Free FE Environmental Practice Test: 55 Questions With Explanations

Answer and explanation

Correct answer: A. 75,600 m³.

Volume is the area under the flow–time curve. With equal steps, the trapezoidal rule is:

V ≈ Δt × [(Q₀ + Q₄)/2 + Q₁ + Q₂ + Q₃]

Convert the time step to seconds so it matches m³/s: Δt = 1 h = 3,600 s.

V ≈ 3,600 s × [(0 + 2)/2 + 4 + 10 + 6] m³/s = 3,600 × 21 = 75,600 m³.

Why the other answers miss

• B (1,260 m³) uses Δt = 60 s, a minute instead of an hour.
• C (79,200 m³) treats all five readings as full one-hour blocks (sum of 22 × 3,600). The end readings only get half weight.
• D (144,000 m³) assumes the 10 m³/s peak lasts all 4 hours.

Review cue: Before you multiply, make the time unit match the flow unit.

Source: OpenStax Calculus Volume 2, §3.6 Numerical Integration, Theorem 3.4, the trapezoidal rule; NIST Guide to the SI (SP 811), Chapter 5: Units Outside the SI, Table 6 (liter, metric ton, minute, hour, day).

Answer and explanation

Correct answer: C. 0.010 mol/L.

Use the quadratic formula with a = 1, b = 0.01, c = −0.0002:

x = [−b ± √(b² − 4ac)] / 2a = [−0.01 ± √(0.0001 + 0.0008)] / 2 = (−0.01 ± 0.03) / 2

The roots are 0.010 and −0.020. A concentration can’t be negative, so x = 0.010 mol/L.

Check: 0.0001 + 0.0001 − 0.0002 = 0. ✓

Why the other answers miss

• A (−0.020) is a real root, but a negative concentration has no physical meaning.
• B (0.005) divides by 4a instead of 2a.
• D (0.020) forgets to divide by 2a, or flips the sign of b.

Review cue: Solve, then ask which root the physical situation allows. Plug it back in to check.

Source: OpenStax Intermediate Algebra 2e, §9.3 Solve Quadratic Equations Using the Quadratic Formula, “Quadratic Formula” box.

Answer and explanation

Correct answer: D. 20 thousand m³.

Set the first derivative to zero to find the critical point:

C′(x) = 2 − 800/x² = 0 → x² = 400 → x = 20 (thousand m³).

Confirm it’s a minimum: C″(x) = 1,600/x³ > 0 for x > 0. The minimum cost is C(20) = 40 + 40 = 80 (thousand dollars per year).

Why the other answers miss

• A (10) is too small: C(10) = 20 + 80 = 100, higher than C(20) = 80.
• B (40) is too large: C(40) = 80 + 20 = 100, also higher than C(20).
• C (400) is x², not x. The square root step was skipped.

Review cue: Write C′(x) = 0, solve, then check the second derivative or nearby values.

Source: OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems, Problem-Solving Strategy: Solving Optimization Problems.

Answer and explanation

Correct answer: B. 10.85 to 13.15 mg/L.

With an unknown population standard deviation and a small sample, use t with n − 1 = 8 degrees of freedom:

x̄ ± t × s/√n = 12.0 ± 2.306 × 1.5/√9 = 12.0 ± 2.306 × 0.5 = 12.0 ± 1.153

Interval: 10.85 to 13.15 mg/L.

Why the other answers miss

• A divides by √8 instead of √9. Degrees of freedom set the t value, not the √n term.
• C uses z = 1.96. With s estimated from 9 samples, t is the right multiplier.
• D divides s by n instead of √n.

Review cue: Standard error is s/√n. Degrees of freedom only pick the t value.

Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.5.2 Confidence Limits for the Mean, confidence-limit formula.

Answer and explanation

Correct answer: 1.6. Accepted range here: 1.55 ≤ answer < 1.65.

b = Σ(x − x̄)(y − ȳ) / Σ(x − x̄)²

x̄ = 2.5 and ȳ = 4.5.

Σ(products) = 8.0 and Σ(x − x̄)² = 2.25 + 0.25 + 0.25 + 2.25 = 5.0.

b = 8.0/5.0 = 1.6. (The intercept is a = 4.5 − 1.6 × 2.5 = 0.5.)

Common wrong answers

• 1.67 is the slope through the first and last points only, (7 − 2)/(4 − 1). Least squares uses every point.
• 0.5 is the intercept, not the slope.

Review cue: Build the deviation table. It keeps the sums organized and catches sign errors.

Source: OpenStax Introductory Statistics 2e, §12.3 The Regression Equation, Least Squares Criteria for Best Fit.

Answer and explanation

Correct answer: D.

NCEES Model Rules §240.15 A.3 says licensees shall notify their employer or client and such other authority as may be appropriate when their professional judgment is overruled and the public’s health, safety, or welfare is endangered. Section A.1 adds that safeguarding the public is the licensee’s first responsibility.

The Model Rules are a template. Each state board adopts its own rules, so check your board’s version for practice.

Why the other answers miss

• A treats employer authority as the end of the matter. A.3 requires notification when public health is at risk.
• B removes her from the job but leaves the danger unreported.
• C protects her, not the public. A private note notifies no one.

Review cue: Ask who is at risk and who must be told. Silence is rarely the answer when public safety is involved.

Source: NCEES Model Rules (revised August 2026), §240.15 Rules of Professional Conduct, §240.15 A.1 and A.3, printed p. 11.

Answer and explanation

Correct answer: A.

Model Rules §240.15 B.6 requires licensees to disclose to their employers or clients all known or potential conflicts of interest or other circumstances that could influence, or appear to influence, their judgment. Owning stock in a bidder is exactly that kind of circumstance. Disclosure lets the client decide how to handle it, for example by assigning another reviewer.

