Wastewater Treatment Plant Operator
Complete Exam Preparation Materials
2026/2027 – Complete Exam-Style Questions
with Correct Answers & Detailed Rationales |
Verified & Reliable
1. A wastewater treatment plant receives an average flow of
2.4 MGD with an influent BOD₅ concentration of 210 mg/L.
Approximately how many pounds of BOD₅ per day enter
the plant?
A. 2,103 lb/day
B. 4,205 lb/day
C. 5,040 lb/day
D. 8,410 lb/day
Rationale: The standard loading equation is lb/day = MGD ×
mg/L × 8.34. Therefore, 2.4 × 210 × 8.34 = approximately 4,205
lb/day. This calculation is useful for evaluating organic loading
and activated-sludge process performance.
2. An aeration basin contains 1.2 million gallons of mixed
liquor at an MLSS concentration of 3,000 mg/L.
Approximately how many pounds of suspended solids are
present in the basin?
,A. 2,504 lb
B. 10,008 lb
C. 30,024 lb
D. 36,000 lb
Rationale: Solids inventory is calculated as volume in MG ×
concentration in mg/L × 8.34. Thus, 1.2 × 3,000 × 8.34 = 30,024
lb of MLSS. This represents the approximate mass of suspended
solids contained in the aeration basin.
3. A secondary treatment process has an influent BOD₅ of
240 mg/L and an effluent BOD₅ of 30 mg/L. What is the
approximate BOD₅ removal efficiency?
A. 75.0%
B. 80.0%
C. 85.0%
D. 87.5%
Rationale: Percent removal = [(influent − effluent) ÷ influent] ×
100. Therefore, [(240 − 30) ÷ 240] × 100 = 87.5%. The result
exceeds the commonly applicable 85% minimum removal
associated with federal secondary-treatment requirements,
although the actual permit must always be consulted.
4. A plant has an aeration volume of 1.8 million gallons and
an average flow of 3.6 MGD. What is the theoretical
hydraulic retention time?
,A. 6 hours
B. 12 hours
C. 18 hours
D. 24 hours
Rationale: HRT in hours = basin volume in MG ÷ flow in MGD ×
24. Thus, 1.8 ÷ 3.6 × 24 = 12 hours. Actual hydraulic
performance can differ because of short-circuiting, mixing
characteristics, and flow variations.
5. A secondary clarifier is receiving 1.5 MGD of RAS while
plant influent flow is 5.0 MGD. What is the RAS rate
expressed as a percentage of influent flow?
A. 15%
B. 20%
C. 30%
D. 33%
Rationale: RAS percentage = RAS flow ÷ influent flow × 100.
Therefore, 1.5 ÷ 5.0 × 100 = 30%. RAS flow should be adjusted
based on solids inventory, settling characteristics, blanket
depth, and overall process conditions rather than percentage
alone.
6. Which condition most directly favors nitrification in an
activated-sludge process?
, A. Very short sludge age and high F/M ratio
B. High dissolved sulfide concentration
C. Anaerobic conditions throughout the aeration basin
D. Adequate dissolved oxygen and sufficient sludge age
Rationale: Nitrifying organisms grow more slowly than ordinary
carbonaceous heterotrophs and therefore generally require
adequate solids retention time. They also require oxygen for
ammonia oxidation. Insufficient DO or excessive wasting can
reduce the nitrifier population and cause ammonia
breakthrough.
7. A plant's effluent ammonia concentration has increased
while effluent nitrate has decreased. Aeration-basin DO is
consistently below 0.5 mg/L. Which operational concern is
most likely?
A. Excessive denitrification
B. Inhibited nitrification
C. Excessive phosphorus precipitation
D. Overchlorination
Rationale: Nitrification is an aerobic process and is sensitive to
low DO. When oxygen becomes limiting, ammonia-oxidizing
and nitrite-oxidizing organisms cannot maintain adequate
activity, resulting in higher ammonia and less nitrate formation.
EPA nutrient-control guidance identifies low DO as a significant
limitation on nitrification.
