Free PE Environmental Practice Test
This free 80-question practice test follows the six knowledge areas and question ranges in the NCEES PE Environmental specifications effective April 2026. Every question is original and unofficial, written by Castleport rather than taken from an NCEES exam, and each one has a worked solution.
Practice questions
Water (Questions 1–24)
Question 1: Reactor type matters
Water · Water engineering applications · Multiple choice
A pollutant decays by first-order kinetics with k = 0.30 per day. A lagoon modeled as a single completely mixed reactor (CSTR) at steady state has a hydraulic retention time of 10 days. Influent concentration is 200 mg/L.
What is the effluent concentration?
Show answer and worked solution
Answer: B. 50 mg/L
Steady-state CSTR mass balance with first-order decay: C = C₀ ÷ (1 + kθ) = 200 ÷ (1 + 0.30 × 10) = 50 mg/L.
Why the other choices are wrong
- A. That's the plug-flow result, C₀·e^(−kθ). Same k and θ, but a plug-flow reactor removes more because nothing short-circuits.
- C. Divides by kθ alone and drops the 1.
- D. Applies one day of 30% decay.
Basis: Derived: steady-state CSTR mass balance with first-order reaction.
Question 2: Ideal settling removal
Water · Water engineering applications · Multiple choice
An ideal rectangular settling basin has an overflow rate of 1.2 m/h. Influent particles are evenly distributed over the inlet depth, and flow is uniform with no short-circuiting. One particle class settles at 0.9 m/h.
What fraction of that particle class is removed?
Show answer and worked solution
Answer: B. 75%
In an ideal basin, particles that settle at least as fast as the overflow rate are fully removed. Slower particles are removed in proportion to their settling velocity: fraction removed = vs ÷ v₀ = 0.9 ÷ 1.2 = 0.75.
The slower particles that start low enough in the water column still reach the floor before the outlet.
Why the other choices are wrong
- A. That's the fraction that escapes.
- C. Full removal needs vs ≥ 1.2 m/h.
- D. Treats the overflow rate as a hard cutoff and ignores particles that enter near the bottom.
Basis: Derived: ideal discrete-settling basin model under the assumptions stated in the stem.
Question 3: Groundwater travel time
Water · Water engineering applications · Multiple choice · SI
A sandy aquifer has hydraulic conductivity K = 25 m/day, effective porosity 0.25, and a hydraulic gradient of 0.004. A conservative tracer is released at an upgradient well.
About how long does it take to reach a monitoring well 200 m downgradient?
Show answer and worked solution
Answer: B. 500 days
Darcy flux: q = Ki = 25 × 0.004 = 0.10 m/day.
Water moves only through connected pores, so the average linear (seepage) velocity is v = q ÷ nₑ = 0.10 ÷ 0.25 = 0.40 m/day.
Travel time = 200 ÷ 0.40 = 500 days.
Why the other choices are wrong
- A. Treats K as a velocity and leaves out the gradient.
- C. Uses the Darcy flux; that spreads flow over the whole cross-section, including solids.
- D. Multiplies by porosity instead of dividing.
Basis: Derived: Darcy's law and seepage velocity, v = Ki ÷ nₑ.
Question 4: Alkalinity as CaCO₃
Water · Water engineering applications · Multiple choice
A water's alkalinity is entirely bicarbonate at 1.2 meq/L. Use 50 mg CaCO₃ per meq.
What is the alkalinity expressed as mg/L as CaCO₃?
Show answer and worked solution
Answer: B. 60 mg/L as CaCO₃
Expressing a constituent "as CaCO₃" means multiplying its milliequivalents by the equivalent weight of CaCO₃: 1.2 × 50 = 60 mg/L as CaCO₃.
Why the other choices are wrong
- A. Reports meq/L as if it were mg/L.
- C. That's the mass of bicarbonate ion itself (1.2 × 61), not the CaCO₃ basis.
- D. Uses CaCO₃'s molar mass (100) instead of its equivalent weight (50).
Basis: Derived: equivalent-weight conversion with the factor given in the stem.
Question 5: Monod growth rate
Water · Water engineering applications · Multiple choice
A heterotrophic culture follows Monod kinetics, μ = μmax·S ÷ (Ks + S), with μmax = 6.0 per day and half-saturation constant Ks = 20 mg/L. The substrate concentration in the reactor is 10 mg/L.
What is the specific growth rate?
Show answer and worked solution
Answer: A. 2.0 per day
μ = 6.0 × 10 ÷ (20 + 10) = 2.0 per day.
Check: when S = Ks, μ = μmax/2 = 3.0. Here S is below Ks, so μ must be below 3.0.
Why the other choices are wrong
- B. Assumes S equals Ks.
- C. Swaps S and Ks in the numerator.
- D. Ignores substrate limitation; μmax is reached only when S ≫ Ks.
Basis: Derived: Monod equation given in the stem.
Question 6: Hypochlorite feed rate
Water · Drinking water · Multiple choice · USCS
A plant treats 2.5 MGD. Jar testing shows a chlorine demand of 3.2 mg/L, and the target free residual is 0.8 mg/L. Chlorine is fed as calcium hypochlorite with 65% available chlorine. Use lb/day = MGD × mg/L × 8.34.
How many pounds of calcium hypochlorite are needed per day?
Show answer and worked solution
Answer: D. 128 lb/day
Dose = demand + residual = 3.2 + 0.8 = 4.0 mg/L.
Pure chlorine needed: 2.5 × 4.0 × 8.34 = 83.4 lb/day.
Only 65% of the product is available chlorine: 83.4 ÷ 0.65 = 128 lb/day.
Why the other choices are wrong
- A. Multiplies by 0.65. A weaker product means you feed more of it, not less.
- B. Pounds of pure chlorine. The question asks for pounds of product.
- C. Leaves out the residual and doses only the demand.
Basis: Florida DEP, Wastewater Treatment Formulas (April 2024), "Pounds" formula; dose = demand + residual given in the stem.
Question 7: Basin detention time
Water · Drinking water · Multiple choice · USCS
A plant's 4.0 MGD flow is split equally between two identical rectangular sedimentation basins. Each basin is 120 ft long, 30 ft wide, and has a 12-ft water depth. Use 7.48 gal/ft³.
What is the theoretical detention time in each basin?
Show answer and worked solution
Answer: C. 3.88 h
Volume of one basin: 120 × 30 × 12 = 43,200 ft³ × 7.48 = 323,136 gal.
Flow to one basin: 4.0 ÷ 2 = 2.0 MGD = 83,333 gal/h.
Detention time = V/Q = 323,136 ÷ 83,333 = 3.88 h.
Why the other choices are wrong
- A. Divides cubic feet by gallons per hour; the 7.48 conversion is missing.
- B. Sends the full 4.0 MGD through one basin.
- D. Doubles the volume instead of halving the flow.
Basis: Derived: theoretical detention time = volume ÷ flow.
Question 8: Filter rate with one out of service
Water · Drinking water · Multiple choice · USCS
A 3.0 MGD plant has four identical gravity filters, each 20 ft × 25 ft. One filter is out of service for backwashing, and the other three share the full plant flow equally.
What is the filtration rate on each operating filter?
Show answer and worked solution
Answer: B. 1.39 gpm/ft²
Plant flow: 3,000,000 ÷ 1,440 = 2,083 gpm.
Operating area: 3 × (20 × 25) = 1,500 ft².
Rate = 2,083 ÷ 1,500 = 1.39 gpm/ft².
Design checks use the reduced number of filters because that's when loading peaks.
Why the other choices are wrong
- A. Uses all four filters.
- C. Uses only two filters.
- D. Puts the whole flow on one filter.
Basis: Derived: filtration rate = flow ÷ operating filter area.
Question 9: Log inactivation from CT
Water · Drinking water · Multiple choice
A disinfection segment achieves a calculated CT of 60 mg·min/L. Under the segment's temperature and pH, the CT required for 3-log Giardia inactivation (CT₉₉.₉) is 90 mg·min/L. Use estimated log inactivation = 3.0 × (CTcalc ÷ CT₉₉.₉).
What log inactivation does the segment achieve?
Show answer and worked solution
Answer: B. 2.0-log
3.0 × (60 ÷ 90) = 2.0-log (99% inactivation).
The ratio scales linearly here because CT₉₉.₉ is the CT for exactly 3 logs.
Why the other choices are wrong
- A. Stops at the CT ratio and skips the × 3.0.
- C. Assumes any CT credit meets the 3-log benchmark.
- D. Inverts the ratio.
Basis: EPA, Disinfection Profiling and Benchmarking Technical Guidance Manual, CT and log-inactivation concepts; ratio equation given in the stem.
Question 10: Distribution main velocity
Water · Drinking water · Multiple choice · USCS
A 12-in. (inside diameter) water main carries 1,500 gpm. Use 448.8 gpm per ft³/s.
What is the average velocity?
Show answer and worked solution
Answer: C. 4.26 ft/s
Q = 1,500 ÷ 448.8 = 3.342 ft³/s.
A = π(1.0 ft)² ÷ 4 = 0.785 ft².
V = Q ÷ A = 3.342 ÷ 0.785 = 4.26 ft/s.
Why the other choices are wrong
- A. Computes the area in square inches but leaves the flow in ft³/s.
- B. Uses the diameter as the radius, quadrupling the area.
- D. Converts gpm to gallons per second and treats them as ft³/s, skipping the 7.48 gal/ft³ step.
Basis: Derived: continuity, V = Q ÷ A.
Question 11: Reverse osmosis concentrate
Water · Drinking water · Multiple choice
A reverse osmosis unit receives 1.25 MGD of feed at 800 mg/L TDS and produces 1.00 MGD of permeate at 16 mg/L TDS. Assume steady state and no other outflows.
What is the TDS of the concentrate?
Show answer and worked solution
Answer: B. 3,936 mg/L
Concentrate flow = 1.25 − 1.00 = 0.25 MGD.
Salt balance: 1.25 × 800 = 1.00 × 16 + 0.25 × Cc → Cc = (1,000 − 16) ÷ 0.25 = 3,936 mg/L.
Why the other choices are wrong
- A. Treats the feed flow as 1.0 MGD (800 ÷ 0.25) instead of 1.25 MGD.
- C. Assumes the permeate has zero TDS.
- D. Adds the permeate salt instead of subtracting it.
Basis: Derived: steady-state flow and mass balance around the membrane.
Question 12: BOD₅ from a dilution test
Water · Wastewater · Numeric entry
A 6.0-mL wastewater sample is diluted to 300 mL in a BOD bottle with unseeded dilution water. Initial DO is 8.6 mg/L and DO after 5 days at 20 °C is 4.1 mg/L. Use BOD₅ = (D₁ − D₂) ÷ P, where P is the decimal fraction of sample in the bottle.
What is the BOD₅, in mg/L? Round to the nearest whole number.
Enter a number in mg/L.
Show answer and worked solution
Answer: 225 mg/L
P = 6.0 ÷ 300 = 0.020.
BOD₅ = (8.6 − 4.1) ÷ 0.020 = 4.5 ÷ 0.020 = 225 mg/L.
Accepted: 224 to 226.
Common mistake: Reporting the 4.5 mg/L DO drop without correcting for dilution, or dividing the initial DO alone (8.6 ÷ 0.02 = 430 mg/L).
Basis: Derived: dilution-correction equation given in the stem.