Why the other answers miss

• B relies on his own sense of objectivity. The rule covers conflicts that could appear to influence judgment, too.
• C is irrelevant. A low bid doesn’t remove the conflict, and the evaluation itself is what’s compromised.
• D acts too late and still leaves the client uninformed during the evaluation.

Review cue: If a reasonable client would want to know, disclose it before you act.

Source: NCEES Model Rules (revised August 2026), §240.15 Rules of Professional Conduct, §240.15 B.6, printed p. 12.

Answer and explanation

Correct answer: C.

Model Rules §240.15 A.4 says licensees shall, to the best of their knowledge, include all relevant and pertinent information in an objective and truthful manner within all professional documents, statements, and testimony. The readings are valid and relevant, so they belong in the report. Context about why they occurred is fine; omission isn’t.

Why the other answers miss

• A treats a manager’s approval as permission to omit relevant data. It isn’t.
• B quietly changes the data. That’s not objective reporting.
• D still leaves out relevant information unless someone asks.

Review cue: Valid and relevant data goes in. Explain it; don’t hide it.

Source: NCEES Model Rules (revised August 2026), §240.15 Rules of Professional Conduct, §240.15 A.4, printed p. 11.

Answer and explanation

Correct answer: B. 1.42.

Put everything in present worth.

PW of benefits = 90,000 × 11.4699 = $1,032,291

PW of costs = 500,000 + 20,000 × 11.4699 = $729,398

Conventional B/C = 1,032,291 / 729,398 = 1.42. Because it’s above 1.0, the project is economically justified at 6%.

Why the other answers miss

• A (0.71) inverts the ratio: costs over benefits.
• C (1.61) is the modified B/C, which subtracts O&M from benefits: (70,000 × 11.4699)/500,000. It’s a valid method, but not the one the question asked for.
• D (2.00) ignores interest: $1.8 million of benefits over $900,000 of costs.

Review cue: Read which B/C convention is asked for. Conventional and modified ratios give different numbers but the same accept/reject decision.

Source: NPTEL, Comparison of Alternatives, Lecture 13: Benefit-cost analysis, conventional and modified B/C ratio; Penn State EME 460, Compound Interest Formulas III, P/F, P/A, and A/P factors.

Answer and explanation

Correct answer: A. $11,932 per year.

EUAC spreads the first cost over the asset’s own life with the capital recovery factor, then adds annual O&M:

EUAC_B = 70,000 × 0.14903 + 1,500 = 10,432 + 1,500 = $11,932 per year.

For comparison, EUAC_A = 40,000 × 0.25046 + 3,000 = $13,018 per year. Pump B is cheaper on an annual basis even though it costs more up front. EUAC works for unequal lives because each pump is assumed to be replaced in kind.

Why the other answers miss

• B ($8,500) ignores interest: 70,000/10 + 1,500.
• C ($13,018) is Pump A’s EUAC, not Pump B’s.
• D ($19,032) uses the 5-year factor for a 10-year pump.

Review cue: Match each factor to that asset’s own life.

Source: Penn State EME 460, Compound Interest Formulas III, P/F, P/A, and A/P factors.

Answer and explanation

Correct answer: D. $159,000.

A future target funded by equal deposits calls for the sinking-fund factor:

A = F × (A/F, 5%, 10) = 2,000,000 × 0.07950 = $159,000 per year (about $159,009 unrounded).

Interest does part of the work, so the deposits are less than $2,000,000/10.

Why the other answers miss

• A ($122,800) finds the present worth of $2 million and divides by 10. That mixes two different conversions.
• B ($200,000) ignores interest.
• C ($259,000) uses the capital recovery factor (A/P), which converts a present amount, not a future one.

Review cue: Name the known (P, F, or A) and the unknown before choosing the factor.

Source: Penn State EME 460, Compound Interest Formulas II, sinking-fund deposit factor (A/F).

Answer and explanation

Correct answer: B. 59,700.

P = 40,000 × e^(0.02 × 20) = 40,000 × e^0.4 = 40,000 × 1.4918 = 59,673 ≈ 59,700.

Why the other answers miss

• A (48,000) applies 2% growth for 10 years arithmetically. It’s the wrong model and the wrong time span.
• C (56,000) is arithmetic (linear) growth: 40,000 × (1 + 0.02 × 20).
• D (59,400) uses annual compounding, 40,000 × 1.02²⁰ = 59,438. That model wasn’t the one given.

Review cue: Continuous, compounded, and arithmetic growth are different models. Use the one stated.

Source: OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay, Rule: Exponential Growth Model.

Answer and explanation

Correct answer: 15,200 m³/day. Accepted range here: 15,199.5 ≤ answer < 15,200.5.

Average day = 25,000 × 380 L/d = 9,500,000 L/d = 9,500 m³/d (1 m³ = 1,000 L).

Maximum day = 9,500 × 1.6 = 15,200 m³/day.

Common wrong answers

• 9,500 is the average day; the peaking factor was left off.
• 15,200,000 is the answer in liters per day, not m³.
• 5,938 divides by 1.6 instead of multiplying.

Review cue: Write the units through every step. L to m³ divides by 1,000.

Source: NIST Guide to the SI (SP 811), Chapter 5: Units Outside the SI, Table 6 (liter, metric ton, minute, hour, day).

Answer and explanation

Correct answer: C. CMFR 0.33; PFR 0.14.

kθ = 0.5 × 4 = 2.

CMFR (steady-state mass balance): C/C₀ = 1/(1 + kθ) = 1/3 = 0.33

PFR (each parcel reacts like a batch): C/C₀ = e^(−kθ) = e^(−2) = 0.14

For positive-order reactions, plug flow removes more than complete mixing at the same residence time. The mixed reactor runs entirely at the low outlet concentration, which slows the reaction.

