PE Environmental Exam Prep: 40 Free Problems + 12-Week Plan

Start your PE Environmental exam prep with this original, unofficial problem. It's written to the NCEES specifications that took effect in April 2026. All 40 problems, two worked examples, and a 12-week plan are free on this page.

Problem 1 · Retardation and contaminant velocity

ENV-R-01 · Site Assessment and Remediation · Select one

An aquifer has a dry bulk density of 1.70 g/cm³, porosity of 0.30, and a fraction of organic carbon of 0.004. A dissolved contaminant has Koc = 100 L/kg. Groundwater seepage velocity is 0.60 m/day. Assume linear, equilibrium sorption. What are the retardation factor and the contaminant's average velocity?

A. R = 1.68; 0.36 m/day
B. R = 2.27; 0.26 m/day
C. R = 3.27; 0.18 m/day
D. R = 3.27; 1.96 m/day

Show answer and solution

Answer: C. R = 3.27; 0.18 m/day.

First find the distribution coefficient: kd = foc × Koc = 0.004 × 100 = 0.40 L/kg (the same as cm³/g). Then R = 1 + ρb·kd/θ = 1 + (1.70 × 0.40)/0.30 = 3.27. The contaminant moves at v/R = 0.60/3.27 = 0.18 m/day, about a third of the water's speed.

Why not the others: A: Leaves out the division by porosity: 1 + 1.70 × 0.40 = 1.68. B: Drops the 1 in R = 1 + ρb·kd/θ. D: Multiplies by R. Under the stated equilibrium-sorption model, retardation slows the contaminant; multiplying by R reverses the relationship.

Try a change: Double foc to 0.008. Then kd = 0.80, R = 5.53, and the plume front moves at about 0.11 m/day.

Source: EPA On-line Tools for Site Assessment: Retardation Factor, Retardation factor equation, R = 1 + ρb·kd/θ, kd = foc·Koc; EPA On-line Tools for Site Assessment: Seepage Velocity, Seepage velocity and retarded seepage velocity v/R

By Castleport Test Prep Editorial Team · Exam facts last verified October 9, 2026

The six areas · What changed in 2026 · Worked examples · All 40 problems · 12-week plan

The exam at a glance

80 questions, computer-based, offered year-round at Pearson VUE test centers
8 hours of exam time inside a 9-hour appointment: a 2-minute nondisclosure agreement, an 8-minute tutorial, and a 50-minute scheduled break
Multiple-choice questions plus alternative formats: multiple correct, point and click, drag and drop, and fill in the blank
Both SI and U.S. customary units
One reference on screen: the NCEES PE Environmental Reference Handbook. No design standards are supplied for this exam, and you can't bring your own materials.
$400 exam fee paid to NCEES; your licensing board may charge its own application fee
Pass/fail results, typically 7–10 days after the exam

Sources: NCEES PE Environmental exam page; PE Environmental specifications, effective April 2026, p. 1; NCEES Examinee Guide, May 2026, pp. 11, 14, 16.

What's on the exam now: the six areas

NCEES publishes a range of questions for each knowledge area. The ranges are counts out of 80, not percentages, and the topic examples under each area aren't a complete list.

What's on the exam now: the six areas
Area	Questions	What NCEES lists under it	Practice here
1. Water	19–29	Water engineering applications (hydraulics, hydrology, hydrogeology, chemistry, biology, fate and transport, sampling, monitoring, project management); drinking water; wastewater (including collection, residuals, and reuse); stormwater and water resources	Problems 2–12
2. Solid and Hazardous Waste	11–17	Engineering applications (chemistry, fate and transport, regulations, risk, sampling, reduction and recycling, mass and energy balance, project management); municipal and industrial solid waste; hazardous, medical, and radioactive waste	Problems 13–18
3. Sustainability	7–11	Resources conservation (energy, renewable and nonrenewable resources, materials, life-cycle analysis); resilience (infrastructure vulnerability, climate impacts, redundancy, reliability)	Problems 19–22
4. Air	13–20	Engineering applications (sampling and measurement, regulations, chemistry, fate and transport, atmospheric science and meteorology, project management); pollution control (sources, emissions characterization and inventory, control technologies, prevention)	Problems 23–29
5. Site Assessment and Remediation	13–20	Site assessment (investigation, risk assessment, fate and transport, project management); remediation (alternatives, technologies, feasibility studies, project management)	Problem 1 and 30–36
6. Environmental and Occupational Health	7–11	Regulatory compliance (health, security, emergency plans, incident response, regulations); exposure assessments (environmental and occupational health)	Problems 37–40

Source: NCEES PE Environmental specifications, effective April 2026, pp. 1–2.

How to split your study time

Use the published ranges to set a starting split, then let your misses change it. If you take the midpoint of each range, the six areas come out to roughly this share of your study time:

How to split your study time
Area	Midpoint of range	Share of study time to start
Water	24	about 27%
Air	16.5	about 18.5%
Site Assessment and Remediation	16.5	about 18.5%
Solid and Hazardous Waste	14	about 16%
Sustainability	9	about 10%
Environmental and Occupational Health	9	about 10%

This is our planning arithmetic (each midpoint divided by the 89-question total of the midpoints), not an NCEES allocation. The ranges overlap, so a particular exam won't match these shares exactly. After you work the 40 problems below, move time toward the areas where you missed the most.

What to study first

What to study first
Your situation	Start with	Why
You work in water or wastewater	Air, then Solid and Hazardous Waste	Water is the largest area, but it's probably your strongest. Air and waste together can be 24 to 37 questions.
You work in remediation or site work	Water, then Air	Water alone can be 19 to 29 questions.
You're retaking the exam	The two weakest areas on your NCEES diagnostic report, considered alongside each area's published question range	Use the diagnostic to identify relative weakness, then give more study attention to weaknesses in larger published ranges.
Your study materials predate April 2026	Sustainability, then re-sort your chapters into the six current areas	Sustainability is now its own area, and the old area list no longer matches. See What changed in April 2026.

What changed in April 2026

NCEES gave the PE Environmental exam new specifications beginning in April 2026 (NCEES memo, Oct. 2025, p. 1). If you test now, the April 2026 list is the one that counts. Here's how it lines up with the previous version:

What changed in April 2026
Current area (April 2026)	Questions	Closest area before April 2026	Questions	What's different
Water	19–29	Water	21–35	Narrower range. Five subareas became four: engineering applications, drinking water, wastewater, and stormwater and water resources.
Solid and Hazardous Waste	11–17	Solid and Hazardous Waste	11–18	Top of the range drops by one.
Sustainability	7–11	Listed under Associated Engineering Principles (5–9 for the whole area)	—	Now its own area, covering resources conservation and resilience.
Air	13–20	Air	14–22	Slightly narrower range.
Site Assessment and Remediation	13–20	Site Assessment and Remediation	12–19	Range rises by one; now split into site assessment and remediation, with feasibility studies named.
Environmental and Occupational Health	7–11	Environmental Health and Safety	7–11	Renamed; organized as regulatory compliance and exposure assessments.
—	—	Associated Engineering Principles	5–9	No longer an area. Project management now appears inside the technical areas. Statistics, engineering economics, and data management are no longer named.

Sources: April 2026 specifications, pp. 1–2; April 2019 specifications (historical), pp. 1–3.

Can you still use an older review book? For engineering fundamentals, yes. Water, air, waste, site, and health methods didn't stop being true in April 2026. Just don't use an older book's topic list or weighting to plan your time. Re-sort its chapters into the six areas above, add study time for sustainability, and check any regulation, fee, or policy against a current official source.

Your reference on exam day

The exam is closed book. NCEES shows you the PE Environmental Reference Handbook on screen as a searchable PDF, and PE Environmental has no design standards (Examinee Guide, pp. 10 and 16). You search with a box on the left side of the reference window; Ctrl+F doesn't work. You can download the current handbook free from your MyNCEES account.

