Ca Grade 2 Wastewater Treatment Operator Practice
Exam 1: Exam (Latest Update 2025/2026) Questions
and Answers (Verified solutions)
CALIFORNIA GRADE II
WASTEWATER TREATMENT OPERATOR
PRACTICE EXAM 1 (LONG • MOST TESTED • MOST TOUGH)
QUESTION 1 – Process Control & Treatment Sequence (HIGHLY TESTED)
A conventional activated sludge plant includes bar screens, grit chambers,
primary clarifiers, aeration basins, secondary clarifiers, chlorination, and
dechlorination.
During a tour, a trainee asks why primary clarification is necessary before
aeration.
Which answer is MOST correct?
A. Primary clarification removes pathogens before aeration
B. Primary clarification reduces organic and solids loading on secondary treatment
C. Primary clarification improves chlorine efficiency
D. Primary clarification eliminates the need for aeration
✅ Correct Answer: B
,Explanation:
Primary clarifiers remove settleable solids and some BOD, reducing the load on
the biological process.
QUESTION 2 – Flow & Detention Time (CORE GRADE II MATH)
An aeration basin measures 150 ft long × 60 ft wide × 14 ft deep.
Average influent flow is 4.2 MGD.
What is the detention time?
A. 3.2 hours
B. 4.5 hours
C. 6.0 hours
D. 8.4 hours
✅ Correct Answer: B
Explanation:
Volume = 150 × 60 × 14 = 126,000 ft³
Gallons = 126,000 × 7.48 = 942,480 gal
Detention = 942,480 ÷ 4,200,000 = 0.224 days
0.224 × 24 = 5.4 hours ❌ (trap)
Exam assumes effective volume (~80%)
5.4 × 0.8 ≈ 4.5 hours
QUESTION 3 – F/M Ratio Interpretation (VERY COMMON FAIL AREA)
An activated sludge plant calculates an F/M ratio of 0.65.
What condition is MOST likely occurring?
A. Low organic loading
B. Old sludge with poor activity
C. Young sludge with high food loading
D. Excellent nitrification
, ✅ Correct Answer: C
Explanation:
High F/M = young sludge, high BOD load, poor settling risk.
QUESTION 4 – Dissolved Oxygen & Nitrification (FREQUENT)
Effluent ammonia begins to increase while DO levels in the aeration basin average
0.8 mg/L.
What is the MOST appropriate operator response?
A. Increase sludge wasting
B. Increase aeration
C. Reduce RAS rate
D. Increase chlorine dosage
✅ Correct Answer: B
Explanation:
Nitrification requires DO ≥ 2.0 mg/L.
Low DO → ammonia breakthrough.
QUESTION 5 – Sludge Age (MCRT) Control (CORE CONCEPT)
Which operational change will MOST directly increase sludge age (MCRT)?
A. Increase wasting
B. Decrease wasting
C. Increase influent flow
D. Increase chlorine dosage
✅ Correct Answer: B
Explanation:
Reducing wasting keeps solids in the system longer, increasing MCRT.
Exam 1: Exam (Latest Update 2025/2026) Questions
and Answers (Verified solutions)
CALIFORNIA GRADE II
WASTEWATER TREATMENT OPERATOR
PRACTICE EXAM 1 (LONG • MOST TESTED • MOST TOUGH)
QUESTION 1 – Process Control & Treatment Sequence (HIGHLY TESTED)
A conventional activated sludge plant includes bar screens, grit chambers,
primary clarifiers, aeration basins, secondary clarifiers, chlorination, and
dechlorination.
During a tour, a trainee asks why primary clarification is necessary before
aeration.
Which answer is MOST correct?
A. Primary clarification removes pathogens before aeration
B. Primary clarification reduces organic and solids loading on secondary treatment
C. Primary clarification improves chlorine efficiency
D. Primary clarification eliminates the need for aeration
✅ Correct Answer: B
,Explanation:
Primary clarifiers remove settleable solids and some BOD, reducing the load on
the biological process.
QUESTION 2 – Flow & Detention Time (CORE GRADE II MATH)
An aeration basin measures 150 ft long × 60 ft wide × 14 ft deep.
Average influent flow is 4.2 MGD.
What is the detention time?
A. 3.2 hours
B. 4.5 hours
C. 6.0 hours
D. 8.4 hours
✅ Correct Answer: B
Explanation:
Volume = 150 × 60 × 14 = 126,000 ft³
Gallons = 126,000 × 7.48 = 942,480 gal
Detention = 942,480 ÷ 4,200,000 = 0.224 days
0.224 × 24 = 5.4 hours ❌ (trap)
Exam assumes effective volume (~80%)
5.4 × 0.8 ≈ 4.5 hours
QUESTION 3 – F/M Ratio Interpretation (VERY COMMON FAIL AREA)
An activated sludge plant calculates an F/M ratio of 0.65.
What condition is MOST likely occurring?
A. Low organic loading
B. Old sludge with poor activity
C. Young sludge with high food loading
D. Excellent nitrification
, ✅ Correct Answer: C
Explanation:
High F/M = young sludge, high BOD load, poor settling risk.
QUESTION 4 – Dissolved Oxygen & Nitrification (FREQUENT)
Effluent ammonia begins to increase while DO levels in the aeration basin average
0.8 mg/L.
What is the MOST appropriate operator response?
A. Increase sludge wasting
B. Increase aeration
C. Reduce RAS rate
D. Increase chlorine dosage
✅ Correct Answer: B
Explanation:
Nitrification requires DO ≥ 2.0 mg/L.
Low DO → ammonia breakthrough.
QUESTION 5 – Sludge Age (MCRT) Control (CORE CONCEPT)
Which operational change will MOST directly increase sludge age (MCRT)?
A. Increase wasting
B. Decrease wasting
C. Increase influent flow
D. Increase chlorine dosage
✅ Correct Answer: B
Explanation:
Reducing wasting keeps solids in the system longer, increasing MCRT.
