10 FE practice problems: activated sludge, with solutions
These ten original problems practice activated sludge, a topic from the FE Environmental exam specification, using the Environmental Engineering, Activated Sludge part of the FE Reference Handbook 10.6. Each problem gives the situation and the values with units. Work it with the handbook PDF open and commit to an answer before you open the solution. Every solution shows the handbook page, the equation, the substitution with units, a size check, and the mistake behind each wrong option, so a wrong pick tells you exactly what to fix.
How to use these problems
Give each problem an honest attempt before you open the solution: write the given values with units, find the equation in the FE Reference Handbook PDF, and commit to an answer. Then compare line by line. If you picked a wrong option, read the note for that option; each one names the mistake that produces it.
The problems
Problem 1 · FE Environmental, Activated sludge
An activated sludge aeration basin has a volume of 13,100 m³. It receives 33,500 m³/d of primary effluent with a BOD5 of 200 mg/L, and the mixed liquor suspended solids (MLSS) concentration is 1,850 mg/L. The food-to-microorganism (F:M) ratio, based on MLSS, is most nearly:
Handbook: Environmental Engineering, Activated Sludge, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (0.276 kg BOD5/(kg MLSS·d))
- Given
- V = 13,100 m³, Q = 33,500 m³/d, S0 = 200 mg/L, XA = 1,850 mg/L
- Find
- F:M ratio (kg BOD5/kg MLSS·d)
- Handbook
- Environmental Engineering, Activated Sludge, Organic loading rate (F:M), page 340
- Equation
F:M = Q0·S0/(Vol·XA)- Substitute
F:M = Q0·S0/(V·XA) = (33,500 m³/d)(200 mg/L)/[(13,100 m³)(1,850 mg/L)] = 0.276 per day(mg/L cancels, so the units are kg BOD5 per kg MLSS per day)
- Result
- 0.276 kg BOD5/(kg MLSS·d), 3 significant figures
- Check
- the hydraulic residence time is V/Q = 9.39 h; F:M has units of 1/day.
- Why the others are wrong
- A: converted MLSS to an assumed MLVSS (0.8) that the problem did not give
- B: inverted the ratio (microorganisms over food)
- D: treated the daily flow as hourly
Problem 2 · FE Environmental, Activated sludge
An aeration basin of 6,900 m³ holds MLSS of 3,050 mg/L. Waste sludge is removed at 206 m³/d with 11,100 mg/L of solids, and the plant discharges 46,500 m³/d of effluent containing 28 mg/L of suspended solids. The solids residence time (sludge age) is most nearly:
Handbook: Environmental Engineering, Activated Sludge, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (5.86 d)
- Given
- V = 6,900 m³, XA = 3,050 mg/L, Qw = 206 m³/d, Xw = 11,100 mg/L, Qe = 46,500 m³/d, Xe = 28 mg/L
- Find
- solids residence time θc (d)
- Handbook
- Environmental Engineering, Activated Sludge (solids residence time), page 339
- Equation
θc = V·XA/(Qw·Xw + Qe·Xe)- Substitute
Solids in the basin: V·XA = (6,900)(3,050) = 21,050,000 m³·mg/LSolids leaving per day: Qw·Xw + Qe·Xe = (206)(11,100) + (46,500)(28) = 3,589,000 m³·mg/(L·d)θc = 21,050,000/3,589,000 = 5.86 d
- Result
- 5.86 d, 3 significant figures
- Check
- θc is the mass of solids in the basin divided by the mass of solids leaving each day; the units reduce to days.
