NJ WASTEWATER TREATMENT OPERATOR
S2 EXAM COMPLETE PRACTICE EXAM
VERIFIED ANSWERS WITH RATIONALES |
2026 EDITION
SECTION 1: ACTIVATED SLUDGE PROCESS CONTROL (20
Questions)
1. A conventional activated sludge plant has an MLSS of 2,500 mg/L
and a settleability (30-min) of 300 mL/L. What is the SVI?
A) 80 mL/g
B) 100 mL/g
C) 120 mL/g
D) 150 mL/g
✅ Correct Answer: C
✅ Rationale: SVI = (Settleability mL/L × 1000) / MLSS mg/L = (300 ×
1000) / 2500 = 120 mL/g. Normal range 80–120; above 150 indicates
bulking.
,2. The food-to-microorganism (F/M) ratio is too high. What should the
operator do?
A) Increase waste activated sludge (WAS) flow
B) Decrease return activated sludge (RAS) flow
C) Increase MLSS by reducing WAS
D) Add more primary effluent
✅ Correct Answer: C
✅ Rationale: High F/M means too much food for microorganisms.
Increasing MLSS (by wasting less) adds more biomass to consume the
food, lowering F/M.
3. Which condition typically causes dispersed growth (turbid
effluent)?
A) Old sludge (high MCRT)
B) Young sludge (low MCRT) with high shear
C) Low DO (<0.5 mg/L)
D) High RAS rate
✅ Correct Answer: B
✅ Rationale: Dispersed growth (pin floc, turbid effluent) occurs when
sludge is too young (low MCRT) or subjected to excessive aeration
shear. Increasing MCRT helps.
,4. A filamentous bulking problem is identified as Type 1701 (low F/M
filaments). The best control strategy is:
A) Add chlorine to return sludge
B) Increase F/M by wasting more sludge
C) Increase DO to 4 mg/L
D) Add polymer to secondary clarifier
✅ Correct Answer: B
✅ Rationale: Low F/M filaments are controlled by increasing F/M
(waste more sludge, reduce MLSS). Chlorine is a temporary band-aid,
not a process fix.
5. What is the typical MCRT (mean cell residence time) range for
conventional activated sludge?
A) 1–3 days
B) 3–8 days
C) 10–20 days
D) 25–40 days
✅ Correct Answer: B
✅ Rationale: Conventional activated sludge: MCRT 3–8 days. Extended
aeration: 20–30+ days. High-rate: 0.5–2 days.
6. The aeration basin DO is 0.3 mg/L. What is the immediate concern?
, A) Nitrification will increase
B) Filamentous organisms may proliferate
C) Sludge will settle faster
D) Phosphorus removal improves
✅ Correct Answer: B
✅ Rationale: Low DO (<1.0 mg/L) favors filamentous organisms like S.
natans and Type 1701, leading to bulking. Maintain DO at 1.5–3.0 mg/L.
7. RAS flow is typically what percentage of influent flow for
conventional activated sludge?
A) 10–20%
B) 25–50%
C) 75–100%
D) 150–200%
✅ Correct Answer: C
✅ Rationale: Conventional plants return 75–100% of influent flow as
RAS. Extended aeration may use 50–75%; high-rate may use 25–50%.
8. A microscopic exam shows many stalked ciliates and rotifers. This
indicates:
A) Young sludge, good settleability
B) Old sludge, stable operation
S2 EXAM COMPLETE PRACTICE EXAM
VERIFIED ANSWERS WITH RATIONALES |
2026 EDITION
SECTION 1: ACTIVATED SLUDGE PROCESS CONTROL (20
Questions)
1. A conventional activated sludge plant has an MLSS of 2,500 mg/L
and a settleability (30-min) of 300 mL/L. What is the SVI?
A) 80 mL/g
B) 100 mL/g
C) 120 mL/g
D) 150 mL/g
✅ Correct Answer: C
✅ Rationale: SVI = (Settleability mL/L × 1000) / MLSS mg/L = (300 ×
1000) / 2500 = 120 mL/g. Normal range 80–120; above 150 indicates
bulking.
,2. The food-to-microorganism (F/M) ratio is too high. What should the
operator do?
A) Increase waste activated sludge (WAS) flow
B) Decrease return activated sludge (RAS) flow
C) Increase MLSS by reducing WAS
D) Add more primary effluent
✅ Correct Answer: C
✅ Rationale: High F/M means too much food for microorganisms.
Increasing MLSS (by wasting less) adds more biomass to consume the
food, lowering F/M.
3. Which condition typically causes dispersed growth (turbid
effluent)?
A) Old sludge (high MCRT)
B) Young sludge (low MCRT) with high shear
C) Low DO (<0.5 mg/L)
D) High RAS rate
✅ Correct Answer: B
✅ Rationale: Dispersed growth (pin floc, turbid effluent) occurs when
sludge is too young (low MCRT) or subjected to excessive aeration
shear. Increasing MCRT helps.
,4. A filamentous bulking problem is identified as Type 1701 (low F/M
filaments). The best control strategy is:
A) Add chlorine to return sludge
B) Increase F/M by wasting more sludge
C) Increase DO to 4 mg/L
D) Add polymer to secondary clarifier
✅ Correct Answer: B
✅ Rationale: Low F/M filaments are controlled by increasing F/M
(waste more sludge, reduce MLSS). Chlorine is a temporary band-aid,
not a process fix.
5. What is the typical MCRT (mean cell residence time) range for
conventional activated sludge?
A) 1–3 days
B) 3–8 days
C) 10–20 days
D) 25–40 days
✅ Correct Answer: B
✅ Rationale: Conventional activated sludge: MCRT 3–8 days. Extended
aeration: 20–30+ days. High-rate: 0.5–2 days.
6. The aeration basin DO is 0.3 mg/L. What is the immediate concern?
, A) Nitrification will increase
B) Filamentous organisms may proliferate
C) Sludge will settle faster
D) Phosphorus removal improves
✅ Correct Answer: B
✅ Rationale: Low DO (<1.0 mg/L) favors filamentous organisms like S.
natans and Type 1701, leading to bulking. Maintain DO at 1.5–3.0 mg/L.
7. RAS flow is typically what percentage of influent flow for
conventional activated sludge?
A) 10–20%
B) 25–50%
C) 75–100%
D) 150–200%
✅ Correct Answer: C
✅ Rationale: Conventional plants return 75–100% of influent flow as
RAS. Extended aeration may use 50–75%; high-rate may use 25–50%.
8. A microscopic exam shows many stalked ciliates and rotifers. This
indicates:
A) Young sludge, good settleability
B) Old sludge, stable operation
