ARIZONA WATER TREATMENT OPERATOR CERTIFICATION EXAM
GRADE 3 ACTUAL EXAM 2026/2027 ACCURATE TEST COMPLETE
APPROVED QUESTIONS AND CORRECT DETAILED ANSWERS WITH
RATIONALES (VERIFIED SOLUTIONS) LATEST UPDATED VERSION
2026 EDITION |GUARANTEED SUCCESS A+ |FULL REVISED EXAM
1. A water treatment plant processes 6.0 MGD. The sedimentation basin is 120 ft.
long, 40 ft. wide, and 15 ft. deep. What is the detention time in hours?
A) 1.2 hours
B) 2.1 hours CORRECT ANSWER – *Rationale: Volume = 120×40×15 =
72,000 ft³ × 7.48 gal/ft³ = 538,560 gal. Flow = 6,000,000 god. Detention time (hr.)
= (538,,000,000) × 24 = 2.15 hours.*
C) 3.4 hours
D) 4.5 hours
2. The filter loading rate is 4 gym/ft². Each filter is 25 ft. × 25 ft. How many filters
are needed to treat 12 MGD?
A) 2 filters
B) 4 filters CORRECT ANSWER – *Rationale: Total area needed = Flow (gym)
/ rate. Flow gym = 12 MGD × 694.4 = 8,333 gyms. Area needed = 8, =
2,083 ft². Each filter area = 625 ft². Number filters = 2, = 3.33, so 4 filters
required for redundancy and operation at 75% capacity each.*
C) 6 filters
D) 8 filters
3. A chemical feed pump delivers 35 lbs./day of chlorine. The plant flow is 2.8
MGD. What is the chlorine dose in mg/L?
A) 1.0 mg/L
B) 1.5 mg/L CORRECT ANSWER – *Rationale: Dose (mg/L) = (lbs./day) /
(Flow MGD × 8.34) = 35 / (2.8 × 8.34) = .35 = 1.5 mg/L.*
,C) 2.0 mg/L
D) 2.5 mg/L
4. What is the required CT value for 99.9% inactivation of Giardia labia at 15°C,
pH 7.0, with free chlorine residual of 1.2 mg/L? (Use CT table: at 15°C, pH 7,
CT=79 for 3-log)
A) 40 gamin/L
B) 79 gamin/L CORRECT ANSWER – *Rationale: EPA CT tables for Giardia
at 15°C, free chlorine, pH 7.0 require CT=79. With residual 1.2 mg/L, required
time = 79/1.2 = 66 minutes. The CT value itself is 79.*
C) 120 gamin/L
D) 150 gamin/L
5. The operator performs a jar test and finds that 18 mg/L of alum produces the
best floc. The plant treats 15 MGD. How many pounds of alum are needed per
day?
A) 1,800 lbs./day
B) 2,252 lbs./day CORRECT ANSWER – *Rationale: 15 MGD × 18 mg/L ×
8.34 = 2,251.8 lbs./day.*
C) 3,000 lbs./day
D) 4,500 lbs./day
6. Which of the following is the primary cause of filter mud balls?
A) Insufficient backwash
B) Inadequate surface wash or air scour CORRECT ANSWER – Rationale:
Mud balls form when floc agglomerates with sand and anthracite due to poor
cleaning, especially lack of surface wash or air scour.
C) High filtration rate
D) Low coagulant dose
7. A finished water sample has a pH of 9.5. What is the approximate hydrogen ion
concentration?
, A) 10⁻⁹.⁵ M
B) 3.16 × 10⁻¹⁰ M CORRECT ANSWER – *Rationale: [H+] = 10⁻pH = 10⁻⁹·⁵ =
3.16 × 10⁻¹⁰ M.*
C) 10⁻⁵ M
D) 9.5 M
8. A filter has a surface area of 400 ft² and operates at a rate of 5 gym/ft². The filter
run time is 48 hours. How many gallons are filtered per run?
A) 2.88 MG
B) 5.76 MG CORRECT ANSWER – *Rationale: Flow rate = 400 × 5 = 2,000
gym. 2,000 gym × 60 min/hr. × 48 hr. = 5,760,000 gal = 5.76 MG.*
C) 8.64 MG
D) 10.0 MG
9. The operator adds 750 lbs. of 65% available calcium hypochlorite. How many
lbs. of pure chlorine does this provide?
A) 265 lbs.
B) 487.5 lbs. CORRECT ANSWER – *Rationale: 750 lbs. × 0.65 = 487.5 lbs.
pure chlorine.*
C) 500 lbs.
D) 650 lbs.
10. Under the Lead and Copper Rule, if more than 10% of tap samples exceed the
action level of 0.015 mg/L for lead, the system must:
A) Issue a boil water advisory
B) Implement corrosion control treatment CORRECT ANSWER – Rationale:
Exceeding lead action level triggers corrosion control study and treatment
optimization (e.g., pH/alkalinity adjustment, orthophosphate).
