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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: To convert gallons to cubic feet, divide by 7.48: 150 gallons ÷ 7.48 gallons/ft³ =
20.05 ft³ ≈ 20.1 ft³. Option A (15.0) incorrectly divides by 10. Option C (1122) mistakenly
multiplies 150 × 7.48 instead of dividing. Option D (1500) is simply 150 × 10 with no
conversion factor. Always verify unit conversions by checking if the answer makes logical
sense—150 gallons should be approximately 20 cubic feet, not over 1,000.



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: Volume = length × width × depth = 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) is the volume in cubic feet, not gallons—a common error of stopping the calculation too
early. Option C (14,960) might use 10 ft depth instead of 8 ft. Option D (20,000) appears to
multiply dimensions incorrectly (20 × 10 × 100).

,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 through unit analysis: 500 gpm × 60 minutes/hour × 24
hours/day = 720,000 gallons/day = 0.72 MGD. Option A (0.5) incorrectly assumes 1,000 gpm =
1 MGD without time conversion. Option C (1.2) might use 2.4 hours instead of 24. Option D
(12) incorrectly calculates 500 × 24 = 12,000 then divides by 1,000 without the minutes
conversion. Remember: 1 MGD = 694.4 gpm (a useful conversion factor to memorize).



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 dosage 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) forgets the 8.34 factor (just multiplies mg/L × MGD).
Option B (20.9) might be 2.5 × 8.34 (forgetting the flow). Option D (208.5) incorrectly
multiplies by 4, perhaps confusing with a different flow rate. The factor 8.34 converts mg/L to
lbs/MG.



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: Rearrange the detention time formula: Volume = Flow × Detention Time. Convert 4
hours to days: 4 ÷ 24 = 0.1667 days. Volume = 1.5 MG/day × 0.1667 day = 0.25 MG = 250,000
gallons. Option B (500,000) would result from 8 hours detention. Option C (750,000) would be
12 hours. Option D (1,000,000) would be 16 hours. Always ensure time units match—flow is in
days, so detention time must be converted to days.



Question 6 A circular clarifier has a diameter of 40 feet and a water depth of 10 feet. What is the
volume in gallons?
A. 9,400 gallons
B. 70,336 gallons
C. 94,000 gallons [CORRECT]
D. 125,600 gallons

Correct Answer: C

Rationale: Calculate area first: Area = π × r² = 3.1416 × (20 ft)² = 1,256.6 ft². Volume = 1,256.6
ft² × 10 ft = 12,566 ft³. Convert to gallons: 12,566 ft³ × 7.48 gal/ft³ = 94,000 gallons. Option A
(9,400) is off by a factor of 10. Option B (70,336) might use diameter instead of radius (40²
instead of 20²). Option D (125,600) is the volume in cubic feet, not gallons.



Question 7 What is the flow velocity in a 12-inch pipe carrying 2.0 cfs (cubic feet per second)?

A. 0.5 ft/sec
B. 1.5 ft/sec
C. 2.55 ft/sec [CORRECT]
D. 4.0 ft/sec

Correct Answer: C

Rationale: Use the continuity equation: Q = A × V, so V = Q ÷ A. Convert 12 inches to 1 foot
diameter, radius = 0.5 ft. Area = π × r² = 3.1416 × (0.5)² = 0.785 ft². Velocity = 2.0 cfs ÷ 0.785 ft²
= 2.55 ft/sec. Option A (0.5) might use diameter as radius. Option B (1.5) is a rough estimate.
Option D (4.0) might use Q × A instead of Q ÷ A. Velocity should typically be 2-5 ft/sec in water
pipes.



Question 8 A chemical feed pump delivers 25 mL/min of a 12% sodium hypochlorite solution.
How many gallons per day is this?

A. 6.3 gallons/day
B. 9.5 gallons/day [CORRECT]

, C. 15.8 gallons/day
D. 25 gallons/day

Correct Answer: B

Rationale: Convert mL/min to gal/day: 25 mL/min × 60 min/hr × 24 hr/day = 36,000 mL/day =
36 L/day. Convert liters to gallons: 36 L ÷ 3.785 L/gal = 9.51 gallons/day. Option A (6.3) might
use 1 L = 1 gal approximation incorrectly. Option C (15.8) might forget the 24-hour conversion.
Option D (25) keeps the original number without conversion. Remember: 1 gallon = 3.785 liters
= 3,785 mL.


Question 9 The surface loading rate (SLR) of a sedimentation basin is calculated as:

A. Flow ÷ Volume
B. Flow ÷ Surface Area [CORRECT]
C. Volume ÷ Flow
D. Surface Area ÷ Flow

Correct Answer: B

Rationale: Surface Loading Rate (also called overflow rate) = Flow Rate ÷ Surface Area,
typically expressed as gpd/ft² or m³/m²/day. It represents the water volume applied per unit of
surface area per day. Option A (Flow ÷ Volume) gives detention time inverse. Option C (Volume
÷ Flow) is detention time. Option D is the inverse of the correct formula. SLR determines
whether particles will settle—typical design values are 300-1,000 gpd/ft² for water treatment.



Question 10 A water treatment plant has a filter with surface area of 500 ft². If the filtration rate
is 4 gpm/ft², what is the total flow rate in gpm?

A. 125 gpm
B. 500 gpm
C. 2,000 gpm [CORRECT]
D. 8,000 gpm

Correct Answer: C

Rationale: Total flow = Filtration rate × Area = 4 gpm/ft² × 500 ft² = 2,000 gpm. Option A (125)
divides instead of multiplies. Option B (500) ignores the rate. Option D (8,000) might double the
calculation incorrectly. Filtration rates typically range from 2-10 gpm/ft² depending on media
type and water quality.

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
39
Escrito en
2025/2026
Tipo
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