Wastewater Treatment Plant Operator
Licensing Exam 2026/2027 – Complete Exam-
Style Questions with Correct Answers &
Detailed Rationales | Verified & Reliable
1. A wastewater plant receives an average flow of 2.40 MGD.
The influent BOD₅ concentration is 210 mg/L. Approximately
how many pounds of BOD₅ enter the plant each day?
A. 2,100 lb/day
B. 3,500 lb/day
C. 4,204 lb/day
D. 5,040 lb/day
The standard loading equation is lb/day = MGD × mg/L × 8.34.
Therefore, 2.40 × 210 × 8.34 = approximately 4,204 lb/day. The
factor 8.34 accounts for the weight of one million gallons of
water at approximately 1,000,000 gallons × 8.34 lb/gal.
2. An activated-sludge process has an MLSS of 3,200 mg/L and
an aeration-basin volume of 1.5 million gallons.
Approximately how many pounds of suspended solids are
contained in the aeration basin?
,A. 20,000 lb
B. 40,040 lb
C. 48,000 lb
D. 60,060 lb
The solids inventory is calculated as MGD × mg/L × 8.34. Thus,
1.5 × 3,200 × 8.34 = 40,032 pounds, or approximately 40,040 lb.
This calculation estimates the total suspended-solids inventory
in the basin.
3. A secondary clarifier begins producing effluent with
increasing turbidity while the sludge blanket remains
unusually high. Which condition should the operator
investigate first?
A. Excessive chlorine residual
B. Low influent alkalinity
C. Solids overloading or inadequate return activated sludge
withdrawal
D. Excessive dissolved oxygen in the final effluent
An elevated sludge blanket combined with deteriorating
effluent clarity strongly suggests that solids are accumulating in
the clarifier faster than they are being removed. Excessive solids
loading, insufficient return activated sludge (RAS) flow, poor
,settling, or a combination of these conditions can cause the
blanket to rise and eventually result in solids washout.
4. A plant has an average flow of 4.0 MGD and an aeration-
basin volume of 1.0 million gallons. What is the approximate
hydraulic retention time (HRT)?
A. 2 hours
B. 6 hours
C. 12 hours
D. 24 hours
HRT in days is basin volume divided by flow: 1.0 MG ÷ 4.0 MGD
= 0.25 day. Multiplying by 24 hours/day gives 6 hours. HRT
describes the theoretical average time wastewater remains in
the basin.
5. Which condition most directly promotes nitrification in a
conventional activated-sludge process?
A. Very short sludge age
B. Absence of dissolved oxygen
C. Adequate sludge age, dissolved oxygen, and suitable
temperature and pH
D. Continuous operation at maximum organic loading
, Nitrifying organisms grow more slowly than typical carbon-
oxidizing organisms and therefore generally require sufficient
solids retention time. They also require oxygen and favorable
pH and temperature conditions. A very short sludge age can
wash nitrifiers out of the system.
6. A plant's aeration dissolved oxygen falls from 2.0 mg/L to
0.3 mg/L while airflow remains unchanged. Influent loading
has increased substantially. What is the most likely
explanation?
A. The clarifier has stopped removing phosphorus
B. Oxygen demand has increased relative to oxygen-transfer
capacity
C. RAS concentration has necessarily decreased to zero
D. Chlorine demand has decreased
An increase in biodegradable organic loading increases the
oxygen demand exerted by microorganisms. If oxygen transfer
does not increase sufficiently, dissolved oxygen can decline even
though the blower airflow remains constant. The operator
should evaluate loading, blower performance, diffuser
condition, and process control.
Licensing Exam 2026/2027 – Complete Exam-
Style Questions with Correct Answers &
Detailed Rationales | Verified & Reliable
1. A wastewater plant receives an average flow of 2.40 MGD.
The influent BOD₅ concentration is 210 mg/L. Approximately
how many pounds of BOD₅ enter the plant each day?
A. 2,100 lb/day
B. 3,500 lb/day
C. 4,204 lb/day
D. 5,040 lb/day
The standard loading equation is lb/day = MGD × mg/L × 8.34.
Therefore, 2.40 × 210 × 8.34 = approximately 4,204 lb/day. The
factor 8.34 accounts for the weight of one million gallons of
water at approximately 1,000,000 gallons × 8.34 lb/gal.
2. An activated-sludge process has an MLSS of 3,200 mg/L and
an aeration-basin volume of 1.5 million gallons.
Approximately how many pounds of suspended solids are
contained in the aeration basin?
,A. 20,000 lb
B. 40,040 lb
C. 48,000 lb
D. 60,060 lb
The solids inventory is calculated as MGD × mg/L × 8.34. Thus,
1.5 × 3,200 × 8.34 = 40,032 pounds, or approximately 40,040 lb.
This calculation estimates the total suspended-solids inventory
in the basin.
3. A secondary clarifier begins producing effluent with
increasing turbidity while the sludge blanket remains
unusually high. Which condition should the operator
investigate first?
A. Excessive chlorine residual
B. Low influent alkalinity
C. Solids overloading or inadequate return activated sludge
withdrawal
D. Excessive dissolved oxygen in the final effluent
An elevated sludge blanket combined with deteriorating
effluent clarity strongly suggests that solids are accumulating in
the clarifier faster than they are being removed. Excessive solids
loading, insufficient return activated sludge (RAS) flow, poor
,settling, or a combination of these conditions can cause the
blanket to rise and eventually result in solids washout.
4. A plant has an average flow of 4.0 MGD and an aeration-
basin volume of 1.0 million gallons. What is the approximate
hydraulic retention time (HRT)?
A. 2 hours
B. 6 hours
C. 12 hours
D. 24 hours
HRT in days is basin volume divided by flow: 1.0 MG ÷ 4.0 MGD
= 0.25 day. Multiplying by 24 hours/day gives 6 hours. HRT
describes the theoretical average time wastewater remains in
the basin.
5. Which condition most directly promotes nitrification in a
conventional activated-sludge process?
A. Very short sludge age
B. Absence of dissolved oxygen
C. Adequate sludge age, dissolved oxygen, and suitable
temperature and pH
D. Continuous operation at maximum organic loading
, Nitrifying organisms grow more slowly than typical carbon-
oxidizing organisms and therefore generally require sufficient
solids retention time. They also require oxygen and favorable
pH and temperature conditions. A very short sludge age can
wash nitrifiers out of the system.
6. A plant's aeration dissolved oxygen falls from 2.0 mg/L to
0.3 mg/L while airflow remains unchanged. Influent loading
has increased substantially. What is the most likely
explanation?
A. The clarifier has stopped removing phosphorus
B. Oxygen demand has increased relative to oxygen-transfer
capacity
C. RAS concentration has necessarily decreased to zero
D. Chlorine demand has decreased
An increase in biodegradable organic loading increases the
oxygen demand exerted by microorganisms. If oxygen transfer
does not increase sufficiently, dissolved oxygen can decline even
though the blower airflow remains constant. The operator
should evaluate loading, blower performance, diffuser
condition, and process control.