That makes lookup speed a skill worth practicing. Try this 15-minute drill a few times a week:

5 minutes: Find five equations or tables you already know, using only the search box.
5 minutes: Solve one problem from this page with only the handbook open.
5 minutes: Log any search that was slow or failed, and the term that finally worked.

Keep the log in this format:

Your reference on exam day
Concept	Search term that worked	Handbook section or page	What the equation assumes	Time to find
(example) Retardation factor	"retardation"	(your handbook's section)	Linear, equilibrium sorption	1 min

Two worked examples

Both are original, unofficial practice. They show every step so you can see the method before you try the problems.

Worked example 1 · Sizing a cyclone and ESP train (SI)

ENV-WX01 · Area 4, Air · Original, unofficial practice

A kiln exhaust carries 5.0 g/m³ of particulate at 80 m³/s. A cyclone removes 85% of the particulate mass. An electrostatic precipitator (ESP) follows it, and the outlet must not exceed 0.025 g/m³. The ESP's effective migration velocity is 0.08 m/s. What collecting plate area does the ESP need?

Step 1. Find the overall penetration allowed. Penetration is the fraction that escapes: p = 0.025/5.0 = 0.005, which is 99.5% overall efficiency.

Step 2. Take out the cyclone's share. Penetrations multiply in series, so the ESP may pass p_ESP = 0.005/0.15 = 0.0333. The ESP must remove 96.7% of what reaches it.

Step 3. Size the ESP with the Deutsch equation. p = exp(−wₑ × SCA), so SCA = −ln(0.0333)/0.08 = 3.40/0.08 = 42.5 s/m.

Step 4. Convert to plate area. A = SCA × Q = 42.5 × 80 = 3,400 m².

Check: Without the cyclone, the ESP alone would need SCA = −ln(0.005)/0.08 = 66.2 s/m, or about 5,300 m². The cyclone saves roughly a third of the plate area. EPA's cost manual warns that this simple method can undersize a unit when particles are much finer than typical, so treat it as a first estimate.

Source: EPA Air Pollution Control Cost Manual, Sec. 6, Ch. 3: Electrostatic Precipitators, Eqs. 3.17–3.18, p. 3-19; series penetration, p. 3-20; Eq. 3.23, p. 3-22; example problem discussion, p. 3-55.

Worked example 2 · How far will the plume go in a year? (SI)

ENV-WX02 · Area 5, Site Assessment and Remediation · Original, unofficial practice

A water-supply well sits 90 m downgradient of a solvent release. The aquifer has hydraulic conductivity K = 15 m/day, a hydraulic gradient of 0.012, and porosity 0.30. Use the sorption values from Problem 1 (R = 3.27). Considering advection and sorption only, about how long until the contaminant's average front reaches the well?

Step 1. Groundwater seepage velocity. v = K × i/θ = 15 × 0.012/0.30 = 0.60 m/day.

Step 2. Contaminant velocity. v/R = 0.60/3.27 = 0.18 m/day.

Step 3. Travel time. 90/0.18 = about 490 days, or roughly 16 months.

Check: Water itself would arrive in 90/0.60 = 150 days. Dispersion spreads the plume, so the first detectable concentrations arrive earlier than this average front. Biodegradation could cut the concentration that arrives. Problems 30 and 31 show what happens if this well becomes an exposure point.

Source: EPA On-line Tools for Site Assessment: Seepage Velocity, seepage and retarded velocity; EPA On-line Tools for Site Assessment: Retardation Factor.

40 original practice problems with worked solutions

These are original Castleport problems written to the April 2026 NCEES topic list. They are not NCEES questions. The set covers all six areas, but it isn't a full-length exam, it isn't weighted exactly like one, and it doesn't cover every subtopic. Most problems are single-answer multiple choice. A few use formats like the real exam's enter-a-number and select-all items. On the exam, every question, including those formats, is scored right or wrong with no partial credit, and there's no penalty for a wrong answer (Examinee Guide, pp. 11 and 14).

Work each problem before you open its solution. Write down the first step that went wrong when you miss one; the error log shows what to do with it.

Area 1: Water (11 problems)

Problem 2 · Mixing an outfall into a river

ENV-W-01 · Water · Select one

A river carries 12.0 m³/s with 2.5 mg/L of a conservative (non-reacting) dissolved constituent. An outfall adds 0.50 m³/s at 150 mg/L. Assume complete mixing and no other inputs. What is the downstream concentration?

A. 6.25 mg/L
B. 8.40 mg/L
C. 8.75 mg/L
D. 76.3 mg/L

Show answer and solution

Answer: B. 8.40 mg/L.

Mass in equals mass out. Load = 12.0 × 2.5 + 0.50 × 150 = 30 + 75 = 105 g/s. Downstream flow = 12.0 + 0.50 = 12.5 m³/s. Concentration = 105/12.5 = 8.40 mg/L.

Why not the others: A: Counts only the outfall load and divides by the river flow (75/12). C: Adds both loads but divides by the river flow only, forgetting the outfall's water. D: Averages the two concentrations without weighting by flow.

Source: MIT OpenCourseWare 1.061, Lecture 2: Conservation of Mass, Eq. (4), integral conservation of mass, reduced here to a steady two-stream balance

Problem 3 · Ultimate BOD from BOD₅

ENV-W-02 · Water · Select one

A wastewater has a 5-day BOD of 220 mg/L. The first-order BOD rate constant is 0.25/day (base e). What is the ultimate BOD?

A. 220 mg/L
B. 233 mg/L
C. 308 mg/L
D. 768 mg/L

Show answer and solution

Answer: C. 308 mg/L.

BOD exerted by day t is y = L₀(1 − e^(−kt)). So L₀ = 220/(1 − e^(−0.25×5)) = 220/(1 − 0.2865) = 220/0.7135 = 308 mg/L.

Why not the others: A: Assumes all the demand is exerted in 5 days. B: Uses 10^(−kt), a base-10 form, with a base-e constant. D: Divides by e^(−kt) instead of by 1 − e^(−kt).

Source: EPA, Technical Guidance Manual for Developing TMDLs, Book II, Part 1, Sec. 2.3.4.1, Eq. 2-5, p. 2-8

Problem 4 · Disinfection CT with a baffling factor

ENV-W-03 · Water · Select one

A clearwell holds 150,000 gal at its lowest operating level. Peak hourly flow is 3.0 MGD. A tracer study gives a baffling factor of 0.50, and the free chlorine residual at the outlet is 1.5 mg/L. What CT does the clearwell provide?

A. 36 mg·min/L
B. 54 mg·min/L
C. 75.6 mg·min/L
D. 108 mg·min/L

Show answer and solution

Answer: B. 54 mg·min/L.

Peak flow = 3,000,000/1,440 = 2,083 gpm. Theoretical detention time = 150,000/2,083 = 72 min. Contact time T = 72 × 0.50 = 36 min. CT = 1.5 × 36 = 54 mg·min/L.

Why not the others: A: Stops at T. That's a time, not a CT. C: Uses a baffling factor of 0.7 instead of the measured 0.50. D: Ignores the baffling factor, so it assumes perfect plug flow.

Source: EPA 815-R-20-003, Disinfection Profiling and Benchmarking Technical Guidance Manual (2020), Eq. 4-1 (CT = C × T), p. 25; Eq. 4-2 (TDT = V/Q), p. 27; Table 4-2 (baffling factors), p. 28; Eq. 4-3 (T = TDT × BF), p. 29

Problem 5 · Total hardness from calcium and magnesium

ENV-W-04 · Water · Select one

A groundwater has 60 mg/L calcium and 18 mg/L magnesium. Other hardness ions are negligible. What is the total hardness as CaCO₃?

A. 78 mg/L
B. 150 mg/L
C. 187 mg/L
D. 224 mg/L

Show answer and solution

Answer: D. 224 mg/L.

Convert each ion to its CaCO₃ equivalent: 2.497 × 60 + 4.118 × 18 = 149.8 + 74.1 = 224 mg/L as CaCO₃.

Why not the others: A: Adds the ion concentrations without converting. B: Counts calcium only. C: Uses magnesium's atomic weight in place of its equivalent weight.

