is the primary purpose of a bar screen at the headworks of a
wastewater treatment plant?
1|Page
,A. Remove dissolved organic matter
B. Remove large debris and screenings
C. Reduce ammonia concentration
D. Disinfect the wastewater
Answer: B. Remove large debris and screenings
Rationale: Bar screens provide preliminary treatment by removing
large objects such as rags, sticks, plastics, and other debris. Removing
these materials protects pumps and downstream equipment from
clogging or mechanical damage.
3. Which treatment process primarily removes settleable solids from
wastewater?
A. Primary sedimentation
B. Chlorination
C. Aeration
D. Ultraviolet disinfection
Answer: A. Primary sedimentation
Rationale: Primary sedimentation allows heavier settleable solids to
settle by gravity while floatable materials are removed by skimming.
This reduces the solids and organic loading placed on downstream
biological treatment.
4. A plant has a flow of 1.5 MGD and a primary clarifier with a surface
area of 10,000 square feet. What is the approximate surface overflow
rate?
A. 100 gpd/ft²
B. 150 gpd/ft²
2|Page
,C. 250 gpd/ft²
D. 1,500 gpd/ft²
Answer: B. 150 gpd/ft²
Rationale: Surface overflow rate equals flow in gallons per day
divided by clarifier surface area. 1.5 MGD = 1,500,000 gallons/day.
1,500,000 ÷ 10,000 = 150 gpd/ft².
5. What is the primary purpose of an aeration basin in an activated-
sludge process?
A. Remove grit
B. Provide oxygen and mixing for biological treatment
C. Disinfect final effluent
D. Remove large floating debris
Answer: B. Provide oxygen and mixing for biological treatment
Rationale: Aeration basins provide an environment where
microorganisms can consume biodegradable organic matter. Aeration
supplies oxygen for aerobic biological activity while mixing keeps
microorganisms and wastewater in contact.
6. Which parameter is commonly used to indicate the amount of
oxygen required to biologically stabilize biodegradable organic
matter?
A. BOD
B. pH
C. TSS
D. Turbidity
3|Page
, Answer: A. BOD
Rationale: Biochemical oxygen demand, or BOD, represents the
amount of dissolved oxygen microorganisms require to biologically
oxidize biodegradable organic material under specified test
conditions.
7. A wastewater sample has a BOD of 240 mg/L and a TSS
concentration of 300 mg/L. What is the BOD-to-TSS ratio?
A. 0.40
B. 0.60
C. 0.80
D. 1.25
Answer: C. 0.80
Rationale: Divide BOD by TSS: 240 ÷ 300 = 0.80. Ratios such as this can
help operators evaluate wastewater characteristics and treatment
performance.
8. What is the main purpose of returning activated sludge from the
secondary clarifier to the aeration basin?
A. Increase chlorine residual
B. Maintain an appropriate concentration of microorganisms in the
biological process
C. Remove grit from the influent
D. Increase primary sludge production
Answer: B. Maintain an appropriate concentration of microorganisms
in the biological process
4|Page
