Louisiana Class IV Wastewater Operator
Exam Practice Questions And Correct
Answers (Verified Answers) Plus
1. A 5 MGD activated sludge plant achieves 85% BOD removal. If the influent BOD is
220 mg/L, what is the effluent BOD concentration in mg/L?
A. 33
B. 37
C. 44
D. 55
Answer: A. 33
Rationale: Effluent BOD = influent BOD × (1 - removal efficiency) = 220 × (1 - 0.85) =
220 × 0.15 = 33 mg/L. Options B, C, and D result from incorrect multiplication or
misapplication of the removal percentage.
2. In a biological nutrient removal process, which condition is essential for achieving
simultaneous nitrification and denitrification?
A. High dissolved oxygen (>2 mg/L) throughout the basin
B. Alternating aerobic and anoxic zones with internal recycle
C. Low SRT (<5 days) and high F/M ratio
D. Addition of external carbon source in aerobic zone
Answer: B. Alternating aerobic and anoxic zones with internal recycle
Rationale: Simultaneous nitrification and denitrification requires alternating aerobic and
anoxic conditions, often achieved with internal recycle to bring nitrate to the anoxic zone.
High DO inhibits denitrification, low SRT washes out nitrifiers, and adding carbon in the
aerobic zone does not promote anoxic denitrification.
Page 1 of 78
,3. A UV disinfection system receives effluent with transmittance of 65%. To achieve
the same log inactivation as at 75% transmittance, how must the UV dose be
adjusted?
A. Decrease dose proportionally to transmittance
B. Increase dose to compensate for reduced transmittance
C. Maintain the same dose since UV dose is independent of transmittance
D. Increase transmittance by adding coagulant
Answer: B. Increase dose to compensate for reduced transmittance
Rationale: Lower transmittance means less UV light penetrates the water, reducing the
effective dose. To achieve the same inactivation, the UV dose must be increased. Option A
is incorrect because dose and transmittance are inversely related; C ignores the effect of
transmittance; D is not a direct adjustment to the UV system.
4. Which polymer characteristic is most critical for effective conditioning of
anaerobically digested sludge prior to belt filter pressing?
A. High molecular weight and moderate charge density
B. Low molecular weight and high charge density
C. Cationic charge with very low molecular weight
D. Anionic charge with high molecular weight
Answer: A. High molecular weight and moderate charge density
Rationale: High molecular weight polymers with moderate charge density effectively
bridge and flocculate digested sludge particles, improving dewaterability. Low molecular
weight or incorrect charge types (anionic for typically negatively charged sludge) result in
poor flocculation and high filtrate solids.
Page 2 of 78
,5. Under Louisiana Department of Environmental Quality (LDEQ) regulations, which
of the following is a Class IV operator's responsibility regarding bypass reporting?
A. Report bypasses only if they exceed 10,000 gallons
B. Notify LDEQ orally within 24 hours and submit written report within 5 days
C. Report bypasses in the monthly discharge monitoring report (DMR) only
D. No reporting required if bypass is within permitted limits
Answer: B. Notify LDEQ orally within 24 hours and submit written report within 5
days
Rationale: LDEQ requires oral notification within 24 hours of a bypass and a written
report within 5 days, regardless of volume, to ensure timely regulatory oversight. Options
A, C, and D underestimate reporting requirements and could lead to violations.
6. A sample has a 5-day BOD of 200 mg/L and a ultimate BOD (BODu) of 300 mg/L.
What is the BOD reaction rate constant (k) at 20°C?
A. 0.10 day¹
B. 0.15 day¹
C. 0.23 day¹
D. 0.35 day¹
Answer: C. 0.23 day¹
Rationale: Using first-order kinetics: BOD5 = BODu (1 - e^(-k*5)). 200 = 300(1 - e^(-5k))
-> 0.667 = 1 - e^(-5k) -> e^(-5k) = 0.333 -> -5k = ln(0.333) -1.099 -> k 0.22 day¹. The
closest is 0.23 day¹. Other options result from calculation errors.
Page 3 of 78
, 7. In anaerobic digestion, which parameter is the most reliable indicator of process
instability?
A. Methane content in biogas above 65%
B. Volatile solids reduction exceeding 50%
C. Volatile fatty acids (VFA) to alkalinity ratio exceeding 0.3
D. pH between 6.8 and 7.4
Answer: C. Volatile fatty acids (VFA) to alkalinity ratio exceeding 0.3
Rationale: A VFA/alkalinity ratio >0.3 indicates acid accumulation and impending souring,
signaling instability. High methane, good VS reduction, and neutral pH are indicators of
stable digestion, not instability.
8. A chlorination system must achieve a CT value of 100 mg-min/L for 3-log virus
inactivation. If the chlorine residual is 2.0 mg/L, what contact time is required?
A. 25 minutes
B. 50 minutes
C. 100 minutes
D. 200 minutes
Answer: B. 50 minutes
Rationale: CT = concentration × time -> time = CT / concentration = .0 = 50
minutes. Options A, C, and D are incorrect because they misapply the division or multiply
instead.