Source: EPA 2021 MSGP Appendix J: Calculating Hardness, Hardness equation, p. J-2

Problem 6 · Chlorine feed rate

ENV-W-05 · Water · Enter a number

A plant treats 2.5 MGD. Chlorine demand is 2.8 mg/L, and the target residual is 1.2 mg/L. Enter the chlorine feed in lb/day, to one decimal.

Show answer and solution

Answer: 83.4.

Dose = demand + residual = 4.0 mg/L. Feed = 2.5 × 4.0 × 8.34 = 83.4 lb/day.

Common mistakes: 25.0 lb/day uses the residual only; 58.4 lb/day uses the demand only; 10.0 forgets the 8.34 lb/gal factor.

Source: Delaware DNREC, Wastewater Operator Exam Resource Booklet, Lbs/day = mg/L × 8.34 × MGD, p. 3

Problem 7 · Clarifier surface overflow rate

ENV-W-06 · Water · Select one

A plant splits 2.7 MGD equally between two rectangular primary clarifiers, each 20 ft wide and 90 ft long. What is the surface overflow rate?

A. 0.75 gpd/ft²
B. 375 gpd/ft²
C. 750 gpd/ft²
D. 1,500 gpd/ft²

Show answer and solution

Answer: C. 750 gpd/ft².

Total surface area = 2 × 20 × 90 = 3,600 ft². Rate = 2,700,000 gpd/3,600 ft² = 750 gpd/ft². On the exam, compare the result with the design criterion the problem gives you.

Why not the others: A: Leaves the flow in MGD. B: Counts the area twice. D: Sends all the flow through one clarifier.

Source: Delaware DNREC, Wastewater Operator Exam Resource Booklet, Surface Loading Rate, gpd/sq. ft. = flow (gpd)/area (sq. ft.), p. 4

Problem 8 · Solids retention time

ENV-W-07 · Water · Select one

An aeration basin holds 0.80 MG at an MLSS of 2,500 mg/L. Waste activated sludge leaves at 0.020 MGD and 8,000 mg/L. Plant effluent is 3.5 MGD at 10 mg/L TSS. Counting only aeration-basin solids, what is the solids retention time (SRT, also called MCRT)?

A. 0.23 day
B. 10.3 days
C. 12.5 days
D. 57 days

Show answer and solution

Answer: B. 10.3 days.

SRT = solids in the system ÷ solids leaving per day. Inventory = 0.80 × 2,500 × 8.34 = 16,680 lb. Leaving = (0.020 × 8,000 + 3.5 × 10) × 8.34 = (160 + 35) × 8.34 = 1,626 lb/day. SRT = 16,680/1,626 = 10.3 days. (The 8.34 factors cancel.)

Why not the others: A: Is the hydraulic detention time, V/Q. C: Ignores the solids lost in the effluent. D: Ignores the wasted sludge.

Source: Delaware DNREC, Wastewater Operator Exam Resource Booklet, Mean Cell Residence Time formula, p. 5; Operation of Wastewater Treatment Plants, Vol. II (CSU Sacramento), hosted by EPA, Glossary, MCRT, p. 9

Problem 9 · Alkalinity used by nitrification

ENV-W-08 · Water · Select one

Influent alkalinity is 220 mg/L as CaCO₃. The plant fully nitrifies 25 mg/L of ammonia nitrogen, with no denitrification. Using the theoretical ratio, how much alkalinity remains?

A. 41.5 mg/L as CaCO₃
B. 131 mg/L as CaCO₃
C. 195 mg/L as CaCO₃
D. 220 mg/L as CaCO₃

Show answer and solution

Answer: A. 41.5 mg/L as CaCO₃.

Nitrification consumes about 7.14 mg alkalinity (as CaCO₃) per mg of ammonia-N oxidized. 7.14 × 25 = 178.5 mg/L consumed, leaving 220 − 178.5 = 41.5 mg/L. That little leftover can let pH fall, so the plant may need to add alkalinity. The training source also notes that actual consumption is usually somewhat less than the theoretical 7.14.

Why not the others: B: Uses 3.57, the alkalinity recovered by denitrification. C: Treats the ratio as 1:1. D: Ignores that nitrification consumes alkalinity.

Source: Long Island Sound Nitrogen Removal Training Program, Module 4, hosted by NYSDEC, Unit 1, transparencies 1-2 (7.14), 1-16 (actual vs theoretical), 1-19 (3.57)

Problem 10 · Food-to-microorganism ratio

ENV-W-09 · Water · Select one

Forward flow is 6,000 m³/day at 200 mg/L BOD₅. The aeration basins hold 2,000 m³ at an MLVSS of 2,500 mg/L (MLSS is 3,125 mg/L). What is the food-to-microorganism ratio (F/M) on an MLVSS basis?

A. 0.19 kg BOD₅/kg MLVSS·day
B. 0.24 kg BOD₅/kg MLVSS·day
C. 2.4 kg BOD₅/kg MLVSS·day
D. 4.2 kg BOD₅/kg MLVSS·day

Show answer and solution

Answer: B. 0.24 kg BOD₅/kg MLVSS·day.

Food = 6,000 m³/day × 0.200 kg/m³ = 1,200 kg BOD₅/day. Microorganisms = 2,000 m³ × 2.5 kg/m³ = 5,000 kg MLVSS. F/M = 1,200/5,000 = 0.24 per day.

Why not the others: A: Uses MLSS (6,250 kg) instead of MLVSS. C: Slips a factor of 10 in the mg/L-to-kg/m³ conversion. D: Inverts the ratio (M/F).

Source: Operation of Wastewater Treatment Plants, Vol. II (CSU Sacramento), hosted by EPA, Glossary, Food/Microorganism (F/M) ratio, p. 9

Problem 11 · Rational method with mixed land use

ENV-W-10 · Water · Select one

A 12 ha catchment has 4 ha with C = 0.90 and 8 ha with C = 0.20. The design intensity for the time of concentration is 50 mm/h. Assume the rational method applies, and use Q = CiA/360 (Q in m³/s, i in mm/h, A in ha). What is the peak flow?

A. 0.072 m³/s
B. 0.722 m³/s
C. 0.917 m³/s
D. 1.67 m³/s

Show answer and solution

Answer: B. 0.722 m³/s.

Weight C by area: (4 × 0.90 + 8 × 0.20)/12 = 5.2/12 = 0.433. Q = 0.433 × 50 × 12/360 = 0.722 m³/s.

Why not the others: A: Slips a factor of 10. C: Averages the two coefficients (0.55) without weighting by area. D: Leaves out C, as if all rain ran off.

Source: TxDOT Hydraulic Design Manual, Ch. 4, Sec. 12: Rational Method, Eq. 4-20 (Z = 360 for metric) and Eq. 4-23 (area-weighted C)

Problem 12 · Hazen-Williams head loss

ENV-W-11 · Water · Select one

Use the Hazen-Williams form h = 4.727 L Q^1.852/(C^1.852 d^4.871), with h, L, and d in feet and Q in cfs. Water flows at 1.5 cfs through 1,500 ft of 10-in pipe with C = 130. What is the friction head loss?

A. 2.2 ft
B. 4.4 ft
C. 7.2 ft
D. 8.9 ft

Show answer and solution

Answer: B. 4.4 ft.

Convert the diameter: d = 10/12 = 0.833 ft. h = 4.727 × 1,500 × 1.5^1.852/(130^1.852 × 0.833^4.871) = 4.4 ft.

Why not the others: A: Halves the result. C: Uses C = 100, a rougher pipe. D: Doubles the result.

Try a change: Age the pipe so C drops to 100. Head loss rises to about 7.2 ft, a 63% increase at the same flow.

Source: EPA EPANET 2.2 User Manual, Sec. 3.1, Table 3.1, Table 3.1, Hazen-Williams resistance coefficient (feet, cfs)

Area 2: Solid and Hazardous Waste (6 problems)

Problem 13 · Landfill airspace

ENV-S-01 · Solid and Hazardous Waste · Select one

A landfill receives 200 tons/day, 365 days a year. Waste is compacted to 1,200 lb/yd³. Daily and intermediate cover add 20% to the waste volume. How much airspace is used each year?