Why the other answers miss

• A swaps the two equations.
• B treats 4 h as a half-life. Nothing in the question says so.
• D is e^(−0.125). It divides k by θ instead of multiplying.

Review cue: Same kθ, two equations. The mixed reactor always leaves more behind for first-order removal.

Source: Stenstrom & Rosso, Fundamentals of Chemical Reactor Theory (UCLA), pp. 4–5: Eq. 16 (CSTR), p. 4; Eq. 22 (plug flow), p. 5.

Answer and explanation

Correct answer: A and C.

A sacrificial anode must be a more active metal than iron, so it corrodes instead of the steel. Magnesium and zinc are both more active than iron and are the standard choices for buried tanks and pipes.

Why the other answers miss

• B (copper) is less active than iron. Connecting it would speed up corrosion of the steel.
• D (lead) is also less active than iron and gives no protection.

Review cue: A sacrificial anode must sit above the protected metal in the activity series.

Source: OpenStax Chemistry 2e, §17.6 Corrosion, sacrificial anodes / cathodic protection.

Answer and explanation

Correct answer: C. 320 mg/L.

Moles of glucose = 300 mg/L ÷ 180 mg/mmol = 1.667 mmol/L.

Moles of O₂ = 6 × 1.667 = 10.0 mmol/L.

ThOD = 10.0 mmol/L × 32 mg/mmol = 320 mg/L.

Shortcut: 300 × (6 × 32)/180 = 320 mg/L.

Why the other answers miss

• A (53 mg/L) uses one O₂ per glucose: 300 × 32/180.
• B (281 mg/L) inverts the mass ratio: 300 × 180/(6 × 32).
• D (1,920 mg/L) applies the 6:1 mole ratio twice: 300 × (6 × 32)/180 × 6.

Review cue: Convert mass to moles, apply the coefficients, then convert back.

Source: OpenStax Chemistry 2e, §4.3 Reaction Stoichiometry, Example 4.10.

Answer and explanation

Correct answer: A. 8.7 mg/L.

Henry’s law: C = kP = 1.3 × 10⁻³ × 0.21 = 2.73 × 10⁻⁴ mol/L.

Convert to mass: 2.73 × 10⁻⁴ mol/L × 32,000 mg/mol = 8.7 mg/L.

The constant changes with temperature, which is why the question supplies it. Colder water holds more oxygen.

Why the other answers miss

• B (0.27) stops at mmol/L (0.273) and labels it mg/L.
• C (4.4) uses 16 g/mol, the mass of one oxygen atom rather than O₂.
• D (41.6) uses 1 atm instead of the 0.21 atm partial pressure of oxygen.

Review cue: Henry’s law uses the gas’s partial pressure, not total pressure.

Source: OpenStax Chemistry 2e, §11.3 Solubility, Henry’s law, Cg = kPg.

Answer and explanation

Correct answer: 10.0 days. Accepted range here: 9.95 ≤ answer < 10.05.

First-order: ln(C/C₀) = −kt.

90% removal leaves C/C₀ = 0.10, so t = ln(10)/k = 2.3026/0.23 = 10.0 days.

Common wrong answers

• 3.0 days is the half-life, 0.693/0.23. That’s 50% removal.
• 0.46 days uses −ln(0.9)/k, the time to remove 10%, not 90%.

Review cue: 90% removal means 10% remaining. Write C/C₀ before you take the log.

Source: OpenStax Chemistry 2e, §12.4 Integrated Rate Laws, first-order integrated rate law and half-life.

Answer and explanation

Correct answer: D. 3.2 × 10⁻⁵ mol/L.

At 25 °C, pH + pOH = 14.00, so pOH = 14.00 − 9.5 = 4.5.

[OH⁻] = 10^(−4.5) = 3.2 × 10⁻⁵ mol/L.

In very dilute water, concentration and activity are close. That’s why the question lets you treat them as equal.

Why the other answers miss

• A is the hydrogen ion concentration, 10^(−9.5).
• B reports pOH as if it were a concentration.
• C treats the pH value as a mantissa instead of an exponent.

Review cue: At 25 °C, pH + pOH = 14. Then undo the log: concentration = 10^(−pOH).

Source: OpenStax Chemistry 2e, §14.2 pH and pOH, pH + pOH = 14.00 at 25 °C.

Answer and explanation

Correct answer: B.

Average daily dose: ADD = C × IR / BW = 0.02 mg/L × 2 L/d / 70 kg = 5.7 × 10⁻⁴ mg/kg-day.

HQ = ADD / RfD = (5.7 × 10⁻⁴) / (3 × 10⁻⁴) = 1.9.

EPA describes exposures at or below the reference level (HQ ≤ 1) as not likely to cause adverse effects. An HQ above 1 flags potential concern that deserves a closer look. EPA also says the HQ is not a probability of illness.

Why the other answers miss

• A reports the dose (ADD) as if it were the hazard quotient.
• C inverts the ratio: RfD/ADD.
• D has the right number but the wrong meaning. HQ is a ratio, not a probability.

Review cue: HQ compares a dose to a reference dose. It’s a screening ratio, not a risk percentage.

Source: EPA ExpoBox, Exposure Assessment Tools by Routes – Ingestion, average daily dose equation; EPA, Risk Assessment for Other Effects (noncancer hazard quotient), final two explanatory paragraphs before References: hazard-quotient definition and interpretation.

Answer and explanation

Correct answer: D. 100%.

OSHA’s dose formula: D = 100 × (C₁/T₁ + C₂/T₂ + …), where C is actual time at a level and T is the allowed time.

D = 100 × (4/8 + 2/4) = 100 × (0.5 + 0.5) = 100%.