Question 13: Ultimate BOD
Water · Wastewater · Multiple choice
A wastewater has BOD₅ = 180 mg/L. Its first-order deoxygenation rate is k = 0.23 per day (base e). Use BODt = L₀(1 − e^(−kt)).
What is the ultimate carbonaceous BOD, L₀?
Show answer and worked solution
Answer: C. 263 mg/L
1 − e^(−0.23 × 5) = 1 − e^(−1.15) = 1 − 0.3166 = 0.6834.
L₀ = 180 ÷ 0.6834 = 263 mg/L.
Sanity check: ultimate BOD must exceed BOD₅, because only part of the demand is exerted in 5 days.
Why the other choices are wrong
- A. Multiplies by 0.6834 instead of dividing. L₀ can't be smaller than BOD₅.
- B. Uses 10^(−kt), treating a base-e rate as base 10.
- D. Divides by e^(−kt), the fraction remaining, instead of the fraction exerted.
Basis: Derived: first-order BOD model given in the stem.
Question 14: Sludge volume index
Water · Wastewater · Numeric entry
A 1-L mixed liquor sample settles to 270 mL in 30 minutes. The mixed liquor suspended solids concentration is 2,700 mg/L. Use SVI (mL/g) = settled volume (mL/L) × 1,000 ÷ MLSS (mg/L).
What is the SVI, in mL/g?
Enter a number in mL/g.
Show answer and worked solution
Answer: 100 mL/g
SVI = 270 × 1,000 ÷ 2,700 = 100 mL/g.
The 1,000 converts mg to g. Each gram of solids occupies 100 mL after 30 minutes of settling.
Accepted: 100 only.
Common mistake: Leaving out the 1,000 gives 0.1, which has no practical meaning as an SVI.
Basis: Derived: equation given in the stem.
Question 15: Primary clarifier overflow rate
Water · Wastewater · Multiple choice · USCS
A 5.0 MGD plant splits flow equally between two circular primary clarifiers, each 70 ft in diameter.
What is the surface overflow rate?
Show answer and worked solution
Answer: C. 650 gpd/ft²
Area of one clarifier: π(70)² ÷ 4 = 3,848 ft². Total area: 7,697 ft².
Overflow rate = 5,000,000 ÷ 7,697 = 650 gpd/ft² (each clarifier gets 2.5 MGD over 3,848 ft², which gives the same answer).
Why the other choices are wrong
- A. Uses πD² without dividing by 4.
- B. Halves the flow and also uses the combined area, counting the split twice.
- D. Puts the full plant flow on one clarifier.
Basis: Derived: overflow rate = flow ÷ surface area.
Question 16: Alkalinity consumed by nitrification
Water · Wastewater · Multiple choice
An activated sludge plant will fully nitrify 25 mg/L of ammonia-N. Influent alkalinity is 200 mg/L as CaCO₃. Use 7.14 mg alkalinity (as CaCO₃) consumed per mg NH₄⁺-N nitrified, and ignore other sources and sinks. The operating target is a residual of at least 50 mg/L as CaCO₃.
What is the residual alkalinity, and is supplemental alkalinity needed?
Show answer and worked solution
Answer: A. 21.5 mg/L; yes, add alkalinity
Alkalinity consumed: 7.14 × 25 = 178.5 mg/L as CaCO₃.
Residual: 200 − 178.5 = 21.5 mg/L, below the 50 mg/L target, so supplemental alkalinity is needed to keep pH from dropping and stalling nitrification.
Why the other choices are wrong
- B. That's the amount consumed, reported as if it were left over.
- C. Nitrification releases acid and consumes alkalinity.
- D. The calculation is right, but it ignores the stated 50 mg/L target.
Basis: Derived: stoichiometric factor and target given in the stem.
Question 17: Gravity sewer full-flow capacity
Water · Wastewater · Multiple choice · USCS
A 15-in. circular gravity sewer is laid at a slope of 0.002 ft/ft with Manning's n = 0.013. Use Manning's equation in U.S. customary units, Q = (1.49/n)·A·R^(2/3)·S^(1/2). For a circular pipe flowing full, R = D/4.
What is the full-flow capacity?
Show answer and worked solution
Answer: C. 2.90 ft³/s
D = 1.25 ft. A = π(1.25)² ÷ 4 = 1.227 ft². R = 1.25 ÷ 4 = 0.3125 ft; R^(2/3) = 0.4605.
Q = (1.49 ÷ 0.013) × 1.227 × 0.4605 × (0.002)^0.5 = 114.6 × 1.227 × 0.4605 × 0.0447 = 2.90 ft³/s.
Full-flow velocity is 2.90 ÷ 1.227 = 2.4 ft/s.
Why the other choices are wrong
- A. Uses S instead of √S.
- B. Uses the SI coefficient 1.0 with U.S. customary inputs.
- D. Uses R = D/2. The hydraulic radius of a full circle is A/P = D/4.
Basis: FHWA HDS-4, Introduction to Highway Hydraulics (2008), §4.3.1 Manning's equation, pp. 4-3 to 4-4.
Question 18: Sludge volume from solids
Water · Wastewater residuals · Multiple choice · USCS
A thickener produces 2,000 lb/day of dry solids at 4.0% solids by weight. The thickened sludge has a specific gravity of 1.02. Use 8.34 lb/gal for water.
What volume of sludge is produced per day?
Show answer and worked solution
Answer: C. 5,880 gal/day
Wet sludge mass: 2,000 ÷ 0.04 = 50,000 lb/day.
Sludge density: 8.34 × 1.02 = 8.51 lb/gal.
Volume: 50,000 ÷ 8.51 = 5,880 gal/day.
Why the other choices are wrong
- A. Multiplies by the solids fraction instead of dividing.
- B. Divides by 7.48 again; that's about 786 ft³, not gallons.
- D. Ignores the specific gravity and uses the density of water.
Basis: Derived: solids mass balance and sludge density.
Question 19: Digester volatile solids reduction
Water · Wastewater residuals · Multiple choice
Raw sludge fed to an anaerobic digester is 72% volatile solids (VS). Digested sludge is 55% VS. Assume fixed (inorganic) solids pass through unchanged. Use VS reduction = (VSin − VSout) ÷ (VSin − VSin × VSout), with fractions as decimals.
What is the volatile solids reduction?
Show answer and worked solution
Answer: C. 52.5%
(0.72 − 0.55) ÷ (0.72 − 0.72 × 0.55) = 0.17 ÷ 0.324 = 0.525.
The equation works because fixed solids are conserved. The digested sludge has a lower VS percentage partly because there's less total sludge, so the drop in percentage points understates the destruction.
Why the other choices are wrong
- A. The difference in percentage points.
- B. Divides that difference by 0.72 alone; it ignores the change in total solids.
- D. The ratio 0.55 ÷ 0.72, which has no meaning here.
Basis: Derived: fixed-solids mass balance; equation given in the stem.
Question 20: Picking the design intensity
Water · Stormwater and water resources · Multiple choice · USCS
An 18-acre drainage area has a runoff coefficient of 0.65 and a time of concentration of 30 minutes. The design-storm IDF data are:
| Duration (min) | 10 | 20 | 30 | 60 |
|---|---|---|---|---|
| Intensity (in./h) | 5.1 | 3.9 | 3.2 | 2.1 |
Use the rational method, Q = CiA (Q in cfs, i in in./h, A in acres).
What is the peak runoff?
Show answer and worked solution
Answer: B. 37.4 cfs
The rational method uses the intensity for a duration equal to the time of concentration: 30 minutes, so i = 3.2 in./h.
Q = 0.65 × 3.2 × 18 = 37.4 cfs.
Why the other choices are wrong
- A. Uses the 60-minute intensity.
- C. Uses the 20-minute intensity.
- D. Uses the 10-minute intensity, the largest in the table, rather than the one that matches tc.
Basis: FHWA HDS-4, Introduction to Highway Hydraulics (2008), §2.5.1, Eq. 2.1 (Ku = 1.0 for U.S. customary units).
Question 21: Daily phosphorus load
Water · Stormwater and water resources · Multiple choice · SI
A river carries 2.5 m³/s with a total phosphorus concentration of 0.12 mg/L. Assume both are constant over the day.
What is the phosphorus load?
Show answer and worked solution
Answer: C. 25.9 kg/day
1 mg/L = 1 g/m³. Load = 2.5 m³/s × 0.12 g/m³ × 86,400 s/day = 25,920 g/day = 25.9 kg/day.
A handy factor: kg/day = m³/s × mg/L × 86.4.
Why the other choices are wrong
- A. Stops at g/s (0.30) and labels it kg/day.
- B. Uses 8.64 instead of 86.4, a decimal slip.
- D. Converts with 3,600 s/h and skips the gram-to-kilogram step.
Basis: Derived: load = flow × concentration with unit conversion.
Question 22: Water quality volume
Water · Stormwater and water resources · Multiple choice · USCS
A 10-acre development site will be 60% impervious. Size the water quality volume with the simple method: runoff coefficient Rv = 0.05 + 0.009 × I (I = percent impervious), and WQv = P × Rv × A, using a 1.0-in. water quality rainfall depth.
What is the water quality volume?
Show answer and worked solution
Answer: B. 21,400 ft³
Rv = 0.05 + 0.009 × 60 = 0.59.
Runoff depth: 1.0 × 0.59 = 0.59 in.
WQv = (0.59 ÷ 12 ft) × 10 ac × 43,560 ft²/ac = 21,400 ft³ (0.49 ac-ft).
Why the other choices are wrong
- A. A decimal slip by a factor of 10.
- C. Uses the full 1.0 in. of rain as runoff (Rv = 1).
- D. Uses 0.59 ft instead of 0.59 in.; the 12 is missing.
Basis: Derived: equations given in the stem.
Question 23: Detention outlet orifice
Water · Stormwater and water resources · Multiple choice · USCS
A detention pond drains through a 6-in. diameter orifice with discharge coefficient C = 0.61. The water surface is 4.0 ft above the orifice centerline, and the outlet is free (not submerged). Use Q = C·A·√(2gh), g = 32.2 ft/s².
What is the discharge?
Show answer and worked solution
Answer: B. 1.92 cfs
A = π(0.5 ft)² ÷ 4 = 0.196 ft².
√(2 × 32.2 × 4.0) = 16.05 ft/s.
Q = 0.61 × 0.196 × 16.05 = 1.92 cfs.
Why the other choices are wrong
- A. Uses half the head.
- C. Leaves out the discharge coefficient (C = 1).
- D. Uses πd² without dividing by 4.
Basis: Derived: orifice equation given in the stem.
Question 24: Reservoir water balance
Water · Stormwater and water resources · Multiple choice · SI
A reservoir with a 4.0 km² surface receives a steady inflow of 3.0 m³/s and releases 2.5 m³/s. Evaporation is 6 mm/day. Ignore rainfall, seepage, and change in surface area.
What is the daily change in storage?
Show answer and worked solution
Answer: A. +19,200 m³/day
Net flow: (3.0 − 2.5) × 86,400 = +43,200 m³/day.
Evaporation: 0.006 m × 4,000,000 m² = 24,000 m³/day.
Change in storage: 43,200 − 24,000 = +19,200 m³/day.
Why the other choices are wrong
- B. Ignores evaporation.
- C. Counts evaporation but forgets the inflow and release.
- D. Uses 6 m/day instead of 6 mm/day.