- Why the others are wrong
- A: left out the solids lost in the effluent (Qe·Xe)
- B: used the waste sludge concentration for the aeration tank
- D: is a hydraulic residence time, not the solids residence time
Problem 3 · FE Environmental, Activated sludge
An aeration basin of 5,400 m³ holds MLSS of 3,200 mg/L. Waste sludge is removed at 124 m³/d with 11,100 mg/L of solids, and the plant discharges 30,000 m³/d of effluent containing 30 mg/L of suspended solids. The solids residence time (sludge age) is most nearly:
Handbook: Environmental Engineering, Activated Sludge, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (7.59 d)
- Given
- V = 5,400 m³, XA = 3,200 mg/L, Qw = 124 m³/d, Xw = 11,100 mg/L, Qe = 30,000 m³/d, Xe = 30 mg/L
- Find
- solids residence time θc (d)
- Handbook
- Environmental Engineering, Activated Sludge (solids residence time), page 339
- Equation
θc = V·XA/(Qw·Xw + Qe·Xe)- Substitute
Solids in the basin: V·XA = (5,400)(3,200) = 17,280,000 m³·mg/LSolids leaving per day: Qw·Xw + Qe·Xe = (124)(11,100) + (30,000)(30) = 2,276,000 m³·mg/(L·d)θc = 17,280,000/2,276,000 = 7.59 d
- Result
- 7.59 d, 3 significant figures
- Check
- θc is the mass of solids in the basin divided by the mass of solids leaving each day; the units reduce to days.
- Why the others are wrong
- B: is a hydraulic residence time, not the solids residence time
- C: left out the solids lost in the effluent (Qe·Xe)
- D: used the aeration-tank MLSS instead of the waste sludge concentration
Problem 4 · FE Environmental, Activated sludge
An aeration basin of 8,200 m³ holds MLSS of 2,100 mg/L. Waste sludge is removed at 265 m³/d with 6,600 mg/L of solids, and the plant discharges 29,000 m³/d of effluent containing 27 mg/L of suspended solids. The solids residence time (sludge age) is most nearly:
Handbook: Environmental Engineering, Activated Sludge, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (6.80 d)
- Given
- V = 8,200 m³, XA = 2,100 mg/L, Qw = 265 m³/d, Xw = 6,600 mg/L, Qe = 29,000 m³/d, Xe = 27 mg/L
- Find
- solids residence time θc (d)
- Handbook
- Environmental Engineering, Activated Sludge (solids residence time), page 339
- Equation
θc = V·XA/(Qw·Xw + Qe·Xe)- Substitute
Solids in the basin: V·XA = (8,200)(2,100) = 17,220,000 m³·mg/LSolids leaving per day: Qw·Xw + Qe·Xe = (265)(6,600) + (29,000)(27) = 2,532,000 m³·mg/(L·d)θc = 17,220,000/2,532,000 = 6.80 d
- Result
- 6.80 d, 3 significant figures
- Check
- θc is the mass of solids in the basin divided by the mass of solids leaving each day; the units reduce to days.
- Why the others are wrong
- A: left out the solids lost in the effluent (Qe·Xe)
- C: used the waste sludge concentration for the aeration tank
- D: is a hydraulic residence time, not the solids residence time
Problem 5 · FE Environmental, Activated sludge
An activated sludge aeration basin has a volume of 2,500 m³. It receives 11,000 m³/d of primary effluent with a BOD5 of 230 mg/L, and the mixed liquor suspended solids (MLSS) concentration is 2,450 mg/L. The food-to-microorganism (F:M) ratio, based on MLSS, is most nearly:
Handbook: Environmental Engineering, Activated Sludge, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (0.413 kg BOD5/(kg MLSS·d))
- Given
- V = 2,500 m³, Q = 11,000 m³/d, S0 = 230 mg/L, XA = 2,450 mg/L
- Find
- F:M ratio (kg BOD5/kg MLSS·d)
- Handbook
- Environmental Engineering, Activated Sludge, Organic loading rate (F:M), page 340
- Equation
F:M = Q0·S0/(Vol·XA)- Substitute
F:M = Q0·S0/(V·XA) = (11,000 m³/d)(230 mg/L)/[(2,500 m³)(2,450 mg/L)] = 0.413 per day(mg/L cancels, so the units are kg BOD5 per kg MLSS per day)
- Result
- 0.413 kg BOD5/(kg MLSS·d), 3 significant figures
- Check
- the hydraulic residence time is V/Q = 5.45 h; F:M has units of 1/day.
