Free FE Environmental practice problems

Ten real questions from our FE Environmental bank — one from each knowledge area, with the same exam-realistic figures and fully worked solutions our members get. No signup, no email. Try each one first, then reveal the solution.

5 with figures10 knowledge areasFull worked solutions

1
medium
A. Fluid statics & dynamics
Water flows steadily through a pipe that contracts from a diameter $D_1 = 200$ mm to $D_2 = 100$ mm. The upstream velocity is $V_1 = 2$ m/s. The velocity in the smaller section is most nearly:
1. 1 m/s
2. 32 m/s
3. 4 m/s
4. 8 m/s
Show the answer & worked solution
Answer: D.8 m/s
Continuity $A_1V_1=A_2V_2$ with $A\propto D^2$ gives $V_2=V_1(D_1/D_2)^2=2\,(200/100)^2=8$ m/s. Distractors: the diameter ratio raised to the FOURTH power ( $V_1(D_1/D_2)^4$ , the area-ratio squaring applied a second time); the DIAMETER ratio used un-squared ( $V_1 D_1/D_2$ , forgetting $A\propto D^2$ ); and the ratio both inverted and un-squared ( $V_1 D_2/D_1$ ).
2
medium
A. Time value of money & equivalence
An engineer borrows $200,000 to be repaid in 15 equal end-of-year payments at 6% per year. The annual payment is most nearly:
1. $12,000
2. $20,593
3. $8,593
4. $13,333
Show the answer & worked solution
Answer: B.$20,593
$(A/P,0.06,15)=\dfrac{i(1+i)^n}{(1+i)^n-1}=0.10296$ , so $A=$ $200,000 $\times\ 0.10296=$ $20,593. Distractors: $P/n$ ignoring interest, the interest-only payment $P\cdot i$ , and the sinking-fund factor (A/F) used in place of A/P.
3
easy
A. Dose-response, exposure & occupational health
The dose-response curve shown is for a noncarcinogenic solvent, a noncarcinogen. Which statement correctly interprets it?
1. Response decreases as dose increases beyond the threshold, so larger doses are progressively safer.
2. Any dose above zero produces a proportional response; there is no threshold and risk rises linearly from the origin.
3. A threshold exists at the NOAEL (about 5 mg/kg·day): no adverse effect occurs below it, and the reference dose (RfD) is obtained by dividing the NOAEL by uncertainty factors.
4. The marked threshold is the LD₅₀ — the dose that is lethal to 50% of the exposed population.
Show the answer & worked solution
Answer: C.A threshold exists at the NOAEL (about 5 mg/kg·day): no adverse effect occurs below it, and the reference dose (RfD) is obtained by dividing the NOAEL by uncertainty factors.
A threshold (noncarcinogen) curve shows ZERO response up to the NOAEL, then an S-shaped rise. Below the NOAEL no adverse effect is expected; the RfD is set as $RfD = NOAEL/(UF\cdot MF)$ . Distractors: the linear-no-threshold (carcinogen) interpretation; an impossible "higher dose is safer" reading; and confusing the NOAEL with the LD₅₀ (a lethality endpoint, not the no-effect level).
4
medium
A. Central tendency, dispersion & distributions
The breakdown voltage of a batch of Zener diodes is normally distributed with mean $\mu=50$ V and standard deviation $\sigma=8$ V. Given $\Phi(0.50)=0.6915$ and $\Phi(1.50)=0.9332$ , the probability that a randomly selected diode breaks down between 54 V and 62 V is most nearly:
1. 0.6247
2. 0.2417
3. 0.7583
4. 0.9332
Show the answer & worked solution
Answer: B.0.2417
$z_1=(54-50)/8=0.50$ and $z_2=(62-50)/8=1.50$ . $P=\Phi(z_2)-\Phi(z_1)=0.9332-0.6915=0.2417$ . Distractors subtract the wrong lower tail ( $1-\Phi(z_1)$ instead of $\Phi(z_1)$ ), ignore the lower limit and report $\Phi(z_2)$ alone, and report the complement $1-P$ .
5
easy
A. Runoff & water budget
An urban catchment of area $A = 12$ ha has a runoff coefficient $C = 0.7$ . During a design storm the rainfall intensity is $i = 90$ mm/h. Using the rational method $Q = 0.00278\,C\,i\,A$ (Q in m³/s, i in mm/h, A in ha), the peak runoff is most nearly:
1. 0.21 m³/s
2. 21 m³/s
3. 2.1 m³/s
4. 3 m³/s
Show the answer & worked solution
Answer: C.2.1 m³/s
$Q = 0.00278\,C\,i\,A = 0.00278(0.7)(90)(12) = 2.102$ m³/s. Distractors: the runoff coefficient $C$ OMITTED (treated as 1); the unit factor a decade too large (0.0278 instead of 0.00278); and the factor derived as $1/3600 = 0.000278$ instead of $1/360 = 0.00278$ (the mm/h→m/s conversion taken a decade too small).
