Practice Test Geeks(WTO) Water Treatment Operator Certification Practice Test
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Math Flashcards

7 cards from real WTO practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 7 Math flashcards as text

  1. A chemical solution is prepared by dissolving 2 lbs of alum in 10 gallons of water. What is the solution concentration in mg/L?

    Answer: 23,957 mg/L

    2 lbs × 453,592 mg/lb ÷ (10 gal × 3,785 mL/gal) = 907,184 ÷ 37,850 = 23,963 mg/L ≈ 23,957 mg/L.

  2. A plant produces 6.5 MGD. The CT value required for Giardia inactivation is 150 mg/L·min. If chlorine residual is 1.5 mg/L, what is the minimum contact time needed?

    Answer: 100 minutes

    CT = C × T → T = 150 ÷ 1.5 = 100 minutes.

  3. A 6-inch diameter pipe carries water at 4 ft/sec. What is the flow rate in gpm?

    Answer: 352 gpm

    Area = π × (0.25 ft)² = 0.196 ft²; flow = 0.196 × 4 = 0.785 cfs × 448.8 = 352 gpm.

  4. An operator needs to raise the fluoride level from 0.4 mg/L to 0.8 mg/L in a 1.5 MGD plant. How many lbs/day of fluoride must be added?

    Answer: 5.0 lb/day

    Dose = 0.8 − 0.4 = 0.4 mg/L; lbs/day = 0.4 × 1.5 × 8.34 = 5.0 lb/day.

  5. A well pump test shows 450 gpm flow with a 25-foot drawdown. What is the specific capacity of the well?

    Answer: 18 gpm/ft

    Specific capacity = 450 gpm ÷ 25 ft = 18 gpm/ft.

  6. A water tower holds 500,000 gallons and must be disinfected using 50 mg/L chlorine solution. How many pounds of chlorine are needed?

    Answer: 208.5 lbs

    lbs = 50 × (500,000 ÷ 1,000,000) × 8.34 = 50 × 0.5 × 8.34 = 208.5 lbs.

  7. A sludge dewatering unit processes 10,000 gallons of sludge at 3% solids content. How many pounds of dry solids are removed?

    Answer: 2,502 lbs

    Dry solids = 10,000 gal × 8.34 lb/gal × 0.03 = 2,502 lbs.