Basis: Derived: continuity, inflow − outflow = change in storage.
Solid and Hazardous Waste (Questions 25–37)
Question 25: Toxicity characteristic codes
Solid and Hazardous Waste · Engineering applications · Multiple choice
A waste's TCLP extract contains benzene 0.62 mg/L, lead 3.1 mg/L, chromium 4.8 mg/L, and methyl ethyl ketone 150 mg/L. No other Table 1 contaminants are detected. The waste is not listed and shows no other characteristic.
Which toxicity characteristic code(s) apply?
Show answer and worked solution
Answer: B. D018 only
Compare each extract concentration with its regulatory level in Table 1. A waste exhibits toxicity at a concentration equal to or greater than the level.
- Benzene (D018): 0.62 ≥ 0.5 mg/L → toxic
- Lead (D008): 3.1 < 5.0 mg/L → not toxic
- Chromium (D007): 4.8 < 5.0 mg/L → not toxic
- Methyl ethyl ketone (D035): 150 < 200 mg/L → not toxic
Why the other choices are wrong
- A. Misses benzene.
- C. Lead is below its 5.0 mg/L level.
- D. Treats any detection as a characteristic. The test is the regulatory level, not detection.
Basis: 40 CFR 261.24 (eCFR), paragraph (a) and Table 1.
Question 26: When the waste is the extract
Solid and Hazardous Waste · Engineering applications · Multiple choice
A generator needs to know whether an aqueous waste exhibits the toxicity characteristic. The waste contains less than 0.5% filterable solids.
Under 40 CFR 261.24, what is treated as the extract for comparison with the Table 1 levels?
Show answer and worked solution
Answer: B. The waste itself, after filtering by the Method 1311 procedure
For a waste with less than 0.5% filterable solids, the rule says the waste itself, after filtering under Method 1311, is the extract. There's nothing meaningful to leach, so the liquid is compared directly with Table 1.
Why the other choices are wrong
- A. There are too few solids for an extraction to be the defined approach.
- C. The toxicity characteristic still applies; only the extract definition changes.
- D. The regulation specifies filtering, not a total digestion.
Basis: 40 CFR 261.24(a) (eCFR).
Question 27: Generator category
Solid and Hazardous Waste · Hazardous, medical, and radioactive waste · Multiple choice
In one calendar month, a facility generates 60 kg of spent solvent and 45 kg of corrosive waste (both non-acute hazardous waste) plus 0.5 kg of acute hazardous waste. It generates no spill-cleanup residues.
Under the federal definitions, what is its generator category for that month?
Show answer and worked solution
Answer: B. Small quantity generator
Add all non-acute hazardous waste generated in the month: 60 + 45 = 105 kg. That's more than 100 kg and less than 1,000 kg.
Acute hazardous waste is 0.5 kg, at or below 1 kg.
Both conditions match the small quantity generator definition. States can be stricter than the federal rule.
Why the other choices are wrong
- A. Looks at each stream separately; 60 kg alone would fit, but the monthly total doesn't.
- C. LQG starts at 1,000 kg of non-acute waste or more than 1 kg of acute waste.
- D. The definitions count the total generated in the calendar month.
Basis: 40 CFR 260.10 (eCFR), definitions of very small, small, and large quantity generator.
Question 28: Which wastes are characteristic?
Solid and Hazardous Waste · Engineering applications · Select all that apply
Each item below is a solid waste that is not listed and not excluded. Select all that exhibit the ignitability or corrosivity characteristic.
Show answer and worked solution
Answer: A, B, E
Credit requires exactly these three.
- A. Corrosive: aqueous with pH ≥ 12.5 (D002).
- B. Ignitable: liquid with flash point below 60 °C (140 °F). The low-alcohol aqueous exception doesn't apply because it contains no water (D001).
- E. Corrosive: aqueous with pH ≤ 2. The limit is inclusive.
Why the other statements are false
- C. pH 2.4 is above the 2.0 limit.
- D. 65 °C is above the 60 °C flash-point limit.
Basis: 40 CFR 261.21 and 261.22 (eCFR), ignitability (a)(1) and corrosivity (a)(1).
Question 29: Universal waste categories
Solid and Hazardous Waste · Hazardous, medical, and radioactive waste · Select all that apply
Under the federal definition of universal waste in 40 CFR 260.10, which of these hazardous wastes can be managed under the universal waste rules? Select all that apply.
Show answer and worked solution
Answer: A, B, C
Credit requires exactly these three.
The federal universal waste categories are batteries, pesticides, mercury-containing equipment, lamps, and aerosol cans. Universal waste rules ease storage and handling for common wastes generated by many businesses. States may add categories.
Why the other statements are false
- D. Spent solvents aren't a universal waste category; they're managed under the full hazardous waste rules.
- E. Contaminated soil isn't a universal waste category.
Basis: 40 CFR 260.10 (eCFR), definition of universal waste.
Question 30: Solvent mass balance
Solid and Hazardous Waste · Engineering applications · Multiple choice · SI
A degreasing operation buys 1,000 kg of solvent in a month and starts and ends the month with the same inventory. Spent solvent shipped off-site contains 520 kg of solvent, and an estimated 30 kg leaves as residue on cleaned parts. Assume everything else evaporates.
What are the month's air emissions of solvent?
Show answer and worked solution
Answer: A. 450 kg
Input − outputs accounted for = emissions: 1,000 − 520 − 30 = 450 kg.
Mass balances like this are a common way to estimate fugitive emissions when they can't be measured directly.
Why the other choices are wrong
- B. Forgets the solvent leaving on parts.
- C. That's the solvent in the shipped waste.
- D. Subtracts only the parts residue.
Basis: Derived: steady-state mass balance.
Question 31: Clay liner seepage
Solid and Hazardous Waste · Engineering applications · Multiple choice · SI
A compacted clay liner is 0.60 m thick with hydraulic conductivity 1 × 10⁻⁷ cm/s. Leachate stands 0.30 m deep on top of it. Assume saturated, steady, one-dimensional flow with atmospheric pressure at the liner's base, so i = (h + L) ÷ L.
What is the seepage rate per hectare of liner?
Show answer and worked solution
Answer: C. 1.30 m³/day
i = (0.30 + 0.60) ÷ 0.60 = 1.5.
K = 1 × 10⁻⁷ cm/s = 1 × 10⁻⁹ m/s, so q = Ki = 1.5 × 10⁻⁹ m/s.
Flow per hectare: 1.5 × 10⁻⁹ × 10,000 m² × 86,400 s/day = 1.30 m³/day.
Why the other choices are wrong
- A. Uses i = h ÷ L = 0.5 and ignores the liner's own thickness.
- B. Uses i = 1 (unit gradient).
- D. Leaves K in cm/s but treats it as m/s, a factor of 100.
Basis: Derived: Darcy's law with the gradient definition given in the stem.
Question 32: Landfill airspace per year
Solid and Hazardous Waste · Municipal and industrial solid waste · Multiple choice · SI
A community of 50,000 people landfills 2.0 kg of municipal solid waste per person per day. Waste is compacted in place to 650 kg/m³. Daily and intermediate cover use one volume of soil for every four volumes of compacted waste.
How much landfill airspace is consumed per year?
Show answer and worked solution
Answer: C. 70,200 m³
Waste mass: 50,000 × 2.0 × 365 = 36,500,000 kg/yr.
Compacted waste volume: 36,500,000 ÷ 650 = 56,154 m³.
A 4:1 waste-to-cover ratio adds one-quarter of the waste volume: 56,154 × 1.25 = 70,200 m³.
Why the other choices are wrong
- A. Multiplies by 0.8, subtracting the cover instead of adding it.
- B. Compacted waste alone; it leaves out cover soil.
- D. Adds half the waste volume as cover (a 2:1 ratio).
Basis: Derived: mass ÷ density, plus cover by the stated volume ratio.
Question 33: Collection truck loads
Solid and Hazardous Waste · Municipal and industrial solid waste · Numeric entry · USCS
A town generates 60 tons (2,000 lb/ton) of municipal solid waste per day. Its compactor trucks have 20-yd³ bodies, and waste is compacted in the truck to 600 lb/yd³. Assume every truck leaves full.
How many truckloads are needed per day?
Enter a number.
Show answer and worked solution
Answer: 10 loads
Waste per day: 60 × 2,000 = 120,000 lb.
Payload per truck: 20 × 600 = 12,000 lb.
Loads: 120,000 ÷ 12,000 = 10.
Accepted: 10 only.
Common mistake: Using 1,000 lb per ton gives 5 loads; a U.S. short ton is 2,000 lb.
Basis: Derived: mass ÷ payload per load.
Question 34: Heating value of mixed waste
Solid and Hazardous Waste · Municipal and industrial solid waste · Multiple choice · USCS
A waste stream is 40% paper (7,200 Btu/lb), 15% plastics (14,000 Btu/lb), 20% food waste (2,000 Btu/lb), and 25% inert material (0 Btu/lb), all by weight on the same as-received basis.
What is the composite heating value?
Show answer and worked solution
Answer: A. 5,380 Btu/lb
Mass-weighted average: 0.40 × 7,200 + 0.15 × 14,000 + 0.20 × 2,000 + 0.25 × 0 = 2,880 + 2,100 + 400 = 5,380 Btu/lb.
Why the other choices are wrong
- B. Simple average of the four values, ignoring mass fractions.
- C. Simple average of the three combustible values, dropping the inerts.
- D. Adds the heating values.
Basis: Derived: mass-weighted average on a consistent basis.
Question 35: Diversion rate
Solid and Hazardous Waste · Municipal and industrial solid waste · Multiple choice
A county's MSW program handles 120,000 tons in a year: 30,000 tons recycled, 12,000 tons composted, and 78,000 tons landfilled. Define diversion rate as tons recycled plus composted divided by total tons generated.
What is the diversion rate?
Show answer and worked solution
Answer: B. 35%
(30,000 + 12,000) ÷ 120,000 = 35%.
Why the other choices are wrong
- A. Counts recycling only.
- C. Divides by tons landfilled instead of total generated.
- D. That's the share landfilled.
Basis: Derived: definition given in the stem.
Question 36: Decay-in-storage time
Solid and Hazardous Waste · Hazardous, medical, and radioactive waste · Multiple choice
A short-lived radionuclide in a waste has a half-life of 8.0 days. How long must it be held for its activity to fall to 1/1,000 of the starting activity? Assume simple exponential decay and no other removal.
Show answer and worked solution
Answer: C. 80 days
The number of half-lives n satisfies (1/2)ⁿ = 1/1,000, so n = log₂(1,000) = 9.97.
Time = 9.97 × 8.0 = 79.7 ≈ 80 days.
A useful anchor: 10 half-lives reduce activity by a factor of 1,024.
Why the other choices are wrong
- A. Three half-lives only reduce activity to 1/8.
- B. Uses ln(1,000) instead of log₂(1,000).
- D. Treats decay as linear in the reduction factor.
Basis: Derived: exponential decay, A = A₀(1/2)^(t/t½). This is a calculation only; whether decay-in-storage is allowed for a waste depends on the applicable license and regulations.
Question 37: Incinerator DRE
Solid and Hazardous Waste · Hazardous, medical, and radioactive waste · Multiple choice
A hazardous waste incinerator is fed 500 kg/h of a principal organic hazardous constituent (POHC). The stack emits 0.030 kg/h of that constituent. The permit requires a destruction and removal efficiency (DRE) of at least 99.99% for it.