That is exactly at the permissible limit. Sound below 80 dBA is left out of this calculation.

Why the other answers miss

• A (50%) counts only the 90 dBA period.
• B (75%) divides the 95 dBA time by 8 h. Each level has its own allowed time.
• C (150%) divides both times by 4 h.

Review cue: Divide time at each level by the time allowed at that level, then add the fractions.

Source: OSHA 29 CFR 1910.95, Occupational noise exposure, Table G-16 and footnote; OSHA 29 CFR 1910.95 Appendix A, Noise Exposure Computation, §I, dose formula.

Answer and explanation

Correct answer: A. 58.9 kPa.

p_gauge = ρgh = 1,000 × 9.81 × 6.0 = 58,860 Pa = 58.9 kPa.

Gauge pressure is measured relative to the atmosphere, so atmospheric pressure is not added.

Why the other answers miss

• B (6.0) drops ρg and reports depth as pressure.
• C (160.2) adds atmospheric pressure (101.3 kPa). That’s absolute pressure.
• D (588.6) is a factor-of-10 slip in the Pa to kPa conversion.

Review cue: Gauge or absolute? Decide before you add 101.3 kPa.

Source: OpenStax University Physics Volume 1, §14.2 Measuring Pressure, gauge vs. absolute pressure; p = hρg.

Answer and explanation

Correct answer: B. 2.29 m.

h_f = f (L/D)(v²/2g) = 0.025 × (200/0.25) × (1.50²/(2 × 9.81))

= 0.025 × 800 × 0.1147 = 2.29 m.

Why the other answers miss

• A (1.53 m) uses v instead of v².
• C (4.59 m) uses the radius (0.125 m) in place of the diameter.
• D (9.17 m) multiplies by 4, as if the 0.025 were a Fanning factor. The question says it’s a Darcy factor.

Review cue: Check which friction factor you were given. Darcy f = 4 × Fanning f.

Source: DOE Fundamentals Handbook: Thermodynamics, Heat Transfer, and Fluid Flow, Vol. 3 (DOE-HDBK-1012/3-92), Fluid Flow module, “Darcy’s Equation,” printed p. 32; NPTEL, Module 6, Chapter 11, Applications of Viscous Flows Through Pipes, Recap, first two bullets, f = 4C_f and Darcy head loss.

Answer and explanation

Correct answer: C. Re ≈ 3.6 × 10⁵; turbulent.

First find velocity from continuity: A = πD²/4 = π(0.30)²/4 = 0.0707 m², so V = Q/A = 0.085/0.0707 = 1.20 m/s.

Re = ρVD/μ = 1,000 × 1.20 × 0.30 / (1.002 × 10⁻³) = 3.6 × 10⁵.

That is far above about 3,000, so the flow is turbulent.

Why the other answers miss

• A leaves out density (Re = VD/μ). Re must be dimensionless.
• B uses πD² as the area, which gives a velocity four times too low.
• D uses the radius instead of the diameter. The regime happens to be right, but the number isn’t.

Review cue: Get V from Q = AV, then use the diameter in Re. Check that the units cancel.

Source: OpenStax University Physics Volume 1, §14.5 Fluid Dynamics, Q = Av; continuity; OpenStax University Physics Volume 1, §14.7 Viscosity and Turbulence, Reynolds number and flow regimes.

Answer and explanation

Correct answer: B. 39.2 kW.

Power delivered to the water: P_w = ρgQH = 1,000 × 9.81 × 0.10 × 30 = 29,430 W.

Efficiency = output ÷ input, so shaft input = 29.43 kW / 0.75 = 39.2 kW.

Why the other answers miss

• A (4.0 kW) leaves out g.
• C (22.1 kW) multiplies by efficiency. The input must be larger than the output.
• D (29.4 kW) is the water power, before efficiency is applied.

Review cue: For this pump, shaft input is larger than water power. Divide by efficiency to get what you must supply.

Source: DOE, Improving Pumping System Performance: A Sourcebook for Industry (2nd ed., 2006), Section 1, System Operating Costs, printed p. 9 (PDF p. 13).

Answer and explanation

Correct answer: 3.53 cfs. Accepted range here: 3.525 ≤ answer < 3.535.

Q = 3.33 × 3.0 × (0.50)^1.5 = 3.33 × 3.0 × 0.3536 = 3.53 cfs.

Common wrong answers

• 5.00 cfs raises h to the first power instead of 1.5.
• 2.50 cfs raises h to the second power instead of 1.5.

Review cue: Weir flow scales with head to the 1.5 power. Small head errors make big flow errors.

Source: USBR Water Measurement Manual, Table A7-3: Standard suppressed rectangular weirs, Q = 3.33Lh^1.5.

Answer and explanation

Correct answer: D. 79 cfs.

Area: A = 10 × 2 = 20 ft². Wetted perimeter: P = 10 + 2 × 2 = 14 ft. Hydraulic radius: R = A/P = 1.429 ft.

Q = (1.486/0.015) × 20 × (1.429)^(2/3) × (0.001)^(1/2)

= 99.07 × 20 × 1.268 × 0.03162 = 79 cfs.

Why the other answers miss

• A (2.5 cfs) uses S instead of √S.
• B (53 cfs) uses the SI coefficient (1.0) with feet.
• C (99 cfs) sets R equal to the depth. In a wide channel R approaches y, but this channel isn’t wide enough.

Review cue: Units decide the coefficient: 1.486 for feet, 1.0 for meters.

Source: USBR Water Measurement Manual, Chapter 13, §6, Slope-Area Method, Manning discharge equation; energy slope and hydraulic radius definitions.

Halfway point. On the real exam, after about half the questions you review and submit the first section, then you can take the optional scheduled break (NCEES Examinee Guide, May 2026, p. 11). This page doesn’t lock anything, so it’s a sensible place to check your pace.