C) Replace all lead pipes immediately
D) Shut down the plant
GRADE 3 ACTUAL EXAM 2026/2027 ACCURATE TEST COMPLETE
APPROVED QUESTIONS AND CORRECT DETAILED ANSWERS WITH
RATIONALES (VERIFIED SOLUTIONS) LATEST UPDATED VERSION
2026 EDITION |GUARANTEED SUCCESS A+ |FULL REVISED EXAM
1. A water treatment plant processes 6.0 MGD. The sedimentation basin is 120 ft.
long, 40 ft. wide, and 15 ft. deep. What is the detention time in hours?
A) 1.2 hours
B) 2.1 hours CORRECT ANSWER – *Rationale: Volume = 120×40×15 =
72,000 ft³ × 7.48 gal/ft³ = 538,560 gal. Flow = 6,000,000 god. Detention time (hr.)
= (538,,000,000) × 24 = 2.15 hours.*
C) 3.4 hours
D) 4.5 hours
2. The filter loading rate is 4 gym/ft². Each filter is 25 ft. × 25 ft. How many filters
are needed to treat 12 MGD?
A) 2 filters
B) 4 filters CORRECT ANSWER – *Rationale: Total area needed = Flow (gym)
/ rate. Flow gym = 12 MGD × 694.4 = 8,333 gyms. Area needed = 8, =
2,083 ft². Each filter area = 625 ft². Number filters = 2, = 3.33, so 4 filters
required for redundancy and operation at 75% capacity each.*
C) 6 filters
D) 8 filters
3. A chemical feed pump delivers 35 lbs./day of chlorine. The plant flow is 2.8
MGD. What is the chlorine dose in mg/L?
A) 1.0 mg/L
B) 1.5 mg/L CORRECT ANSWER – *Rationale: Dose (mg/L) = (lbs./day) /
(Flow MGD × 8.34) = 35 / (2.8 × 8.34) = .35 = 1.5 mg/L.*
,C) 2.0 mg/L
D) 2.5 mg/L
4. What is the required CT value for 99.9% inactivation of Giardia labia at 15°C,
pH 7.0, with free chlorine residual of 1.2 mg/L? (Use CT table: at 15°C, pH 7,
CT=79 for 3-log)
A) 40 gamin/L
B) 79 gamin/L CORRECT ANSWER – *Rationale: EPA CT tables for Giardia
at 15°C, free chlorine, pH 7.0 require CT=79. With residual 1.2 mg/L, required
time = 79/1.2 = 66 minutes. The CT value itself is 79.*
C) 120 gamin/L
D) 150 gamin/L
5. The operator performs a jar test and finds that 18 mg/L of alum produces the
best floc. The plant treats 15 MGD. How many pounds of alum are needed per
day?
A) 1,800 lbs./day
B) 2,252 lbs./day CORRECT ANSWER – *Rationale: 15 MGD × 18 mg/L ×
8.34 = 2,251.8 lbs./day.*
C) 3,000 lbs./day
D) 4,500 lbs./day
6. Which of the following is the primary cause of filter mud balls?
A) Insufficient backwash
B) Inadequate surface wash or air scour CORRECT ANSWER – Rationale:
Mud balls form when floc agglomerates with sand and anthracite due to poor
cleaning, especially lack of surface wash or air scour.
C) High filtration rate
D) Low coagulant dose
7. A finished water sample has a pH of 9.5. What is the approximate hydrogen ion
concentration?
, A) 10⁻⁹.⁵ M
B) 3.16 × 10⁻¹⁰ M CORRECT ANSWER – *Rationale: [H+] = 10⁻pH = 10⁻⁹·⁵ =
3.16 × 10⁻¹⁰ M.*
C) 10⁻⁵ M
D) 9.5 M
8. A filter has a surface area of 400 ft² and operates at a rate of 5 gym/ft². The filter
run time is 48 hours. How many gallons are filtered per run?
A) 2.88 MG
B) 5.76 MG CORRECT ANSWER – *Rationale: Flow rate = 400 × 5 = 2,000
gym. 2,000 gym × 60 min/hr. × 48 hr. = 5,760,000 gal = 5.76 MG.*
C) 8.64 MG
D) 10.0 MG
9. The operator adds 750 lbs. of 65% available calcium hypochlorite. How many
lbs. of pure chlorine does this provide?
A) 265 lbs.
B) 487.5 lbs. CORRECT ANSWER – *Rationale: 750 lbs. × 0.65 = 487.5 lbs.
pure chlorine.*
C) 500 lbs.
D) 650 lbs.
10. Under the Lead and Copper Rule, if more than 10% of tap samples exceed the
action level of 0.015 mg/L for lead, the system must:
A) Issue a boil water advisory
B) Implement corrosion control treatment CORRECT ANSWER – Rationale:
Exceeding lead action level triggers corrosion control study and treatment
optimization (e.g., pH/alkalinity adjustment, orthophosphate).
C) Replace all lead pipes immediately
D) Shut down the plant