What is the DRE, and does it meet the permit?
Show answer and worked solution
Answer: B. 99.994%; meets
DRE = (Win − Wout) ÷ Win × 100 = (500 − 0.030) ÷ 500 × 100 = 99.994%, which exceeds 99.99%.
Count nines carefully: 0.030 ÷ 500 = 6 × 10⁻⁵ escapes.
Why the other choices are wrong
- A. Drops a digit; 0.06% would escape at this DRE.
- C. Adds an extra nine; that would allow only 0.003 kg/h out.
- D. The fraction emitted, not the DRE.
Basis: Derived: DRE definition; the 99.99% requirement is stated in the stem.
Sustainability (Questions 38–45)
Question 38: Life cycle assessment phases
Sustainability · Resources conservation · Ordering
Put the four phases of a life cycle assessment in order, first to last.
- Life cycle impact assessment
- Interpretation
- Goal and scope definition
- Life cycle inventory analysis
Show answer and worked solution
Answer: 1. Goal and scope definition → 2. Life cycle inventory analysis → 3. Life cycle impact assessment → 4. Interpretation
Goal and scope set the functional unit and system boundaries. Inventory quantifies inputs and outputs. Impact assessment converts that inventory into impact categories. Interpretation evaluates the results against the goal and scope.
In practice the phases loop back on each other, but this is their defined order. Credit requires the full order.
Basis: ISO 14040:2006, Life cycle assessment: Principles and framework, abstract listing the four phases.
Question 39: Blower upgrade payback
Sustainability · Resources conservation · Multiple choice · USD
An aeration blower draws 150 kW continuously, 8,760 h/yr. A $95,000 upgrade cuts its power draw by 18%. Electricity costs $0.11/kWh. Ignore demand charges, maintenance changes, and discounting.
What is the simple payback period?
Show answer and worked solution
Answer: B. 3.65 years
Energy saved: 150 × 0.18 × 8,760 = 236,520 kWh/yr.
Cost saved: 236,520 × 0.11 = $26,017/yr.
Payback = 95,000 ÷ 26,017 = 3.65 years.
Why the other choices are wrong
- A. Uses the blower's full energy cost as the savings, not 18% of it.
- C. Assumes 8 h/day of operation instead of continuous.
- D. Uses $0.011/kWh, a decimal slip.
Basis: Derived: simple payback = first cost ÷ annual savings.
Question 40: Annual pumping energy
Sustainability · Resources conservation · Multiple choice · SI
A pump delivers 0.050 m³/s of water against a total dynamic head of 40 m, 8,760 h/yr. The combined pump-and-motor (wire-to-water) efficiency is 70%. Use ρ = 1,000 kg/m³ and g = 9.81 m/s².
How much electrical energy does it use per year?
Show answer and worked solution
Answer: C. 245,500 kWh
Water power: ρgQH = 1,000 × 9.81 × 0.050 × 40 = 19,620 W = 19.62 kW.
Electrical input: 19.62 ÷ 0.70 = 28.03 kW.
Annual energy: 28.03 × 8,760 = 245,500 kWh.
Why the other choices are wrong
- A. That's the power in kW, not annual energy.
- B. Uses water power and skips the efficiency.
- D. Divides by 0.70 twice.
Basis: DOE/Hydraulic Institute, Pump Life Cycle Costs (2001), input-power equation with pump and motor efficiency, p. 6.
Question 41: Biogas to electricity
Sustainability · Resources conservation · Multiple choice · SI
An anaerobic digester produces 2,000 m³/day of biogas that is 62% methane by volume. Use a methane lower heating value of 35.8 MJ/m³ and an engine-generator electrical efficiency of 38%. Use 3.6 MJ per kWh.
How much electricity does the engine produce per day?
Show answer and worked solution
Answer: A. 4,690 kWh/day
Methane: 2,000 × 0.62 = 1,240 m³/day.
Fuel energy: 1,240 × 35.8 = 44,392 MJ/day.
Electricity: 44,392 × 0.38 ÷ 3.6 = 4,690 kWh/day.
Why the other choices are wrong
- B. Treats all the biogas as methane.
- C. Skips the engine efficiency.
- D. Leaves the answer in MJ and labels it kWh.
Basis: Derived: energy balance with the values given in the stem.
Question 42: Water loss reduction
Sustainability · Resources conservation · Multiple choice · USCS
A utility's system input is 10.0 MGD and its non-revenue water (input minus billed authorized consumption) is 18% of input. A leak-reduction program cuts non-revenue water to 12% of input with no change in billed use.
How much water does the program save per year?
Show answer and worked solution
Answer: C. 219 MG
Reduction: (0.18 − 0.12) × 10.0 = 0.60 MGD.
Annual savings: 0.60 × 365 = 219 MG.
Why the other choices are wrong
- A. The daily savings, not annual.
- B. Bases the percentages on billed use (8.2 MGD) instead of system input.
- D. Uses the full 18% as if all non-revenue water were eliminated.
Basis: Derived: definitions given in the stem.
Question 43: Annualized life-cycle cost
Sustainability · Resources conservation · Multiple choice · USD
A treatment upgrade costs $500,000 now and $25,000 per year to operate. Use a 20-year life, a 6% interest rate, no salvage value, and the capital recovery factor (A/P) = i(1 + i)ⁿ ÷ [(1 + i)ⁿ − 1].
What is its equivalent annual cost?
Show answer and worked solution
Answer: D. $68,600
(A/P, 6%, 20) = 0.06 × 1.06²⁰ ÷ (1.06²⁰ − 1) = 0.08718.
Annualized capital: 500,000 × 0.08718 = $43,592.
Equivalent annual cost: 43,592 + 25,000 = $68,600.
Why the other choices are wrong
- A. Capital only; it drops the operating cost.
- B. Straight-line capital ($25,000) with no interest, plus O&M.
- C. Simple interest on the capital plus O&M; it never recovers the principal.
Basis: Derived: capital recovery factor given in the stem.
Question 44: Lift station reliability
Sustainability · Resilience · Multiple choice
A lift station has two identical pumps in parallel; either one alone can carry the design flow. Each pump is available 95% of the time. The station also depends on a single standby generator that is available 97% of the time. Assume failures are independent and the station needs the generator and at least one pump.
What is the probability the station is available?
Show answer and worked solution
Answer: C. 0.968
Parallel pumps fail only if both fail: 1 − (0.05)² = 0.9975.
The generator is in series with the pump pair: 0.9975 × 0.97 = 0.968.
The redundancy lesson: the duplicated pumps are no longer the weak link; the single generator is.
Why the other choices are wrong
- A. Treats the two pumps as if both were required (series).
- B. Counts only one pump, ignoring the redundancy.
- D. The pump pair alone; it leaves out the generator.
Basis: Derived: probability of independent components in series and parallel.
Question 45: Updated rainfall, bigger risk
Sustainability · Resilience · Multiple choice
A culvert was designed for a storm with a 1% annual exceedance probability (AEP). Updated rainfall data show that the same storm depth now has a 4% AEP. Assume each year is independent.
What is the chance that storm depth is exceeded at least once during the culvert's remaining 30-year life?
Show answer and worked solution
Answer: C. 70.6%
P(at least once) = 1 − (1 − AEP)ⁿ = 1 − 0.96³⁰ = 1 − 0.294 = 70.6%.
Under the original 1% AEP it was 1 − 0.99³⁰ = 26.0%. The design storm hasn't changed size, but the odds of meeting it have nearly tripled.
Why the other choices are wrong
- A. Uses the original 1% AEP.
- B. That's the chance of no exceedance.
- D. Multiplies 4% × 30. Probabilities can't exceed 100%.
Basis: Derived: probability of at least one exceedance in n independent years.
Air (Questions 46–59)
Question 46: ppm to µg/m³
Air · Engineering applications · Numeric entry
An ambient SO₂ reading is 0.075 ppm by volume. Convert to µg/m³ at 25 °C and 1 atm. Use a molar volume of 24.45 L/mol and a molecular weight of 64.07 g/mol for SO₂.
Round to the nearest whole number.
Enter a number in µg/m³.
Show answer and worked solution
Answer: 197 µg/m³
µg/m³ = ppm × MW ÷ 24.45 × 1,000 = 0.075 × 64.07 ÷ 24.45 × 1,000 = 196.5 ≈ 197 µg/m³.
Accepted: 195 to 198.
Common mistake: Using 22.4 L/mol (0 °C) gives 215 µg/m³. Match the molar volume to the stated temperature.
Basis: Derived: ideal-gas conversion with the constants given in the stem.
Question 47: Atmospheric stability
Air · Engineering applications · Multiple choice
A morning sounding shows air temperature falling 5 °C per km of altitude. The dry adiabatic lapse rate is 9.8 °C per km. Treat the air as dry.
How should the atmosphere be classified, and what does that mean for a plume?
Show answer and worked solution
Answer: C. Stable; vertical mixing is suppressed
The environment cools more slowly (5 °C/km) than a rising dry parcel cools (9.8 °C/km). A displaced parcel ends up colder and denser than its surroundings and sinks back. That's a subadiabatic, stable atmosphere, which limits vertical dispersion.
Why the other choices are wrong
- A. Unstable needs the environment to cool faster than 9.8 °C/km.
- B. Neutral means the two rates are equal.
- D. Superadiabatic is the same thing as unstable; looping plumes go with strong instability.
Basis: Derived: parcel comparison using the lapse rates given in the stem.
Question 48: Wind speed at stack height
Air · Engineering applications · Multiple choice · SI
Wind speed measured at 10 m is 3.0 m/s. For the stability class at the site, use the power law u₂ = u₁(z₂/z₁)^p with p = 0.25.
What is the wind speed at an 80-m stack top?
Show answer and worked solution
Answer: C. 5.0 m/s
u₂ = 3.0 × (80 ÷ 10)^0.25 = 3.0 × 8^0.25 = 3.0 × 1.682 = 5.0 m/s.
Why the other choices are wrong
- A. Assumes wind speed doesn't change with height.
- B. Uses p = 1/7 instead of the given 0.25.
- D. Scales linearly with height (p = 1).
Basis: Derived: power-law profile given in the stem.
Question 49: Ground-level plume concentration
Air · Engineering applications · Multiple choice
A stack emits 100 g/s of a gas with an effective stack height of 60 m. At the receptor distance, σy = 80 m and σz = 45 m, and the wind speed at stack height is 4.0 m/s. Use the ground-level, plume-centerline Gaussian equation with full ground reflection:
C = Q ÷ (π σy σz u) × exp[−H² ÷ (2σz²)]
What is the concentration?
Show answer and worked solution
Answer: B. 909 µg/m³
Q ÷ (π σy σz u) = 100 ÷ (π × 80 × 45 × 4.0) = 2.21 × 10⁻³ g/m³.
exp[−60² ÷ (2 × 45²)] = exp(−0.889) = 0.411.
C = 2.21 × 10⁻³ × 0.411 = 9.09 × 10⁻⁴ g/m³ = 909 µg/m³.
Why the other choices are wrong
- A. Uses 2π in the denominator. That's the form without ground reflection; the given equation already includes it.
- C. Leaves out the stack-height term, as if the source were at ground level.
- D. Converts g/m³ to µg/m³ with 10³ instead of 10⁶.
Basis: Derived: equation given in the stem.