Answer and explanation

Correct answer: A. 52,250 kJ.

Heat absorbed by the water: Q = mcΔT = 500 × 4.18 × (35 − 15) = 41,800 kJ.

Fuel energy = useful heat ÷ efficiency = 41,800 / 0.80 = 52,250 kJ.

Why the other answers miss

• B (33,440 kJ) multiplies by the efficiency. The fuel must supply more than the water absorbs.
• C (41,800 kJ) is the heat the water absorbs, before efficiency.
• D (91,440 kJ) uses the final temperature, 35 °C, as ΔT (and then divides by 0.80).

Review cue: ΔT is final minus initial. Then divide by efficiency to get the input.

Source: OpenStax University Physics Volume 2, §1.4 Heat Transfer, Specific Heat, and Calorimetry, Eq. 1.5, Q = mcΔT.

Answer and explanation

Correct answer: C. 651 m³/min.

At constant pressure and molar flow rate, volumetric flow is proportional to absolute temperature (Q₁/T₁ = Q₂/T₂):

Q₂ = 1,000 × 293/450 = 651 m³/min.

Cooling the gas shrinks its volume, so the answer must be below 1,000.

Why the other answers miss

• A (112) uses Celsius temperatures (19.85/176.85). Gas laws need kelvin.
• B (1,000) ignores the temperature change.
• D (1,536) inverts the ratio, which would make cooled gas expand.

Review cue: Absolute temperatures only, and sanity-check the direction of the change.

Source: OpenStax University Physics Volume 2, §2.1 Molecular Model of an Ideal Gas, The Gas Laws.

Answer and explanation

Correct answer: D. 2.04 in.

Potential maximum retention: S = 1,000/CN − 10 = 1,000/80 − 10 = 2.5 in.

Initial abstraction: Ia = 0.2 × 2.5 = 0.5 in.

Runoff: Q = (P − 0.2S)² / (P + 0.8S) = (4.0 − 0.5)² / (4.0 + 2.0) = 12.25/6.0 = 2.04 in.

Why the other answers miss

• A (2.46) leaves out the initial abstraction: P²/(P + S).
• B (2.50) reports S, the retention, instead of runoff.
• C (3.50) stops at P − Ia. Rain beyond Ia isn’t all runoff.

Review cue: Find S, then Ia, then Q. For this storm, runoff is less than P − Ia.

Source: USDA SCS, Urban Hydrology for Small Watersheds (TR-55), June 1986, Chapter 2, Eqs. 2-1 to 2-4.

Answer and explanation

Correct answer: B. 18 cfs.

Q = CiA = 0.6 × 2.5 in/h × 12 ac = 18 cfs.

In U.S. units, 1 acre-in/h ≈ 1.008 cfs, so the conversion factor is taken as 1. The rational method gives a peak rate, not a runoff volume.

Why the other answers miss

• A (1.5) multiplies only C × i, leaving out the area.
• C (7.2) multiplies only C × A, leaving out the intensity.
• D (30) leaves out the runoff coefficient. Not all rain runs off.

Review cue: Q = CiA. Count three factors before you finish.

Source: TxDOT Hydraulic Design Manual, Chapter 4, Section 12: Rational Method, Eq. 4-20.

Answer and explanation

Correct answer: A. +140,000 m³.

ΔS = inflows − outflows.

Rain on the surface: 0.050 m × 2.0 × 10⁶ m² = 100,000 m³ (in).

Evaporation: 0.080 m × 2.0 × 10⁶ m² = 160,000 m³ (out).

ΔS = 1,200,000 + 100,000 − 1,000,000 − 160,000 = +140,000 m³.

Why the other answers miss

• B (−60,000) counts only rain and evaporation, leaving out the stream flows.
• C (+200,000) counts only stream inflow and release.
• D (+460,000) adds evaporation instead of subtracting it.

Review cue: List every in and out with a sign before adding. Convert depth to volume with the surface area.

Source: USGS Water-Supply Paper 2425, Technical Aspects of Wetlands, Hydrologic Processes in Wetlands, Figure 18 water-budget equation.

Answer and explanation

Correct answer: C. 4.5 mg/L.

Flow-weighted mass balance:

C = (Q_r C_r + Q_e C_e)/(Q_r + Q_e) = (10 × 2 + 1 × 30)/(10 + 1) = 50/11 = 4.5 mg/L.

The answer must fall between 2 and 30 mg/L, and closer to the larger stream’s value.

Why the other answers miss

• A (5.0) divides the load by 10 instead of the total flow, 11.
• B (16) is a simple average that ignores the very different flows.
• D (32) adds the concentrations.

Review cue: Mix loads, not concentrations. Then divide by total flow.

Source: EPA Region VIII Mixing Zones and Dilution Policy (Dec. 1994, updated Sept. 1995), Appendix A, Eq. A-9 and Figure A-1, with the fraction of upstream flow mixed equal to 1.

Answer and explanation

Correct answer: D. 263 mg/L.

BOD_t = L₀(1 − e^(−kt)), so L₀ = BOD₅/(1 − e^(−0.23 × 5)) = 180/(1 − e^(−1.15)) = 180/(1 − 0.317) = 180/0.683 = 263 mg/L.

Ultimate BOD must be larger than BOD₅, because only part of the demand is exerted in 5 days.

Why the other answers miss

• A (123) multiplies by 0.683 instead of dividing.
• B (180) assumes all the demand is exerted in 5 days.
• C (568) divides by e^(−kt) (0.317), the fraction remaining, instead of the fraction exerted.

Review cue: For this first-order model and positive BOD, ultimate BOD is larger than BOD₅. Check your answer against that.

