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10 FE practice problems: health risk: dose and hazard, with solutions

By Dr. Mohamed Elansary, PhD Environmental EngineeringUpdated

These ten original problems practice health risk: dose and hazard, a topic from the FE Environmental exam specification, using the Safety, Exposure and Risk Assessment/Toxicology 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, Health risk: dose and hazard

A resident drinks 2.0 L/day of water containing 0.271 mg/L of a noncarcinogenic chemical, 350 days/year for 9 years. Body weight = 80 kg; for a noncarcinogen the averaging time equals the exposure duration (365 days/year). The reference dose is 0.0003 mg/kg·day. The hazard quotient is most nearly:

  • A 22.6
  • B 903
  • C 21.7
  • D 2.78

Handbook: Safety, Exposure and Risk Assessment/Toxicology, FE Reference Handbook 10.6

Show the worked solution
Answer
C (21.7)
Given
CW = 0.271 mg/L, IR = 2.0 L/day, EF = 350 days/year, ED = 9 years, BW = 80 kg, days = 365 days/year, RfD = 0.0003 mg/kg·day
Find
hazard quotient
Handbook
Safety, Exposure (ingestion in drinking water) and Risk Assessment/Toxicology (noncarcinogens, RfD), pages 26 and 29
Equation
CDI = (CW)(IR)(EF)(ED)/[(BW)(AT)]; HQ = CDI/RfD (the hazard index sums HQs)
Substitute
  1. AT = ED × 365 = 9 × 365 = 3,285 days
  2. CDI = (0.271)(2.0)(350)(9)/[(80)(3,285)] = 6.497 × 10⁻³ mg/(kg·day)
  3. HQ = CDI/RfD = 6.497 × 10⁻³/0.0003 = 21.7
Result
21.7, 3 significant figures
Check
HQ > 1, so adverse effects are a concern for this one chemical.
Why the others are wrong
  • A: assumed exposure every day (EF left out)
  • B: divided the water concentration by the RfD (no intake calculation)
  • D: averaged over 70 years (the cancer convention) instead of the exposure duration

Problem 2 · FE Environmental, Health risk: dose and hazard

A resident drinks 2.0 L/day of water containing 0.123 mg/L of a noncarcinogenic chemical, 350 days/year for 6 years. Body weight = 15 kg; for a noncarcinogen the averaging time equals the exposure duration (365 days/year). The reference dose is 0.005 mg/kg·day. The hazard quotient is most nearly:

  • A 3.28
  • B 0.270
  • C 24.6
  • D 3.15

Handbook: Safety, Exposure and Risk Assessment/Toxicology, FE Reference Handbook 10.6

Show the worked solution
Answer
D (3.15)
Given
CW = 0.123 mg/L, IR = 2.0 L/day, EF = 350 days/year, ED = 6 years, BW = 15 kg, days = 365 days/year, RfD = 0.005 mg/kg·day
Find
hazard quotient
Handbook
Safety, Exposure (ingestion in drinking water) and Risk Assessment/Toxicology (noncarcinogens, RfD), pages 26 and 29
Equation
CDI = (CW)(IR)(EF)(ED)/[(BW)(AT)]; HQ = CDI/RfD (the hazard index sums HQs)
Substitute
  1. AT = ED × 365 = 6 × 365 = 2,190 days
  2. CDI = (0.123)(2.0)(350)(6)/[(15)(2,190)] = 1.573 × 10⁻² mg/(kg·day)
  3. HQ = CDI/RfD = 1.573 × 10⁻²/0.005 = 3.15
Result
3.15, 3 significant figures
Check
HQ > 1, so adverse effects are a concern for this one chemical.
Why the others are wrong
  • A: assumed exposure every day (EF left out)
  • B: averaged over 70 years (the cancer convention) instead of the exposure duration
  • C: divided the water concentration by the RfD (no intake calculation)