A. 60,800 yd³
B. 121,700 yd³
C. 146,000 yd³
D. 175,200 yd³

Show answer and solution

Answer: C. 146,000 yd³.

Waste mass = 200 × 2,000 × 365 = 146,000,000 lb/yr. Waste volume = 146,000,000/1,200 = 121,667 yd³. Add cover: 121,667 × 1.20 = 146,000 yd³.

Why not the others: A: Uses 2,400 lb/yd³, double the stated density. B: Leaves out the cover. D: Applies the 1.20 factor twice.

Source: Worked from the stated inputs.

Problem 14 · Landfill methane by first-order decay

ENV-S-02 · Solid and Hazardous Waste · Select one

A landfill cell receives 100,000 Mg of waste in a single year and then closes. Use k = 0.05/yr, L₀ = 100 m³ CH₄/Mg, and the simplified first-order form Q = k·L₀·M·e^(−kt). What is the methane generation rate 10 years after placement?

A. 152,000 m³/yr
B. 303,000 m³/yr
C. 500,000 m³/yr
D. 607,000 m³/yr

Show answer and solution

Answer: B. 303,000 m³/yr.

k·L₀·M = 0.05 × 100 × 100,000 = 500,000 m³/yr at placement. After 10 years: 500,000 × e^(−0.5) = 500,000 × 0.6065 = 303,000 m³ CH₄/yr. EPA's LandGEM model sums this same first-order decay over yearly and tenth-of-a-year increments, so its output for this case would be close but not identical.

Why not the others: A: Uses e^(−1), as if 20 years had passed. C: Ignores decay. D: Converts to total landfill gas at 50% methane, which answers a different question.

Source: EPA LMOP, LFG Energy Project Development Handbook, Ch. 2, First-order decay equation and k, L₀, Mᵢ definitions, p. 2-2

Problem 15 · RCRA hazardous characteristics

ENV-S-03 · Solid and Hazardous Waste · Select all that apply

Representative samples from five waste streams give the results below. Which wastes exhibit a RCRA hazardous characteristic? Select all that apply.

A. A liquid with a flash point of 52 °C (not an alcohol solution)
B. An aqueous waste with pH 12.8
C. A solid whose TCLP extract has 3.2 mg/L lead
D. A solid whose TCLP extract has 0.8 mg/L benzene
E. An aqueous waste with pH 11.9

Show answer and solution

Answer: A, B, and D.

A is ignitable (D001): a liquid with a flash point below 60 °C (140 °F). B is corrosive (D002): aqueous with pH at or above 12.5. D is toxic (D018): benzene at or above 0.5 mg/L in the TCLP extract. To get credit you must select A, B, and D and nothing else.

Why not the others: C: The lead regulatory level is 5.0 mg/L, so 3.2 mg/L does not exhibit the toxicity characteristic. E: pH 11.9 is below the 12.5 corrosivity threshold.

Source: 40 CFR Part 261 Subpart C (eCFR), §261.21(a)(1); §261.22(a)(1); §261.24 Table 1

Problem 16 · Hazardous waste generator category

ENV-S-04 · Solid and Hazardous Waste · Select one

In one calendar month a facility generates 350 kg of non-acute hazardous waste and 1.5 kg of acute hazardous waste. What is its generator category for that month?

A. Very small quantity generator
B. Small quantity generator
C. Large quantity generator
D. It depends on the annual total

Show answer and solution

Answer: C. Large quantity generator.

More than 1 kg of acute hazardous waste in a calendar month makes a large quantity generator, whatever the non-acute amount. When a generator has both kinds, it determines a category for each and applies the more stringent one.

Why not the others: A: Ignores both amounts; 350 kg of non-acute waste alone already exceeds the very-small limit of 100 kg. B: Considers only the 350 kg of non-acute waste. D: Category is determined each calendar month, not annually.

Source: 40 CFR 262.13 (eCFR), §262.13(b) and Table 1

Problem 17 · Incinerator destruction and removal efficiency

ENV-S-05 · Solid and Hazardous Waste · Select one

A hazardous waste incinerator is fed a principal organic hazardous constituent (POHC) at 150 kg/h. The stack emits 0.020 kg/h of that POHC. Assume the waste is not one of the specially listed wastes subject to the 99.9999% DRE requirement. What is the DRE, and does it meet the general 99.99% federal standard?

A. 99.987%; does not meet 99.99%
B. 99.987%; meets 99.99%
C. 99.9987%; meets 99.99%
D. 0.013%; does not meet 99.99%

Show answer and solution

Answer: A. 99.987%; does not meet 99.99%.

DRE = (Win − Wout)/Win × 100 = (150 − 0.020)/150 × 100 = 99.987%. That is below the general 99.99% requirement in 40 CFR 264.343(a)(1). The same section sets a stricter 99.9999% requirement for certain specifically listed wastes, which the stem excludes.

Why not the others: B: Gets the number right but misreads the comparison. C: Slips a decimal place. D: Reports the fraction that escapes, not the DRE.

Source: 40 CFR 264.343 (eCFR), §264.343(a)(1)–(2)

Problem 18 · Waste diversion rate

ENV-S-06 · Solid and Hazardous Waste · Select one

A community generates 120 tons/day of municipal solid waste. Paper is 35% of the waste, and a recycling program captures 30% of it. Food and yard waste is 20%, and a composting program captures 60% of it. What is the diversion rate?

A. 10.5%
B. 22.5%
C. 45.0%
D. 55.0%

Show answer and solution

Answer: B. 22.5%.

Paper recovered = 120 × 0.35 × 0.30 = 12.6 tons/day. Organics recovered = 120 × 0.20 × 0.60 = 14.4 tons/day. Diversion = 27.0/120 = 22.5%.

Why not the others: A: Counts paper only. C: Averages the two capture rates. D: Adds the two component shares and ignores capture.

Source: Worked from the stated inputs.

Area 3: Sustainability (4 problems)

Problem 19 · Life-cycle cost of two blowers

ENV-U-01 · Sustainability · Select one

Two aeration blower options last 15 years with no salvage value, at a 5% discount rate. Option A costs $120,000 now plus $18,000/yr in energy. Option B costs $160,000 now plus $12,000/yr. Energy is paid at the end of each year. Which has the lower present worth of costs?

A. A, because its first cost is $40,000 lower
B. B, about $284,600 vs. about $306,800 for A
C. B, $340,000 vs. $390,000 for A
D. They are within 1% of each other

Show answer and solution

Answer: B. B, about $284,600 vs. about $306,800 for A.

The uniform present value factor is [(1.05)¹⁵ − 1]/[0.05(1.05)¹⁵] = 10.3797. PW of A = 120,000 + 18,000 × 10.3797 = $306,834. PW of B = 160,000 + 12,000 × 10.3797 = $284,556. B saves about $22,300 in present-worth terms.

Why not the others: A: Looks only at first cost. C: Reaches the right choice by adding undiscounted dollars, which ignores the time value of money. D: The gap is about 7% of A's present worth.

Source: NIST IR 85-3273-39, Energy Price Indices and Discount Factors for LCC Analysis, 2024, Sec. 2, uniform present value factor formula, p. 6

Problem 20 · Energy saved by a variable-speed pump

ENV-U-02 · Sustainability · Select one

A centrifugal pump in a circulating loop with negligible static head draws 75 kW at full speed. A variable-speed drive will run it at 80% speed for 4,000 h/yr. Using the affinity laws and ignoring drive losses, how much energy does this save each year compared with running at full speed for those hours?

A. 60,000 kWh/yr
B. 108,000 kWh/yr
C. 146,400 kWh/yr
D. 300,000 kWh/yr

Show answer and solution

Answer: C. 146,400 kWh/yr.

