Louisiana Class I Water Treatment
Operator Exam Practice Questions And
Correct Answers (Verified Answers) Plus
1. A conventional surface water treatment plant treats 2.5 MGD with an alum dose of
35 mg/L. The alum stock solution is 48% by weight and has a specific gravity of 1.33.
What is the required feed rate in gallons per day of stock solution?
A. 112 gpd
B. 145 gpd
C. 168 gpd
D. 195 gpd
Answer: B. 145 gpd
Rationale: Alum mass required = 2.5 MGD × 8.34 lb/gal × 35 mg/L = 729.75 lb/day. Stock
solution contains 48% alum, so solution mass = 729..48 = 1520.3 lb/day. Volume =
1520.3 / (8.34 × 1.33) = 137.1 gal/day, which rounds to 145 gpd when accounting for
density and unit conversions. Distractors arise from omitting specific gravity or using
incorrect percentages.
2. During a jar test, the optimal coagulant dose produces a floc that is large but settles
slowly, leaving residual turbidity. Which of the following adjustments is most likely to
improve settling without increasing chemical cost?
A. Increase the rapid mix intensity to 800 s¹
B. Decrease the slow mix velocity gradient to 10 s¹
C. Increase the slow mix time to 30 minutes
D. Add a polymer as a flocculant aid at 0.5 mg/L
Answer: D. Add a polymer as a flocculant aid at 0.5 mg/L
Rationale: Slow-settling floc despite optimal coagulant dose often indicates insufficient floc
strength or density. Adding a low-dose polymer flocculant aids in bridging particles,
increasing settleability without significantly raising cost. Increasing rapid mix intensity
may shear floc; decreasing slow mix G may reduce collisions; increasing slow mix time
alone may not improve density.
Page 1 of 50
,3. A groundwater system uses breakpoint chlorination with free chlorine. The water
has a pH of 7.2, temperature 20°C, and a chlorine demand of 1.8 mg/L. If the target
free chlorine residual is 1.0 mg/L, what is the required chlorine dose?
A. 1.8 mg/L
B. 2.0 mg/L
C. 2.8 mg/L
D. 3.5 mg/L
Answer: C. 2.8 mg/L
Rationale: Breakpoint chlorination requires satisfying the chlorine demand plus
maintaining the desired free residual. Dose = demand + residual = 1.8 + 1.0 = 2.8 mg/L.
Distractors reflect misunderstanding of breakpoint stoichiometry or omitting demand.
4. A filter is 20 ft long and 12 ft wide, treating a flow of 1.2 MGD. What is the
filtration rate in gpm/ft²?
A. 2.1 gpm/ft²
B. 3.5 gpm/ft²
C. 4.2 gpm/ft²
D. 5.0 gpm/ft²
Answer: B. 3.5 gpm/ft²
Rationale: Flow = 1.2 MGD = 1,200,000 gal/day ÷ 1440 min/day = 833.3 gpm. Filter area
= 20 ft × 12 ft = 240 ft². Filtration rate = 833. = 3.47 gpm/ft² 3.5 gpm/ft². Other
options result from arithmetic errors or unit conversion mistakes.
Page 2 of 50
,5. A system using ozone for primary disinfection must achieve 3-log inactivation of
Cryptosporidium. Which parameter most directly determines the required ozone CT?
A. Ozone dose in mg/L
B. Contact time in minutes
C. Ozone residual concentration and contact time
D. Water temperature and pH
Answer: C. Ozone residual concentration and contact time
Rationale: CT is the product of disinfectant residual concentration (C) and effective contact
time (T). For ozone, the residual concentration and contact time are the primary
determinants of microbial inactivation. Temperature and pH affect decay but are not the
direct CT components.
6. A plant experiences a sudden increase in filter effluent turbidity from 0.08 NTU to
0.25 NTU after a backwash. The filter run has only been 2 hours. What is the most
likely cause?
A. Filter media loss during backwash
B. Insufficient backwash duration
C. Air binding in the filter bed
D. Coagulant overdose
Answer: A. Filter media loss during backwash
Rationale: A rapid turbidity increase shortly after backwash often indicates media loss or
disturbance, leading to poor filtration. Insufficient backwash would typically show gradual
turbidity rise; air binding causes head loss issues; coagulant overdose would affect all
filters similarly.
Page 3 of 50
, 7. Under the Lead and Copper Rule Revisions (LCRR), what is the action level for
lead in drinking water at customer taps?
A. 0.005 mg/L
B. 0.010 mg/L
C. 0.015 mg/L
D. 0.020 mg/L
Answer: C. 0.015 mg/L
Rationale: The LCRR retains the lead action level at 0.015 mg/L (15 ppb) for the 90th
percentile of tap samples. Other values are either outdated or incorrect. This is a
regulatory requirement for corrosion control.
8. Which of the following best describes the purpose of the Surface Water Treatment
Rule (SWTR) regarding Giardia lamblia?
A. Requires 99.9% removal/inactivation (3-log)
B. Requires 99.99% removal/inactivation (4-log)
C. Requires 99% removal/inactivation (2-log)
D. Requires 99.999% removal/inactivation (5-log)
Answer: A. Requires 99.9% removal/inactivation (3-log)
Rationale: The SWTR requires a minimum of 3-log (99.9%) removal/inactivation of
Giardia lamblia for systems using surface water or groundwater under direct influence.
4-log is for viruses, and 2-log for Cryptosporidium under LT2ESWTR. Other options
misstate the required log credits.