- Why the others are wrong
- A: treated the daily flow as hourly
- C: inverted the ratio (microorganisms over food)
- D: converted MLSS to an assumed MLVSS (0.8) that the problem did not give
Problem 6 · FE Environmental, Activated sludge
An activated sludge aeration basin has a volume of 4,100 m³. It receives 16,000 m³/d of primary effluent with a BOD5 of 175 mg/L, and the mixed liquor suspended solids (MLSS) concentration is 1,900 mg/L. The food-to-microorganism (F:M) ratio, based on MLSS, is most nearly:
Handbook: Environmental Engineering, Activated Sludge, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (0.359 kg BOD5/(kg MLSS·d))
- Given
- V = 4,100 m³, Q = 16,000 m³/d, S0 = 175 mg/L, XA = 1,900 mg/L
- Find
- F:M ratio (kg BOD5/kg MLSS·d)
- Handbook
- Environmental Engineering, Activated Sludge, Organic loading rate (F:M), page 340
- Equation
F:M = Q0·S0/(Vol·XA)- Substitute
F:M = Q0·S0/(V·XA) = (16,000 m³/d)(175 mg/L)/[(4,100 m³)(1,900 mg/L)] = 0.359 per day(mg/L cancels, so the units are kg BOD5 per kg MLSS per day)
- Result
- 0.359 kg BOD5/(kg MLSS·d), 3 significant figures
- Check
- the hydraulic residence time is V/Q = 6.15 h; F:M has units of 1/day.
- Why the others are wrong
- A: converted MLSS to an assumed MLVSS (0.8) that the problem did not give
- B: inverted the ratio (microorganisms over food)
- C: treated the daily flow as hourly
Problem 7 · FE Environmental, Activated sludge
An activated sludge aeration basin has a volume of 4,100 m³. It receives 23,600 m³/d of primary effluent with a BOD5 of 185 mg/L, and the mixed liquor suspended solids (MLSS) concentration is 2,400 mg/L. The food-to-microorganism (F:M) ratio, based on MLSS, is most nearly:
Handbook: Environmental Engineering, Activated Sludge, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (0.444 kg BOD5/(kg MLSS·d))
- Given
- V = 4,100 m³, Q = 23,600 m³/d, S0 = 185 mg/L, XA = 2,400 mg/L
- Find
- F:M ratio (kg BOD5/kg MLSS·d)
- Handbook
- Environmental Engineering, Activated Sludge, Organic loading rate (F:M), page 340
- Equation
F:M = Q0·S0/(Vol·XA)- Substitute
F:M = Q0·S0/(V·XA) = (23,600 m³/d)(185 mg/L)/[(4,100 m³)(2,400 mg/L)] = 0.444 per day(mg/L cancels, so the units are kg BOD5 per kg MLSS per day)
- Result
- 0.444 kg BOD5/(kg MLSS·d), 3 significant figures
- Check
- the hydraulic residence time is V/Q = 4.17 h; F:M has units of 1/day.
- Why the others are wrong
- A: is the volumetric organic loading (kg BOD5/m³·d), not F:M
- C: inverted the ratio (microorganisms over food)
- D: converted MLSS to an assumed MLVSS (0.8) that the problem did not give
Problem 8 · FE Environmental, Activated sludge
An aeration basin of 6,100 m³ holds MLSS of 2,150 mg/L. Waste sludge is removed at 279 m³/d with 9,400 mg/L of solids, and the plant discharges 7,300 m³/d of effluent containing 9 mg/L of suspended solids. The solids residence time (sludge age) is most nearly:
Handbook: Environmental Engineering, Activated Sludge, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (4.88 d)
- Given
- V = 6,100 m³, XA = 2,150 mg/L, Qw = 279 m³/d, Xw = 9,400 mg/L, Qe = 7,300 m³/d, Xe = 9 mg/L
- Find
- solids residence time θc (d)
- Handbook
- Environmental Engineering, Activated Sludge (solids residence time), page 339
- Equation
θc = V·XA/(Qw·Xw + Qe·Xe)- Substitute
Solids in the basin: V·XA = (6,100)(2,150) = 13,120,000 m³·mg/LSolids leaving per day: Qw·Xw + Qe·Xe = (279)(9,400) + (7,300)(9) = 2,688,000 m³·mg/(L·d)θc = 13,120,000/2,688,000 = 4.88 d
- Result
- 4.88 d, 3 significant figures
- Check
- θc is the mass of solids in the basin divided by the mass of solids leaving each day; the units reduce to days.