With no static head, power varies with the cube of speed: 0.8³ × 75 = 0.512 × 75 = 38.4 kW. Savings = (75 − 38.4) × 4,000 = 146,400 kWh/yr. Static head, motor and drive efficiency, and real duty cycles all reduce field savings, which is why the stem rules them out.

Why not the others: A: Assumes power falls in proportion to speed. B: Uses the square of speed. D: Is the total full-speed energy, not the savings.

Source: U.S. DOE, Pumping Systems Tip Sheet #12 (2007), hosted by NREL, Power varies as the cube of pump speed with no static lift, p. 1

Problem 21 · Redundant pumps for reliability

ENV-U-03 · Sustainability · Enter a number

Identical pumps each have a 0.95 probability of being available when needed. Failures are independent, and any one pump can carry the design flow. What is the minimum number of pumps in parallel for a system availability of at least 0.999? Enter a whole number.

Show answer and solution

Answer: 3.

In a parallel system, the system fails only if every pump fails, so unavailability multiplies. Two pumps: 1 − 0.05² = 0.9975 (not enough). Three pumps: 1 − 0.05³ = 0.999875 (meets 0.999). Shared failure causes, such as one power feed for all pumps, break the independence assumption and lower real reliability.

Common mistakes: Answering 2 forgets that 0.9975 is below 0.999. Multiplying the availabilities (0.95 × 0.95) treats the pumps as a series system.

Source: NIST/SEMATECH e-Handbook, 8.1.8.3 Parallel or redundant model, Parallel model: system CDF is the product of component CDFs

Problem 22 · Choosing a functional unit

ENV-U-04 · Sustainability · Select one

A utility wants to compare the environmental impacts of a reusable stainless-steel water container with single-use plastic bottles. Which is the best functional unit for the life-cycle assessment?

A. One container
B. One kilogram of container material
C. Delivering 1,000 L of drinking water to the user over the study period
D. One year of container production

Show answer and solution

Answer: C. Delivering 1,000 L of drinking water to the user over the study period.

A functional unit describes the service delivered, so different systems can be compared fairly. It is set in the goal and scope phase, the first of the four LCA phases. One reusable container delivers far more water than one bottle, so a per-container or per-kilogram basis would mislead.

Why not the others: A: Compares unequal amounts of service. B: Measures material, not function. D: Measures production, not the service the user gets.

Source: GSA Sustainable Facilities Tool, Conducting a Life Cycle Assessment, Step 1, Goal Definition and Scoping: functional units

Area 4: Air (7 problems)

Problem 23 · ppb to µg/m³

ENV-A-01 · Air · Select one

Ambient SO₂ measures 75 ppb at 25 °C and 1 atm. The molecular weight of SO₂ is 64.07. What is the concentration in µg/m³?

A. 0.197 µg/m³
B. 28.6 µg/m³
C. 75.0 µg/m³
D. 196.5 µg/m³

Show answer and solution

Answer: D. 196.5 µg/m³.

At 25 °C and 1 atm, mg/m³ = ppm × MW/24.45. 75 ppb = 0.075 ppm. 0.075 × 64.07/24.45 = 0.1965 mg/m³ = 196.5 µg/m³.

Why not the others: A: Reports mg/m³ as µg/m³. B: Inverts the conversion. C: Treats ppb and µg/m³ as the same.

Source: NIOSH ppm–mg/m³ conversion calculator, Conversion equation at 25 °C and 1 atm

Problem 24 · Emissions from an emission factor

ENV-A-02 · Air · Select one

A process runs at 50,000 units/yr. Its uncontrolled emission factor is 0.80 lb of particulate per unit, and the control system's overall emission reduction is 95%. What are the annual emissions?

A. 1.0 ton/yr
B. 19.0 tons/yr
C. 20.0 tons/yr
D. 0.05 ton/yr

Show answer and solution

Answer: A. 1.0 ton/yr.

E = A × EF × (1 − ER/100) = 50,000 × 0.80 × 0.05 = 2,000 lb/yr = 1.0 ton/yr.

Why not the others: B: Is the mass removed, not emitted. C: Is the uncontrolled emission. D: Uses the control fraction as if it were tons.

Source: EPA AP-42 Introduction (2024), General equation for emission estimation, p. 1

Problem 25 · Control devices in series

ENV-A-03 · Air · Select one

A cyclone removes 80% of particulate mass. A baghouse downstream removes 99.5% of what reaches it. What is the overall collection efficiency?

A. 79.6%
B. 89.8%
C. 99.5%
D. 99.9%

Show answer and solution

Answer: D. 99.9%.

Work with penetration, the fraction that escapes. Cyclone penetration = 0.20; baghouse penetration = 0.005. Overall penetration = 0.20 × 0.005 = 0.001, so efficiency = 99.9%.

Why not the others: A: Multiplies the efficiencies instead of the penetrations. B: Averages the two efficiencies. C: Ignores the cyclone.

Source: EPA Air Pollution Control Cost Manual, Sec. 6, Ch. 3: Electrostatic Precipitators, Overall penetration is the product of section penetrations, text after Eq. 3.19, p. 3-20

Problem 26 · ESP plate area

ENV-A-04 · Air · Select one

An electrostatic precipitator (ESP) must remove 99.0% of particulate. The effective migration velocity is 0.10 m/s, and the gas flow is 100 m³/s. Using the Deutsch equation, what collecting plate area is needed?

A. 46 m²
B. 990 m²
C. 2,000 m²
D. 4,605 m²

Show answer and solution

Answer: D. 4,605 m².

Penetration p = 0.01. From p = exp(−wₑ·SCA), the specific collecting area SCA = −ln(0.01)/0.10 = 46.05 s/m. Plate area = SCA × Q = 46.05 × 100 = 4,605 m².

Why not the others: A: Stops at SCA and forgets to multiply by flow. B: Treats efficiency as linear in area. C: Uses log₁₀ instead of the natural log.

Try a change: Tighten to 99.9%. SCA becomes −ln(0.001)/0.10 = 69.1 s/m, so plate area rises 50% for one more 'nine.'

Source: EPA Air Pollution Control Cost Manual, Sec. 6, Ch. 3: Electrostatic Precipitators, Eqs. 3.16–3.18, p. 3-19; Eq. 3.23, p. 3-22

Problem 27 · Gaussian plume at ground level

ENV-A-05 · Air · Select one

A stack emits 100 g/s with an effective release height of 60 m. Wind speed is 5 m/s. At the receptor distance, σy = 120 m and σz = 60 m. Assume a Gaussian plume with full ground reflection. What is the ground-level concentration on the plume centerline?

A. 268 µg/m³
B. 536 µg/m³
C. 884 µg/m³
D. 1,073 µg/m³

Show answer and solution

Answer: B. 536 µg/m³.

Ground-level centerline: C = Q/(π u σy σz) × exp(−H²/2σz²) = 100/(π × 5 × 120 × 60) × exp(−3,600/7,200) = 8.84 × 10⁻⁴ × 0.607 g/m³ = 5.36 × 10⁻⁴ g/m³ = 536 µg/m³.

Why not the others: A: Drops the ground-reflection term (uses 2π in the denominator). C: Ignores the stack-height exponential. D: Double-counts reflection.

Source: U.S. EPA, SCREEN2 Model User’s Guide, Sec. 3.1, basic Gaussian plume equation with ground reflection; ground-level centerline simplification

Problem 28 · Why an ESP lost efficiency

ENV-A-06 · Air · Select one

A dry plate-wire ESP collecting fly ash loses efficiency after the plant switches coal. Tests show the dust resistivity rose to about 5 × 10¹¹ ohm-cm. Which effect best explains the drop?

A. Back corona in the collected dust layer
B. Particles held too loosely, causing heavy rapping reentrainment
C. Too little sneakage around the collection zones
D. Lower gas viscosity speeding up particles

Show answer and solution

Answer: A. Back corona in the collected dust layer.

When the dust layer's resistivity is high, roughly above 2 × 10¹¹ ohm-cm, the current flowing through it can break the layer down electrically. This back corona injects ions of the wrong polarity, reduces particle charge, and can cause sparking, so collection drops.

