CALIFORNIA WASTEWATER TREATMENT OPERATOR GRADE V
PRACTICE EXAM QUESTIONS AND CORRECT ANSWERS (VERIFIED
ANSWERS) WITH RATIONALES 2026 | MCQs |INSTANT DOWNLOAD
PDF VERSION LATEST UPDATE
SECTION 1: ACTIVATED SLUDGE AND BIOLOGICAL TREATMENT
(Questions 1-30)
Question 1
Which of the following is the primary purpose of activated sludge in wastewater
treatment?
A. To filter out solids
B. To biologically remove organic matter
C. To disinfect the effluent
D. To increase pH
Answer: B
Rationale: Activated sludge uses microorganisms to consume and break down
organic pollutants, reducing BOD and improving water quality.[reference:13]
Question 2
In wastewater treatment, BOD stands for:
A. Basic Oxidation Demand
B. Biological Oxygen Deficiency
C. Biochemical Oxygen Demand
D. Biological Oxidation Discharge
Answer: C
Rationale: BOD measures the amount of oxygen microorganisms need to
decompose
organic matter in water over a specific time, typically 5 days
(BOD5).[reference:14]
Question 3
1
,In an activated sludge system, the primary purpose of returning activated sludge
(RAS) is to:
A. Increase oxygen transfer
B. Maintain microorganisms in the aeration tank
C. Remove nutrients from the effluent
D. Increase chlorine demand
Answer: B
Rationale: Returning activated sludge keeps a high concentration of
microorganisms in the aeration tank so they can continue breaking down organic
waste.[reference:15]
Question 4
Which factor most significantly affects the growth rate of microorganisms in an
aeration tank?
A. Temperature
B. pH
C. Dissolved oxygen
D. All of the above
Answer: D
Rationale: Microbial growth depends on temperature, pH, dissolved oxygen, and
nutrient availability; all these factors must be optimized for effective
treatment.[reference:16]
Question 5
The Food-to-Microorganism (F/M) ratio is calculated as:
A. Influent BOD ÷ Aeration tank volume
B. Influent BOD ÷ Mixed liquor suspended solids (MLSS)
C. Effluent BOD ÷ Influent BOD
D. Return sludge flow ÷ Influent flow
Answer: B
2
,Rationale: F/M ratio = (Influent BOD × Flow) ÷ (Aeration tank volume × MLSS).
It indicates the amount of food available per unit of microorganisms.
Question 6
A low F/M ratio in an activated sludge system typically results in:
A. Pinpoint floc and poor settling
B. Filamentous bulking
C. High oxygen demand
D. Excessive foam production
Answer: A
Rationale: Low F/M ratios (starved conditions) can lead to pinpoint floc
formation and poor solids settling.
Question 7
Sludge Volume Index (SVI) is calculated as:
A. Settled sludge volume (mL/L) ÷ MLSS (mg/L) × 1000
B. MLSS ÷ Settled sludge volume × 1000
C. Influent flow ÷ Return sludge flow
D. Aeration tank volume ÷ Clarifier surface area
Answer: A
Rationale: SVI = (Settled sludge volume in mL/L after 30 minutes ÷ MLSS in mg/L)
× 1000. Normal SVI ranges from 50-150 mL/g.
Question 8
An SVI value greater than 150 mL/g typically indicates:
A. Good settling sludge
B. Bulking sludge
C. Pinpoint floc
D. Low MLSS concentration
Answer: B
3
, Rationale: SVI > 150 mL/g indicates poor settling and potential bulking
conditions, often caused by filamentous bacteria overgrowth.
Question 9
Which microorganism is primarily responsible for nitrification?
A. Escherichia coli
B. Nitrosomonas and Nitrobacter
C. Bacillus subtilis
D. Streptococcus
Answer: B
Rationale: Nitrifying bacteria include Nitrosomonas (ammonia to nitrite) and
Nitrobacter (nitrite to nitrate).[reference:17]
Question 10
The process of converting ammonia to nitrite is performed by:
A. Nitrobacter
B. Nitrosomonas
C. Pseudomonas
D. Zoogloea
Answer: B
Rationale: Nitrosomonas oxidizes ammonia (NH3) to nitrite (NO2-). Nitrobacter
then oxidizes nitrite to nitrate (NO3-).[reference:18]
Question 11
Aeration in an activated sludge system serves which primary function?