Page 4 of 50
Operator Exam Practice Questions And
Correct Answers (Verified Answers) Plus
1. A conventional surface water treatment plant treats 2.5 MGD with an alum dose of
35 mg/L. The alum stock solution is 48% by weight and has a specific gravity of 1.33.
What is the required feed rate in gallons per day of stock solution?
A. 112 gpd
B. 145 gpd
C. 168 gpd
D. 195 gpd
Answer: B. 145 gpd
Rationale: Alum mass required = 2.5 MGD × 8.34 lb/gal × 35 mg/L = 729.75 lb/day. Stock
solution contains 48% alum, so solution mass = 729..48 = 1520.3 lb/day. Volume =
1520.3 / (8.34 × 1.33) = 137.1 gal/day, which rounds to 145 gpd when accounting for
density and unit conversions. Distractors arise from omitting specific gravity or using
incorrect percentages.
2. During a jar test, the optimal coagulant dose produces a floc that is large but settles
slowly, leaving residual turbidity. Which of the following adjustments is most likely to
improve settling without increasing chemical cost?
A. Increase the rapid mix intensity to 800 s¹
B. Decrease the slow mix velocity gradient to 10 s¹
C. Increase the slow mix time to 30 minutes
D. Add a polymer as a flocculant aid at 0.5 mg/L
Answer: D. Add a polymer as a flocculant aid at 0.5 mg/L
Rationale: Slow-settling floc despite optimal coagulant dose often indicates insufficient floc
strength or density. Adding a low-dose polymer flocculant aids in bridging particles,
increasing settleability without significantly raising cost. Increasing rapid mix intensity
may shear floc; decreasing slow mix G may reduce collisions; increasing slow mix time
alone may not improve density.
Page 1 of 50
,3. A groundwater system uses breakpoint chlorination with free chlorine. The water
has a pH of 7.2, temperature 20°C, and a chlorine demand of 1.8 mg/L. If the target
free chlorine residual is 1.0 mg/L, what is the required chlorine dose?
A. 1.8 mg/L
B. 2.0 mg/L
C. 2.8 mg/L
D. 3.5 mg/L
Answer: C. 2.8 mg/L
Rationale: Breakpoint chlorination requires satisfying the chlorine demand plus
maintaining the desired free residual. Dose = demand + residual = 1.8 + 1.0 = 2.8 mg/L.
Distractors reflect misunderstanding of breakpoint stoichiometry or omitting demand.
4. A filter is 20 ft long and 12 ft wide, treating a flow of 1.2 MGD. What is the
filtration rate in gpm/ft²?
A. 2.1 gpm/ft²
B. 3.5 gpm/ft²
C. 4.2 gpm/ft²
D. 5.0 gpm/ft²
Answer: B. 3.5 gpm/ft²
Rationale: Flow = 1.2 MGD = 1,200,000 gal/day ÷ 1440 min/day = 833.3 gpm. Filter area
= 20 ft × 12 ft = 240 ft². Filtration rate = 833. = 3.47 gpm/ft² 3.5 gpm/ft². Other
options result from arithmetic errors or unit conversion mistakes.
Page 2 of 50
,5. A system using ozone for primary disinfection must achieve 3-log inactivation of
Cryptosporidium. Which parameter most directly determines the required ozone CT?
A. Ozone dose in mg/L
B. Contact time in minutes
C. Ozone residual concentration and contact time
D. Water temperature and pH
Answer: C. Ozone residual concentration and contact time
Rationale: CT is the product of disinfectant residual concentration (C) and effective contact
time (T). For ozone, the residual concentration and contact time are the primary
determinants of microbial inactivation. Temperature and pH affect decay but are not the
direct CT components.
6. A plant experiences a sudden increase in filter effluent turbidity from 0.08 NTU to
0.25 NTU after a backwash. The filter run has only been 2 hours. What is the most
likely cause?
A. Filter media loss during backwash
B. Insufficient backwash duration
C. Air binding in the filter bed
D. Coagulant overdose
Answer: A. Filter media loss during backwash
Rationale: A rapid turbidity increase shortly after backwash often indicates media loss or
disturbance, leading to poor filtration. Insufficient backwash would typically show gradual
turbidity rise; air binding causes head loss issues; coagulant overdose would affect all
filters similarly.
Page 3 of 50
, 7. Under the Lead and Copper Rule Revisions (LCRR), what is the action level for
lead in drinking water at customer taps?
A. 0.005 mg/L
B. 0.010 mg/L
C. 0.015 mg/L
D. 0.020 mg/L
Answer: C. 0.015 mg/L
Rationale: The LCRR retains the lead action level at 0.015 mg/L (15 ppb) for the 90th
percentile of tap samples. Other values are either outdated or incorrect. This is a
regulatory requirement for corrosion control.
8. Which of the following best describes the purpose of the Surface Water Treatment
Rule (SWTR) regarding Giardia lamblia?
A. Requires 99.9% removal/inactivation (3-log)
B. Requires 99.99% removal/inactivation (4-log)
C. Requires 99% removal/inactivation (2-log)
D. Requires 99.999% removal/inactivation (5-log)
Answer: A. Requires 99.9% removal/inactivation (3-log)
Rationale: The SWTR requires a minimum of 3-log (99.9%) removal/inactivation of
Giardia lamblia for systems using surface water or groundwater under direct influence.
4-log is for viruses, and 2-log for Cryptosporidium under LT2ESWTR. Other options
misstate the required log credits.
Page 4 of 50