Complete Exam Preparation Materials
2026/2027 – Complete Exam-Style Questions
with Correct Answers & Detailed Rationales |
Verified & Reliable
1. A wastewater treatment plant receives an average flow of
2.4 MGD with an influent BOD₅ concentration of 210 mg/L.
Approximately how many pounds of BOD₅ per day enter
the plant?
A. 2,103 lb/day
B. 4,205 lb/day
C. 5,040 lb/day
D. 8,410 lb/day
Rationale: The standard loading equation is lb/day = MGD ×
mg/L × 8.34. Therefore, 2.4 × 210 × 8.34 = approximately 4,205
lb/day. This calculation is useful for evaluating organic loading
and activated-sludge process performance.
2. An aeration basin contains 1.2 million gallons of mixed
liquor at an MLSS concentration of 3,000 mg/L.
Approximately how many pounds of suspended solids are
present in the basin?
,A. 2,504 lb
B. 10,008 lb
C. 30,024 lb
D. 36,000 lb
Rationale: Solids inventory is calculated as volume in MG ×
concentration in mg/L × 8.34. Thus, 1.2 × 3,000 × 8.34 = 30,024
lb of MLSS. This represents the approximate mass of suspended
solids contained in the aeration basin.
3. A secondary treatment process has an influent BOD₅ of
240 mg/L and an effluent BOD₅ of 30 mg/L. What is the
approximate BOD₅ removal efficiency?
A. 75.0%
B. 80.0%
C. 85.0%
D. 87.5%
Rationale: Percent removal = [(influent − effluent) ÷ influent] ×
100. Therefore, [(240 − 30) ÷ 240] × 100 = 87.5%. The result
exceeds the commonly applicable 85% minimum removal
associated with federal secondary-treatment requirements,
although the actual permit must always be consulted.
4. A plant has an aeration volume of 1.8 million gallons and
an average flow of 3.6 MGD. What is the theoretical
hydraulic retention time?
,A. 6 hours
B. 12 hours
C. 18 hours
D. 24 hours
Rationale: HRT in hours = basin volume in MG ÷ flow in MGD ×
24. Thus, 1.8 ÷ 3.6 × 24 = 12 hours. Actual hydraulic
performance can differ because of short-circuiting, mixing
characteristics, and flow variations.
5. A secondary clarifier is receiving 1.5 MGD of RAS while
plant influent flow is 5.0 MGD. What is the RAS rate
expressed as a percentage of influent flow?
A. 15%
B. 20%
C. 30%
D. 33%
Rationale: RAS percentage = RAS flow ÷ influent flow × 100.
Therefore, 1.5 ÷ 5.0 × 100 = 30%. RAS flow should be adjusted
based on solids inventory, settling characteristics, blanket
depth, and overall process conditions rather than percentage
alone.
6. Which condition most directly favors nitrification in an
activated-sludge process?
, A. Very short sludge age and high F/M ratio
B. High dissolved sulfide concentration
C. Anaerobic conditions throughout the aeration basin
D. Adequate dissolved oxygen and sufficient sludge age
Rationale: Nitrifying organisms grow more slowly than ordinary
carbonaceous heterotrophs and therefore generally require
adequate solids retention time. They also require oxygen for
ammonia oxidation. Insufficient DO or excessive wasting can
reduce the nitrifier population and cause ammonia
breakthrough.
7. A plant's effluent ammonia concentration has increased
while effluent nitrate has decreased. Aeration-basin DO is
consistently below 0.5 mg/L. Which operational concern is
most likely?
A. Excessive denitrification
B. Inhibited nitrification
C. Excessive phosphorus precipitation
D. Overchlorination
Rationale: Nitrification is an aerobic process and is sensitive to
low DO. When oxygen becomes limiting, ammonia-oxidizing
and nitrite-oxidizing organisms cannot maintain adequate
activity, resulting in higher ammonia and less nitrate formation.
EPA nutrient-control guidance identifies low DO as a significant
limitation on nitrification.
