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Level 3 · Advanced sample questions

Math and Calculations Practice Questions for Level 3 · Advanced

5 free sample questions on math and calculations for the Level 3 · Advanced wastewater treatment operator exam. Try each one, then open the answer to check your work.

  1. Question 1

    A pump delivers 900 gpm against 80 ft of head. Pump efficiency is 75%, and motor efficiency is 90%. What is the motor electrical input expressed in horsepower?

    1. 24.2 hp
    2. 20.2 hp
    3. 26.9 hp
    4. 12.3 hp
    Show the answer

    Answer: C. 26.9 hp

    The pump requires 24.2 brake hp at the shaft. Accounting for 90% motor efficiency gives an electrical input equivalent to 26.9 hp. Calculation: (900 × 80) ÷ (3,960 × 0.75 × 0.90) = 26.9 hp.

  2. Question 2

    A single-phase AC circuit supplying a blower operates at 480 volts and 40 amps with a power factor of 0.85. What is the power demand?

    1. 16.32 kW
    2. 19.20 kW
    3. 13.87 kW
    4. 22.59 kW
    Show the answer

    Answer: A. 16.32 kW

    Multiply volts, amps, and power factor, then convert watts to kW. Calculation: 480 × 40 × 0.85 = 16,320 W = 16.32 kW.

  3. Question 3

    A plant treats 1.0 MGD. The active chemical dose is 4.0 mg/L. The feed solution is 50% active and has a density of 1,200 mg/mL. What feed pump setting is needed?

    1. 4.63 mL/min
    2. 35.05 mL/min
    3. 8.76 mL/min
    4. 17.52 mL/min
    Show the answer

    Answer: D. 17.52 mL/min

    The feed rate must account for flow, dose, chemical density, active strength, and minutes per day. Calculation: (1.0 × 4.0 × 3.785 × 1,000,000) ÷ (1,200 × 0.50 × 1,440) = 17.52 mL/min.

  4. Question 4

    A centrifuge feed sludge is 4.0% TS. The cake is 25.0% TS, and the centrate is 0.2% TSS. What is the solids capture?

    1. 95.57%
    2. 95.77%
    3. 99.20%
    4. 95.00%
    Show the answer

    Answer: B. 95.77%

    The calculation corrects both the feed and cake values for solids leaving in the centrate. Calculation: [25.0 × (4.0 − 0.2)] ÷ [4.0 × (25.0 − 0.2)] × 100 = 95.77%.

  5. Question 5

    A DAF thickener receives 6,000 lb of solids per day over 400 ft². What is the solids loading rate?

    1. 1.5 lb/day/ft²
    2. 6.7 lb/day/ft²
    3. 15.0 lb/day/ft²
    4. 24.0 lb/day/ft²
    Show the answer

    Answer: C. 15.0 lb/day/ft²

    Divide the daily solids load by the thickener surface area. Calculation: 6,000 ÷ 400 = 15.0 lb/day/ft².

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Other Level 3 · Advanced Topics

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