Problem 3 · FE Environmental, Health risk: dose and hazard

A resident drinks 2.0 L/day of water containing 0.189 mg/L of a noncarcinogenic chemical, 350 days/year for 9 years. Body weight = 15 kg; for a noncarcinogen the averaging time equals the exposure duration (365 days/year). The reference dose is 0.003 mg/kg·day. The hazard quotient is most nearly:

  • A 8.05
  • B 63.0
  • C 1.04
  • D 8.40

Handbook: Safety, Exposure and Risk Assessment/Toxicology, FE Reference Handbook 10.6

Show the worked solution
Answer
A (8.05)
Given
CW = 0.189 mg/L, IR = 2.0 L/day, EF = 350 days/year, ED = 9 years, BW = 15 kg, days = 365 days/year, RfD = 0.003 mg/kg·day
Find
hazard quotient
Handbook
Safety, Exposure (ingestion in drinking water) and Risk Assessment/Toxicology (noncarcinogens, RfD), pages 26 and 29
Equation
CDI = (CW)(IR)(EF)(ED)/[(BW)(AT)]; HQ = CDI/RfD (the hazard index sums HQs)
Substitute
  1. AT = ED × 365 = 9 × 365 = 3,285 days
  2. CDI = (0.189)(2.0)(350)(9)/[(15)(3,285)] = 2.416 × 10⁻² mg/(kg·day)
  3. HQ = CDI/RfD = 2.416 × 10⁻²/0.003 = 8.05
Result
8.05, 3 significant figures
Check
HQ > 1, so adverse effects are a concern for this one chemical.
Why the others are wrong
  • B: divided the water concentration by the RfD (no intake calculation)
  • C: averaged over 70 years (the cancer convention) instead of the exposure duration
  • D: assumed exposure every day (EF left out)

Problem 4 · FE Environmental, Health risk: dose and hazard

A resident drinks 2.0 L/day of water containing 0.087 mg/L of a carcinogen, 350 days/year for 24 years. Body weight = 15 kg; averaging time = 70 years (365 days/year); cancer slope factor = 0.05 (mg/kg·day)⁻¹. The lifetime excess cancer risk is most nearly:

  • A 5.80 × 10⁻⁴
  • B 1.91 × 10⁻⁴
  • C 1.99 × 10⁻⁴
  • D 5.56 × 10⁻⁴

Handbook: Safety, Exposure and Risk Assessment/Toxicology, FE Reference Handbook 10.6

Show the worked solution
Answer
B (1.91 × 10⁻⁴)
Given
CW = 0.087 mg/L, IR = 2.0 L/day, EF = 350 days/year, ED = 24 years, BW = 15 kg, AT_yr = 70 years, days = 365 days/year, CSF = 0.05
Find
lifetime excess cancer risk
Handbook
Safety, Exposure (ingestion in drinking water) and Risk Assessment/Toxicology (carcinogens), pages 25 and 29
Equation
CDI = (CW)(IR)(EF)(ED)/[(BW)(AT)]; Risk = CDI × CSF
Substitute
  1. CDI = (CW)(IR)(EF)(ED)/[(BW)(AT)] = (0.087)(2.0)(350)(24)/[(15)(25,550)] = 3.814 × 10⁻³ mg/(kg·day)
  2. Risk = CDI × CSF = 3.814 × 10⁻³ × 0.05 = 1.91 × 10⁻⁴
Result
1.91 × 10⁻⁴, 3 significant figures
Check
risk is a probability (no units); carcinogens are averaged over a lifetime, so AT = 70 × 365 days.
Why the others are wrong
  • A: left out EF·ED/AT (treated it as lifetime daily exposure)
  • C: assumed exposure every day (EF left out)
  • D: averaged over the exposure duration instead of the 70-year lifetime