Page 4 of 78
Exam Practice Questions And Correct
Answers (Verified Answers) Plus
1. A 5 MGD activated sludge plant achieves 85% BOD removal. If the influent BOD is
220 mg/L, what is the effluent BOD concentration in mg/L?
A. 33
B. 37
C. 44
D. 55
Answer: A. 33
Rationale: Effluent BOD = influent BOD × (1 - removal efficiency) = 220 × (1 - 0.85) =
220 × 0.15 = 33 mg/L. Options B, C, and D result from incorrect multiplication or
misapplication of the removal percentage.
2. In a biological nutrient removal process, which condition is essential for achieving
simultaneous nitrification and denitrification?
A. High dissolved oxygen (>2 mg/L) throughout the basin
B. Alternating aerobic and anoxic zones with internal recycle
C. Low SRT (<5 days) and high F/M ratio
D. Addition of external carbon source in aerobic zone
Answer: B. Alternating aerobic and anoxic zones with internal recycle
Rationale: Simultaneous nitrification and denitrification requires alternating aerobic and
anoxic conditions, often achieved with internal recycle to bring nitrate to the anoxic zone.
High DO inhibits denitrification, low SRT washes out nitrifiers, and adding carbon in the
aerobic zone does not promote anoxic denitrification.
Page 1 of 78
,3. A UV disinfection system receives effluent with transmittance of 65%. To achieve
the same log inactivation as at 75% transmittance, how must the UV dose be
adjusted?
A. Decrease dose proportionally to transmittance
B. Increase dose to compensate for reduced transmittance
C. Maintain the same dose since UV dose is independent of transmittance
D. Increase transmittance by adding coagulant
Answer: B. Increase dose to compensate for reduced transmittance
Rationale: Lower transmittance means less UV light penetrates the water, reducing the
effective dose. To achieve the same inactivation, the UV dose must be increased. Option A
is incorrect because dose and transmittance are inversely related; C ignores the effect of
transmittance; D is not a direct adjustment to the UV system.
4. Which polymer characteristic is most critical for effective conditioning of
anaerobically digested sludge prior to belt filter pressing?
A. High molecular weight and moderate charge density
B. Low molecular weight and high charge density
C. Cationic charge with very low molecular weight
D. Anionic charge with high molecular weight
Answer: A. High molecular weight and moderate charge density
Rationale: High molecular weight polymers with moderate charge density effectively
bridge and flocculate digested sludge particles, improving dewaterability. Low molecular
weight or incorrect charge types (anionic for typically negatively charged sludge) result in
poor flocculation and high filtrate solids.
Page 2 of 78
,5. Under Louisiana Department of Environmental Quality (LDEQ) regulations, which
of the following is a Class IV operator's responsibility regarding bypass reporting?
A. Report bypasses only if they exceed 10,000 gallons
B. Notify LDEQ orally within 24 hours and submit written report within 5 days
C. Report bypasses in the monthly discharge monitoring report (DMR) only
D. No reporting required if bypass is within permitted limits
Answer: B. Notify LDEQ orally within 24 hours and submit written report within 5
days
Rationale: LDEQ requires oral notification within 24 hours of a bypass and a written
report within 5 days, regardless of volume, to ensure timely regulatory oversight. Options
A, C, and D underestimate reporting requirements and could lead to violations.
6. A sample has a 5-day BOD of 200 mg/L and a ultimate BOD (BODu) of 300 mg/L.
What is the BOD reaction rate constant (k) at 20°C?
A. 0.10 day¹
B. 0.15 day¹
C. 0.23 day¹
D. 0.35 day¹
Answer: C. 0.23 day¹
Rationale: Using first-order kinetics: BOD5 = BODu (1 - e^(-k*5)). 200 = 300(1 - e^(-5k))
-> 0.667 = 1 - e^(-5k) -> e^(-5k) = 0.333 -> -5k = ln(0.333) -1.099 -> k 0.22 day¹. The
closest is 0.23 day¹. Other options result from calculation errors.
Page 3 of 78
, 7. In anaerobic digestion, which parameter is the most reliable indicator of process
instability?
A. Methane content in biogas above 65%
B. Volatile solids reduction exceeding 50%
C. Volatile fatty acids (VFA) to alkalinity ratio exceeding 0.3
D. pH between 6.8 and 7.4
Answer: C. Volatile fatty acids (VFA) to alkalinity ratio exceeding 0.3
Rationale: A VFA/alkalinity ratio >0.3 indicates acid accumulation and impending souring,
signaling instability. High methane, good VS reduction, and neutral pH are indicators of
stable digestion, not instability.
8. A chlorination system must achieve a CT value of 100 mg-min/L for 3-log virus
inactivation. If the chlorine residual is 2.0 mg/L, what contact time is required?
A. 25 minutes
B. 50 minutes
C. 100 minutes
D. 200 minutes
Answer: B. 50 minutes
Rationale: CT = concentration × time -> time = CT / concentration = .0 = 50
minutes. Options A, C, and D are incorrect because they misapply the division or multiply
instead.
Page 4 of 78