Source: Rowan University CEE study guide: Oxygen Demand, Biological Oxygen Demand section, PDF p. 2, BOD_t = L₀(1 − e^(−kt)).

Answer and explanation

Correct answer: B. 60 m³/day.

Darcy’s law: Q = KiA.

A = 50 m × 20 m = 1,000 m².

Q = 15 × 0.004 × 1,000 = 60 m³/day.

Porosity doesn’t enter the discharge calculation. It matters for velocity (next question).

Why the other answers miss

• A (0.06) is the Darcy flux q = Ki in m/day, not a flow rate.
• C (240) divides by porosity. That belongs in seepage velocity, not discharge.
• D (15,000) multiplies K by area and leaves out the gradient.

Review cue: Darcy discharge uses the full cross-section and no porosity.

Source: The Groundwater Project, Hydrogeologic Properties of Earth Materials, §4.1 Darcy’s Law, Eqs. 16–18.

Answer and explanation

Correct answer: 833 days. Accepted range here: 832.5 ≤ answer < 833.5.

Darcy flux: q = Ki = 15 × 0.004 = 0.06 m/day.

Average linear (seepage) velocity: v = q/nₑ = 0.06/0.25 = 0.24 m/day.

Travel time: t = 200/0.24 = 833 days.

Common wrong answers

• 3,333 days uses the Darcy flux as the travel velocity. Water only moves through the pore space, so it travels faster than q.
• 13.3 days treats K as a velocity without the gradient or porosity.

Review cue: Water travels through pores, so divide q by effective porosity before computing time.

Source: The Groundwater Project, Hydrogeologic Properties of Earth Materials, §4.1 Darcy’s Law, Eqs. 16–18.

Answer and explanation

Correct answer: C. 275 m²/day.

Thiem: T = Q ln(r₂/r₁) / [2π(s₁ − s₂)]

T = 1,500 × ln(100/10) / [2π × (3.0 − 1.0)] = 1,500 × 2.303 / 12.57 = 275 m²/day.

Why the other answers miss

• A (119) uses log₁₀ without the matching conversion factor.
• B (550) uses π instead of 2π.
• D (1,727) leaves out the 2π term entirely.

Review cue: This form of the Thiem equation uses ln and 2π. Check which log your equation form calls for.

Source: MIT OpenCourseWare 1.72 Groundwater Hydrology, Lecture 8, p. 1, assumptions; p. 2, Thiem equation.

Answer and explanation

Correct answer: A. 3.67.

R = 1 + (ρb/n)Kd = 1 + (1.6/0.30) × 0.5 = 1 + 2.67 = 3.67.

The units cancel because 1 mL = 1 cm³. The contaminant front moves at about 1/3.67 of the groundwater velocity. Sorption slows the plume; it doesn’t destroy the contaminant.

Why the other answers miss

• B (1.09) swaps bulk density and porosity.
• C (1.24) multiplies by porosity instead of dividing.
• D (2.67) leaves out the leading 1.

Review cue: For this sorption model, R is at least 1. Check that the units of ρb·Kd cancel.

Source: EPA On-line Tools for Site Assessment Calculation: Retardation Factor, retardation equation.

Answer and explanation

Correct answer: D. 92%.

Track what passes through. Fraction remaining after both = (1 − 0.60)(1 − 0.80) = 0.40 × 0.20 = 0.08.

Overall removal = 1 − 0.08 = 92%.

Why the other answers miss

• A (48%) multiplies the removal efficiencies instead of the fractions remaining.
• B (70%) averages the two efficiencies.
• C (140%) adds them. Removal can’t exceed 100%.

Review cue: Multiply the fractions that pass through, then subtract from 1.

Source: DOE Fundamentals Handbook: Thermodynamics, Heat Transfer, and Fluid Flow, Vol. 3 (DOE-HDBK-1012/3-92), printed p. 10, Conservation of Mass; the calculation above applies that principle to the supplied removal fractions.

Answer and explanation

Correct answer: A. 108 mg·min/L.

Theoretical detention time: TDT = V/Q = 500/4,000 d = 0.125 d = 180 min.

Effective contact time: T = TDT × baffling factor = 180 × 0.5 = 90 min. (EPA calls this T₁₀, the time by which only 10% of the water has passed through.)

CT = C × T = 1.2 × 90 = 108 mg·min/L.

This is a calculation only. Whether it meets a required inactivation level depends on the target organism, temperature, and pH, which aren’t given.

Why the other answers miss

• B (0.15) uses C × V/Q in days and also omits the baffling factor.
• C (90) is T alone, before multiplying by the residual.
• D (216) uses the full theoretical detention time and skips the baffling factor.

Review cue: CT uses effective contact time, not V/Q. Apply the baffling factor and work in minutes.

Source: EPA Disinfection Profiling and Benchmarking Technical Guidance Manual (EPA 815-R-20-003, 2020), §§4.2–4.4, printed pp. 24–28 (PDF pp. 34–38): CT = C × T; TDT = V/Q; T = TDT × BF.

Answer and explanation

Correct answer: C. 0.27 d⁻¹.

F/M = (Q × S₀)/(V × X), with X as MLVSS, used here as the biomass measure:

F/M = (10,000 × 200)/(3,000 × 2,500) = 2,000,000/7,500,000 = 0.27 d⁻¹ (kg BOD per kg MLVSS per day).

The mg/L and m³ units cancel, leaving per day.

Why the other answers miss

• A (0.22) uses MLSS instead of the MLVSS the question specifies. MLVSS is the volatile fraction of the suspended solids.
• B (0.67) is the volumetric organic loading in kg BOD/(m³·d). It leaves out the biomass.
• D (3.75) inverts the ratio.

Review cue: F is the BOD load per day; M is the biomass in the basin. Divide load by mass.