Problem 5 · FE Environmental, Health risk: dose and hazard

A resident drinks 2.0 L/day of water containing 0.034 mg/L of a noncarcinogenic chemical, 350 days/year for 9 years. Body weight = 15 kg; for a noncarcinogen the averaging time equals the exposure duration (365 days/year). The reference dose is 0.02 mg/kg·day. The hazard quotient is most nearly:

  • A 0.217
  • B 1.70
  • C 0.227
  • D 0.0279

Handbook: Safety, Exposure and Risk Assessment/Toxicology, FE Reference Handbook 10.6

Show the worked solution
Answer
A (0.217)
Given
CW = 0.034 mg/L, IR = 2.0 L/day, EF = 350 days/year, ED = 9 years, BW = 15 kg, days = 365 days/year, RfD = 0.02 mg/kg·day
Find
hazard quotient
Handbook
Safety, Exposure (ingestion in drinking water) and Risk Assessment/Toxicology (noncarcinogens, RfD), pages 26 and 29
Equation
CDI = (CW)(IR)(EF)(ED)/[(BW)(AT)]; HQ = CDI/RfD (the hazard index sums HQs)
Substitute
  1. AT = ED × 365 = 9 × 365 = 3,285 days
  2. CDI = (0.034)(2.0)(350)(9)/[(15)(3,285)] = 4.347 × 10⁻³ mg/(kg·day)
  3. HQ = CDI/RfD = 4.347 × 10⁻³/0.02 = 0.217
Result
0.217, 3 significant figures
Check
HQ < 1, so adverse effects are not expected for this one chemical.
Why the others are wrong
  • B: divided the water concentration by the RfD (no intake calculation)
  • C: assumed exposure every day (EF left out)
  • D: averaged over 70 years (the cancer convention) instead of the exposure duration

Problem 6 · FE Environmental, Health risk: dose and hazard

A resident drinks 2.0 L/day of water containing 0.003 mg/L of a carcinogen, 350 days/year for 9 years. Body weight = 70 kg; averaging time = 70 years (365 days/year); cancer slope factor = 0.15 (mg/kg·day)⁻¹. The lifetime excess cancer risk is most nearly:

  • A 1.29 × 10⁻⁵
  • B 1.23 × 10⁻⁵
  • C 7.05 × 10⁻⁵
  • D 1.59 × 10⁻⁶

Handbook: Safety, Exposure and Risk Assessment/Toxicology, FE Reference Handbook 10.6

Show the worked solution
Answer
D (1.59 × 10⁻⁶)
Given
CW = 0.003 mg/L, IR = 2.0 L/day, EF = 350 days/year, ED = 9 years, BW = 70 kg, AT_yr = 70 years, days = 365 days/year, CSF = 0.15
Find
lifetime excess cancer risk
Handbook
Safety, Exposure (ingestion in drinking water) and Risk Assessment/Toxicology (carcinogens), pages 25 and 29
Equation
CDI = (CW)(IR)(EF)(ED)/[(BW)(AT)]; Risk = CDI × CSF
Substitute
  1. CDI = (CW)(IR)(EF)(ED)/[(BW)(AT)] = (0.003)(2.0)(350)(9)/[(70)(25,550)] = 1.057 × 10⁻⁵ mg/(kg·day)
  2. Risk = CDI × CSF = 1.057 × 10⁻⁵ × 0.15 = 1.59 × 10⁻⁶
Result
1.59 × 10⁻⁶, 3 significant figures
Check
risk is a probability (no units); carcinogens are averaged over a lifetime, so AT = 70 × 365 days.
Why the others are wrong
  • A: left out EF·ED/AT (treated it as lifetime daily exposure)
  • B: averaged over the exposure duration instead of the 70-year lifetime
  • C: divided by the slope factor instead of multiplying

Problem 7 · FE Environmental, Health risk: dose and hazard

A resident drinks 2.0 L/day of water containing 0.236 mg/L of a noncarcinogenic chemical, 350 days/year for 6 years. Body weight = 15 kg; for a noncarcinogen the averaging time equals the exposure duration (365 days/year). The reference dose is 0.003 mg/kg·day. The hazard quotient is most nearly:

