T2 water treatment math worksheet
California T2 water treatment operator exam · 10 problems
Name: ______________________________ Date: ______________
Use the CA Exam Conversion Sheet (revised 4/30/2026) and a calculator.
1. A liquid chemical weighs 10.9 lbs per gallon. What is its specific gravity?
Chemical feed and solutions
- A) 0.175
- B) 1.31
- C) 1.46
- D) 0.765
2. A flow transmitter is scaled 4 mA = 0 GPM and 20 mA = 2,600 GPM. The SCADA input reads 6.9 mA. What flow does that represent?
Unit conversions and pressure
- A) 377 GPM
- B) 471 GPM
- C) 1,120 GPM
- D) 897 GPM
3. A 6-inch well casing holds a 435-ft column of water. How many pounds of 65% HTH are needed to dose the water in the casing at 25 mg/L?
Chlorination and CT
- A) 0.819 lbs
- B) 0.410 lbs
- C) 0.205 lbs
- D) 0.308 lbs
4. A 17 ft × 18 ft filter runs all day at 3.2 GPM/ft². How many gallons does it produce per day?
Filtration and removal
- A) 58,800 GPD
- B) 161,000 GPD
- C) 10,500,000 GPD
- D) 1,410,000 GPD
5. A round clearwell is 55 ft in diameter with water 11 ft deep. Flow through it is 3,250 GPM. What is the detention time in minutes?
Flow, velocity and detention
- A) 90.2 min
- B) 120 min
- C) 60.1 min
- D) 240 min
6. A cylindrical storage tank is 40 ft in diameter with water 14 ft deep. How many gallons does it hold?
Area and volume
- A) 17,600 gal
- B) 132,000 gal
- C) 3,290 gal
- D) 526,000 gal
7. Before pumping, the water level in a well is 28 ft below ground. After pumping for several hours it stabilizes at 127 ft below ground. What is the drawdown?
Wells, pumps and power
- A) 77.5 ft
- B) 99.0 ft
- C) 155 ft
- D) 127 ft
8. An operator dissolves 95 lbs of alum in 85 gallons of water. What is the percent strength of the solution by weight?
Chemical feed and solutions
- A) 5.90%
- B) 11.8%
- C) 3.93%
- D) 8.85%
9. The water surface in a reservoir is 121 ft above a pressure gauge on a closed (static) main. What should the gauge read?
Unit conversions and pressure
- A) 26.2 psi
- B) 52.4 psi
- C) 16.0 psi
- D) 39.3 psi
10. A chlorinator fed 247 lbs of chlorine in one day to a flow of 6.3 MGD. What was the dose in mg/L?
Chlorination and CT
- A) 3.53 mg/L
- B) 2.35 mg/L
- C) 4.70 mg/L
- D) 0.213 mg/L
Answer key: T2 water treatment math worksheet
1. B) 1.31
1. Specific Gravity = Chemical Weight (lbs/gal) ÷ 8.34 [Sheet p.1, Solutions: "Specific Gravity (SG) = Chemical Weight (Lbs/Gal) / 8.34"] = 10.9 ÷ 8.34 = 1.307. 2. Answer: 1.31 (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Solutions: "Specific Gravity (SG) = Chemical Weight (Lbs/Gal) / 8.34"
2. B) 471 GPM
1. Process Units = (Live Signal − 4 mA) × Process Range ÷ 16 mA [Sheet p.1, Electrical/SCADA: "Process Units = (Live Signal (mA) - 4 mA) x Process Range / 16 mA"]. 2. (6.9 − 4) = 2.9 mA; 2.9 × 2,600 ÷ 16 = 471.25 GPM. 3. Answer: 471 GPM (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Electrical/SCADA: "Process Units = (Live Signal (mA) - 4 mA) x Process Range / 16 mA"
3. C) 0.205 lbs