- Why the others are wrong
- A: used the aeration-tank MLSS instead of the waste sludge concentration
- C: left out the solids lost in the effluent (Qe·Xe)
- D: is a hydraulic residence time, not the solids residence time
Problem 9 · FE Environmental, Activated sludge
An activated sludge aeration basin has a volume of 4,700 m³. It receives 16,600 m³/d of primary effluent with a BOD5 of 280 mg/L, and the mixed liquor suspended solids (MLSS) concentration is 3,500 mg/L. The food-to-microorganism (F:M) ratio, based on MLSS, is most nearly:
Handbook: Environmental Engineering, Activated Sludge, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (0.283 kg BOD5/(kg MLSS·d))
- Given
- V = 4,700 m³, Q = 16,600 m³/d, S0 = 280 mg/L, XA = 3,500 mg/L
- Find
- F:M ratio (kg BOD5/kg MLSS·d)
- Handbook
- Environmental Engineering, Activated Sludge, Organic loading rate (F:M), page 340
- Equation
F:M = Q0·S0/(Vol·XA)- Substitute
F:M = Q0·S0/(V·XA) = (16,600 m³/d)(280 mg/L)/[(4,700 m³)(3,500 mg/L)] = 0.283 per day(mg/L cancels, so the units are kg BOD5 per kg MLSS per day)
- Result
- 0.283 kg BOD5/(kg MLSS·d), 3 significant figures
- Check
- the hydraulic residence time is V/Q = 6.80 h; F:M has units of 1/day.
- Why the others are wrong
- A: inverted the ratio (microorganisms over food)
- B: converted MLSS to an assumed MLVSS (0.8) that the problem did not give
- C: treated the daily flow as hourly
Problem 10 · FE Environmental, Activated sludge
An activated sludge aeration basin has a volume of 3,600 m³. It receives 19,100 m³/d of primary effluent with a BOD5 of 200 mg/L, and the mixed liquor suspended solids (MLSS) concentration is 3,550 mg/L. The food-to-microorganism (F:M) ratio, based on MLSS, is most nearly:
Handbook: Environmental Engineering, Activated Sludge, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (0.299 kg BOD5/(kg MLSS·d))
- Given
- V = 3,600 m³, Q = 19,100 m³/d, S0 = 200 mg/L, XA = 3,550 mg/L
- Find
- F:M ratio (kg BOD5/kg MLSS·d)
- Handbook
- Environmental Engineering, Activated Sludge, Organic loading rate (F:M), page 340
- Equation
F:M = Q0·S0/(Vol·XA)- Substitute
F:M = Q0·S0/(V·XA) = (19,100 m³/d)(200 mg/L)/[(3,600 m³)(3,550 mg/L)] = 0.299 per day(mg/L cancels, so the units are kg BOD5 per kg MLSS per day)
- Result
- 0.299 kg BOD5/(kg MLSS·d), 3 significant figures
- Check
- the hydraulic residence time is V/Q = 4.52 h; F:M has units of 1/day.
- Why the others are wrong
- A: converted MLSS to an assumed MLVSS (0.8) that the problem did not give
- B: treated the daily flow as hourly
- C: inverted the ratio (microorganisms over food)
A new problem posts every day inside the lab, with the full worked solution the same evening.
Frequently asked questions
Where is activated sludge in the FE Reference Handbook?
Look in the Environmental Engineering, Activated Sludge part of FE Reference Handbook 10.6. Each solution gives the exact page, so you can practice finding it in the PDF the way you will on exam day.
Are these real FE exam questions?
No. They are original problems generated from the handbook formulas and checked by code. Real exam questions are confidential, and sharing them breaks the NCEES agreement every examinee accepts.
How do I check my answer before opening the solution?
Check the units of your result and whether its size makes sense for the situation. Each solution ends with the same kind of check, so you can compare your habit with ours.
Sources
- NCEES FE Environmental CBT exam specifications (PDF). Retrieved October 3, 2026.
- NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.