FE Environmental practice problems with solutions
Below are five FE Environmental practice problems in exam style, each with a full step-by-step solution and the FE Reference Handbook 10.6 page. Give yourself about 3 minutes per problem, the same pace as the real 110-question exam.
Last reviewed October 10, 2026 by Engineer Exam Lab. Exam facts link to NCEES. Problems are from our practice packs, written by a PhD environmental engineer, and every numeric answer was checked by a second, independent calculation.
How to use these problems
Open the FE Reference Handbook 10.6, set a timer for 15 minutes, and work all five before you read a solution. That is about the pace of the real exam: 110 questions in 5 hours 20 minutes, or 2.9 minutes each. Then check each step and the handbook page we cite, and note the section you had to search for.
The five problems come from five different NCEES FE Environmental knowledge areas. The question ranges for each area are on the FE Environmental topics guide.
Five FE Environmental practice problems
Problem 1 · FE Environmental · Mathematics · Multiple choice
For f(x) = 3x³ + 6x² + 5x - 2, the slope of the curve at x = 4 is most nearly:
Show the worked solution
Answer: C) 197
Equation: d(axⁿ)/dx = n·a·xⁿ⁻¹; the slope is f′(x0)
- f′(x) = 9x² + 12x + 5
- f′(4) = 3(3)(4)² + 2(6)(4) + (5) = 197
Why the wrong choices are wrong:
- A is the second derivative f″(x0).
- B is f(x0), the function value, not the slope.
- D multiplied by each power but did not lower it.
Handbook: FE Reference Handbook 10.6, Mathematics, Differential Calculus (the derivative), p. 47.
Problem 2 · FE Environmental · Fundamental Principles · Multiple choice
A contaminant enters an ideal completely mixed flow reactor (CMFR) at 145 mg/L and decays by a first-order reaction with k = 3.9/d. The hydraulic residence time is 4 h. At steady state, the effluent concentration is most nearly:
Show the worked solution
Answer: D) 87.9 mg/L
Equation: Ct = Co/(1 + kθ) (first order, steady state)
- θ = 4 h = 4/24 d = 0.1667 d; kθ = 3.9 × 0.1667 = 0.6500
- Ct = 145/(1 + 0.6500) = 87.9 mg/L
Why the wrong choices are wrong:
- A is the amount removed, not the effluent concentration.
- B used θ in hours with k per day.
- C used Co(1 - kθ), a zero-order-like straight line.
Handbook: FE Reference Handbook 10.6, Environmental Engineering, Steady-State Reactor Parameters (first order), p. 331.
Problem 3 · FE Environmental · Fluid Mechanics and Hydraulics · Multiple choice
A solid rectangular block 5 ft tall with a specific gravity of 0.75 floats upright in seawater (SG = 1.025). The depth of the block below the water line is most nearly:
Show the worked solution
Answer: B) 3.66 ft
Equation: Buoyant force = weight of displaced fluid; γf·Vsub = γb·V
- Floating: buoyant force = weight, so γf·A·y = γb·A·H
- y = H·SGb/SGf = 5 ft × 0.75/1.025 = 3.66 ft
Why the wrong choices are wrong:
- A multiplied the two specific gravities instead of dividing.
- C took the fluid as fresh water (SG 1.00).
- D divided the height by the block's specific gravity instead of multiplying.
Handbook: FE Reference Handbook 10.6, Fluid Mechanics, Archimedes Principle and Buoyancy, p. 184.
Problem 4 · FE Environmental · Groundwater, Soils, and Sediments · Fill in the blank
A well fully penetrates a confined aquifer 19 m thick with K = 28 m/d. At steady pumping, observation wells 20 m and 60 m from the well show piezometric heads of 36.4 m and 37.7 m above the aquifer bottom. Find the pumping rate, in m³/d.
Show the worked solution
Answer: 3,960 m³/d
Equation: Q = 2πT(h2 - h1)/ln(r2/r1), T = Kb
- T = Kb = 28 × 19 = 532 m²/d
- Q = 2πT(h2 - h1)/ln(r2/r1) = 2π(532)(37.7 - 36.4)/ln(60/20) = 3,960 m³/d
Handbook: FE Reference Handbook 10.6, Civil Engineering, Groundwater, Thiem Equation (confined aquifer), p. 299.
Problem 5 · FE Environmental · Water and Wastewater · Fill in the blank
A circular clarifier 45 ft in diameter treats a flow of 1.96 MGD. Find the overflow rate in gpd/ft².
Show the worked solution
Answer: 1,230 gpd/ft²
Equation: vo = Q/A_surface, with A = πD²/4
- A = πD²/4 = π(45 ft)²/4 = 1,590 ft²
- Q = 1.96 MGD = 1,960,000 gal/d
- vo = Q/A = (1,960,000 gal/d)/(1,590 ft²) = 1,230 gpd/ft²
Handbook: FE Reference Handbook 10.6, Environmental Engineering, Clarifier (overflow rate), p. 345.
What to look for when you check your work
- Units. Wrong choices are often built from a unit slip: hours vs days, mg/L vs kg/d.
- The right equation form. CMFR vs plug flow, first order vs zero order. The handbook lists both side by side.
- Time spent finding the page. If you needed more than a minute to find the equation, practice searching the handbook PDF for that section.
Want 100 more FE Environmental problems like these, plus a timed 110-question mock? $39, one time. Every solution shows each step and the handbook 10.6 page.
What this page doesn't cover
Five problems can't cover 15 knowledge areas. The NCEES FE Environmental interactive practice exam (50 questions per volume, written from retired exam items) is the closest look at real questions. Our pack adds 100 problems by area and a full 110-question mock.
Questions
How many questions are on the FE Environmental exam?
110 questions, with 5 hours 20 minutes of exam time inside a 6-hour appointment, according to NCEES.
Are these real FE exam questions?
No. They are original practice problems written in the style of the exam. Real exam questions are confidential.
Which handbook version do the page numbers use?
FE Reference Handbook 10.6. Check which edition applies to your exam date in your MyNCEES account.
How long should each problem take?
About 3 minutes on average. The exam gives 320 minutes for 110 questions, roughly 2.9 minutes each.
Related guides
- FE Environmental exam topics and question counts
- FE exam sample questions by type
- How to take an FE practice exam
- FE Environmental practice pack ($39)
Engineer Exam Lab is not affiliated with or endorsed by NCEES. The problems here are original practice problems, not questions from a real exam.