Source: EPA, Field Manual for Performance Evaluation and Troubleshooting at Municipal Wastewater Treatment Facilities (January 1978), printed pp. 66–67, Food to Microorganism (F/M) Ratio and calculation Step 4; EPA, Development of a Biological Simulation Monitor for Joint Municipal/Industrial Treatment Systems, §5, printed p. 8, paragraph (a) MLVSS.

Answer and explanation

Correct answer: B. 15.5 days.

SRT = solids in the system ÷ solids leaving per day.

In system: 3,000 m³ × 3,000 g/m³ = 9,000,000 g.

Leaving: waste 60 × 8,000 = 480,000 g/d, plus effluent 10,000 × 10 = 100,000 g/d, for 580,000 g/d.

SRT = 9,000,000 / 580,000 = 15.5 days.

Why the other answers miss

• A (0.3) is the hydraulic residence time, V/Q.
• C (18.8) ignores solids lost in the effluent.
• D (50) treats the waste sludge as if it were at MLSS concentration and ignores effluent solids.

Review cue: Count every path solids take out of the system, including the effluent.

Source: EPA, Field Manual for Performance Evaluation and Troubleshooting at Municipal Wastewater Treatment Facilities (January 1978), printed p. 68, Step 6, symbolic SRT inventory/loss-rate equation.

Answer and explanation

Correct answer: D. 194 mg/L as CaCO₃.

Calcium: 48 × 2.50 = 120 mg/L as CaCO₃.

Magnesium: 18 × 4.12 = 74 mg/L as CaCO₃.

Total hardness = 194 mg/L as CaCO₃.

The factors are ratios of equivalent weights: CaCO₃ (50.0) ÷ Ca²⁺ (20.0) ≈ 2.50, and 50.0 ÷ Mg²⁺ (12.15) ≈ 4.12.

Why the other answers miss

• A (66) adds the ion masses without converting to a common basis.
• B (74) counts magnesium only.
• C (120) counts calcium only.

Review cue: Convert every ion to the same basis before adding.

Source: EPA, National Functional Guidelines for Inorganic Superfund Data Review (August 2014), printed p. 42, Hardness (Total) in Aqueous/Water Samples, H = 2.497[Ca] + 4.118[Mg]; this item uses the supplied rounded factors.

Answer and explanation

Correct answer: 84%. Accepted range here: 83.5 ≤ answer < 84.5.

Dry solids mass is m_s = ρVw, where ρ is bulk sludge density and w is dry-solids mass fraction. The dry solids don’t change, and density is equal, so volume is inversely proportional to the solids fraction:

V₂/V₁ = 4%/25% = 0.16.

Volume reduction = 1 − 0.16 = 84%.

Common wrong answers

• 16% is the fraction of volume that remains, not the reduction.
• 21% is the difference in solids percentage (25 − 4). It isn’t a volume change.

Review cue: Hold the dry solids constant. Volume scales with 1/(solids fraction).

Source: DOE Fundamentals Handbook: Thermodynamics, Heat Transfer, and Fluid Flow, Vol. 3 (DOE-HDBK-1012/3-92), printed p. 2, density as mass per volume; printed p. 10, Conservation of Mass.

Answer and explanation

Correct answer: A and B.

Conventional nitrogen removal is a two-stage process.

• Nitrification (aerobic): bacteria oxidize ammonia to nitrite, then nitrite to nitrate. This converts nitrogen from one form to another but doesn’t remove it.
• Denitrification (anoxic): bacteria use nitrate in place of oxygen and convert it to nitrogen gas, which leaves the water.

Why the other answers miss

• C is wrong because nitrification ends at nitrate. Removing nitrogen as a gas takes denitrification.
• D is wrong because denitrification happens under anoxic conditions, where nitrate, not dissolved oxygen, accepts the electrons.

Review cue: Nitrification changes the form; denitrification removes the nitrogen.

Source: EPA Municipal Nutrient Removal Technologies Reference Document (EPA 832-R-08-006, 2008), §2.2.1; EPA Distribution System Issue Paper: Nitrification (2002), pp. 3 and 9.

Answer and explanation

Correct answer: C. 1,000 kg/yr.

E = A × EF × (1 − ER/100) = 50,000 × 0.4 × (1 − 0.95) = 20,000 × 0.05 = 1,000 kg/yr.

The question says capture is complete, so the overall reduction equals the device efficiency.

Why the other answers miss

• A (2,000) uses 90% instead of 95%.
• B (19,000) is the mass removed, not the mass emitted.
• D (20,000) is the uncontrolled emission rate.

Review cue: Emitted = uncontrolled × (1 − efficiency). Check whether you found what’s emitted or what’s removed.

Source: EPA AP-42 Introduction, p. 1, general emission equation.

Answer and explanation

Correct answer: A. 131 µg/m³.

mg/m³ = ppm × MW / 24.45 = 0.05 × 64.06 / 24.45 = 0.131 mg/m³.

Convert to micrograms: 0.131 × 1,000 = 131 µg/m³.

Why the other answers miss

• B (0.131) is in mg/m³ but labeled µg/m³.
• C (71) uses 22.4 L/mol (0 °C) and the molecular weight of S alone (32).
• D (143) uses 22.4 L/mol, the molar volume at 0 °C, while the gas is at 25 °C.

Review cue: Match the molar volume to the stated temperature and pressure.

Source: NIOSH Occupational Exposure Banding: ppm–mg/m³ calculator, conversion at 25 °C and 1 atm.

Answer and explanation

Correct answer: B. 1.05 mg/m³.

Ventilation flow: Q = ACH × V = 0.60 × 300 = 180 m³/h.

Steady state, well mixed: C_in = C_out + S/Q = 0.05 + 180/180 = 1.05 mg/m³.

