TEXAS WASTEWATER CLASS A EXAM 2025 –
PRACTICE QUESTIONS WITH ANSWERS &
RATIONALES
SECTION 1: REGULATORY AGENCIES & BASIC
CONCEPTS (Questions 1–25)
Question 1: In an activated sludge system, the "food-to-microorganism"
(F/M) ratio is a key control parameter. Which of the following correctly
describes the effect of an F/M ratio that is too high?
A) Pin floc formation and high effluent TSS
B) Filamentous bulking and poor settling
C) Zoogleal bulking and high SVI
D) Reduced sludge production and low oxygen demand
Correct Answer: B) Filamentous bulking and poor settling
Rationale: A high F/M ratio (more food than microorganisms) promotes the
growth of filamentous organisms, leading to bulking sludge, poor settling,
and high SVI. Low F/M ratios can cause pin floc (A). Zoogleal bulking (C) is
associated with high carbohydrate loading. Reduced sludge production (D)
occurs with low F/M ratios.
,Question 2: The sludge volume index (SVI) is calculated using which
formula?
A) (Settled sludge volume in mL/L) ÷ MLSS in mg/L × 1,000
B) (MLSS in mg/L) ÷ Settled sludge volume in mL/L × 1,000
C) (Settled sludge volume in mL) ÷ MLSS in mg/L × 100
D) (MLSS in mg/L) ÷ Settled sludge volume in mL × 100
Correct Answer: A) (Settled sludge volume in mL/L) ÷ MLSS in mg/L ×
1,000
Rationale: SVI is calculated as the settled sludge volume (after 30 minutes in
a 1,000 mL graduated cylinder, expressed in mL/L) divided by the MLSS
concentration (in mg/L), multiplied by 1,000. An SVI below 100 indicates
good settling, while an SVI above 150 indicates bulking sludge.
Question 3: A "Racetrack" is a common term used to describe which type
of activated sludge treatment process?
A) Conventional plug flow
B) Complete mix
C) Oxidation ditch
D) Sequencing batch reactor
Correct Answer: C) Oxidation ditch
Rationale: An oxidation ditch is a type of extended aeration activated sludge
system that is commonly shaped like a "racetrack" (oval or circular with a
continuous channel). The mixed liquor circulates continuously around the
track, providing biological treatment with extended aeration.
,Question 4: The mixed liquor suspended solids (MLSS) concentration in an
aeration basin is typically maintained by controlling:
A) The return activated sludge (RAS) rate
B) The waste activated sludge (WAS) rate
C) Both RAS and WAS rates
D) The influent flow rate
Correct Answer: C) Both RAS and WAS rates
Rationale: MLSS concentration is controlled by both the return activated
sludge (RAS) rate (which returns solids to the aeration basin) and the waste
activated sludge (WAS) rate (which removes excess solids from the system).
Adjusting these rates allows operators to maintain the desired MLSS and
sludge age.
Question 5: The sludge age (mean cell residence time) in an extended
aeration activated sludge system is typically:
A) 1–3 days
B) 3–10 days
C) 10–25 days
D) 25–40 days
Correct Answer: C) 10–25 days
Rationale: Extended aeration systems operate with sludge ages of 10–25
days, which is longer than conventional activated sludge (3–10 days). This
longer sludge age results in more complete oxidation of organic matter, less
waste sludge production, and improved effluent quality.
, Question 6: The waste activated sludge (WAS) removal rate should be
increased when:
A) The MLSS concentration is too low
B) The sludge age is too high
C) The F/M ratio is too low
D) The effluent TSS is increasing
Correct Answer: B) The sludge age is too high
Rationale: Increasing the WAS rate reduces the sludge age (mean cell
residence time). When the sludge age is too high, the sludge becomes old and
can form pin floc, leading to high effluent TSS. Increasing WAS removes older
sludge and rejuvenates the microbial population.
Question 7: In the activated sludge process, the term "utilization" refers to:
A) The process where food supply is changed into a liquid state for use
B) The settling of solids in the clarifier
C) The return of sludge to the aeration basin
D) The wasting of excess sludge
Correct Answer: A) The process where food supply is changed into a
liquid state for use
Rationale: Utilization is the process by which microorganisms convert organic
matter (food) into cellular material, energy, and byproducts. This is the
fundamental biological process in activated sludge treatment where BOD is
removed as microorganisms metabolize the organic matter.
Question 8: A clarifier will concentrate return sludge and waste sludge
about how many times the solids concentration of the mixed liquor?
