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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


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

A. 15.0 ft³ B. 20.1 ft³ [CORRECT] C. 1122 ft³ D. 1500 ft³

Correct Answer: B

Rationale: Calculation: 150 gallons ÷ 7.48 gallons/ft³ = 20.05 ft³ ≈ 20.1 ft³

Step-by-step:

1. Set up conversion: gallons ÷ gallons per cubic foot = cubic feet

2. 150 ÷ 7.48 = 20.05

3. Round to one decimal place: 20.1 ft³

Why Others Are Wrong:

• A (15.0): Incorrectly divides by 10 instead of 7.48

• C (1122): Multiplies 150 × 7.48 instead of dividing—common error

• D (1500): Multiplies by 10, confusing units

Practical Tip: Always verify whether answer should be larger or smaller than original. Cubic
feet are larger than gallons, so dividing gives smaller number.



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: Step 1: Calculate volume in cubic feet Volume = Length × Width × Depth = 20 ft ×
10 ft × 8 ft = 1,600 ft³

Step 2: Convert to gallons 1,600 ft³ × 7.48 gal/ft³ = 11,968 gallons

Why Others Are Wrong:

• A (1,600): This is cubic feet, not gallons—forgetting conversion factor

• C (14,960): Uses 10 ft depth instead of 8 ft (20 × 10 × 10 × 7.48)

• D (20,000): Incorrect formula or unit confusion

Practical Tip: Always check units. If answer seems too small (1,600 for a large tank), you likely
forgot the 7.48 conversion.



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: Step 1: Convert gpm to gallons per day 500 gpm × 60 minutes/hour × 24 hours/day =
720,000 gallons/day

Step 2: Convert to MGD 720,000 gallons ÷ 1,000,000 = 0.72 MGD

Shortcut formula: gpm × 1.44 = MGD (500 × 1.44 = 720, then ÷ 1000 = 0.72)

Why Others Are Wrong:

• A (0.5): Divides 500 by 1000 without time conversion

• C (1.2): Incorrect multiplier (500 × 2.4)

• D (12): Multiplies 500 × 24 but keeps in thousands, not millions

Practical Tip: Remember: 1 MGD = 694.4 gpm (1,000,000 ÷ 1440 minutes/day). 500 gpm is
less than 694, so answer must be less than 1.0 MGD.



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: Formula: lbs/day = mg/L × MGD × 8.34

Calculation: 2.5 mg/L × 2.5 MGD × 8.34 = 6.25 × 8.34 = 52.125 lbs/day ≈ 52.1 lbs

Step-by-step:

1. Multiply concentration × flow: 2.5 × 2.5 = 6.25

2. Multiply by conversion factor: 6.25 × 8.34 = 52.125

Why Others Are Wrong:

• A (5.2): Multiplies mg/L × MGD only, forgetting 8.34

• B (20.9): Uses only mg/L × 8.34 (2.5 × 8.34), missing MGD

• D (208.5): Multiplies by 4 instead of 8.34, or dosage error

Practical Tip: The 8.34 factor converts mg/L-MGD to pounds (8.34 lbs/gallon). Always include
it for chemical dosage calculations.



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: Formula: Detention Time = Volume ÷ Flow Rate Rearranged: Volume = Flow Rate
× Detention Time

Calculation:

1. Convert flow to gallons per hour: 1.5 MGD = 1,500,000 gallons/day ÷ 24 hours = 62,500
gallons/hour

2. Volume = 62,500 gallons/hour × 4 hours = 250,000 gallons

Alternative: 1.5 MGD × (4/24) days = 1.5 × 0.1667 = 0.25 MG = 250,000 gallons

Why Others Are Wrong:

• B (500,000): 8 hours detention time

• C (750,000): 12 hours detention time

• D (1,000,000): 16 hours detention time or forgetting to convert days to hours

Practical Tip: Detention time calculations often require time unit conversions. Always ensure
flow and time units match.

, Question 6 A circular clarifier has a diameter of 40 feet and a depth of 12 feet. What is the
volume in gallons when full?

A. 75,400 gallons B. 112,900 gallons [CORRECT] C. 150,800 gallons D. 451,500 gallons

Correct Answer: B

Rationale: Step 1: Calculate area of circle Area = π × r² = 3.1416 × (20 ft)² = 3.1416 × 400 =
1,256.6 ft²

Step 2: Calculate volume in cubic feet Volume = Area × Depth = 1,256.6 ft² × 12 ft = 15,079.6
ft³

Step 3: Convert to gallons 15,079.6 ft³ × 7.48 gal/ft³ = 112,795 gallons ≈ 112,900 gallons

Why Others Are Wrong:

• A (75,400): Uses diameter instead of radius (π × 40² = 5,027 ft² area, then wrong volume)

• C (150,800): Uses r = 40 ft instead of 20 ft (double radius = 4× volume)

• D (451,500): Multiplies by 30 instead of 7.48, or uses wrong formula

Practical Tip: Remember: radius = diameter ÷ 2. Using diameter instead of radius is the most
common error in circular tank calculations.



Question 7 What is the surface loading rate (overflow rate) of a circular clarifier 50 feet in
diameter with a flow of 1.0 MGD?

A. 407 gpd/ft² B. 509 gpd/ft² [CORRECT] C. 636 gpd/ft² D. 1,018 gpd/ft²

Correct Answer: B

Rationale: Formula: Surface Loading Rate = Flow ÷ Surface Area

Step 1: Calculate surface area Area = π × r² = 3.1416 × (25 ft)² = 3.1416 × 625 = 1,963.5 ft²

Step 2: Convert flow to gallons per day 1.0 MGD = 1,000,000 gallons/day

Step 3: Calculate loading rate 1,000,000 gpd ÷ 1,963.5 ft² = 509.3 gpd/ft² ≈ 509 gpd/ft²

Why Others Are Wrong:

• A (407): Uses diameter instead of radius (area = 2,454 ft²)

• C (636): Uses 20 ft radius (area = 1,257 ft²)

• D (1,018): Uses half the area or doubles the flow

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
46
Escrito en
2025/2026
Tipo
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