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Colorado WW Operator D Exam Actual Exam QUESTIONS AND ANSWERS 2026 | Complete Solution Guide A+ Graded | Pass Guaranteed - A+ Graded

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Colorado WW Operator D Exam Actual Exam QUESTIONS AND ANSWERS 2026 | Complete Solution Guide A+ Graded | Pass Guaranteed - A+ Graded

Institución
Colorado WW Operator D
Grado
Colorado WW Operator D

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Colorado WW Operator D Exam Actual
Exam QUESTIONS AND ANSWERS 2026 |
Complete Solution Guide A+ Graded | Pass
Guaranteed - A+ Graded

Section 1: Basic Mathematics and Calculations

Questions 1-15



Question 1 Convert 150 gallons to cubic feet. (1 cubic foot = 7.48 gallons)

A. 15.0 ft³ B. 20.1 ft³ [CORRECT] C. 1,122 ft³ D. 1,500 ft³

Correct Answer: B

Rationale: To convert gallons to cubic feet, divide by 7.48: 150 gallons ÷ 7.48 gal/ft³ = 20.05 ft³
≈ 20.1 ft³.

• Option A (15.0): Incorrect, likely from dividing by 10 instead of 7.48.

• Option C (1,122): Incorrect, results from multiplying 150 × 7.48 instead of dividing.

• Option D (1,500): Incorrect, results from multiplying 150 × 10.



Question 2 A rectangular tank is 20 feet long, 10 feet wide, and has a water depth of 8 feet. What
is the volume of water in gallons? (1 ft³ = 7.48 gallons)
A. 1,600 gallons B. 11,968 gallons [CORRECT] C. 14,960 gallons D. 20,000 gallons

Correct Answer: B

Rationale: First calculate volume in cubic feet: 20 ft × 10 ft × 8 ft = 1,600 ft³. Then convert to
gallons: 1,600 ft³ × 7.48 gal/ft³ = 11,968 gallons.

• Option A (1,600): Incorrect, this is the volume in cubic feet, not gallons.

• Option C (14,960): Incorrect, likely using 10 ft depth instead of 8 ft (20 × 10 × 10 ×
7.48).

• Option D (20,000): Incorrect, appears to be a rough estimate without proper calculation.

,Question 3 A pump delivers 500 gallons per minute (gpm). How many million gallons per day
(MGD) does this represent?

A. 0.5 MGD B. 0.72 MGD [CORRECT] C. 1.2 MGD D. 12 MGD

Correct Answer: B

Rationale: Convert gpm to MGD: 500 gpm × 60 minutes/hour × 24 hours/day = 720,000
gallons/day = 0.72 MGD.

• Option A (0.5): Incorrect, might be from 500/1000 without time conversion.
• Option C (1.2): Incorrect, possibly from 500 × 2.4 (incorrect conversion factor).

• Option D (12): Incorrect, from 500 × 24 = 12,000 then misplaced decimal.



Question 4 A treatment plant treats 2.5 MGD. The chlorine dosage is 2.5 mg/L. How many
pounds of chlorine are used per day? (Use formula: lbs/day = mg/L × MGD × 8.34)

A. 5.2 lbs B. 20.9 lbs C. 52.1 lbs [CORRECT] D. 208.5 lbs

Correct Answer: C

Rationale: Using the formula: lbs/day = 2.5 mg/L × 2.5 MGD × 8.34 = 6.25 × 8.34 = 52.125 ≈
52.1 lbs/day.

• Option A (5.2): Incorrect, missing the 8.34 factor (2.5 × 2.5 = 6.25, misplaced decimal).

• Option B (20.9): Incorrect, might be 2.5 × 8.34 (missing MGD factor).

• Option D (208.5): Incorrect, possibly multiplied by 4 or used wrong decimal placement.



Question 5 The detention time in a clarifier is 4 hours. If the flow rate is 1.5 MGD, what is the
volume of the clarifier in gallons?

A. 250,000 gallons [CORRECT] B. 500,000 gallons C. 750,000 gallons D. 1,000,000 gallons

Correct Answer: A

Rationale: Detention time = Volume ÷ Flow rate. Rearranged: Volume = Flow rate × Detention
time. Convert 4 hours to days: 4 ÷ 24 = 0.1667 days. Volume = 1.5 MGD × 0.1667 days = 0.25
million gallons = 250,000 gallons.

• Option B (500,000): Incorrect, would require 8 hours detention time.

, • Option C (750,000): Incorrect, would require 12 hours detention time.

• Option D (1,000,000): Incorrect, would require 16 hours detention time.



Question 6 What is the surface area of a circular clarifier with a diameter of 40 feet?

A. 1,256 ft² [CORRECT] B. 2,512 ft² C. 125.6 ft² D. 160 ft²

Correct Answer: A

Rationale: Area = π × r² where r = diameter/2 = 20 ft. Area = 3.1416 × (20)² = 3.1416 × 400 =
1,256.6 ≈ 1,257 ft² (closest to 1,256).

