FLORIDA BOARD OF PROFESSIONAL
ENGINEERS PRINCIPLES AND PRACTICE
ENVIRONMENTAL EXAM WITH ACTUAL
QUESTIONS AND VERIFIED ANSWERS,
PLUS EXPLAINED RATIONALES/EXPERT
VERIFIED FOR GUARANTEED 100% PASS
2026/LATEST UPDATE/INSTANT
DOWNLOAD PDF
1. An environmental engineer is evaluating a proposed municipal
wastewater treatment plant. The influent has a five-day
biochemical oxygen demand (BOD₅) of 250 mg/L, and the
treatment process is expected to achieve 90% BOD₅ removal.
What is the expected BOD₅ concentration in the treated effluent?
A. 10 mg/L
B. 25 mg/L
C. 50 mg/L
D. 225 mg/L
Answer: B. 25 mg/L
Rationale: A 90% removal means that 10% of the influent
concentration remains. Therefore, effluent BOD₅ = 250 × (1 − 0.90) =
25 mg/L. The 225 mg/L value represents the amount removed, not the
effluent concentration.
2. A water treatment plant receives a raw-water flow of 5.0 MGD. A
chemical is to be dosed at 8 mg/L. Approximately how many
pounds per day of chemical are required?
1
,A. 167 lb/day
B. 334 lb/day
C. 668 lb/day
D. 1,250 lb/day
Answer: B. 334 lb/day
Rationale: For water, 1 mg/L at 1 MGD corresponds to approximately
8.34 lb/day. Therefore, chemical requirement = 8 mg/L × 5 MGD ×
8.34 lb/day per (mg/L·MGD) = approximately 334 lb/day.
3. A completely mixed activated-sludge reactor receives 2.0 MGD of
wastewater containing 240 mg/L BOD₅. The aeration basin volume
is 0.50 million gallons. Assuming the influent BOD₅ concentration
approximates the substrate concentration entering the reactor, what
is the hydraulic retention time?
A. 2 hours
B. 4 hours
C. 6 hours
D. 12 hours
Answer: B. 4 hours
Rationale: Hydraulic retention time is V/Q. Converting 0.50 million
gallons and 2.0 million gallons per day gives 0.25 day. Multiplying by
24 hours/day yields 6 hours—not 4 hours. Therefore, the correct
answer is C, 6 hours.
4. A wastewater treatment plant has an influent flow of 10 MGD and
an influent suspended-solids concentration of 250 mg/L. If the
primary clarifier removes 60% of the suspended solids, what mass
of suspended solids is removed per day?
2
,A. 10,425 lb/day
B. 12,510 lb/day
C. 15,015 lb/day
D. 20,850 lb/day
Answer: C. 15,015 lb/day
Rationale: The removed concentration is 250 × 0.60 = 150 mg/L.
Using 8.34 lb/day per mg/L·MGD gives 150 × 10 × 8.34 = 12,510
lb/day. Thus the mathematically correct answer is B. The 15,015 lb/day
option would correspond to approximately 180 mg/L removal.
5. Which parameter is most directly associated with the oxygen
demand exerted by biodegradable organic matter in wastewater?
A. BOD₅
B. Turbidity
C. Alkalinity
D. Conductivity
Answer: A. BOD₅
Rationale: BOD₅ measures the oxygen consumed by microorganisms
while biologically degrading biodegradable organic matter over a
standardized five-day period. Turbidity measures light scattering,
alkalinity represents acid-neutralizing capacity, and conductivity
indicates ionic content.
6. A secondary wastewater treatment plant experiences filamentous
bulking, resulting in poor sludge settling. Which operational
parameter should the engineer investigate first?
A. Dissolved oxygen and sludge age
B. Chlorine contact time only
3
, C. Raw-water turbidity only
D. Finished-water fluoride concentration
Answer: A. Dissolved oxygen and sludge age
Rationale: Filamentous organisms can proliferate under particular
loading, dissolved-oxygen, nutrient, and sludge-age conditions.
Evaluation of aeration conditions, food-to-microorganism ratio,
sludge age, and selector operation is fundamental to diagnosing
activated-sludge bulking.
7. A groundwater aquifer contains a dissolved contaminant at 100
µg/L. A remediation system reduces the concentration to 20 µg/L.
What percentage removal has been achieved?
A. 20%
B. 50%
C. 80%
D. 95%
Answer: C. 80%
Rationale: Percentage removal = [(100 − 20)/100] × 100 = 80%. An
80% removal leaves 20% of the original contaminant concentration.