Question 50: Urban box model
Air · Engineering applications · Multiple choice · SI
An urban area 20 km long in the wind direction has a uniform area emission rate of 4 × 10⁻⁶ g/(m²·s). Wind speed is 3 m/s, the mixing height is 500 m, and incoming air is clean. Use the steady-state box model, C = qL ÷ (uH).
What is the concentration?
Show answer and worked solution
Answer: C. 53 µg/m³
C = (4 × 10⁻⁶ × 20,000) ÷ (3 × 500) = 0.08 ÷ 1,500 = 5.33 × 10⁻⁵ g/m³ = 53 µg/m³.
Why the other choices are wrong
- A. Leaves out the box length L.
- B. Halves the result with no basis.
- D. Uses the mixing height in km instead of m.
Basis: Derived: box model given in the stem.
Question 51: Combustion air with excess air
Air · Engineering applications · Multiple choice
Methane burns completely: CH₄ + 2O₂ → CO₂ + 2H₂O. Air is 21% O₂ by volume. The burner runs at 10% excess air.
How many moles of air are supplied per mole of methane?
Show answer and worked solution
Answer: C. 10.5
Stoichiometric O₂: 2 mol per mol CH₄. Stoichiometric air: 2 ÷ 0.21 = 9.52 mol.
With 10% excess: 9.52 × 1.10 = 10.5 mol air.
Why the other choices are wrong
- A. Moles of oxygen with excess; the 79% nitrogen is missing.
- B. Stoichiometric air with no excess.
- D. Applies 20% excess air.
Basis: Derived: combustion stoichiometry.
Question 52: NOx mass emission rate
Air · Pollution control · Multiple choice · USCS
A stack's dry standard flow is 25,000 dscfm, and NOx measures 120 ppmvd, reported as NO₂ (MW 46.01). Use a standard molar volume of 385.3 ft³/lb-mol.
What is the NOx emission rate?
Show answer and worked solution
Answer: C. 21.5 lb/h
Volume of NOx: 120 × 10⁻⁶ × 25,000 × 60 = 180 ft³/h.
Moles: 180 ÷ 385.3 = 0.467 lb-mol/h.
Mass as NO₂: 0.467 × 46.01 = 21.5 lb/h.
Why the other choices are wrong
- A. Leaves out the × 60 min/h; that's lb/min.
- B. Uses the molecular weight of NO (30.01). The reporting basis is NO₂.
- D. Multiplies by 60 twice.
Basis: Derived: ideal-gas mole conversion with the constants given in the stem.
Question 53: Emission inventory with control
Air · Pollution control · Multiple choice · USCS
A boiler burns 2.0 million gallons of fuel oil per year. Use an SO₂ emission factor of 71 lb per 1,000 gal burned, and a scrubber with 90% control efficiency.
What are the annual controlled SO₂ emissions?
Show answer and worked solution
Answer: A. 7.1 tons/yr
Uncontrolled: (2,000,000 ÷ 1,000) × 71 = 142,000 lb/yr.
Controlled: 142,000 × (1 − 0.90) = 14,200 lb/yr ÷ 2,000 = 7.1 tons/yr.
Why the other choices are wrong
- B. Converts with 1,000 lb per ton.
- C. Uncontrolled emissions.
- D. Applies the factor per gallon instead of per 1,000 gallons.
Basis: Derived: emissions = activity × emission factor × (1 − control efficiency).
Question 54: Coating substitution
Air · Pollution control · Multiple choice · USCS
A plant uses 10,000 gal/yr of a solventborne coating containing 3.5 lb VOC/gal. It switches to a waterborne coating at 1.2 lb VOC/gal with the same usage. Assume all VOC in the coating is emitted.
How much does the switch reduce VOC emissions?
Show answer and worked solution
Answer: B. 11.5 tons/yr
Reduction: 10,000 × (3.5 − 1.2) = 23,000 lb/yr ÷ 2,000 = 11.5 tons/yr.
This is pollution prevention: the VOC is never created, so no control device is needed to capture it.
Why the other choices are wrong
- A. Emissions after the switch, not the reduction.
- C. Emissions before the switch.
- D. Reports pounds as tons.
Basis: Derived: emissions = usage × VOC content.
Question 55: Control devices in series
Air · Pollution control · Multiple choice
A particulate stream of 2,000 kg/h passes through a cyclone (80% efficient) and then a baghouse (99% efficient on what reaches it).
What are the outlet emission rate and the overall control efficiency?
Show answer and worked solution
Answer: A. 4.0 kg/h; 99.8%
After the cyclone: 2,000 × (1 − 0.80) = 400 kg/h.
After the baghouse: 400 × (1 − 0.99) = 4.0 kg/h.
Overall: 1 − (0.20 × 0.01) = 99.8%.
Why the other choices are wrong
- B. Applies only the baghouse to the full inlet.
- C. Stops after the cyclone.
- D. Cuts the result in half with no basis.
Basis: Derived: penetration through devices in series multiplies.
Question 56: Baghouse bag count
Air · Pollution control · Numeric entry · USCS
A baghouse handles 60,000 acfm at a net air-to-cloth ratio of 3.0 ft/min. Each bag is a cylinder 8 in. in diameter and 10 ft long; count only the side area. Ignore compartments taken offline for cleaning.
What is the minimum number of bags?
Enter a number.
Show answer and worked solution
Answer: 955 bags
Required net cloth area: 60,000 ÷ 3.0 = 20,000 ft².
Area per bag: π × (8/12) × 10 = 20.94 ft².
Bags: 20,000 ÷ 20.94 = 954.9 → round up to 955.
Accepted: 955 only.
Common mistake: Using the radius (4 in.) in πDL gives 1,910 bags. Rounding down to 954 leaves the cloth slightly short of the required area.
Basis: Derived: air-to-cloth ratio = gas flow ÷ cloth area.
Question 57: Electrostatic precipitator efficiency
Air · Pollution control · Multiple choice · SI
An electrostatic precipitator has 5,000 m² of collecting plate area and treats 150 m³/s of gas. The effective migration velocity is 0.08 m/s. Use the Deutsch-Anderson equation, η = 1 − exp(−wA ÷ Q).
What is the collection efficiency?
Show answer and worked solution
Answer: C. 93.1%
wA ÷ Q = 0.08 × 5,000 ÷ 150 = 2.67.
η = 1 − e^(−2.67) = 1 − 0.069 = 93.1%.
Why the other choices are wrong
- A. Uses Q in m³/min (9,000) with w in m/s.
- B. That's the penetration, the fraction escaping.
- D. Doubles the plate area.
Basis: Derived: Deutsch-Anderson equation given in the stem.
Question 58: Carbon bed service life
Air · Pollution control · Multiple choice · SI
An activated carbon adsorber holds 2,000 kg of carbon with a working capacity of 0.25 kg VOC per kg carbon. The inlet VOC load is 12 kg/h, and the bed captures it all until breakthrough.
How long until the bed must be regenerated or replaced?
Show answer and worked solution
Answer: B. 41.7 h
Capacity: 2,000 × 0.25 = 500 kg VOC.
Service time: 500 ÷ 12 = 41.7 h.
Why the other choices are wrong
- A. A decimal slip.
- C. Ignores the working capacity and divides the carbon mass by the load.
- D. Multiplies capacity by load instead of dividing.
Basis: Derived: service time = working capacity × carbon mass ÷ VOC load.
Question 59: NOx control technology
Air · Pollution control · Multiple choice
A boiler needs NOx control. The proposed device injects ammonia or urea into the flue gas upstream of a catalyst bed, where NOx is reduced to nitrogen and water.
What technology is this?
Show answer and worked solution
Answer: A. Selective catalytic reduction (SCR)
SCR injects a nitrogen-based reagent (ammonia or urea) and passes the gas over a catalyst, which reduces NOx to molecular nitrogen and water vapor. The catalyst lets the reaction run at lower, wider temperature ranges than the non-catalytic version and typically achieves higher removal.
Why the other choices are wrong
- B. SNCR uses the same reagents but no catalyst.
- C. A combustion modification that limits NOx formation; it doesn't treat flue gas.
- D. Wet scrubbers mainly target acid gases such as SO₂ and particulates, not NOx reduction with ammonia.
Basis: EPA Air Pollution Control Technology Fact Sheet: Selective Catalytic Reduction, EPA-452/F-03-032, description of technology and comparison with SNCR.
Site Assessment and Remediation (Questions 60–73)
Question 60: Is the exposure pathway complete?
Site Assessment and Remediation · Site assessment · Multiple choice
Which scenario describes an incomplete exposure pathway?
Show answer and worked solution
Answer: A
An exposure pathway has five elements: a contaminant source, environmental fate and transport, an exposure point, an exposure route, and a potentially exposed population. In A there's no exposure point: nobody drinks the groundwater and it doesn't reach surface water. The contaminant is a nonvolatile metal, so vapor intrusion isn't a route.
An incomplete pathway now doesn't mean no future risk; a new well in the plume would complete it.
Why the other choices are wrong
- B. Soil is the medium and the exposure point; incidental ingestion is the route.
- C. Drinking the well water completes the pathway.
- D. Fish accumulate PCBs; eating them is the route.
Basis: ATSDR, Overview of the Exposure Pathway Evaluation, five elements of an exposure pathway.
Question 61: Three-point gradient
Site Assessment and Remediation · Site assessment · Numeric entry
Three monitoring wells in a confined aquifer are at these plan coordinates (m) with these heads:
| Well | x (m) | y (m) | Head (m) |
|---|---|---|---|
| MW-1 | 0 | 0 | 100.0 |
| MW-2 | 100 | 0 | 99.0 |
| MW-3 | 0 | 100 | 99.5 |
Assume the potentiometric surface is a plane. What is the magnitude of the hydraulic gradient? Round to four decimal places.
Enter a number in m/m.
Show answer and worked solution
Answer: 0.0112
Along x: (99.0 − 100.0) ÷ 100 = −0.010. Along y: (99.5 − 100.0) ÷ 100 = −0.005.
Magnitude: √(0.010² + 0.005²) = 0.0112.
Flow runs down the gradient, toward +x and +y, about 27° from the x-axis.
Accepted: 0.0111 to 0.0113.
Common mistake: Adding the two components (0.015) or using only one of them (0.010 or 0.005). The gradient is a vector.
Basis: Derived: gradient of a planar head surface.
Question 62: Retarded contaminant velocity
Site Assessment and Remediation · Site assessment · Multiple choice · SI
Groundwater flows with an average linear (seepage) velocity of 0.50 m/day. A dissolved organic contaminant has Koc = 100 L/kg. The aquifer has fraction organic carbon foc = 0.004, dry bulk density 1.7 kg/L, and porosity 0.30. Assume linear, reversible sorption. Use Kd = Koc × foc and R = 1 + ρb·Kd ÷ n.
How fast does the contaminant front move?
Show answer and worked solution
Answer: A. 0.153 m/day
Kd = 100 × 0.004 = 0.40 L/kg.
R = 1 + 1.7 × 0.40 ÷ 0.30 = 3.27.
Contaminant velocity = 0.50 ÷ 3.27 = 0.153 m/day.
Why the other choices are wrong
- B. Leaves out the bulk density in R.
- C. Ignores sorption; that's the water's velocity.
- D. Multiplies by R. Sorption slows the contaminant down.
Basis: Derived: linear-sorption retardation equations given in the stem.
