NJ WASTEWATER TREATMENT OPERATOR S2 EXAM COMPLETE PRACTICE EXAM VERIFIED ANSWERS WITH RATIONALES | 2026 EDITION

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