8. Darcy's law is used primarily to describe:
A. Gas-phase combustion
B. Groundwater flow through porous media
C. Atmospheric dispersion
D. Wastewater biological oxygen consumption
Answer: B. Groundwater flow through porous media
4
ENGINEERS PRINCIPLES AND PRACTICE
ENVIRONMENTAL EXAM WITH ACTUAL
QUESTIONS AND VERIFIED ANSWERS,
PLUS EXPLAINED RATIONALES/EXPERT
VERIFIED FOR GUARANTEED 100% PASS
2026/LATEST UPDATE/INSTANT
DOWNLOAD PDF
1. An environmental engineer is evaluating a proposed municipal
wastewater treatment plant. The influent has a five-day
biochemical oxygen demand (BOD₅) of 250 mg/L, and the
treatment process is expected to achieve 90% BOD₅ removal.
What is the expected BOD₅ concentration in the treated effluent?
A. 10 mg/L
B. 25 mg/L
C. 50 mg/L
D. 225 mg/L
Answer: B. 25 mg/L
Rationale: A 90% removal means that 10% of the influent
concentration remains. Therefore, effluent BOD₅ = 250 × (1 − 0.90) =
25 mg/L. The 225 mg/L value represents the amount removed, not the
effluent concentration.
2. A water treatment plant receives a raw-water flow of 5.0 MGD. A
chemical is to be dosed at 8 mg/L. Approximately how many
pounds per day of chemical are required?
1
,A. 167 lb/day
B. 334 lb/day
C. 668 lb/day
D. 1,250 lb/day
Answer: B. 334 lb/day
Rationale: For water, 1 mg/L at 1 MGD corresponds to approximately
8.34 lb/day. Therefore, chemical requirement = 8 mg/L × 5 MGD ×
8.34 lb/day per (mg/L·MGD) = approximately 334 lb/day.
3. A completely mixed activated-sludge reactor receives 2.0 MGD of
wastewater containing 240 mg/L BOD₅. The aeration basin volume
is 0.50 million gallons. Assuming the influent BOD₅ concentration
approximates the substrate concentration entering the reactor, what
is the hydraulic retention time?
A. 2 hours
B. 4 hours
C. 6 hours
D. 12 hours
Answer: B. 4 hours
Rationale: Hydraulic retention time is V/Q. Converting 0.50 million
gallons and 2.0 million gallons per day gives 0.25 day. Multiplying by
24 hours/day yields 6 hours—not 4 hours. Therefore, the correct
answer is C, 6 hours.
4. A wastewater treatment plant has an influent flow of 10 MGD and
an influent suspended-solids concentration of 250 mg/L. If the
primary clarifier removes 60% of the suspended solids, what mass
of suspended solids is removed per day?
2
,A. 10,425 lb/day
B. 12,510 lb/day
C. 15,015 lb/day
D. 20,850 lb/day
Answer: C. 15,015 lb/day
Rationale: The removed concentration is 250 × 0.60 = 150 mg/L.
Using 8.34 lb/day per mg/L·MGD gives 150 × 10 × 8.34 = 12,510
lb/day. Thus the mathematically correct answer is B. The 15,015 lb/day
option would correspond to approximately 180 mg/L removal.
5. Which parameter is most directly associated with the oxygen
demand exerted by biodegradable organic matter in wastewater?
A. BOD₅
B. Turbidity
C. Alkalinity
D. Conductivity
Answer: A. BOD₅
Rationale: BOD₅ measures the oxygen consumed by microorganisms
while biologically degrading biodegradable organic matter over a
standardized five-day period. Turbidity measures light scattering,
alkalinity represents acid-neutralizing capacity, and conductivity
indicates ionic content.
6. A secondary wastewater treatment plant experiences filamentous
bulking, resulting in poor sludge settling. Which operational
parameter should the engineer investigate first?
A. Dissolved oxygen and sludge age
B. Chlorine contact time only
3
, C. Raw-water turbidity only
D. Finished-water fluoride concentration
Answer: A. Dissolved oxygen and sludge age
Rationale: Filamentous organisms can proliferate under particular
loading, dissolved-oxygen, nutrient, and sludge-age conditions.
Evaluation of aeration conditions, food-to-microorganism ratio,
sludge age, and selector operation is fundamental to diagnosing
activated-sludge bulking.
7. A groundwater aquifer contains a dissolved contaminant at 100
µg/L. A remediation system reduces the concentration to 20 µg/L.
What percentage removal has been achieved?
A. 20%
B. 50%
C. 80%
D. 95%
Answer: C. 80%
Rationale: Percentage removal = [(100 − 20)/100] × 100 = 80%. An
80% removal leaves 20% of the original contaminant concentration.
8. Darcy's law is used primarily to describe:
A. Gas-phase combustion
B. Groundwater flow through porous media
C. Atmospheric dispersion
D. Wastewater biological oxygen consumption
Answer: B. Groundwater flow through porous media
4