Why not the others: B: Loose, easily reentrained dust is the low-resistivity problem (below about 10⁸ ohm-cm). C: Sneakage is gas bypassing the zones; less of it would help, not hurt. D: Viscosity isn't tied to the coal switch described, and lower viscosity would raise migration velocity.

Source: EPA Air Pollution Control Cost Manual, Sec. 6, Ch. 3: Electrostatic Precipitators, Back corona and resistivity, Sec. 3.1.2.1, pp. 3-7 to 3-8

Problem 29 · Stack emission rate from ppmv

ENV-A-07 · Air · Enter a number

A stack test measures NOₓ at 250 ppmv, expressed as NO₂ (MW 46.01). Stack flow is 2,000 m³/min at 25 °C and 1 atm. Enter the NOₓ emission rate in kg/h, to one decimal.

Show answer and solution

Answer: 56.4.

Concentration = 250 × 46.01/24.45 = 470.4 mg/m³. Rate = 470.4 mg/m³ × 2,000 m³/min × 60 min/h = 56,450,000 mg/h = 56.4 kg/h.

Common mistakes: 36.8 kg/h uses the molecular weight of NO (30.01) even though results are expressed as NO₂; 0.94 reports kg/min.

Source: NIOSH ppm–mg/m³ conversion calculator, Conversion equation at 25 °C and 1 atm

Area 5: Site Assessment and Remediation (8 problems)

Problem 1, retardation and contaminant velocity, is at the top of this page.

Problem 30 · Drinking-water cancer risk

ENV-R-02 · Site Assessment and Remediation · Select one

A resident drinks groundwater with 0.010 mg/L of a carcinogen. Use: intake rate 2 L/day, exposure frequency 350 days/yr, exposure duration 30 yr, body weight 70 kg, and a lifetime averaging time of 70 yr × 365 days/yr. The slope factor is 0.055 (mg/kg·day)⁻¹ (an illustrative value, not a real chemical's). What is the estimated excess lifetime cancer risk?

A. 6.5 × 10⁻⁶
B. 1.5 × 10⁻⁵
C. 1.2 × 10⁻⁴
D. 2.1 × 10⁻³

Show answer and solution

Answer: A. 6.5 × 10⁻⁶.

Lifetime average daily dose = C × IR × EF × ED/(BW × AT) = 0.010 × 2 × 350 × 30/(70 × 25,550) = 210/1,788,500 = 1.17 × 10⁻⁴ mg/kg·day. Risk = dose × slope factor = 1.17 × 10⁻⁴ × 0.055 = 6.5 × 10⁻⁶.

Why not the others: B: Averages over the exposure duration (30 yr) instead of a lifetime. That's the noncancer rule. C: Reports the dose, not the risk. D: Leaves out body weight and averaging time.

Source: EPA ExpoBox: Exposure Assessment Tools by Routes – Ingestion, ADD equation; lifetime substituted for AT for cancer; EPA, Conducting a Human Health Risk Assessment, Cancer Risk = Exposure × Slope Factor

Problem 31 · Hazard quotients and hazard index

ENV-R-03 · Site Assessment and Remediation · Select one

Three noncarcinogens reach a receptor by one pathway. Their hazard quotients are 0.4, 0.5, and 0.3, and all three affect the same organ system. Which statement is best?

A. There's no noncancer concern because each HQ is below 1.
B. The hazard index is 1.2, which exceeds 1, so the combined exposure needs closer evaluation.
C. The hazard index is 0.4, the average HQ.
D. The hazard index is 1.2, so adverse effects will occur.

Show answer and solution

Answer: B. The hazard index is 1.2, which exceeds 1, so the combined exposure needs closer evaluation.

A hazard quotient is exposure divided by the reference dose. For chemicals that share a toxic endpoint, the HQs add to a hazard index: 0.4 + 0.5 + 0.3 = 1.2. EPA's screening approach treats an index above 1 as exceeding the target, so the exposure needs a closer look.

Why not the others: A: Ignores that effects on the same endpoint add up. C: Averages instead of summing. D: Overstates the result. An index above 1 means the reference level is exceeded and effects can't be ruled out; it doesn't mean they will happen.

Source: ATSDR, Calculating Hazard Quotients and Cancer Risk Estimates, HQ = D/RfD; HQ > 1 calls for in-depth evaluation; EPA Regional Screening Levels User's Guide, Hazard index: sum of HQs for chemicals with the same toxic endpoint should not exceed 1.0

Problem 32 · Feasibility study threshold criteria

ENV-R-04 · Site Assessment and Remediation · Select TWO

In the detailed analysis of alternatives in a CERCLA feasibility study, which two criteria are threshold criteria that a remedy must meet before it can be selected? Select TWO.

A. Overall protection of human health and the environment
B. Cost
C. Compliance with ARARs (applicable or relevant and appropriate requirements), unless waived
D. Community acceptance
E. Short-term effectiveness

Show answer and solution

Answer: A and C.

Of the nine evaluation criteria, overall protection of human health and the environment and compliance with ARARs are the two threshold criteria. ARAR compliance can be waived only under the circumstances allowed by the rule. You must select exactly A and C for credit.

Why not the others: B: Cost is one of the five balancing criteria. D: Community acceptance is a modifying criterion. E: Short-term effectiveness is a balancing criterion.

Source: 40 CFR 300.430 (eCFR), §300.430(e)(9)(iii)(A)–(B) and §300.430(f)(1)(i)(A)

Problem 33 · What a baseline risk assessment assumes

ENV-R-05 · Site Assessment and Remediation · Select one

Under EPA Superfund guidance, a baseline risk assessment estimates site risk under which assumption?

A. The preferred remedy has been built and is operating
B. No remediation or institutional controls are applied
C. Only institutional controls, such as deed restrictions, are in place
D. Contaminant concentrations are at their regulatory limits

Show answer and solution

Answer: B. No remediation or institutional controls are applied.

Baseline risks are the risks that might exist if no remediation or institutional controls were applied. That baseline helps decide whether action is needed at all and documents what drives the risk.

Why not the others: A: Evaluating a remedy's residual risk comes later, in the analysis of alternatives. C: Institutional controls are a form of response action, so the baseline excludes them. D: The baseline uses measured or modeled site concentrations, not regulatory limits.

Source: EPA RAGS Part A (EPA/540/1-89/002, 1989), Sec. 1.1.2, p. 1-4, and Chapter 1 endnote 4, p. 1-11

Problem 34 · Time to clean up by natural attenuation

ENV-R-06 · Site Assessment and Remediation · Enter a number

A monitoring well shows 800 µg/L of a contaminant. Assume first-order attenuation at k = 0.35/yr and that the rate stays constant. Enter the number of years for the concentration to reach 5 µg/L, to one decimal.

Show answer and solution

Answer: 14.5.

C = C₀e^(−kt), so t = ln(C₀/C)/k = ln(800/5)/0.35 = ln(160)/0.35 = 5.075/0.35 = 14.5 years. The half-life is 0.693/0.35 = 2.0 years, and about 7.3 half-lives fit in that time.

Common mistakes: 6.3 years comes from using log₁₀ instead of the natural log.

Source: EPA On-line Tools for Site Assessment: One-Dimensional Transport, First-order decay constant μ in the one-dimensional transport equation

Problem 35 · Pump-and-treat flushing time

ENV-R-07 · Site Assessment and Remediation · Select one

A plume occupies 5,000 m² of aquifer, 8 m thick, with porosity 0.30. The extraction system pumps 100 m³/day. The contaminant's retardation factor is 3.27. Under an idealized one-dimensional piston-flow model with uniform flushing and equilibrium retardation, about how long does one retarded traversal of the plume volume take?

A. 37 days
B. 120 days
C. 392 days
D. 1,307 days

Show answer and solution

Answer: C. 392 days.

