T1 water treatment math worksheet
California T1 water treatment operator exam · 10 problems
Name: ______________________________ Date: ______________
Use the CA Exam Conversion Sheet (revised 4/30/2026) and a calculator.
1. A tank holds 780 gallons of a 25% solution with a specific gravity of 1.27. How many pounds of active chemical does it contain?
Chemical feed and solutions
- A) 1,850 lbs
- B) 2,070 lbs
- C) 248 lbs
- D) 1,630 lbs
2. A hardness test reports 16.5 grains per gallon (gpg). What is the hardness in mg/L?
Unit conversions and pressure
- A) 282 mg/L
- B) 564 mg/L
- C) 138 mg/L
- D) 0.965 mg/L
3. The chlorine demand of a 7.4 MGD supply is 1 mg/L and the target free residual is 0.5 mg/L. How many lbs/day of chlorine must be fed?
Chlorination and CT
- A) 30.9 lbs/day
- B) 92.6 lbs/day
- C) 61.7 lbs/day
- D) 185 lbs/day
4. A storage tank gains 910,000 gallons in 19 hours while nothing is drawn from it. What is the average inflow in GPM?
Flow, velocity and detention
- A) 13.3 GPM
- B) 33.3 GPM
- C) 399 GPM
- D) 798 GPM
5. What is the cross-sectional area, in square feet, of a 24-inch diameter pipe?
Area and volume
- A) 12.6 ft²
- B) 37.7 ft²
- C) 452 ft²
- D) 3.14 ft²
6. A well yields 1,720 GPM. The static water level is 142 ft below ground and the pumping water level is 249 ft below ground. What is the specific capacity?
Wells, pumps and power
- A) 32.2 GPM/ft
- B) 12.1 GPM/ft
- C) 6.91 GPM/ft
- D) 16.1 GPM/ft
7. A chemical metering pump is calibrated at 520 mL/min. How many gallons per day does it feed?
Chemical feed and solutions
- A) 198 GPD
- B) 8.24 GPD
- C) 23.7 GPD
- D) 99.0 GPD
8. A treatment plant produces 1.7 MGD. What is the average flow in gallons per minute (GPM)?
Unit conversions and pressure
- A) 1,180 GPM
- B) 1,770 GPM
- C) 2,450 GPM
- D) 1,480 GPM
9. A 6-inch well casing holds a 500-ft column of water. How many pounds of 70% HTH are needed to dose the water in the casing at 25 mg/L?
Chlorination and CT
- A) 0.437 lbs
- B) 0.0292 lbs
- C) 0.153 lbs
- D) 0.219 lbs
10. Water flows through a 14-inch main at 3.1 ft/sec. What is the flow in GPM?
Flow, velocity and detention
- A) 24.8 GPM
- B) 1,490 GPM
- C) 745 GPM
- D) 1,270 GPM
Answer key: T1 water treatment math worksheet
1. B) 2,070 lbs
1. (lbs) Chemical = SG × 8.34 × Solution (Gal) × Solution (%/100) [Sheet p.1, Solutions: "(lbs) Chemical = SG x 8.34 x Solution (Gal) x Solution (%/100)"]. 2. = 1.27 × 8.34 × 780 × 0.25 = 2,065 lbs. 3. Answer: 2,070 lbs (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Solutions: "(lbs) Chemical = SG x 8.34 x Solution (Gal) x Solution (%/100)"
2. A) 282 mg/L
1. 1 gpg = 17.1 mg/L [Sheet p.1, Units and Conversion Factors, row 18: "1 gpg = 17.1 mg/L"]. 2. Hardness = 16.5 × 17.1 = 282.15 mg/L. 3. Answer: 282 mg/L (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Units and Conversion Factors, row 18: "1 gpg = 17.1 mg/L"
3. B) 92.6 lbs/day