Question 63: Comparing a UCL to a cleanup level
Site Assessment and Remediation · Site assessment · Multiple choice
Eight confirmation soil samples from an excavation have a mean of 40 mg/kg and a standard deviation of 12 mg/kg. The cleanup decision rule compares the one-sided 95% upper confidence limit of the mean with a cleanup level of 50 mg/kg. Use UCL = x̄ + t·s ÷ √n with t = 1.895 (7 degrees of freedom), and assume the data are approximately normal.
What is the UCL, and does the area meet the cleanup level?
Show answer and worked solution
Answer: C. 48.0 mg/kg; meets
UCL = 40 + 1.895 × 12 ÷ √8 = 40 + 8.04 = 48.0 mg/kg, below 50 mg/kg, so the area meets the rule.
The UCL protects against declaring success just because the sample mean happened to come out low.
Why the other choices are wrong
- A. Uses the mean alone; the decision rule calls for the UCL.
- B. Divides by n instead of √n.
- D. Leaves out √n; that's the mean plus t standard deviations.
Basis: Derived: one-sided t-based UCL with the values given in the stem.
Question 64: Child soil ingestion intake
Site Assessment and Remediation · Site assessment · Multiple choice
A residential yard has 400 mg/kg of a noncarcinogenic metal in surface soil. For a child receptor use soil ingestion 200 mg/day, exposure frequency 350 days/yr, exposure duration 6 years, and body weight 15 kg. For noncancer effects, set the averaging time equal to the exposure duration.
Use intake = (C × IR × CF × EF × ED) ÷ (BW × AT), with CF = 10⁻⁶ kg/mg.
What is the chronic daily intake?
Show answer and worked solution
Answer: B. 0.0051 mg/kg-day
Intake = (400 × 200 × 10⁻⁶ × 350 × 6) ÷ (15 × 6 × 365) = 168 ÷ 32,850 = 0.0051 mg/kg-day.
Why the other choices are wrong
- A. Averages over a 70-year lifetime. That's the cancer convention; noncancer intake is averaged over the exposure duration.
- C. Assumes exposure every day of the year (EF = 365).
- D. Leaves out the 10⁻⁶ kg/mg conversion.
Basis: EPA, Regional Screening Levels User's Guide, averaging time for carcinogens versus noncarcinogens; intake equation given in the stem.
Question 65: Drinking water cancer risk
Site Assessment and Remediation · Site assessment · Multiple choice
A residential well contains a carcinogen at 0.005 mg/L. Use ingestion 2 L/day, 350 days/yr for 30 years, body weight 70 kg, and a 70-year averaging time for cancer. The oral slope factor is 0.055 (mg/kg-day)⁻¹.
Use CDI = (C × IR × EF × ED) ÷ (BW × AT) and risk = CDI × SF.
What is the excess lifetime cancer risk, and how does it compare with the 10⁻⁶ to 10⁻⁴ range in the National Contingency Plan?
Show answer and worked solution
Answer: A. 3.2 × 10⁻⁶; within the range
CDI = (0.005 × 2 × 350 × 30) ÷ (70 × 70 × 365) = 5.87 × 10⁻⁵ mg/kg-day.
Risk = 5.87 × 10⁻⁵ × 0.055 = 3.2 × 10⁻⁶.
That's inside the NCP's 10⁻⁴ to 10⁻⁶ range of acceptable exposure levels for carcinogens. Where a site lands in that range is a risk-management decision, not the end of the analysis.
Why the other choices are wrong
- B. Averages over the 30-year exposure duration. Cancer risk is averaged over a lifetime.
- C. Skips the frequency and duration adjustment, as if exposure were every day for a lifetime.
- D. A units slip of 1,000.
Basis: EPA, Conducting a Human Health Risk Assessment (cancer risk = exposure × slope factor); EPA, Regional Screening Levels User's Guide (lifetime averaging for carcinogens); 40 CFR 300.430(e)(2)(i)(A)(2) (eCFR) (10⁻⁴ to 10⁻⁶ range).
Question 66: Hazard index
Site Assessment and Remediation · Site assessment · Numeric entry
A receptor's chronic daily intakes are 0.004 mg/kg-day of Chemical X (RfD 0.003 mg/kg-day) and 0.0006 mg/kg-day of Chemical Y (RfD 0.002 mg/kg-day). Both affect the same target organ.
What is the hazard index? Round to two decimal places.
Enter a number.
Show answer and worked solution
Answer: 1.63
Hazard quotient = intake ÷ RfD.
- X: 0.004 ÷ 0.003 = 1.33
- Y: 0.0006 ÷ 0.002 = 0.30
HI = 1.33 + 0.30 = 1.63. An HI above 1 flags potential concern for noncancer effects; it's not a probability.
Accepted: 1.62 to 1.64.
Common mistake: Multiplying intakes by RfDs (as with slope factors) or averaging the HQs (0.82).
Basis: EPA AirToxScreen Glossary of Terms, definitions of hazard quotient and hazard index.
Question 67: CERCLA threshold criteria
Site Assessment and Remediation · Remediation · Select all that apply
Under the National Contingency Plan, which remedy-selection criteria are threshold criteria that every alternative must meet to be eligible for selection? Select all that apply.
Show answer and worked solution
Answer: A, B
Credit requires exactly these two.
The nine criteria fall into three groups:
- Threshold (must pass): overall protection; compliance with ARARs.
- Primary balancing (traded off): long-term effectiveness and permanence; reduction of toxicity, mobility, or volume through treatment; short-term effectiveness; implementability; cost.
- Modifying: state acceptance; community acceptance.
Why the other statements are false
- C. A primary balancing criterion.
- D. A modifying criterion.
- E. A primary balancing criterion.
Question 68: Containment versus treatment
Site Assessment and Remediation · Remediation · Multiple choice
At a Superfund site, a large volume of soil poses a relatively low long-term threat, and treatment has been shown to be impracticable. Under the NCP's stated expectations, what does EPA generally expect to use for this waste?
Show answer and worked solution
Answer: B. Engineering controls such as containment
The NCP's expectations reserve treatment for principal threats (liquids, highly toxic, highly mobile material) and say EPA expects engineering controls such as containment for waste that poses a relatively low long-term threat or where treatment is impracticable. Institutional controls supplement those measures.
Why the other choices are wrong
- A. Treatment is the expectation for principal threats, and here it's impracticable.
- C. Institutional controls shouldn't be the sole remedy unless active measures aren't practicable.
- D. Low long-term threat isn't the same as no unacceptable risk; nothing in the stem supports no action.
Question 69: Capture zone width
Site Assessment and Remediation · Remediation · Multiple choice · SI
A single extraction well fully penetrates a confined aquifer 20 m thick with hydraulic conductivity 15 m/day. The regional hydraulic gradient is 0.002. The well pumps 600 m³/day. For a uniform regional flow, the capture zone's maximum width far upgradient is W = Q ÷ (b·K·i).
What is the maximum capture zone width?
Show answer and worked solution
Answer: C. 1,000 m
Regional flow per unit width: b·K·i = 20 × 15 × 0.002 = 0.60 m²/day.
W = 600 ÷ 0.60 = 1,000 m. The well captures exactly the regional flow passing through that width.
Why the other choices are wrong
- A. Divides by π; that's the stagnation-point distance downgradient.
- B. The width along the line through the well, half the maximum.
- D. Doubles the width with no basis.
Basis: Derived: equation given in the stem (uniform-flow capture zone).
Question 70: Pump-and-treat duration
Site Assessment and Remediation · Remediation · Multiple choice · SI
A dissolved plume occupies a pore volume of 50,000 m³. A pump-and-treat system extracts 200 m³/day from the plume. Treat the contaminant's retardation factor as R = 2.0, and assume cleanup requires flushing 3 pore volumes of the retarded contaminant, so time = (PV count × R × pore volume) ÷ Q.
About how long will cleanup take?
Show answer and worked solution
Answer: C. 1,500 days
Time = 3 × 2.0 × 50,000 ÷ 200 = 1,500 days (about 4.1 years).
Retardation is why pump-and-treat often runs far longer than the time to pump one plume volume (250 days here).
Why the other choices are wrong
- A. Flushes 2 pore volumes and ignores retardation.
- B. Flushes 3 pore volumes but ignores retardation.
- D. Doubles the answer with no basis.
Basis: Derived: simplified flushing estimate given in the stem.
Question 71: SVE pore volume exchange
Site Assessment and Remediation · Remediation · Multiple choice · SI
A soil vapor extraction (SVE) system treats a zone 30 m × 30 m × 4 m deep with air-filled porosity 0.25. It extracts 2.0 m³/min of soil gas, all from that zone.
How many air-filled pore volumes are exchanged per day?
Show answer and worked solution
Answer: C. 3.2
Air-filled pore volume: 30 × 30 × 4 × 0.25 = 900 m³.
Daily extraction: 2.0 × 1,440 = 2,880 m³.
Exchanges: 2,880 ÷ 900 = 3.2 per day.
Why the other choices are wrong
- A. That's the hourly rate (120 ÷ 900).
- B. Divides by the total soil volume instead of the air-filled pore volume.
- D. Applies the porosity twice.
Basis: Derived: pore volume = bulk volume × air-filled porosity.
Question 72: Dimensionless Henry's constant
Site Assessment and Remediation · Remedial technologies · Numeric entry
For an air stripping design, trichloroethylene has H = 0.0103 atm·m³/mol at 25 °C. Use R = 8.205 × 10⁻⁵ atm·m³/(mol·K).
What is the dimensionless Henry's constant (Cgas ÷ Cwater)? Round to three decimal places.
Enter a number.
Show answer and worked solution
Answer: 0.421
H′ = H ÷ (RT) = 0.0103 ÷ (8.205 × 10⁻⁵ × 298.15) = 0.421.
Accepted: 0.41 to 0.43.
Common mistake: Using T in °C (25) instead of kelvin, or inverting the ratio (2.37).
Basis: Derived: ideal-gas conversion of Henry's constant.
Question 73: Oxygen demand for bioremediation
Site Assessment and Remediation · Remedial technologies · Multiple choice
Benzene is mineralized aerobically: C₆H₆ + 7.5 O₂ → 6 CO₂ + 3 H₂O. Use molar masses C₆H₆ = 78.11 g/mol and O₂ = 32.00 g/mol, and ignore oxygen used for cell growth.
How many grams of oxygen are needed per gram of benzene?
Show answer and worked solution
Answer: C. 3.07 g O₂/g
7.5 × 32.00 ÷ 78.11 = 3.07 g O₂ per g benzene.
It's why oxygen delivery often limits aerobic bioremediation: water saturated with air holds only around 8–9 mg/L O₂.
Why the other choices are wrong
- A. Inverts the ratio.
- B. Uses the molar mass of atomic oxygen (16).
- D. That's the mole ratio, not the mass ratio.
Basis: Derived: reaction stoichiometry given in the stem.
Environmental and Occupational Health (Questions 74–80)
Question 74: Noise exposure TWA
Environmental and Occupational Health · Exposure assessments · Multiple choice
A worker's 8-hour shift: 4 h at 95 dBA, 2 h at 90 dBA, and 2 h at 85 dBA. Compute the dose using OSHA's Appendix A method and Table G-16a (reference durations: 95 dBA = 4 h, 90 dBA = 8 h, 85 dBA = 16 h), then convert with TWA = 16.61 log₁₀(D/100) + 90.