6
medium
A. Codes of ethics, liability & licensure
An engineer preparing the switchgear specification for a hospital project owns a 20 percent interest in one of the three manufacturers likely to bid on the equipment. Under the NCEES Model Rules of Professional Conduct, the engineer should:
1. sell the ownership interest quietly after the contract is awarded to eliminate the conflict
2. write the specification around generic performance requirements so the ownership has no practical effect
3. disclose the ownership interest to the client in writing before continuing with the specification
4. continue normally, since the bids will be evaluated against objective criteria the client controls
Show the answer & worked solution
Answer: C.disclose the ownership interest to the client in writing before continuing with the specification
The Model Rules require prompt written disclosure of any business interest that could influence professional judgment; the informed client then decides how to proceed. Distractors: drafting around the interest conceals the conflict instead of disclosing it; objective bid criteria do not cure an undisclosed stake in a bidder who helped shape the spec; divesting after award removes the benefit only after the influence has already operated.
7
easy
A. Stoichiometry, equilibrium & acid-base
An acidic mountain stream has a hydrogen-ion activity of $[\text{H}^+] = 2\times10^{-5}$ mol/L. Its pH is most nearly:
1. 9.3
2. 5.3
3. 4.7
4. 5
Show the answer & worked solution
Answer: C.4.7
$\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(2\times10^{-5}) = 4.7$ . Distractors: the POWER of ten read off alone with the mantissa ignored (pH taken as the exponent magnitude 5); the mantissa log ADDED instead of subtracted ( $-e+\log_{10}m$ rather than $-e-\log_{10}m$ ); and the pOH computed in place of pH ( $14-\text{pH}$ ).
8
easy
A. Darcy's law & hydrogeology
A confined sand aquifer carries flow between two wells that are $L = 60$ m apart with a head difference of $\Delta h = 3$ m. The hydraulic conductivity is $K = 12$ m/d and the flow cross-section is $A = 50$ m². The volumetric discharge is most nearly:
1. 0.6 m³/d
2. 12000 m³/d
3. 1800 m³/d
4. 30 m³/d
Show the answer & worked solution
Answer: D.30 m³/d
$i = \Delta h/L = 3/60 = 0.05$ , so $Q = K\,i\,A = 12\cdot0.05\cdot50 = 30$ m³/d. Distractors: the cross-sectional area OMITTED ( $K\,i$ , which is the Darcy velocity, not a discharge); the gradient INVERTED ( $L/\Delta h$ used for $i$ ); and the head drop $\Delta h$ used directly as the gradient (the division by the flow-path length $L$ dropped).
9
medium
A. Analytic geometry, algebra & trigonometry
The roots of the quadratic equation $x^2 - 7x + 10 = 0$ are most nearly:
1. x = 2.5 and x = 1
2. x = 5 and x = 2
3. x = -2 and x = -5
4. x = 8.22 and x = -1.22
Show the answer & worked solution
Answer: B.x = 5 and x = 2
$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{7\pm\sqrt{49-40}}{2}=\dfrac{7\pm3}{2}$ , so $x_1=5$ and $x_2=2$ . Distractors flip the sign of $-b$ (using $+b$ in the numerator), use $b^2+4ac$ for the discriminant, and divide by $4a$ instead of $2a$ (half the true roots).
10
easy
A. Solid waste management & landfills
A municipal collection system serves a population of 85,000 people, each generating 1.9 kg of municipal solid waste per person per day. The total MSW generation rate is most nearly:
1. 62.1 tonnes/d
2. 162 tonnes/d
3. 162000 tonnes/d
4. 58900 tonnes/d
Show the answer & worked solution
Answer: B.162 tonnes/d
Generation $= \text{pop}\times r = 85,000\times1.9 = 161,500$ kg/d $= 161.5$ tonnes/d. Distractors: the kg→tonne conversion DROPPED (÷1000 omitted, 1000× high); the daily rate reported on an ANNUAL basis (×365); and a per-HOUSEHOLD basis (population divided by ~2.6 persons/household before applying the per-capita rate).

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