T3 water treatment math worksheet
California T3 water treatment operator exam · 10 problems
Name: ______________________________ Date: ______________
Use the CA Exam Conversion Sheet (revised 4/30/2026) and a calculator.
1. A chemical metering pump is calibrated at 645 mL/min. How many gallons per day does it feed?
Chemical feed and solutions
- A) 3,520 GPD
- B) 245 GPD
- C) 929 GPD
- D) 10.2 GPD
2. A chlorine solution is 11.5% strength. What is its concentration in mg/L?
Unit conversions and pressure
- A) 959,000 mg/L
- B) 57,500 mg/L
- C) 230,000 mg/L
- D) 115,000 mg/L
3. A required CT is 77 mg·min/L and the free chlorine residual is 1.65 mg/L. What contact time is needed?
Chlorination and CT
- A) 46.7 min
- B) 0.0214 min
- C) 0.778 min
- D) 23.4 min
4. A treatment process removes 91.3% of Giardia cysts. What log removal is that?
Filtration and removal
- A) 1.06 log
- B) 0.353 log
- C) 0.795 log
- D) 0.530 log
5. A 4-inch main carries 225 GPM. What is the velocity in ft/sec?
Flow, velocity and detention
- A) 5.75 ft/sec
- B) 11.5 ft/sec
- C) 345 ft/sec
- D) 8.63 ft/sec
6. A cylindrical storage tank is 95 ft in diameter with water 16 ft deep. How many gallons does it hold?
Area and volume
- A) 3,390,000 gal
- B) 113,000 gal
- C) 8,930 gal
- D) 848,000 gal
7. A well yields 590 GPM. The static water level is 89 ft below ground and the pumping water level is 171 ft below ground. What is the specific capacity?
Wells, pumps and power
- A) 3.45 GPM/ft
- B) 14.4 GPM/ft
- C) 6.63 GPM/ft
- D) 7.20 GPM/ft
8. A 1.2 MGD system doses dry soda ash at 30 mg/L. It costs $0.23 per lb. What is the chemical cost per day?
Chemical feed and solutions
- A) $69.1 per day
- B) $300 per day
- C) $138 per day
- D) $1,310 per day
9. A flow transmitter is scaled 4 mA = 0 GPM and 20 mA = 3,200 GPM. The SCADA input reads 8.9 mA. What flow does that represent?
Unit conversions and pressure
- A) 1,230 GPM
- B) 980 GPM
- C) 1,420 GPM
- D) 1,780 GPM
10. A required CT is 131 mg·min/L and the free chlorine residual is 1.2 mg/L. What contact time is needed?
Chlorination and CT
- A) 109 min
- B) 1.82 min
- C) 157 min
- D) 130 min
Answer key: T3 water treatment math worksheet
1. B) 245 GPD
1. GPD = (Feed, mL/min × 1,440 min/day) ÷ (1,000 mL/L × 3.785 L/gal) [Sheet p.1, Solutions: "GPD = (Feed, ml/min. x 1,440 min/day) / (1,000 ml/L x 3.785 L/gal)"]. 2. = 645 × 1,440 ÷ 3,785 = 928,800 ÷ 3,785 = 245.4 GPD. 3. Answer: 245 GPD (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Solutions: "GPD = (Feed, ml/min. x 1,440 min/day) / (1,000 ml/L x 3.785 L/gal)"
2. D) 115,000 mg/L
1. 1 % = 10,000 mg/L [Sheet p.1, Units and Conversion Factors, row 1: "1 % = 10,000 mg/L"]. 2. 11.5% × 10,000 = 115,000 mg/L. 3. Answer: 115,000 mg/L.