What is the 8-hour TWA?
Show answer and worked solution
Answer: C. 92.3 dBA
D = 100 × (4/4 + 2/8 + 2/16) = 100 × 1.375 = 137.5%.
TWA = 16.61 × log₁₀(1.375) + 90 = 92.3 dBA.
Why the other choices are wrong
- A. The criterion level itself, not this worker's exposure.
- B. Drops the 85 dBA period (D = 125%). Appendix A counts levels from 80 to 130 dBA.
- D. The highest level, not a time-weighted average.
Basis: 29 CFR 1910.95 (eCFR), Appendix A, I(1)(ii) and I(2); Table G-16a.
Question 75: What an 87 dBA TWA triggers
Environmental and Occupational Health · Regulatory compliance · Multiple choice
Monitoring shows a general-industry employee's 8-hour TWA noise exposure is 87 dBA. The employee has a valid baseline audiogram and no standard threshold shift. Under 29 CFR 1910.95, which statement is correct?
Show answer and worked solution
Answer: A
87 dBA is at or above the 85 dBA action level, so the hearing conservation program applies: monitoring, notification, audiometric testing, training, and hearing protectors made available. It's below the 90 dBA permissible exposure level, so the controls requirement in paragraph (b) isn't triggered.
Why the other choices are wrong
- B. Engineering or administrative controls are required above the Table G-16 limits (90 dBA for 8 hours).
- C. The 85 dBA action level has its own requirements.
- D. Required use at this level applies only to employees without a baseline audiogram or with a standard threshold shift.
Basis: 29 CFR 1910.95 (eCFR), paragraphs (b)(1), (c)(1), (g)(1), (i)(1), and (i)(2).
Question 76: Adding two noise sources
Environmental and Occupational Health · Exposure assessments · Multiple choice
Two machines each produce 90 dB at a worker's position when run alone. The sources are incoherent. Use L = 10 log₁₀(Σ 10^(Li/10)).
What is the combined sound level when both run?
Show answer and worked solution
Answer: B. 93 dB
L = 10 log₁₀(2 × 10⁹) = 10 × 9.30 = 93 dB.
Doubling the sound energy adds about 3 dB, not 90.
Why the other choices are wrong
- A. Ignores the second source.
- C. Adds 6 dB, which corresponds to four equal sources.
- D. Adds decibels arithmetically; they're logarithmic.
Basis: Derived: logarithmic addition of incoherent sources, equation given in the stem.
Question 77: Distance from a point source
Environmental and Occupational Health · Exposure assessments · Multiple choice
A radiation survey meter reads 40 mR/h at 2 m from a small (point) gamma source with no shielding.
Using the inverse square law, what is the expected reading at 8 m?
Show answer and worked solution
Answer: B. 2.5 mR/h
Dose rate falls with the square of distance: 40 × (2 ÷ 8)² = 40 ÷ 16 = 2.5 mR/h.
Distance is one of the simplest protective measures: quadrupling it cuts the rate to one-sixteenth.
Why the other choices are wrong
- A. Uses the cube of the distance ratio.
- C. Scales linearly with distance.
- D. Halves the reading as if the distance had only doubled linearly.
Basis: Derived: inverse square law for a point source.
Question 78: Confined space atmosphere
Environmental and Occupational Health · Regulatory compliance · Multiple choice
Pre-entry testing of a general-industry tank gives four readings, each taken alone. Under 29 CFR 1910.146, which one by itself makes the atmosphere a hazardous atmosphere?
Show answer and worked solution
Answer: A. Oxygen at 19.0% by volume
The standard defines a hazardous atmosphere to include oxygen below 19.5% or above 23.5%, and flammable gas in excess of 10% of its LFL. At 19.0%, oxygen is deficient.
The standard also sets the testing order: oxygen first, then combustible gases, then toxics.
Why the other choices are wrong
- B. 6% of LFL is below the 10% threshold.
- C. 22.0% is within 19.5–23.5%.
- D. An atmosphere becomes hazardous on this basis only if exposure could exceed the limit.
Basis: 29 CFR 1910.146(b) and (d)(5)(iii) (eCFR), definition of hazardous atmosphere; testing order.
Question 79: Hierarchy of controls
Environmental and Occupational Health · Regulatory compliance · Multiple choice
Workers degrease parts in an open tank of a toxic chlorinated solvent. Eliminating the cleaning step isn't possible. Which option ranks highest on the hierarchy of controls?
Show answer and worked solution
Answer: D. Switch to an aqueous cleaner that does the job
The hierarchy, most to least effective: elimination, substitution, engineering controls, administrative controls, PPE. With elimination ruled out, substitution ranks highest, provided the substitute doesn't introduce new risks.
Why the other choices are wrong
- A. PPE, the last line of defense.
- B. An administrative control.
- C. An engineering control: good, but it still leaves the hazard in place.
Basis: NIOSH, Hierarchy of Controls, overview and substitution section.
Question 80: Mixture exposure
Environmental and Occupational Health · Exposure assessments · Multiple choice
A worker's 8-hour TWA exposures to three hypothetical contaminants with additive effects are: Substance A 120 ppm (limit 200 ppm), Substance B 40 ppm (limit 100 ppm), and Substance C 300 ppm (limit 1,000 ppm). Use OSHA's mixture formula, Em = C₁/L₁ + C₂/L₂ + C₃/L₃.
What is Em, and is the exposure acceptable?
Show answer and worked solution
Answer: B. 1.3; exceeds the limit
Em = 120/200 + 40/100 + 300/1,000 = 0.60 + 0.40 + 0.30 = 1.3.
The standard says Em must not exceed 1. Each substance is under its own limit, but the mixture isn't.
Why the other choices are wrong
- A. Averages the three ratios instead of adding them.
- C. The individual checks pass, but the mixture rule still applies.
- D. Adding concentrations of different substances has no meaning.
Basis: 29 CFR 1910.1000(d)(2)(i) (OSHA).
Answer key
Show / hide the complete answer key
| Q | Answer | Knowledge area | Topic |
|---|---|---|---|
| 1 | B. 50 mg/L | Water | Reactor type matters |
| 2 | B. 75% | Water | Ideal settling removal |
| 3 | B. 500 days | Water | Groundwater travel time |
| 4 | B. 60 mg/L as CaCO₃ | Water | Alkalinity as CaCO₃ |
| 5 | A. 2.0 per day | Water | Monod growth rate |
| 6 | D. 128 lb/day | Water | Hypochlorite feed rate |
| 7 | C. 3.88 h | Water | Basin detention time |
| 8 | B. 1.39 gpm/ft² | Water | Filter rate with one out of service |
| 9 | B. 2.0-log | Water | Log inactivation from CT |
| 10 | C. 4.26 ft/s | Water | Distribution main velocity |
| 11 | B. 3,936 mg/L | Water | Reverse osmosis concentrate |
| 12 | 225 mg/L | Water | BOD₅ from a dilution test |
| 13 | C. 263 mg/L | Water | Ultimate BOD |
| 14 | 100 mL/g | Water | Sludge volume index |
| 15 | C. 650 gpd/ft² | Water | Primary clarifier overflow rate |
| 16 | A. 21.5 mg/L; yes, add alkalinity | Water | Alkalinity consumed by nitrification |
| 17 | C. 2.90 ft³/s | Water | Gravity sewer full-flow capacity |
| 18 | C. 5,880 gal/day | Water | Sludge volume from solids |
| 19 | C. 52.5% | Water | Digester volatile solids reduction |
| 20 | B. 37.4 cfs | Water | Picking the design intensity |
| 21 | C. 25.9 kg/day | Water | Daily phosphorus load |
| 22 | B. 21,400 ft³ | Water | Water quality volume |
| 23 | B. 1.92 cfs | Water | Detention outlet orifice |
| 24 | A. +19,200 m³/day | Water | Reservoir water balance |
| 25 | B. D018 only | Solid and Hazardous Waste | Toxicity characteristic codes |
| 26 | B. The waste itself, after filtering by the Method 1311 procedure | Solid and Hazardous Waste | When the waste is the extract |
| 27 | B. Small quantity generator | Solid and Hazardous Waste | Generator category |
| 28 | A, B, E | Solid and Hazardous Waste | Which wastes are characteristic? |
| 29 | A, B, C | Solid and Hazardous Waste | Universal waste categories |
| 30 | A. 450 kg | Solid and Hazardous Waste | Solvent mass balance |
| 31 | C. 1.30 m³/day | Solid and Hazardous Waste | Clay liner seepage |
| 32 | C. 70,200 m³ | Solid and Hazardous Waste | Landfill airspace per year |
| 33 | 10 loads | Solid and Hazardous Waste | Collection truck loads |
| 34 | A. 5,380 Btu/lb | Solid and Hazardous Waste | Heating value of mixed waste |
| 35 | B. 35% | Solid and Hazardous Waste | Diversion rate |
| 36 | C. 80 days | Solid and Hazardous Waste | Decay-in-storage time |
| 37 | B. 99.994%; meets | Solid and Hazardous Waste | Incinerator DRE |
| 38 | Goal and scope → Inventory → Impact assessment → Interpretation | Sustainability | Life cycle assessment phases |
| 39 | B. 3.65 years | Sustainability | Blower upgrade payback |
| 40 | C. 245,500 kWh | Sustainability | Annual pumping energy |
| 41 | A. 4,690 kWh/day | Sustainability | Biogas to electricity |
| 42 | C. 219 MG | Sustainability | Water loss reduction |
| 43 | D. $68,600 | Sustainability | Annualized life-cycle cost |
| 44 | C. 0.968 | Sustainability | Lift station reliability |
| 45 | C. 70.6% | Sustainability | Updated rainfall, bigger risk |
| 46 | 197 µg/m³ | Air | ppm to µg/m³ |
| 47 | C. Stable; vertical mixing is suppressed | Air | Atmospheric stability |
| 48 | C. 5.0 m/s | Air | Wind speed at stack height |
| 49 | B. 909 µg/m³ | Air | Ground-level plume concentration |
| 50 | C. 53 µg/m³ | Air | Urban box model |
| 51 | C. 10.5 | Air | Combustion air with excess air |
| 52 | C. 21.5 lb/h | Air | NOx mass emission rate |
| 53 | A. 7.1 tons/yr | Air | Emission inventory with control |
| 54 | B. 11.5 tons/yr | Air | Coating substitution |
| 55 | A. 4.0 kg/h; 99.8% | Air | Control devices in series |
| 56 | 955 bags | Air | Baghouse bag count |
| 57 | C. 93.1% | Air | Electrostatic precipitator efficiency |
| 58 | B. 41.7 h | Air | Carbon bed service life |
| 59 | A. Selective catalytic reduction (SCR) | Air | NOx control technology |
| 60 | A (no exposure point) | Site Assessment and Remediation | Is the exposure pathway complete? |
| 61 | 0.0112 | Site Assessment and Remediation | Three-point gradient |
| 62 | A. 0.153 m/day | Site Assessment and Remediation | Retarded contaminant velocity |
| 63 | C. 48.0 mg/kg; meets | Site Assessment and Remediation | Comparing a UCL to a cleanup level |
| 64 | B. 0.0051 mg/kg-day | Site Assessment and Remediation | Child soil ingestion intake |
| 65 | A. 3.2 × 10⁻⁶; within the range | Site Assessment and Remediation | Drinking water cancer risk |
| 66 | 1.63 | Site Assessment and Remediation | Hazard index |
| 67 | A, B | Site Assessment and Remediation | CERCLA threshold criteria |
| 68 | B. Engineering controls such as containment | Site Assessment and Remediation | Containment versus treatment |
| 69 | C. 1,000 m | Site Assessment and Remediation | Capture zone width |
| 70 | C. 1,500 days | Site Assessment and Remediation | Pump-and-treat duration |
| 71 | C. 3.2 | Site Assessment and Remediation | SVE pore volume exchange |
| 72 | 0.421 | Site Assessment and Remediation | Dimensionless Henry's constant |
| 73 | C. 3.07 g O₂/g | Site Assessment and Remediation | Oxygen demand for bioremediation |
| 74 | C. 92.3 dBA | Environmental and Occupational Health | Noise exposure TWA |
| 75 | A. Hearing conservation program applies | Environmental and Occupational Health | What an 87 dBA TWA triggers |
| 76 | B. 93 dB | Environmental and Occupational Health | Adding two noise sources |
| 77 | B. 2.5 mR/h | Environmental and Occupational Health | Distance from a point source |
| 78 | A. Oxygen at 19.0% by volume | Environmental and Occupational Health | Confined space atmosphere |
| 79 | D. Switch to an aqueous cleaner that does the job | Environmental and Occupational Health | Hierarchy of controls |
| 80 | B. 1.3; exceeds the limit | Environmental and Occupational Health | Mixture exposure |
Score your attempt
Give yourself one point per question. A select-all or ordering question counts only when it's fully correct. A numeric entry counts when it falls inside the accepted range shown with its solution. NCEES scores its own items the same all-or-nothing way.