A. Mixing only
B. Oxygen transfer for microbial respiration and mixing
C. Temperature control
D. pH adjustment
Answer: B
4
PRACTICE EXAM QUESTIONS AND CORRECT ANSWERS (VERIFIED
ANSWERS) WITH RATIONALES 2026 | MCQs |INSTANT DOWNLOAD
PDF VERSION LATEST UPDATE
SECTION 1: ACTIVATED SLUDGE AND BIOLOGICAL TREATMENT
(Questions 1-30)
Question 1
Which of the following is the primary purpose of activated sludge in wastewater
treatment?
A. To filter out solids
B. To biologically remove organic matter
C. To disinfect the effluent
D. To increase pH
Answer: B
Rationale: Activated sludge uses microorganisms to consume and break down
organic pollutants, reducing BOD and improving water quality.[reference:13]
Question 2
In wastewater treatment, BOD stands for:
A. Basic Oxidation Demand
B. Biological Oxygen Deficiency
C. Biochemical Oxygen Demand
D. Biological Oxidation Discharge
Answer: C
Rationale: BOD measures the amount of oxygen microorganisms need to
decompose
organic matter in water over a specific time, typically 5 days
(BOD5).[reference:14]
Question 3
1
,In an activated sludge system, the primary purpose of returning activated sludge
(RAS) is to:
A. Increase oxygen transfer
B. Maintain microorganisms in the aeration tank
C. Remove nutrients from the effluent
D. Increase chlorine demand
Answer: B
Rationale: Returning activated sludge keeps a high concentration of
microorganisms in the aeration tank so they can continue breaking down organic
waste.[reference:15]
Question 4
Which factor most significantly affects the growth rate of microorganisms in an
aeration tank?
A. Temperature
B. pH
C. Dissolved oxygen
D. All of the above
Answer: D
Rationale: Microbial growth depends on temperature, pH, dissolved oxygen, and
nutrient availability; all these factors must be optimized for effective
treatment.[reference:16]
Question 5
The Food-to-Microorganism (F/M) ratio is calculated as:
A. Influent BOD ÷ Aeration tank volume
B. Influent BOD ÷ Mixed liquor suspended solids (MLSS)
C. Effluent BOD ÷ Influent BOD
D. Return sludge flow ÷ Influent flow
Answer: B
2
,Rationale: F/M ratio = (Influent BOD × Flow) ÷ (Aeration tank volume × MLSS).
It indicates the amount of food available per unit of microorganisms.
Question 6
A low F/M ratio in an activated sludge system typically results in:
A. Pinpoint floc and poor settling
B. Filamentous bulking
C. High oxygen demand
D. Excessive foam production
Answer: A
Rationale: Low F/M ratios (starved conditions) can lead to pinpoint floc
formation and poor solids settling.
Question 7
Sludge Volume Index (SVI) is calculated as:
A. Settled sludge volume (mL/L) ÷ MLSS (mg/L) × 1000
B. MLSS ÷ Settled sludge volume × 1000
C. Influent flow ÷ Return sludge flow
D. Aeration tank volume ÷ Clarifier surface area
Answer: A
Rationale: SVI = (Settled sludge volume in mL/L after 30 minutes ÷ MLSS in mg/L)
× 1000. Normal SVI ranges from 50-150 mL/g.
Question 8
An SVI value greater than 150 mL/g typically indicates:
A. Good settling sludge
B. Bulking sludge
C. Pinpoint floc
D. Low MLSS concentration
Answer: B
3
, Rationale: SVI > 150 mL/g indicates poor settling and potential bulking
conditions, often caused by filamentous bacteria overgrowth.
Question 9
Which microorganism is primarily responsible for nitrification?
A. Escherichia coli
B. Nitrosomonas and Nitrobacter
C. Bacillus subtilis
D. Streptococcus
Answer: B
Rationale: Nitrifying bacteria include Nitrosomonas (ammonia to nitrite) and
Nitrobacter (nitrite to nitrate).[reference:17]
Question 10
The process of converting ammonia to nitrite is performed by:
A. Nitrobacter
B. Nitrosomonas
C. Pseudomonas
D. Zoogloea
Answer: B
Rationale: Nitrosomonas oxidizes ammonia (NH3) to nitrite (NO2-). Nitrobacter
then oxidizes nitrite to nitrate (NO3-).[reference:18]
Question 11
Aeration in an activated sludge system serves which primary function?
A. Mixing only
B. Oxygen transfer for microbial respiration and mixing
C. Temperature control
D. pH adjustment
Answer: B
4