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Units and Conversion Factors, row 1: "1 % = 10,000 mg/L"
3. A) 46.7 min
1. Time (min) = C × T ÷ Chlorine Residual [Sheet p.2, C x T Calculations: "Time (min) = C x T / Chlorine Residual (mg/L)"] = 77 ÷ 1.65 = 46.67 min. 2. Answer: 46.7 min (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.2, C x T Calculations: "Time (min) = C x T / Chlorine Residual (mg/L)"
4. A) 1.06 log
1. Log Removal = log(1.0 − %Removal/100) × (−1) [Sheet p.2, C x T Calculations: "Log Removal = 1.0 - (% Removal/100) x Log Key x (-1)" (i.e. -log10(1 - %Removal/100))]. 2. 1.0 − 0.913 = 0.087; log(0.087) = -1.06; × (−1) = 1.06 log. 3. Answer: 1.06 log (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.2, C x T Calculations: "Log Removal = 1.0 - (% Removal/100) x Log Key x (-1)" (i.e. -log10(1 - %Removal/100))
5. A) 5.75 ft/sec
1. Q = 225 GPM ÷ 448.8 GPM/CFS = 0.5013 ft³/sec [Sheet p.1, Units and Conversion Factors, row 7: "1 CFS = 448.8 GPM"]. 2. A = .785 × (4 ÷ 12)² = 0.08722 ft² [Sheet p.1, Area and Volume: "Area of Circle = .785 X D2"]. 3. V = Q ÷ A [Sheet p.1, Flow & Velocity: "V (Ft/Sec) = Q (Ft3/Sec) / A (Ft2)"] = 0.5013 ÷ 0.08722 = 5.748 ft/sec. 4. Answer: 5.75 ft/sec (rounded to 3 significant figures). 5. If you used 694.4 GPM per MGD (Units row 29) and 1.55 CFS per MGD (row 28) instead of 448.8 GPM per CFS (row 7), you get 5.76 ft/sec. The sheet prints both factors, so either answer is right; pick the option closest to yours.
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Units and Conversion Factors, row 7: "1 CFS = 448.8 GPM"; p.1, Area and Volume: "Area of Circle = .785 X D2"; p.1, Flow & Velocity: "V (Ft/Sec) = Q (Ft3/Sec) / A (Ft2)"
6. D) 848,000 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 × 95 × 95 × 16 × 7.48 = 847,888 gal. 3. Answer: 848,000 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. D) 7.20 GPM/ft
1. Drawdown = 171 − 89 = 82 ft. 2. Specific Capacity = Well Yield ÷ Drawdown [Sheet p.1, Pumping: "Specific Capacity (GPM/Ft) = Well Yield (GPM) / Drawdown (Ft)"] = 590 ÷ 82 = 7.195 GPM/ft. 3. Answer: 7.20 GPM/ft (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Pumping: "Specific Capacity (GPM/Ft) = Well Yield (GPM) / Drawdown (Ft)"
8. A) $69.1 per day
1. lbs/day = MGD × mg/L × 8.34 (the sheet's "ppm or mg/L" line solved for lbs) [Sheet p.2, Chemical Dosage Calculations: "ppm or mg/L = lbs / (MG x 8.34)"] = 1.2 × 30 × 8.34 = 300.24 lbs/day. 2. Cost/Day = Lbs/Day × Cost/lb [Sheet p.1, Other: "Cost/Day ($) = Lbs/Day x Cost/lb ($)"] = 300.24 × $0.23 = $69.0552 per day. 3. Answer: $69.1 per day (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.2, Chemical Dosage Calculations: "ppm or mg/L = lbs / (MG x 8.34)"; p.1, Other: "Cost/Day ($) = Lbs/Day x Cost/lb ($)"
9. B) 980 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. (8.9 − 4) = 4.9 mA; 4.9 × 3,200 ÷ 16 = 980 GPM. 3. Answer: 980 GPM.
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"
10. A) 109 min
1. Time (min) = C × T ÷ Chlorine Residual [Sheet p.2, C x T Calculations: "Time (min) = C x T / Chlorine Residual (mg/L)"] = 131 ÷ 1.2 = 109.2 min. 2. Answer: 109 min (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.2, C x T Calculations: "Time (min) = C x T / Chlorine Residual (mg/L)"