| Knowledge area | Questions | Your correct |
|---|---|---|
| Water | 1–24 | ___ / 24 |
| Solid and Hazardous Waste | 25–37 | ___ / 13 |
| Sustainability | 38–45 | ___ / 8 |
| Air | 46–59 | ___ / 14 |
| Site Assessment and Remediation | 60–73 | ___ / 14 |
| Environmental and Occupational Health | 74–80 | ___ / 7 |
| Total | 1–80 | ___ / 80 |
What your score means
Your score is your raw result on these 80 original questions. It isn't an NCEES scaled score, and no percentage here converts to a pass or fail.
NCEES counts correct answers with no deduction for wrong ones, converts that count to a scaled score to adjust for small differences in difficulty between exam forms, and doesn't publish the passing score. Results come back as pass or fail, and examinees who fail get a diagnostic report by knowledge area. (NCEES Examinee Guide, May 2026, Section 6)
Use the area rows to decide what to study, not to grade yourself. With 7 to 24 questions per area, one or two misses can swing a percentage a lot, especially in Sustainability and Environmental and Occupational Health.
How to use this practice test
To learn, work it untimed and open each solution right after you answer. Log why you missed each one, not just that you did.
To simulate the exam, give yourself 8 hours for all 80 questions, about 6 minutes each, and hold the solutions until you finish. Use only an NCEES-approved calculator (the list is under "Calculator Policy" on the NCEES exams page) and the PE Environmental Reference Handbook from your MyNCEES account, since that's all you'll have on exam day. A break around Question 40 approximates the exam's two-section structure, though the actual section boundary is after approximately half the questions and the real exam mixes topics rather than grouping them by area like this page does.
Before a retake, wait long enough that you're solving rather than remembering the answers. Change the numbers in the questions you missed and rework them.
Review your misses
- Rework each missed question on paper before rereading the solution.
- Name the cause using the table below. The fix depends on the cause.
- For a lookup miss, find the equation or table in the reference handbook and note the search term that got you there. On exam day you'll search a PDF, so the search term is what's worth remembering.
- Practice more problems in any area where the misses share a cause.
| If the miss was… | Do this next |
|---|---|
| Method or concept | Restate the governing principle (mass balance, reactor model, regulatory test) before calculating again. |
| Reference lookup | Find the equation or rule in the handbook or the regulation and write down your search term. |
| Units | Write the starting and target units and convert before substituting. |
| Regulatory threshold | Check whether the limit is inclusive (≥, ≤) or exclusive (>, <). Several questions here turn on that. |
| Arithmetic | Recompute from unrounded intermediate values and sanity-check the magnitude. |
How this test maps to the exam
The NCEES column is the official number of questions per knowledge area on each 80-question exam. Our counts sit inside every official range and total 80. That split is our editorial choice, not the distribution of any particular exam form.
| Knowledge area | NCEES questions (of 80) | Questions here | Question numbers |
|---|---|---|---|
| 1. Water | 19–29 | 24 | 1–24 |
| 2. Solid and Hazardous Waste | 11–17 | 13 | 25–37 |
| 3. Sustainability | 7–11 | 8 | 38–45 |
| 4. Air | 13–20 | 14 | 46–59 |
| 5. Site Assessment and Remediation | 13–20 | 14 | 60–73 |
| 6. Environmental and Occupational Health | 7–11 | 7 | 74–80 |
| Total | 80 | 80 |
Source: NCEES PE Environmental CBT Exam Specifications, effective beginning April 2026.
This set samples every knowledge area, not every topic in each. It doesn't test, for example, coagulant dosing, membrane fouling, landfill gas generation, leachate collection design, stack sampling methods, ambient air quality standards, emergency response planning, or respirator selection. Those are gaps in this set, not signs they matter less on the exam.
The real exam also uses point-and-click and drag-and-drop items. This set uses multiple choice, numeric entry, select-all, and one ordering question, and it doesn't reproduce the exam's on-screen interface.
PE Environmental exam format
| Item | PE Environmental |
|---|---|
| Questions | 80 |
| Time | 9-hour appointment: 2-minute nondisclosure agreement, 8-minute tutorial, 8-hour exam, 50-minute scheduled break |
| Sections | Two. After about half the questions you review and submit them, and you can't return to them. |
| Item types | Multiple choice plus alternative item types (multiple correct, point and click, drag and drop, fill in the blank), all scored right or wrong. Unscored pretest items are mixed in and not marked. |
| Units | Both SI and U.S. customary |
| References | The NCEES PE Environmental Reference Handbook, on screen as a searchable PDF. No design standards are supplied for this exam. Ctrl+F doesn't work; use the reference panel's search box. |
| Calculator | One NCEES-approved calculator; a TI-30XS is also available on screen |
| Where and when | NCEES-approved Pearson test centers, year-round |
| Fee | $400 to NCEES; some licensing boards charge a separate application fee |
| Results | Pass or fail, typically 7–10 days after the exam |
| Retakes | One attempt per quarterly testing window and no more than three in 12 months under NCEES policy; some boards are stricter |
Sources: NCEES PE Environmental exam page; NCEES Examinee Guide, May 2026, Sections 1–4 and 6 and the exam testing dates and format summary; NCEES computer-based testing.
What changed in April 2026
If your study material predates April 2026, check it against this map. The question count and appointment length didn't change.
| Before April 2026 | Beginning April 2026 |
|---|---|
| Water 21–35 | Water 19–29 (stormwater and water resources now one subarea) |
| Air 14–22 | Air 13–20 |
| Solid and Hazardous Waste 11–18 | Solid and Hazardous Waste 11–17 |
| Site Assessment and Remediation 12–19 | Site Assessment and Remediation 13–20 |
| Environmental Health and Safety 7–11 | Environmental and Occupational Health 7–11 |
| Associated Engineering Principles 5–9 | No longer a separate area |
| — | Sustainability 7–11 (new: resources conservation and resilience) |
Sources: specifications effective beginning April 2026; prior specifications, April 2019 exams.
Other things candidates ask
Is this the same as the PE Civil: Water Resources and Environmental exam? No. That's a module of PE Civil with its own specifications and design standards. If that's your exam, use the NCEES PE Civil exam page to open the Water Resources and Environmental specifications and design standards for your test date.
Is there an official practice exam? Yes. NCEES sells an 80-question PE Environmental practice exam built to the April 2026 specifications, $59.95 as of October 9, 2026, through MyNCEES exam prep. Choose the edition labeled for exams beginning April 2026. Corrections to NCEES practice exams are posted on its exam prep errata page.
What's the pass rate? NCEES reports 73% for 415 first-time takers and 54% for 118 repeat takers, for January–June 2026 examinees under NCEES member boards (table last updated July 2026). That window includes exams taken under both the old and new specifications. Pass rates describe groups of examinees; they don't tell you what raw score you need. (NCEES PE Environmental page, PE exam pass rates table)
Where do I get the reference handbook? Log in to MyNCEES. NCEES provides it to account holders and doesn't sell printed copies.
Do I need board approval first? Each licensing board sets eligibility, and some require a separate application before you can test. Check your board's process before you pay NCEES.
Next steps for registering and testing:
- How to register for an NCEES exam
- NCEES exam accommodations
- Understanding NCEES exam results
- NCEES exam retake policy
Sources and independence
Official exam sources
- NCEES: PE Environmental exam page
- NCEES: PE Environmental CBT Exam Specifications, effective beginning April 2026
- NCEES: PE Environmental CBT Exam Specifications, April 2019 exams (prior version)
- NCEES Examinee Guide, May 2026
- NCEES: Computer-based testing and item types
- NCEES: PE Environmental e-book Practice Exam (April 2026 specifications)
- NCEES: Exam prep errata
Regulations and technical sources used in the solutions
- 40 CFR Part 261, Subpart C: Characteristics of Hazardous Waste (eCFR)
- 40 CFR 260.10: Definitions (eCFR)
- 40 CFR 300.430: Remedial investigation/feasibility study and selection of remedy (eCFR)
- 29 CFR 1910.95: Occupational noise exposure (eCFR)
- 29 CFR 1910.146: Permit-required confined spaces (eCFR)
- 29 CFR 1910.1000: Air contaminants (OSHA)
- EPA: Conducting a Human Health Risk Assessment
- EPA: Regional Screening Levels User's Guide
- EPA: AirToxScreen Glossary of Terms
- ATSDR: Overview of the Exposure Pathway Evaluation
- EPA Air Pollution Control Technology Fact Sheet: Selective Catalytic Reduction
- NIOSH: Hierarchy of Controls
- ISO 14040:2006, Life cycle assessment: Principles and framework
- FHWA HDS-4, Introduction to Highway Hydraulics (2008)
- DOE/Hydraulic Institute, Pump Life Cycle Costs (2001)
- EPA: Disinfection Profiling and Benchmarking Technical Guidance Manual
- Florida DEP, Wastewater Treatment Formulas (April 2024)
We recomputed every answer and every wrong-answer path, and traced each regulatory threshold and stated principle to the source linked with its question. Questions marked "Derived" rest on the equation and assumptions given in the question itself. That's a source and arithmetic check, not a professional engineering review.
Castleport Test Prep Editorial Team. Last verified October 9, 2026: NCEES exam format, fee, April 2026 specifications, practice exam listing, pass-rate table, and the regulatory thresholds and definitions used in the solutions.
Castleport Test Prep is an independent exam prep publisher and is not affiliated with, endorsed by, or approved by NCEES. Exam and credential names are used to identify their subjects; trademarks belong to their respective owners. These practice questions are original and unofficial, not actual or past exam questions. Using this resource doesn't guarantee a passing result or licensure.
Drafting was AI-assisted. No licensed-engineer review has been completed.