Texas Water Class A Operator License Exam | TCEQ
Advanced Water Operator Practice Exam, QUESTIONs &
Detailed Rationales (2026/2027)
QUESTION 1
In a reverse osmosis (RO) membrane system treating high-total-
dissolved-solids (TDS) groundwater, if the feed flow rate is 400 GPM and
the concentrate (reject) flow rate is 100 GPM, what is the recovery rate
of the system?
• A. 25%
• B. 50%
• C. 75%
• D. 80%
Correct Answer: C. 75%
Detailed Rationale: Recovery rate is calculated as (Permeate Flow /
Feed Flow) × 100. Permeate flow = Feed minus Concentrate (400 −
100 = 300 GPM). Recovery = (300/400) × 100 = 75%.
QUESTION 2
According to TCEQ 30 TAC Chapter 290, what is the maximum design
filtration rate for conventional rapid gravity dual-media filters treating
surface water under standard operating conditions without special state
pilot study authorization?
• A. 2.0 gpm/sq ft
, • B. 4.0 gpm/sq ft
• C. 6.0 gpm/sq ft
• D. 10.0 gpm/sq ft
Correct Answer: B. 4.0 gpm/sq ft
Detailed Rationale: TCEQ design criteria limit conventional rapid gravity
filtration rates to a maximum of 4.0 gpm/sq ft under standard
conditions unless specific high-rate filtration demonstration studies are
approved.
QUESTION 3
A disinfection basin operating at peak hourly flow results in a baffled
contact time (𝑇10) of 45 minutes. The free chlorine residual measured at
the end of the basin is 1.5 mg/L at a pH of 7.2 and water temperature of
15°C. What is the calculated CT value?
• A. 30 mg-min/L
• B. 45.0 mg-min/L
• C. 67.5 mg-min/L
• D. 90.0 mg-min/L
Correct Answer: C. 67.5 mg-min/L
Detailed Rationale: CT is calculated using the formula CT = 𝐶 × 𝑇10,
where 𝐶 is the disinfectant residual (1.5 mg/L) and 𝑇10 is the effective
contact time (45 min). Calculation: 1.5 × 45 = 67.5 mg-min/L.
QUESTION 4
,What is the primary mechanism by which an Advanced Oxidation
Process (AOP) utilizing ultraviolet (UV) light and hydrogen peroxide
(𝐻2𝑂2) destroys recalcitrant organic contaminants and micropollutants?
• A. Direct thermal boiling of target molecules
• B. Generation of highly reactive hydroxyl radicals (⋅ 𝑂𝐻)
• C. Complete reduction of sulfate ions
• D. Coagulation with aluminum sulfate
Correct Answer: B. Generation of highly reactive hydroxyl radicals (⋅
𝑂𝐻)
Detailed Rationale: Photolysis of hydrogen peroxide under UV
irradiation produces hydroxyl radicals (⋅ 𝑂𝐻), which are powerful, non-
selective oxidizers that break down stable organic molecules.
QUESTION 5
An anion exchange resin system utilized for nitrate removal in
groundwater is typically regenerated using which chemical solution?
• A. Sodium hydroxide (NaOH)
• B. Sodium chloride (NaCl) brine
• C. Sulfuric acid (𝐻2𝑆𝑂4)
• D. Potassium permanganate (𝐾𝑀𝑛𝑂4)
Correct Answer: B. Sodium chloride (NaCl) brine
Detailed Rationale: Strong-base anion exchange resins selective for
nitrate removal are regenerated by passing a high-concentration
sodium chloride brine through the resin bed to displace nitrate ions.
, QUESTION 6
A circular clarifier has a diameter of 80 feet and treats a flow of 3.5
MGD. What is the surface overflow rate in gallons per day per square
foot (gpd/sq ft)?
• A. 450 gpd/sq ft
• B. 698 gpd/sq ft
• C. 850 gpd/sq ft
• D. 1,200 gpd/sq ft
Correct Answer: B. 698 gpd/sq ft
Detailed Rationale: Surface Area = 𝜋 × 𝑟2 = 3.1416 × (40)2 =
5,026.5 sq ft. Overflow Rate = Flow / Area = 3,500,000 gpd/
5,026.5 sq ft = 696.3 ≈ 698 gpd/sq ft.
QUESTION 7
In water treatment coagulation, what does reducing the zeta potential
of colloidal particles toward zero accomplish?
• A. It increases mutual electrostatic repulsion, keeping particles
suspended indefinitely.
• B. It neutralizes the surface charge, allowing Van der Waals
attractive forces to agglomerate particles into micro-flocs.
• C. It raises finished water pH above 10.0.
• D. It eliminates all dissolved gases.
Correct Answer: B. It neutralizes the surface charge, allowing Van der
Waals attractive forces to agglomerate particles into micro-flocs.