Why the other answers miss

• A (0.65) divides the source by room volume instead of ventilation flow.
• C (0.95) subtracts the outdoor concentration. Outdoor air brings the pollutant in.
• D (1.00) leaves out the outdoor background.

Review cue: Convert air changes to a flow, then add the outdoor background.

Source: EPA, Introduction to Indoor Air Quality: A Reference Manual (EPA/400/3-91/003), §2.1, “Effect of Indoor Sources”.

Answer and explanation

Correct answer: A, C, D, E, G, and I.

EPA sets NAAQS for six criteria pollutants: carbon monoxide, lead, nitrogen dioxide, ozone, particle pollution (PM), and sulfur dioxide.

Why the other answers miss

• B (carbon dioxide) is a greenhouse gas but not a criteria pollutant.
• F (benzene) is regulated as a hazardous air pollutant, not under NAAQS.
• H (radon) is mainly an indoor-air and soil-gas concern, with no NAAQS.

Review cue: Six criteria pollutants: CO, Pb, NO₂, O₃, PM, SO₂.

Source: EPA, NAAQS Table, six criteria pollutants; EPA, Overview of Greenhouse Gases, opening list, carbon dioxide; EPA, Initial List of Hazardous Air Pollutants with Modifications, benzene, CAS 71432; EPA, RadTown Radon Activity 3: Indoor Radon Levels, activity steps 2–3, soil/rock origin and indoor accumulation.

Answer and explanation

Correct answer: A and B.

• A is ignitable (D001): a liquid with a flash point below 60 °C (140 °F).
• B is corrosive (D002): an aqueous waste with pH ≤ 2.

The stated results for the other two fall short of the relevant regulatory thresholds.

Why the other answers miss

• C does not meet the pH criterion for corrosivity. The high-pH threshold is pH ≥ 12.5, and 11.5 is below it.
• D does not meet the toxicity-characteristic threshold for benzene. Its benzene concentration, 0.3 mg/L, is below the 0.5 mg/L regulatory level.

Review cue: Compare each result to its exact threshold, and note which side of the threshold counts.

Source: eCFR, 40 CFR Part 261 Subpart C — Characteristics of Hazardous Waste, §261.20(a); §261.21(a)(1); §261.22(a)(1)–(2); §261.24(a), Table 1, benzene.

Answer and explanation

Correct answer: D.

The TCLP extract (6.2 mg/L) exceeds the lead regulatory level (5.0 mg/L), so the waste exhibits the toxicity characteristic. Its EPA hazardous waste number is D008.

Why the other answers miss

• A names a characteristic the waste doesn’t have.
• B is wrong because the toxicity characteristic is not limited to liquid wastes. TCLP evaluates contaminant mobility in liquid, solid, and multiphasic wastes.
• C is wrong because lead appears on the toxicity characteristic list (D008).

Review cue: Compare the extract concentration to the regulatory level, then name the matching D code.

Source: eCFR, 40 CFR Part 261 Subpart C — Characteristics of Hazardous Waste, §261.20(a); §261.24(a), Table 1, lead 5.0 mg/L and D008; EPA, SW-846 Test Method 1311: Toxicity Characteristic Leaching Procedure, opening paragraph, liquid, solid, and multiphasic wastes.

Answer and explanation

Correct answer: A. 91,250 m³/yr.

Daily volume = mass ÷ density = 200 t/d ÷ 0.8 t/m³ = 250 m³/d.

Annual volume = 250 × 365 = 91,250 m³/yr.

Why the other answers miss

• B (250) is the daily volume, labeled as annual.
• C (58,400) multiplies by density instead of dividing.
• D (73,000) is the annual mass in tons, not volume.

Review cue: Volume = mass ÷ density. Check the time basis at the end.

Source: EPA Environmental Geophysics: Density, definition.

Answer and explanation

Correct answer: C. 99.992%; meets it.

DRE = (W_in − W_out)/W_in × 100% = (1,000 − 0.08)/1,000 × 100% = 99.992%.

99.992% ≥ 99.99%, so it meets the requirement.

Why the other answers miss

• A misplaces a decimal: 0.08/1,000 is 0.008%, not 0.08%.
• B has the right DRE but compares it wrongly. 99.992 is larger than 99.99.
• D reports the fraction emitted, not the fraction destroyed and removed.

Review cue: Write all the nines out and line up the decimals before comparing.

Source: eCFR, 40 CFR 264.343 Performance standards (incinerators), (a)(1).

Answer and explanation

Correct answer: B. 3,117 kg.

Carbon burned: 1,000 × 0.85 = 850 kg.

C + O₂ → CO₂, so each 12 kg of carbon makes 44 kg of CO₂.

CO₂ = 850 × 44/12 = 3,117 kg.

Why the other answers miss

• A (850) is the mass of carbon, not CO₂.
• C (2,267) is the mass of oxygen consumed (850 × 32/12), not the CO₂ produced.
• D (3,667) treats the fuel as 100% carbon.

Review cue: Find the carbon mass first, then scale by 44/12.

Source: OpenStax Chemistry 2e, §4.3 Reaction Stoichiometry, Example 4.10.

Answer and explanation

Correct answer: D. 34%.

Efficiency = energy out ÷ energy in = 3,412 Btu / 10,000 Btu = 0.341 = 34%.

A lower heat rate means a more efficient plant.

Why the other answers miss

• A (3.4%) misplaces the decimal.
• B (66%) is the fraction lost as heat, not the efficiency.
• C (293%) inverts the ratio. Efficiency can’t exceed 100%.

Review cue: Heat rate is input per unit of output. Efficiency is its inverse, in matching units.

Source: EIA FAQ: What is the efficiency of different types of power plants?, heat rate and 3,412 Btu/kWh.