COLORADO WW OPERATOR D EXAM ACTUAL EXAM
QUESTIONS WITH COMPLETE SOLUTION GUIDE (A+
GRADED 100% VERIFIED) LATEST VERSION 2025!!
Question 1
What is the formula for calculating the F/M ratio (Food to Microorganism
ratio)?
A) BOD, lbs/day x MLVSS, lbs
B) MLVSS, lbs / BOD, lbs/day
C) BOD, lbs/day / MLVSS, lbs
D) MLVSS, lbs / Volume, MG
E) BOD, mg/L / MLVSS, mg/L
Correct Answer: C) BOD, lbs/day / MLVSS, lbs
Rationale: The F/M ratio is a key process control parameter in
activated sludge, indicating the amount of food available to the
microorganisms.
Question 2
In a conventional activated sludge plant, where is the oxygen demand
typically highest?
A) At the end of the aeration tank
B) In the secondary clarifier
C) At the primary clarifier effluent
D) At the head end of the tank
E) In the return activated sludge line
Correct Answer: D) At the head end of the tank
Rationale: At the head end, the microorganisms are first exposed to
,the raw wastewater (food), leading to a high metabolic rate and
thus high oxygen consumption.
Question 3
Air requirements in an activated sludge tank are governed by:
A) Only BOD loading
B) Only volatile suspended solids concentration
C) Only primary effluent suspended solids
D) BOD and the desired removal efficiency, volatile suspended solids
concentration in the aerator, and suspended solids concentrations
of the primary effluent.
E) Temperature and pH only
Correct Answer: D) BOD and the desired removal efficiency, volatile
suspended solids concentration in the aerator, and suspended solids
concentrations of the primary effluent.
Rationale: All these factors contribute to the metabolic activity of the
microorganisms and thus their oxygen needs.
Question 4
To operate an activated sludge plant effectively, an operator does not need
to be a skilled microbiologist. Rather, the operator needs only to recognize
major groups of organisms such as:
A) Specific bacterial species, viruses, fungi
B) Coliform bacteria, pathogenic bacteria
C) Filamentous bacteria, rotifers, protozoa
, D) Algae, nematodes, worms
E) Yeast, molds, giardia
Correct Answer: C) Filamentous bacteria, rotifers, protozoa
Rationale: These are key indicator organisms whose presence and
abundance provide valuable clues about the health and
performance of the activated sludge process.
Question 5
In addition to high energy costs, excess Dissolved Oxygen (DO) levels in an
aeration tank can cause:
A) Increased BOD removal
B) Improved settling
C) Reduced filamentous growth
D) Damage to floc particles due to turbulence
E) Increased nitrification
Correct Answer: D) Damage to floc particles due to turbulence
Rationale: Excessive aeration can lead to high turbulence and shear,
which can break apart the delicate microbial floc, resulting in poor
settling in the clarifiers.
Question 6
Which of the following are key factors in activated sludge nitrification? Check
all that apply.
A) MCRT (Mean Cell Residence Time)
B) Alkalinity and pH
QUESTIONS WITH COMPLETE SOLUTION GUIDE (A+
GRADED 100% VERIFIED) LATEST VERSION 2025!!
Question 1
What is the formula for calculating the F/M ratio (Food to Microorganism
ratio)?
A) BOD, lbs/day x MLVSS, lbs
B) MLVSS, lbs / BOD, lbs/day
C) BOD, lbs/day / MLVSS, lbs
D) MLVSS, lbs / Volume, MG
E) BOD, mg/L / MLVSS, mg/L
Correct Answer: C) BOD, lbs/day / MLVSS, lbs
Rationale: The F/M ratio is a key process control parameter in
activated sludge, indicating the amount of food available to the
microorganisms.
Question 2
In a conventional activated sludge plant, where is the oxygen demand
typically highest?
A) At the end of the aeration tank
B) In the secondary clarifier
C) At the primary clarifier effluent
D) At the head end of the tank
E) In the return activated sludge line
Correct Answer: D) At the head end of the tank
Rationale: At the head end, the microorganisms are first exposed to
,the raw wastewater (food), leading to a high metabolic rate and
thus high oxygen consumption.
Question 3
Air requirements in an activated sludge tank are governed by:
A) Only BOD loading
B) Only volatile suspended solids concentration
C) Only primary effluent suspended solids
D) BOD and the desired removal efficiency, volatile suspended solids
concentration in the aerator, and suspended solids concentrations
of the primary effluent.
E) Temperature and pH only
Correct Answer: D) BOD and the desired removal efficiency, volatile
suspended solids concentration in the aerator, and suspended solids
concentrations of the primary effluent.
Rationale: All these factors contribute to the metabolic activity of the
microorganisms and thus their oxygen needs.
Question 4
To operate an activated sludge plant effectively, an operator does not need
to be a skilled microbiologist. Rather, the operator needs only to recognize
major groups of organisms such as:
A) Specific bacterial species, viruses, fungi
B) Coliform bacteria, pathogenic bacteria
C) Filamentous bacteria, rotifers, protozoa
, D) Algae, nematodes, worms
E) Yeast, molds, giardia
Correct Answer: C) Filamentous bacteria, rotifers, protozoa
Rationale: These are key indicator organisms whose presence and
abundance provide valuable clues about the health and
performance of the activated sludge process.
Question 5
In addition to high energy costs, excess Dissolved Oxygen (DO) levels in an
aeration tank can cause:
A) Increased BOD removal
B) Improved settling
C) Reduced filamentous growth
D) Damage to floc particles due to turbulence
E) Increased nitrification
Correct Answer: D) Damage to floc particles due to turbulence
Rationale: Excessive aeration can lead to high turbulence and shear,
which can break apart the delicate microbial floc, resulting in poor
settling in the clarifiers.
Question 6
Which of the following are key factors in activated sludge nitrification? Check
all that apply.
A) MCRT (Mean Cell Residence Time)
B) Alkalinity and pH