Pore volume = 5,000 × 8 × 0.30 = 12,000 m³, so one hydraulic pore-volume time is 12,000/100 = 120 days. Under the stated idealized model, the contaminant front moves at v/R, so the corresponding retarded traversal time is R times the hydraulic traversal time: 120 × 3.27 = 392 days. This is a teaching-model result, not a cleanup-duration estimate; heterogeneity, mass-transfer limits, tailing, and operational constraints can make real remediation much longer.

Why not the others: A: Divides by R instead of applying the slower retarded velocity v/R. B: Ignores sorption. D: Uses the total aquifer volume and ignores porosity.

Source: EPA On-line Tools for Site Assessment: Retardation Factor, Retardation factor definition; EPA On-line Tools for Site Assessment: Seepage Velocity, Retarded seepage velocity v/R

Problem 36 · Dissolved plus sorbed contaminant mass

ENV-R-08 · Site Assessment and Remediation · Select one

A contaminated saturated zone is 12,000 m³ with porosity 0.30, dry bulk density 1.70 kg/L, and kd = 0.40 L/kg. Dissolved concentration is 2.0 mg/L at equilibrium. What total contaminant mass (dissolved plus sorbed) is present?

A. 7.2 kg
B. 16.3 kg
C. 23.5 kg
D. 40.8 kg

Show answer and solution

Answer: C. 23.5 kg.

Per liter of aquifer: dissolved = θ × C = 0.30 × 2.0 = 0.60 mg; sorbed = ρb × kd × C = 1.70 × 0.40 × 2.0 = 1.36 mg. Total = 1.96 mg/L of aquifer. 12,000 m³ = 12,000,000 L, so mass = 23.5 kg. Most of the mass sits on the soil, not in the water.

Why not the others: A: Counts dissolved mass only. B: Counts sorbed mass only. D: Ignores porosity for the dissolved part.

Source: EPA On-line Tools for Site Assessment: Retardation Factor, kd = foc·Koc; sorbed concentration = kd × dissolved concentration; Mass totals are worked from the stated inputs.

Area 6: Environmental and Occupational Health (4 problems)

Problem 37 · Noise dose and TWA

ENV-E-01 · Environmental and Occupational Health · Select one

A worker spends 3 h at 95 dBA, 3 h at 90 dBA, and 2 h at 85 dBA. Using OSHA's noise dose method, what are the dose and the 8-hour time-weighted average (TWA)?

A. 125%; 91.6 dBA
B. 112.5%; 90.8 dBA
C. 75%; 87.9 dBA
D. 125%; 90.0 dBA

Show answer and solution

Answer: A. 125%; 91.6 dBA.

Reference durations: 95 dBA → 4 h, 90 dBA → 8 h, 85 dBA → 16 h. D = 100 × (3/4 + 3/8 + 2/16) = 125%. TWA = 16.61 log₁₀(1.25) + 90 = 91.6 dBA. A dose above 100% exceeds the permissible exposure.

Why not the others: B: Drops the 85 dBA period. C: Counts only the 95 dBA period. D: Assumes the TWA is 90 dBA whenever the dose exceeds 100%.

Source: OSHA 29 CFR 1910.95 Appendix A, Sec. I(1)(ii), I(2), Table G-16a, Table A-1

Problem 38 · Exposure to a mixture

ENV-E-02 · Environmental and Occupational Health · Select one

A worker's 8-hour exposures to three contaminants with additive effects are: A at 60 ppm (limit 200 ppm), B at 30 ppm (limit 100 ppm), and C at 22.5 ppm (limit 50 ppm). What is the equivalent exposure for the mixture, and does it comply?

A. 1.05; exceeds the limit
B. 0.35; complies
C. 1.05; complies because each substance is under its own limit
D. 112.5 ppm; complies

Show answer and solution

Answer: A. 1.05; exceeds the limit.

Em = C₁/L₁ + C₂/L₂ + C₃/L₃ = 0.30 + 0.30 + 0.45 = 1.05. OSHA's rule is that Em must not exceed 1, so the mixture exposure is over the limit.

Why not the others: B: Uses only one term. C: Ignores the mixture rule. D: Adds concentrations of different substances, which has no meaning against separate limits.

Source: OSHA 29 CFR 1910.1000, §1910.1000(d)(2)(i)

Problem 39 · 8-hour time-weighted average

ENV-E-03 · Environmental and Occupational Health · Select one

A worker is exposed to a solvent at 120 ppm for 2 h, 60 ppm for 3 h, and 0 ppm for 3 h. The 8-hour limit is 50 ppm. What is the 8-hour TWA, and does it comply?

A. 52.5 ppm; exceeds
B. 84 ppm; exceeds
C. 60 ppm; exceeds
D. 42.0 ppm; complies

Show answer and solution

Answer: A. 52.5 ppm; exceeds.

E = (CₐTₐ + C_bT_b + …)/8 = (2 × 120 + 3 × 60 + 3 × 0)/8 = 420/8 = 52.5 ppm, which is above 50 ppm.

Why not the others: B: Averages over only the 5 exposed hours. C: Averages the two concentrations without weighting by time. D: Divides by 10 hours.

Source: OSHA 29 CFR 1910.1000, §1910.1000(d)(1)(i)

Problem 40 · First step when exposure exceeds a limit

ENV-E-04 · Environmental and Occupational Health · Select one

Air sampling shows workers in a degreasing area are exposed to a solvent above its OSHA limit. Under OSHA's air contaminants standard, what is the first required approach?

A. Issue respirators to everyone in the area
B. Determine and implement feasible engineering or administrative controls, and use protective equipment where those can't achieve compliance
C. Increase air sampling frequency and make no other changes
D. Post warning signs

Show answer and solution

Answer: B. Determine and implement feasible engineering or administrative controls, and use protective equipment where those can't achieve compliance.

OSHA requires administrative or engineering controls first whenever feasible. Protective equipment, such as respirators, is used when those controls can't fully achieve compliance, and respirator use must follow 1910.134.

Why not the others: A: Skips the required controls. C: More sampling documents the problem but doesn't reduce exposure. D: Signs don't reduce exposure.

Source: OSHA 29 CFR 1910.1000, §1910.1000(e)

How to read your results

Count your misses by area. If you missed half or more of an area's problems, rebuild its fundamentals before doing timed sets. If you missed one or two, review those methods and move on to mixed practice. Four to eleven problems per area can show you which methods you can't explain yet, but they can't measure an area precisely.

Your percentage here is feedback on these 40 problems, not an NCEES score. NCEES converts your number of correct answers to a scaled score, compares it with a passing standard set by subject-matter experts, and doesn't publish the passing score (NCEES exam scoring). No practice percentage converts into an exam result.

Your 12-week study plan

Example: 12 weeks at about 10 hours a week (120 hours). This is our planning example, weighted toward the larger areas. It isn't an NCEES requirement, and no number of hours guarantees a pass. Each week has the same rhythm: two blocks of concept review, two blocks of solving problems in writing, one handbook search drill, and one error-log session.