1. Volume = .785 × D² × Height × 7.48 [Sheet p.1, Area and Volume: "Cylinder Volume (Gal) = .785 x D2 x Height or Length (Ft) x 7.48"], with D = 6 in ÷ 12 [12 in = 1 ft]. 2. = .785 × (6 ÷ 12)² × 435 × 7.48 = 638.6 gal. 3. MG = 638.6 ÷ 1,000,000 = 0.0006386 MG [Sheet p.1, Unit Definition: "MG Million Gallons" (1 MG = 1,000,000 gal)]. 4. HTH Solid (lbs) = MG × mg/L × 8.34 ÷ (Strength/100) [Sheet p.1, Chlorination: "HTH Solid (lbs) = MG x mg/L x 8.34 / (Strength (%)/100)"] = 0.0006386 × 25 × 8.34 ÷ 0.65 = 0.2048 lbs. 5. Answer: 0.205 lbs (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Area and Volume: "Cylinder Volume (Gal) = .785 x D2 x Height or Length (Ft) x 7.48"; p.1, Unit Definition: "MG Million Gallons" (1 MG = 1,000,000 gal); p.1, Chlorination: "HTH Solid (lbs) = MG x mg/L x 8.34 / (Strength (%)/100)"
4. D) 1,410,000 GPD
1. Filter area = 17 × 18 = 306 ft² [Sheet p.1, Area and Volume: "Area of Rectangle = Length x Width"]. 2. Daily Filter Production = Filter Area × (GPM/ft²) × 1440 [Sheet p.2, Filtration: "Daily Filter Production (GPD) = Filter Area (Ft2) x (GPM/Ft2) x 1440"] = 306 × 3.2 × 1440 = 1,410,048 GPD. 3. Answer: 1,410,000 GPD (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Area and Volume: "Area of Rectangle = Length x Width"; p.2, Filtration: "Daily Filter Production (GPD) = Filter Area (Ft2) x (GPM/Ft2) x 1440"
5. C) 60.1 min
1. Volume = .785 × 55² × 11 × 7.48 = 195,384 gal [Sheet p.1, Area and Volume: "Cylinder Volume (Gal) = .785 x D2 x Height or Length (Ft) x 7.48"]. 2. DT (Min) = Volume (Gal) ÷ Flow (GPM) [Sheet p.1, Sedimentation: "DT (Min) = Volume (Gal) / Flow (GPM)"] = 195,384 ÷ 3,250 = 60.12 min. 3. Answer: 60.1 min (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Area and Volume: "Cylinder Volume (Gal) = .785 x D2 x Height or Length (Ft) x 7.48"; p.1, Sedimentation: "DT (Min) = Volume (Gal) / Flow (GPM)"
6. B) 132,000 gal
1. Cylinder Volume (Gal) = .785 × D² × Height × 7.48 [Sheet p.1, Area and Volume: "Cylinder Volume (Gal) = .785 x D2 x Height or Length (Ft) x 7.48"]. 2. = .785 × 40 × 40 × 14 × 7.48 = 131,528 gal. 3. Answer: 132,000 gal (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Area and Volume: "Cylinder Volume (Gal) = .785 x D2 x Height or Length (Ft) x 7.48"
7. B) 99.0 ft
1. Drawdown = pumping water level − static water level (the denominator of [Sheet p.1, Pumping: "Specific Capacity (GPM/Ft) = Well Yield (GPM) / Drawdown (Ft)"]). 2. = 127 − 28 = 99 ft. 3. Answer: 99.0 ft.
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Pumping: "Specific Capacity (GPM/Ft) = Well Yield (GPM) / Drawdown (Ft)"
8. B) 11.8%
1. Water weight = 85 gal × 8.34 lbs/gal = 708.9 lbs [Sheet p.1, Units and Conversion Factors, row 16: "1 Gal (H2O) = 8.34 lbs"]. 2. Solution total weight = 95 + 708.9 = 803.9 lbs. 3. Chemical in Solution (%) = Dry Chemical × 100 ÷ Solution Total Weight [Sheet p.1, Solutions: "Chemical in Solution (%) = Dry Chemical (Lbs) x 100 / Solution Total Weight (lbs)"] = 95 × 100 ÷ 803.9 = 11.82%. 4. Answer: 11.8% (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Units and Conversion Factors, row 16: "1 Gal (H2O) = 8.34 lbs"; p.1, Solutions: "Chemical in Solution (%) = Dry Chemical (Lbs) x 100 / Solution Total Weight (lbs)"
9. B) 52.4 psi
1. 1 Ft (H₂O) = 0.433 PSI [Sheet p.1, Units and Conversion Factors, row 12: "1 Ft (H2O) = 0.433 PSI"]. 2. Pressure = 121 ft × 0.433 psi/ft = 52.393 psi. 3. Answer: 52.4 psi (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Units and Conversion Factors, row 12: "1 Ft (H2O) = 0.433 PSI"
10. C) 4.70 mg/L
1. mg/L = lbs ÷ (MG × 8.34) [Sheet p.1, Solutions: "mg/L = lbs / (MG x 8.34)"] = 247 ÷ (6.3 × 8.34) = 247 ÷ 52.542 = 4.701 mg/L. 2. Answer: 4.70 mg/L (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Solutions: "mg/L = lbs / (MG x 8.34)"