• Option B (2,512): Incorrect, used diameter instead of radius (π × 40² ÷ 4 or calculation
error).

• Option C (125.6): Incorrect, used circumference formula (π × d) instead of area.

• Option D (160): Incorrect, simple multiplication (40 × 4).



Question 7 A chemical feed pump delivers 12% sodium hypochlorite solution. If the pump
delivers 50 gallons per day of solution, how many pounds of pure chlorine are delivered?
(Assume 8.34 lbs/gallon for water; 12% solution is slightly heavier, but use 8.34 for calculation)

A. 4.2 lbs B. 50.0 lbs C. 50.0 lbs of solution D. 50.0 lbs × 0.12 = 6.0 lbs, but closest is Need to
recalculate

Correct Answer: 50 gal/day × 8.34 lbs/gal × 0.12 = 50.04 lbs solution × 0.12 = 6.0 lbs pure
chlorine. (If options don't match, closest correct calculation)

Let me provide proper options:

A. 4.2 lbs B. 5.0 lbs [CORRECT] C. 50.0 lbs D. 100.4 lbs

Correct Answer: B
Rationale: 50 gallons/day × 8.34 lbs/gallon × 0.12 (12% concentration) = 50.04 lbs solution ×
0.12 = 6.0 lbs ≈ 5.0 lbs (or exact calculation: 50 × 8.34 × 0.12 = 50.04). Actually 50 × 8.34 = 417
lbs solution × 0.12 = 50.04 lbs. Wait - let me recalculate: 50 gallons × 8.34 lbs/gal = 417 lbs of
solution. 417 × 0.12 = 50.04 lbs of pure chlorine.

Corrected options should be: A. 4.2 lbs B. 50.0 lbs [CORRECT] C. 417 lbs D. 500 lbs

Correct Answer: B (50.0 lbs)

, Rationale: 50 gallons/day × 8.34 lbs/gallon = 417 lbs of 12% solution. 417 lbs × 0.12 = 50.04
lbs of pure chlorine.

• Option A (4.2): Incorrect, wrong decimal placement.

• Option C (417): Incorrect, this is total solution weight, not pure chlorine.

• Option D (500): Incorrect, rough estimate without proper calculation.



Question 8 What is the flow velocity in a 12-inch pipe flowing at 2.0 MGD?

A. 0.5 ft/sec B. 1.0 ft/sec C. 2.0 ft/sec D. 4.9 ft/sec [CORRECT]

Correct Answer: D

Rationale: First convert MGD to cfs: 2.0 MGD × 1.547 cfs/MGD = 3.094 cfs. Pipe area = π ×
(0.5 ft)² = 0.785 ft². Velocity = Flow ÷ Area = 3.094 ÷ 0.785 = 3.94 ft/sec ≈ 4.9 ft/sec (using
standard conversion: V = 0.408 × Q/D² where Q in gpm, D in inches; or 2.0 MGD = 1,389 gpm;
V = 0.408 × 1,389/144 = 3.93 ft/sec).

Actually using Q = AV: 2 MGD = 2.78 cfs. Area = 0.785 ft². V = 2.78/0.785 = 3.54 ft/sec.

Given variations, the closest standard answer is typically around 4.9 ft/sec for these parameters,
or let me recalculate with exact values: 12" pipe = 1 ft diameter, area = 0.785 ft². 2 MGD =
2,000,000 gal/day ÷ 648,000 gal/day/cfs = 3.086 cfs. V = 3.086/0.785 = 3.93 ft/sec.

Options should be: A. 0.5 ft/sec B. 1.0 ft/sec C. 2.0 ft/sec
D. 3.9 ft/sec [CORRECT]

Correct Answer: D

Rationale: Convert 2.0 MGD to cubic feet per second: 2.0 MGD × 1.547 cfs/MGD = 3.094 cfs.
Calculate pipe cross-sectional area: 12-inch diameter = 1 foot; Area = π × (0.5 ft)² = 0.785 ft².
Velocity = Flow ÷ Area = 3.094 cfs ÷ 0.785 ft² = 3.94 ft/sec ≈ 3.9 ft/sec.
• Option A (0.5): Incorrect, significantly underestimated.

• Option B (1.0): Incorrect, underestimated.

• Option C (2.0): Incorrect, close but still underestimated.



Question 9 A sedimentation basin has a surface loading rate of 800 gpd/ft². If the flow is 0.5
MGD, what is the surface area required?

A. 525 ft² B. 625 ft² [CORRECT] C. 800 ft² D. 1,600 ft²

Escuela, estudio y materia

Institución
Colorado WW Operator D
Grado
Colorado WW Operator D

Información del documento

Subido en
20 de febrero de 2026
Número de páginas
32
Escrito en
2025/2026
Tipo
Examen
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