1. Dosage = Demand + Residual [Sheet p.1, Chlorination: "Dosage (mg/L) = Demand (mg/L) + Residual (mg/L)"] = 1 + 0.5 = 1.5 mg/L. 2. lbs/day = MGD × mg/L × 8.34 [Sheet p.1, Chlorination: "Gas (lbs) = MG x mg/L x 8.34"] = 7.4 × 1.5 × 8.34 = 92.574 lbs/day. 3. Answer: 92.6 lbs/day (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Chlorination: "Dosage (mg/L) = Demand (mg/L) + Residual (mg/L)"; p.1, Chlorination: "Gas (lbs) = MG x mg/L x 8.34"
4. D) 798 GPM
1. Time in minutes = 19 hrs × 60 min/hr = 1,140 min [Sheet p.1, Units and Conversion Factors, rows 8-9: 1 day = 24 hrs = 1440 min, so 1440 / 24 = 60 min/hr]. 2. GPM = Volume ÷ Time [Sheet p.1, Flow & Velocity: "GPM = Volume (Gal) / Time (Min)"] = 910,000 ÷ 1,140 = 798.2 GPM. 3. Answer: 798 GPM (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Units and Conversion Factors, rows 8-9: 1 day = 24 hrs = 1440 min, so 1440 / 24 = 60 min/hr; p.1, Flow & Velocity: "GPM = Volume (Gal) / Time (Min)"
5. D) 3.14 ft²
1. Convert the diameter to feet: D = 24 in ÷ 12 in/ft = 2 ft [12 in = 1 ft]. 2. Area = .785 × D² [Sheet p.1, Area and Volume: "Area of Circle = .785 X D2"] = .785 × (24 ÷ 12)² = 3.14 ft². 3. Answer: 3.14 ft².
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Area and Volume: "Area of Circle = .785 X D2"
6. D) 16.1 GPM/ft
1. Drawdown = 249 − 142 = 107 ft. 2. Specific Capacity = Well Yield ÷ Drawdown [Sheet p.1, Pumping: "Specific Capacity (GPM/Ft) = Well Yield (GPM) / Drawdown (Ft)"] = 1,720 ÷ 107 = 16.07 GPM/ft. 3. Answer: 16.1 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)"
7. A) 198 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. = 520 × 1,440 ÷ 3,785 = 748,800 ÷ 3,785 = 197.8 GPD. 3. Answer: 198 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)"
8. A) 1,180 GPM
1. 1 MGD = 694.4 GPM [Sheet p.1, Units and Conversion Factors, row 29: "1 MGD = 694.4 GPM"]. 2. GPM = 1.7 MGD × 694.4 = 1,180.48 GPM. 3. Answer: 1,180 GPM (rounded to 3 significant figures).
Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Units and Conversion Factors, row 29: "1 MGD = 694.4 GPM"
9. D) 0.219 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 × 500 × 7.48 = 733.975 gal. 3. MG = 733.975 ÷ 1,000,000 = 0.000733975 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)"] = 0.000733975 × 25 × 8.34 ÷ 0.7 = 0.2186 lbs. 5. Answer: 0.219 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)"
10. B) 1,490 GPM
1. A = .785 × D² [Sheet p.1, Area and Volume: "Area of Circle = .785 X D2"] = .785 × (14 ÷ 12)² = 1.068 ft² [12 in = 1 ft]. 2. Q = A × V [Sheet p.1, Flow & Velocity: "(Q) Flow (Ft3/Sec) = (A) Area (Ft2) x (V) Velocity (Ft/Sec)"] = 1.068 × 3.1 = 3.3108 ft³/sec. 3. 1 CFS = 448.8 GPM [Sheet p.1, Units and Conversion Factors, row 7: "1 CFS = 448.8 GPM"]: 3.3108 × 448.8 = 1,486 GPM. 4. Answer: 1,490 GPM (rounded to 3 significant figures). 5. If you used 1.55 CFS per MGD (Units row 28) and 694.4 GPM per MGD (row 29) instead of 448.8 GPM per CFS (row 7), you get 1,480 GPM. 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, Area and Volume: "Area of Circle = .785 X D2"; p.1, Flow & Velocity: "(Q) Flow (Ft3/Sec) = (A) Area (Ft2) x (V) Velocity (Ft/Sec)"; p.1, Units and Conversion Factors, row 7: "1 CFS = 448.8 GPM"