PRACTICE QUESTIONS WITH ANSWERS &
RATIONALES
SECTION 1: REGULATORY AGENCIES & BASIC
CONCEPTS (Questions 1–25)
Question 1: In an activated sludge system, the "food-to-microorganism"
(F/M) ratio is a key control parameter. Which of the following correctly
describes the effect of an F/M ratio that is too high?
A) Pin floc formation and high effluent TSS
B) Filamentous bulking and poor settling
C) Zoogleal bulking and high SVI
D) Reduced sludge production and low oxygen demand
Correct Answer: B) Filamentous bulking and poor settling
Rationale: A high F/M ratio (more food than microorganisms) promotes the
growth of filamentous organisms, leading to bulking sludge, poor settling,
and high SVI. Low F/M ratios can cause pin floc (A). Zoogleal bulking (C) is
associated with high carbohydrate loading. Reduced sludge production (D)
occurs with low F/M ratios.
,Question 2: The sludge volume index (SVI) is calculated using which
formula?
A) (Settled sludge volume in mL/L) ÷ MLSS in mg/L × 1,000
B) (MLSS in mg/L) ÷ Settled sludge volume in mL/L × 1,000
C) (Settled sludge volume in mL) ÷ MLSS in mg/L × 100
D) (MLSS in mg/L) ÷ Settled sludge volume in mL × 100
Correct Answer: A) (Settled sludge volume in mL/L) ÷ MLSS in mg/L ×
1,000
Rationale: SVI is calculated as the settled sludge volume (after 30 minutes in
a 1,000 mL graduated cylinder, expressed in mL/L) divided by the MLSS
concentration (in mg/L), multiplied by 1,000. An SVI below 100 indicates
good settling, while an SVI above 150 indicates bulking sludge.
Question 3: A "Racetrack" is a common term used to describe which type
of activated sludge treatment process?
A) Conventional plug flow
B) Complete mix
C) Oxidation ditch
D) Sequencing batch reactor
Correct Answer: C) Oxidation ditch
Rationale: An oxidation ditch is a type of extended aeration activated sludge
system that is commonly shaped like a "racetrack" (oval or circular with a
continuous channel). The mixed liquor circulates continuously around the
track, providing biological treatment with extended aeration.
,Question 4: The mixed liquor suspended solids (MLSS) concentration in an
aeration basin is typically maintained by controlling:
A) The return activated sludge (RAS) rate
B) The waste activated sludge (WAS) rate
C) Both RAS and WAS rates
D) The influent flow rate
Correct Answer: C) Both RAS and WAS rates
Rationale: MLSS concentration is controlled by both the return activated
sludge (RAS) rate (which returns solids to the aeration basin) and the waste
activated sludge (WAS) rate (which removes excess solids from the system).
Adjusting these rates allows operators to maintain the desired MLSS and
sludge age.
Question 5: The sludge age (mean cell residence time) in an extended
aeration activated sludge system is typically:
A) 1–3 days
B) 3–10 days
C) 10–25 days
D) 25–40 days
Correct Answer: C) 10–25 days
Rationale: Extended aeration systems operate with sludge ages of 10–25
days, which is longer than conventional activated sludge (3–10 days). This
longer sludge age results in more complete oxidation of organic matter, less
waste sludge production, and improved effluent quality.
, Question 6: The waste activated sludge (WAS) removal rate should be
increased when:
A) The MLSS concentration is too low
B) The sludge age is too high
C) The F/M ratio is too low
D) The effluent TSS is increasing
Correct Answer: B) The sludge age is too high
Rationale: Increasing the WAS rate reduces the sludge age (mean cell
residence time). When the sludge age is too high, the sludge becomes old and
can form pin floc, leading to high effluent TSS. Increasing WAS removes older
sludge and rejuvenates the microbial population.
Question 7: In the activated sludge process, the term "utilization" refers to:
A) The process where food supply is changed into a liquid state for use
B) The settling of solids in the clarifier
C) The return of sludge to the aeration basin
D) The wasting of excess sludge
Correct Answer: A) The process where food supply is changed into a
liquid state for use
Rationale: Utilization is the process by which microorganisms convert organic
matter (food) into cellular material, energy, and byproducts. This is the
fundamental biological process in activated sludge treatment where BOD is
removed as microorganisms metabolize the organic matter.
Question 8: A clarifier will concentrate return sludge and waste sludge
about how many times the solids concentration of the mixed liquor?