  • A 78.7
  • B 10.1
  • C 10.5
  • D 0.862

Handbook: Safety, Exposure and Risk Assessment/Toxicology, FE Reference Handbook 10.6

Show the worked solution
Answer
B (10.1)
Given
CW = 0.236 mg/L, IR = 2.0 L/day, EF = 350 days/year, ED = 6 years, BW = 15 kg, days = 365 days/year, RfD = 0.003 mg/kg·day
Find
hazard quotient
Handbook
Safety, Exposure (ingestion in drinking water) and Risk Assessment/Toxicology (noncarcinogens, RfD), pages 26 and 29
Equation
CDI = (CW)(IR)(EF)(ED)/[(BW)(AT)]; HQ = CDI/RfD (the hazard index sums HQs)
Substitute
  1. AT = ED × 365 = 6 × 365 = 2,190 days
  2. CDI = (0.236)(2.0)(350)(6)/[(15)(2,190)] = 3.017 × 10⁻² mg/(kg·day)
  3. HQ = CDI/RfD = 3.017 × 10⁻²/0.003 = 10.1
Result
10.1, 3 significant figures
Check
HQ > 1, so adverse effects are a concern for this one chemical.
Why the others are wrong
  • A: divided the water concentration by the RfD (no intake calculation)
  • C: assumed exposure every day (EF left out)
  • D: averaged over 70 years (the cancer convention) instead of the exposure duration

Problem 8 · FE Environmental, Health risk: dose and hazard

A resident drinks 2.0 L/day of water containing 0.013 mg/L of a carcinogen, 350 days/year for 9 years. Body weight = 80 kg; averaging time = 70 years (365 days/year); cancer slope factor = 0.55 (mg/kg·day)⁻¹. The lifetime excess cancer risk is most nearly:

  • A 1.71 × 10⁻⁴
  • B 2.20 × 10⁻⁵
  • C 2.30 × 10⁻⁵
  • D 7.29 × 10⁻⁵

Handbook: Safety, Exposure and Risk Assessment/Toxicology, FE Reference Handbook 10.6

Show the worked solution
Answer
B (2.20 × 10⁻⁵)
Given
CW = 0.013 mg/L, IR = 2.0 L/day, EF = 350 days/year, ED = 9 years, BW = 80 kg, AT_yr = 70 years, days = 365 days/year, CSF = 0.55
Find
lifetime excess cancer risk
Handbook
Safety, Exposure (ingestion in drinking water) and Risk Assessment/Toxicology (carcinogens), pages 25 and 29
Equation
CDI = (CW)(IR)(EF)(ED)/[(BW)(AT)]; Risk = CDI × CSF
Substitute
  1. CDI = (CW)(IR)(EF)(ED)/[(BW)(AT)] = (0.013)(2.0)(350)(9)/[(80)(25,550)] = 4.007 × 10⁻⁵ mg/(kg·day)
  2. Risk = CDI × CSF = 4.007 × 10⁻⁵ × 0.55 = 2.20 × 10⁻⁵
Result
2.20 × 10⁻⁵, 3 significant figures
Check
risk is a probability (no units); carcinogens are averaged over a lifetime, so AT = 70 × 365 days.
Why the others are wrong
  • A: averaged over the exposure duration instead of the 70-year lifetime
  • C: assumed exposure every day (EF left out)
  • D: divided by the slope factor instead of multiplying

Problem 9 · FE Environmental, Health risk: dose and hazard

A resident drinks 2.0 L/day of water containing 0.079 mg/L of a carcinogen, 350 days/year for 24 years. Body weight = 70 kg; averaging time = 70 years (365 days/year); cancer slope factor = 0.015 (mg/kg·day)⁻¹. The lifetime excess cancer risk is most nearly:

  • A 1.16 × 10⁻⁵
  • B 3.25 × 10⁻⁵
  • C 3.39 × 10⁻⁵
  • D 1.11 × 10⁻⁵

Handbook: Safety, Exposure and Risk Assessment/Toxicology, FE Reference Handbook 10.6

Show the worked solution
Answer
D (1.11 × 10⁻⁵)
Given
CW = 0.079 mg/L, IR = 2.0 L/day, EF = 350 days/year, ED = 24 years, BW = 70 kg, AT_yr = 70 years, days = 365 days/year, CSF = 0.015
Find
lifetime excess cancer risk
Handbook
Safety, Exposure (ingestion in drinking water) and Risk Assessment/Toxicology (carcinogens), pages 25 and 29
Equation
CDI = (CW)(IR)(EF)(ED)/[(BW)(AT)]; Risk = CDI × CSF
Substitute
  1. CDI = (CW)(IR)(EF)(ED)/[(BW)(AT)] = (0.079)(2.0)(350)(24)/[(70)(25,550)] = 7.421 × 10⁻⁴ mg/(kg·day)
  2. Risk = CDI × CSF = 7.421 × 10⁻⁴ × 0.015 = 1.11 × 10⁻⁵
Result
1.11 × 10⁻⁵, 3 significant figures
Check
risk is a probability (no units); carcinogens are averaged over a lifetime, so AT = 70 × 365 days.
Why the others are wrong
  • A: assumed exposure every day (EF left out)
  • B: averaged over the exposure duration instead of the 70-year lifetime
  • C: left out EF·ED/AT (treated it as lifetime daily exposure)

Problem 10 · FE Environmental, Health risk: dose and hazard

A resident drinks 2.0 L/day of water containing 0.040 mg/L of a carcinogen, 350 days/year for 6 years. Body weight = 70 kg; averaging time = 70 years (365 days/year); cancer slope factor = 0.0015 (mg/kg·day)⁻¹. The lifetime excess cancer risk is most nearly:

  • A 1.47 × 10⁻⁷
  • B 1.64 × 10⁻⁶
  • C 1.71 × 10⁻⁶
  • D 1.41 × 10⁻⁷

Handbook: Safety, Exposure and Risk Assessment/Toxicology, FE Reference Handbook 10.6

Show the worked solution
Answer
D (1.41 × 10⁻⁷)
Given
CW = 0.040 mg/L, IR = 2.0 L/day, EF = 350 days/year, ED = 6 years, BW = 70 kg, AT_yr = 70 years, days = 365 days/year, CSF = 0.0015
Find
lifetime excess cancer risk
Handbook
Safety, Exposure (ingestion in drinking water) and Risk Assessment/Toxicology (carcinogens), pages 25 and 29
Equation
CDI = (CW)(IR)(EF)(ED)/[(BW)(AT)]; Risk = CDI × CSF
Substitute
  1. CDI = (CW)(IR)(EF)(ED)/[(BW)(AT)] = (0.040)(2.0)(350)(6)/[(70)(25,550)] = 9.393 × 10⁻⁵ mg/(kg·day)
  2. Risk = CDI × CSF = 9.393 × 10⁻⁵ × 0.0015 = 1.41 × 10⁻⁷
Result
1.41 × 10⁻⁷, 3 significant figures
Check
risk is a probability (no units); carcinogens are averaged over a lifetime, so AT = 70 × 365 days.
Why the others are wrong
  • A: assumed exposure every day (EF left out)
  • B: averaged over the exposure duration instead of the 70-year lifetime
  • C: left out EF·ED/AT (treated it as lifetime daily exposure)

A new problem posts every day inside the lab, with the full worked solution the same evening.

Frequently asked questions

Where is health risk: dose and hazard in the FE Reference Handbook?

Look in the Safety, Exposure and Risk Assessment/Toxicology 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

  1. NCEES FE Environmental CBT exam specifications (PDF). Retrieved October 3, 2026.
  2. NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.