Advanced Water Operator Practice Exam, QUESTIONs &
Detailed Rationales (2026/2027)
QUESTION 1
In a reverse osmosis (RO) membrane system treating high-total-
dissolved-solids (TDS) groundwater, if the feed flow rate is 400 GPM and
the concentrate (reject) flow rate is 100 GPM, what is the recovery rate
of the system?
• A. 25%
• B. 50%
• C. 75%
• D. 80%
Correct Answer: C. 75%
Detailed Rationale: Recovery rate is calculated as (Permeate Flow /
Feed Flow) × 100. Permeate flow = Feed minus Concentrate (400 −
100 = 300 GPM). Recovery = (300/400) × 100 = 75%.
QUESTION 2
According to TCEQ 30 TAC Chapter 290, what is the maximum design
filtration rate for conventional rapid gravity dual-media filters treating
surface water under standard operating conditions without special state
pilot study authorization?
• A. 2.0 gpm/sq ft
, • B. 4.0 gpm/sq ft
• C. 6.0 gpm/sq ft
• D. 10.0 gpm/sq ft
Correct Answer: B. 4.0 gpm/sq ft
Detailed Rationale: TCEQ design criteria limit conventional rapid gravity
filtration rates to a maximum of 4.0 gpm/sq ft under standard
conditions unless specific high-rate filtration demonstration studies are
approved.
QUESTION 3
A disinfection basin operating at peak hourly flow results in a baffled
contact time (𝑇10) of 45 minutes. The free chlorine residual measured at
the end of the basin is 1.5 mg/L at a pH of 7.2 and water temperature of
15°C. What is the calculated CT value?
• A. 30 mg-min/L
• B. 45.0 mg-min/L
• C. 67.5 mg-min/L
• D. 90.0 mg-min/L
Correct Answer: C. 67.5 mg-min/L
Detailed Rationale: CT is calculated using the formula CT = 𝐶 × 𝑇10,
where 𝐶 is the disinfectant residual (1.5 mg/L) and 𝑇10 is the effective
contact time (45 min). Calculation: 1.5 × 45 = 67.5 mg-min/L.
QUESTION 4
,What is the primary mechanism by which an Advanced Oxidation
Process (AOP) utilizing ultraviolet (UV) light and hydrogen peroxide
(𝐻2𝑂2) destroys recalcitrant organic contaminants and micropollutants?
• A. Direct thermal boiling of target molecules
• B. Generation of highly reactive hydroxyl radicals (⋅ 𝑂𝐻)
• C. Complete reduction of sulfate ions
• D. Coagulation with aluminum sulfate
Correct Answer: B. Generation of highly reactive hydroxyl radicals (⋅
𝑂𝐻)
Detailed Rationale: Photolysis of hydrogen peroxide under UV
irradiation produces hydroxyl radicals (⋅ 𝑂𝐻), which are powerful, non-
selective oxidizers that break down stable organic molecules.
QUESTION 5
An anion exchange resin system utilized for nitrate removal in
groundwater is typically regenerated using which chemical solution?
• A. Sodium hydroxide (NaOH)
• B. Sodium chloride (NaCl) brine
• C. Sulfuric acid (𝐻2𝑆𝑂4)
• D. Potassium permanganate (𝐾𝑀𝑛𝑂4)
Correct Answer: B. Sodium chloride (NaCl) brine
Detailed Rationale: Strong-base anion exchange resins selective for
nitrate removal are regenerated by passing a high-concentration
sodium chloride brine through the resin bed to displace nitrate ions.
, QUESTION 6
A circular clarifier has a diameter of 80 feet and treats a flow of 3.5
MGD. What is the surface overflow rate in gallons per day per square
foot (gpd/sq ft)?
• A. 450 gpd/sq ft
• B. 698 gpd/sq ft
• C. 850 gpd/sq ft
• D. 1,200 gpd/sq ft
Correct Answer: B. 698 gpd/sq ft
Detailed Rationale: Surface Area = 𝜋 × 𝑟2 = 3.1416 × (40)2 =
5,026.5 sq ft. Overflow Rate = Flow / Area = 3,500,000 gpd/
5,026.5 sq ft = 696.3 ≈ 698 gpd/sq ft.
QUESTION 7
In water treatment coagulation, what does reducing the zeta potential
of colloidal particles toward zero accomplish?
• A. It increases mutual electrostatic repulsion, keeping particles
suspended indefinitely.
• B. It neutralizes the surface charge, allowing Van der Waals
attractive forces to agglomerate particles into micro-flocs.
• C. It raises finished water pH above 10.0.
• D. It eliminates all dissolved gases.
Correct Answer: B. It neutralizes the surface charge, allowing Van der
Waals attractive forces to agglomerate particles into micro-flocs.