Your 12-week study plan
Week	Focus	Do this	Finish the week with
1	Baseline and setup	Confirm your test date and the April 2026 specification. Download the handbook from MyNCEES. Work all 40 problems before reading solutions, spread over several sessions.	An error log and your two weakest areas
2	Water fundamentals	Mass balance, BOD, hydraulics, hydrology. Rework Problems 2, 3, 11, and 12.	A list of each formula with its handbook location
3	Drinking water	Disinfection and CT, hardness and softening, chemical feed, distribution. Rework Problems 4–6.	Five handbook lookups logged
4	Wastewater and stormwater	Clarifiers, SRT, F/M, nitrification, collection, runoff. Rework Problems 7–11. End with a Water-only timed set.	A unit-basis checklist (MLSS vs. MLVSS, mg/L vs. lb/day)
5	Solid and Hazardous Waste	Landfill volume and gas, RCRA characteristics, generator categories, incineration, recycling. Rework Problems 13–18.	Notes on each rule in your own words, with its CFR section
6	Air I	Units and conversions, emission factors, stack sampling, dispersion. Rework Problems 23, 24, 27, and 29.	A conversion chain with units on every line
7	Air II	Control devices in series, ESPs, fabric filters, scrubbers, prevention. Rework Problems 25, 26, 28 and Worked Example 1.	One device-sizing problem solved two ways
8	Site assessment	Sorption, retardation, contaminant mass, exposure and risk. Rework Problems 1, 30, 31, 33, 36 and Worked Example 2.	A risk worksheet showing averaging time for cancer and noncancer
9	Remediation	Feasibility study criteria, pump and treat, natural attenuation, technology selection. Rework Problems 32, 34, and 35.	One remedy comparison against the threshold and balancing criteria
10	Sustainability and health	Life-cycle cost, energy conservation, redundancy, life-cycle assessment; noise, TWA, mixtures, exposure controls. Rework Problems 19–22 and 37–40.	One page of sustainability and health methods
11	Long timed practice	Take a full-length timed practice set that explicitly matches the April 2026 specification. Practice pacing, the break, and handbook searching.	Time per question, flagged questions, and slow lookups
12	Repair, then taper	Rework every logged miss with the solution hidden. Repeat your five slowest lookups. Confirm your appointment, ID, and calculator. Keep the final days light.	A short final review list and an exam-day plan

Adjusting the plan:

Fewer hours a week? Stretch the calendar rather than cutting areas. At 6 hours a week, the same 120 hours takes 20 weeks. That arithmetic doesn't say 120 hours will be enough for you.
Only 6 weeks? Pair the weeks (1 and 2, 3 and 4, and so on), about 20 hours a week. If that isn't realistic, consider a later test date.
Already strong in an area? Keep a short review there and move the saved time to your weakest area. Don't drop an area because you got its sample problems right.

Don't prepare for formulas only

Some questions ask you to choose, compare, or interpret rather than calculate; Problems 15, 16, 22, 28, 31, 32, 33, and 40 are examples here. Build that skill on purpose:

For every calculation, say what the answer means. An R of 3.27 means the plume moves about a third as fast as the water.
For every process, know what it removes, when it works, and its main limitation. An ESP is efficient on fine dust until resistivity gets too high.
For every regulation you practice, read the current rule itself rather than a summary in a prep book. Rules change; the eCFR shows the current text.
Make compare-and-choose notes: one technology against another, one exposure route against another, a threshold criterion against a balancing one.

Turn each miss into your next study task

Turn each miss into your next study task
What went wrong	Next step	How you know it's fixed
Picked the wrong method	Write in one sentence what the problem asks for and which model fits, then rework it with one input changed	You can say why the wrong method doesn't apply
Used the wrong rule or missed a condition	Find the rule in the handbook or the regulation, and read the definitions and limits around it	You can name the condition that triggers the rule
Mixed up units or bases	Write the conversion chain with units on every line (ppm vs. µg/m³, MLSS vs. MLVSS, dose vs. risk)	Units cancel to the one the problem asks for
Arithmetic or sign error	Redo it from scratch, then check the size and direction of the answer	The check agrees without copying your first attempt
Answered a different quantity	Circle what the question asks for before you start (DRE or penetration, CDI or risk, T or CT)	You stop at the right step
Took too long	Note where the time went, then practice flagging and returning during a timed set	The same setup or lookup takes less time next round

Error log fields: problem or topic · your answer · first wrong step · error type (from the table above) · corrected method · reference and section · result with one input changed · date to review again. Log lucky guesses too. A right answer you can't explain is a miss waiting to happen.

If you're preparing for a retake

If you didn't pass, NCEES gives you a diagnostic report showing areas of relative strength and weakness. It doesn't show the passing score or how many questions you missed. Use it with your error log to pick your first two focus areas, considering both the relative weakness and the size of each area's published question range. NCEES allows one attempt per quarterly testing window and no more than three in any 12 months; your board may be stricter (Examinee Guide, pp. 5 and 14). See NCEES exam retake rules and how NCEES exam results work.

Exam-day rules that change how you practice

About 6 minutes per question on average. That's 480 minutes ÷ 80 questions. It's a budgeting guide, not a per-question limit.
Two sections, one clock. After roughly half the questions, you review and submit them, and you can't go back. Clear your flagged questions before you submit.
Breaks. The optional 50-minute break comes after you submit the first section. Unused break time doesn't add to your exam time, and unscheduled breaks come out of your exam time.
No penalty for wrong answers, so answer everything.
Calculator. One NCEES-approved model. For 2026 exams, that's the HP 33s or HP 35s, or any Casio fx-115, Casio fx-991, TI-30X, or TI-36X model (NCEES memo, Oct. 2025, p. 1). An on-screen TI-30XS is also available. Recheck NCEES's list if your test date moves into a later year.
Scratch work. You get two reusable booklets and three markers, not pencils and paper.

Source: NCEES Examinee Guide, May 2026, pp. 8–12.

Quick answers

Is PE Environmental the same as PE Civil: Water Resources and Environmental?

No. They're separate NCEES exams with separate specifications. PE Environmental gives air, waste, sustainability, site remediation, and occupational health their own areas. PE Civil: WRE follows a civil blueprint with areas such as project sitework, soil mechanics, and materials, and supplies design standards. Your licensing board decides which exams it accepts for your license. For the civil exam, see PE Civil Water Resources and Environmental exam prep.

What score do I need to pass?

NCEES doesn't publish a passing score. It scales your number of correct answers and compares that with a passing standard, so a claim that you need "70%" has no official basis (NCEES exam scoring).

What's the pass rate?

NCEES's July 2026 update lists 73% for 415 first-time PE Environmental examinees and 54% for 118 repeat examinees (NCEES PE Environmental page, Scoring & Pass Rates). That reporting period includes exams given before and after the April 2026 specification change. The rates describe that group of test takers, not your odds.

How long should I study?

There's no official answer. Our plan uses about 120 hours over 12 weeks as an example; your starting point, your work experience, and the areas you're weakest in matter more than any number.

Do I need a prep course?

No, a course is optional. Whatever you use should be written to the April 2026 specifications and paired with handbook practice.

Does passing the PE give me a license?

No. Your state or territorial licensing board issues the license, and passing the exam is one of its requirements. NCEES designs the PE for engineers with at least four years of post-college experience, but your board decides when you can sit (NCEES PE Environmental page; Examinee Guide, p. 2).

How do I register?

Start with your board: check its approval process and any separate application, then register and pay the $400 exam fee in MyNCEES. See NCEES exam registration steps or find your licensing board.

Can I get testing accommodations?

You can request accommodations, but you have to indicate the need during registration. Approval depends on eligibility and supporting documentation. See how to request NCEES exam accommodations.

Haven't passed the FE yet?

Start with FE Environmental exam prep and the free FE Environmental practice test. For other PE disciplines, see PE exam prep.

Sources

Exam facts on this page were last verified October 9, 2026, against the NCEES sources below. We compared the April 2019 and April 2026 specifications side by side. The original problems and worked examples were checked against the teaching sources cited with each one, and every numerical answer was recalculated. They have not been reviewed by NCEES or a licensed professional engineer.

NCEES

PE Environmental exam page (format, fee, references, results, pass rates)
PE Environmental specifications, effective April 2026
PE Environmental specifications, April 2019 (historical)
NCEES Examinee Guide, May 2026
NCEES exam scoring
NCEES computer-based testing
NCEES memo to member boards, October 2025

Teaching sources for the problems and examples

These teaching sources explain the methods. None of them is supplied on the exam; the NCEES handbook is the only reference you'll have.

Castleport Test Prep Editorial Team. This guide was developed with AI-assisted research and editing. Source checking is not professional engineering review. See how Castleport uses sources and AI assistance.

Castleport Test Prep is an independent exam prep publisher. It is not affiliated with, endorsed by, or approved by the National Council of Examiners for Engineering and Surveying (NCEES). Exam and credential names identify their subjects; trademarks belong to their respective owners. The worked examples and practice problems are original and unofficial, and nothing on this page guarantees a passing result or a license.