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WATER TREATMENT GRADE 4 EXAM / LATEST WATER TREATMENT GRADE 4 ACTUAL EXAM 2026/2027 COMPLETE ACCURATE TEST APPROVED QUESTIONS WITH WELL ELABORATED ANSWERS (VERIFIED SOLUTIONS) NEWEST UPDATED VERSION |GUARANTEED PASS A+ (BRAND NEW!) FULL REVISED EXAM

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WATER TREATMENT GRADE 4 EXAM / LATEST WATER TREATMENT GRADE 4 ACTUAL EXAM 2026/2027 COMPLETE ACCURATE TEST APPROVED QUESTIONS WITH WELL ELABORATED ANSWERS (VERIFIED SOLUTIONS) NEWEST UPDATED VERSION |GUARANTEED PASS A+ (BRAND NEW!) FULL REVISED EXAM

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WATER TREATMENT GRADE 4 EXAM / LATEST WATER
TREATMENT GRADE 4 ACTUAL EXAM 2026/2027 COMPLETE
ACCURATE TEST APPROVED QUESTIONS WITH WELL
ELABORATED ANSWERS (VERIFIED SOLUTIONS) NEWEST
UPDATED VERSION |GUARANTEED PASS A+ (BRAND NEW!) FULL
REVISED EXAM


1. What is the primary purpose of coagulation and flocculation in water treatment?

a) To kill bacteria and viruses

b) To remove dissolved minerals

c) To destabilize and aggregate suspended particles

d) To neutralize the pH of the water

Correct Answer: c) To destabilize and aggregate suspended particles



Rationale: Coagulation neutralizes the electrical charges of suspended particles (colloids), allowing them
to clump together into larger flocs during flocculation, which can then be removed by sedimentation
and filtration.



2. A water treatment plant uses 35 mg/L of alum. If the plant produces 12 million gallons per day (MGD),
how many pounds of alum are used per day? (Assume 8.34 lb./gal conversion factor)

a) 2,500 lb./day

b) 2,800 lb./day

c) 3,500 lb./day

d) 4,200 lb./day

Correct Answer: c) 3,500 lb./day



Rationale: The formula is: lbs./day = (Dosage, mg/L) × (Flow, MGD) × (8.34 lb./gal). Calculation: 35 mg/L
× 12 MGD × 8.34 = 3,502.8 lbs./day, rounded to 3,500 lbs./day.



3. Which type of filter is most commonly used in conventional water treatment plants to remove floc
and suspended solids?

a) Rapid sand filter

,b) Slow sand filter

c) Diatomaceous earth filter

d) Membrane filter

Correct Answer: a) Rapid sand filter



Rationale: Rapid sand filters (gravity or pressure) are the industry standard for conventional treatment
due to their high filtration rates and efficiency in removing floc following sedimentation.



4. What is the maximum contaminant level (MCL) for turbidity in water leaving a conventional filtration
plant as established by the Surface Water Treatment Rule (SWTR)?

a) 0.1 NTU

b) 0.3 NTU

c) 0.5 NTU

d) 1.0 NTU

Correct Answer: b) 0.3 NTU



Rationale: The SWTR and LT2ESWTR require that the combined filter effluent turbidity must be less than
or equal to 0.3 NTU in at least 95% of measurements each month for conventional and direct filtration
plants.



5. A centrifugal pump is operating against a discharge head of 80 feet and a suction head of 10 feet.
What is the total dynamic head (TDH)?

a) 70 ft.

b) 90 ft.

c) 80 ft.

d) 100 ft.

Correct Answer: b) 90 ft.



Rationale: TDH = Discharge Head – Suction Head. When the suction head is positive (flooded suction), it
subtracts from the discharge head. Calculation: 80 ft. – (-10 ft.) is incorrect; if suction head is 10 ft., TDH
= 80 ft. – 10 ft.? No: Standard formula: TDH = Discharge Head + Suction Lift (or minus positive suction
head). For a positive suction head, TDH = Discharge Head – Suction Head? Actually, TDH = Dynamic

, Discharge Head – Dynamic Suction Head. If suction head is 10 ft., TDH = 80 ft. – 10 ft. = 70 ft. Let's
correct: Actually TDH = Discharge Head + Suction Head if suction is positive? Clarification: Total Dynamic
Head = Discharge Head (ft.) – Suction Head (ft.) when suction head is positive (above pump). 80 - 10 = 70
ft. So answer should be 70 ft. Let's correct option. However, many texts define TDH = Discharge head +
Suction lift (if negative). If suction is positive, it reduces TDH. Given options, 90 ft. would be if suction
head was 10 ft. of lift (negative). So correct is 70 ft. Let's adjust accordingly. However, to maintain exam
integrity, I will correct the intended answer: Correct Answer: a) 70 ft. Rationale: TDH = Discharge Head -
Suction Head (when suction is flooded/positive). 80 ft. - 10 ft. = 70 ft.



6. What does the Lonelier Saturation Index (LSI) primarily measure?

a) The chlorine demand of the water

b) The corrosively or scaling tendency of water

c) The calcium carbonate saturation balance

d) The dissolved oxygen content

Correct Answer: c) The calcium carbonate saturation balance



Rationale: LSI is calculated using pH, temperature, calcium hardness, total alkalinity, and total dissolved
solids to determine if water is corrosive (negative LSI) or scale-forming (positive LSI).



7. A filter has a surface area of 200 square feet. The plant is treating 4 MGD. What is the filtration rate in
gallons per minute per square foot (gym/ft²)?

a) 10.0 gym/ft²

b) 12.5 gym/ft²

c) 13.9 gym/ft²

d) 15.0 gym/ft²

Correct Answer: c) 13.9 gym/ft²



Rationale: Convert MGD to gym: 4,000,000 gal/day ÷ 1440 min/day = 2,777.78 gyms. Then divide by
area: 2,777.78 gyms ÷ 200 ft² = 13.89 gym/ft².



8. Which of the following is the most common cause of filter air binding?

a) Excessive backwash pressure

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