Free PE Civil Transportation Practice Test
Here are 40 original, unofficial practice questions covering all 10 NCEES PE Civil: Transportation knowledge areas, each with a worked solution. Every answer holds whether you test before or after the April 2027 HCM and MUTCD change, and there's no signup.
Practice questions
Question 1 of 40
Vertical Design · Multiple choice · TRN-01
A crest vertical curve joins a +3.0% grade to a −2.0% grade on a road with a 55-mph design speed. Use a design stopping sight distance of 495 ft, a driver eye height of 3.5 ft, and an object height of 2.0 ft. Assume the sight distance is shorter than the curve. What is the minimum curve length?
Show answer and explanation
Answer: C — 568 ft
A = |+3.0 − (−2.0)| = 5.0 (percent).
For S < L on a crest curve with h₁ = 3.5 ft and h₂ = 2.0 ft, L = AS² ÷ 2,158. (The 2,158 comes from 200(√3.5 + √2.0)².)
L = 5.0 × 495² ÷ 2,158 = 567.7 ft, so 568 ft.
Check the assumption: S = 495 ft is less than L = 568 ft, so the S < L form was the right one.
Why the other answers miss: A (438 ft) uses 2,800, the constant for passing sight distance, where the object height is 3.5 ft. B (558 ft) uses the S > L form, 2S − 2,158/A, which doesn't apply once L comes out longer than S. D (575 ft) uses the sag headlight equation, AS² ÷ (400 + 3.5S), on a crest curve.
Source: TxDOT Roadway Design Manual §4.11.1, Table 4-23; MDT Road Design Manual, Appendix K worked calculations
Question 2 of 40
Traffic Engineering · Multiple choice · TRN-02
A traffic count records 360, 420, 480, and 340 vehicles in four consecutive 15-minute periods. What is the peak-hour factor (PHF) for this hour, to two decimal places?
Show answer and explanation
Answer: B — 0.83
Hourly volume V = 360 + 420 + 480 + 340 = 1,600 veh.
Peak 15-minute count = 480 veh, so the peak rate is 4 × 480 = 1,920 veh/h.
PHF = V ÷ (4 × V₁₅) = 1,600 ÷ 1,920 = 0.833, so 0.83.
Why the other answers miss: A (0.25) puts the peak 15-minute count in the numerator instead of the hourly total. C (1.00) uses the average 15-minute count (400) in the denominator, which always returns 1.00. D (1.20) flips the ratio. PHF can never exceed 1.00.
Source: FHWA, Traffic Data for Highway Engineering, §§5-1a and 5-2b
Question 3 of 40
Project Management · Multiple choice · TRN-03
A rectangular pavement area 900 m long and 7.2 m wide needs a uniform 40-mm layer removed. The unit price is $35 per cubic meter. Ignore bulking, waste, and mobilization. What is the estimated removal cost?
Show answer and explanation
Answer: D — $9,072
Depth: 40 mm = 0.040 m.
Volume = 900 m × 7.2 m × 0.040 m = 259.2 m³.
Cost = 259.2 m³ × $35/m³ = $9,072.
Why the other answers miss: A ($907.20) converts 40 mm to 0.004 m, a factor-of-ten slip. B ($259.20) reports the volume as if it were dollars. C ($226,800) prices the surface area and forgets the depth.
Source: Original calculation from the values given.
Question 4 of 40
Horizontal Design · Fill in the blank · TRN-04
A simple circular curve has a radius of 1,200 ft and a central angle of 32°. The PI is at station 45+30.00. There are no spirals or station equations. What is the PT station? Enter it in feet (for example, 4,856.11 for 48+56.11), to the nearest 0.01 ft.
Your answer: ______ ft (station in feet)
Show answer and explanation
Answer: 48+56.11
Tangent: T = R tan(Δ/2) = 1,200 × tan 16° = 344.09 ft.
Arc length: L = πRΔ/180 = π × 1,200 × 32 ÷ 180 = 670.21 ft.
PC = PI − T = 4,530.00 − 344.094 = 4,185.906 → 41+85.91.
PT = PC + L = 4,185.906 + 670.206 = 4,856.112 → 48+56.11. (Keep full precision until the last step; rounding T and L first can shift the last digit.)
Common mistakes: The common slip is PT = PI + T = 48+74.09. Stationing runs along the curve, so you go back T from the PI to the PC, then forward along the arc length L.
Source: MDT Road Design Manual, Appendix K worked calculations
Question 5 of 40
Traffic Control Design · Fill in the blank · TRN-05
A work zone closes one 12-ft lane on a road where the speed used for taper design is 35 mph. Using the MUTCD taper-length formulas, what minimum merging taper length is required? Enter the nearest foot.
Your answer: ______ ft
Show answer and explanation
Answer: 245 ft
At 40 mph or less, L = WS² ÷ 60.
L = 12 × 35² ÷ 60 = 12 × 1,225 ÷ 60 = 245 ft.
A merging taper needs at least L.
Common mistakes: Using L = WS (the 45-mph-and-up formula) gives 420 ft. Check the speed branch first.
Edition note: The formulas are the same in both editions; only the section and table numbers change (6C.08 → 6B.08).
Source: MUTCD 2009, §6C.08 and Tables 6C-3, 6C-4; MUTCD 11th Edition, Part 6, §6B.08 and Tables 6B-3, 6B-4
Question 6 of 40
Drainage · Fill in the blank · TRN-06
A drainage area has 3.2 acres of pavement (C = 0.90) and 4.8 acres of turf (C = 0.25). The design rainfall intensity for a duration equal to the time of concentration is 4.2 in/hr. Using the Rational Method, what is the peak discharge? Enter to the nearest 0.1 cfs.
Your answer: ______ cfs
Show answer and explanation
Answer: 17.1 cfs
Total area = 3.2 + 4.8 = 8.0 ac.
Area-weighted C = (0.90 × 3.2 + 0.25 × 4.8) ÷ 8.0 = (2.88 + 1.20) ÷ 8.0 = 0.51.
Q = CiA = 0.51 × 4.2 × 8.0 = 17.1 cfs (the U.S.-unit conversion factor is 1).
Common mistakes: Averaging the two C values without weighting by area (0.575) gives 19.3 cfs. Weight by area.
Source: TxDOT Hydraulic Design Manual Ch. 4 §12, Eqs. 4-20 and 4-23
Question 7 of 40
Drainage · Multiple choice · TRN-07
A 24-in reinforced concrete pipe (n = 0.013) is laid at 0.5%. It must carry 17.1 cfs. Using Manning's equation for full flow, can it carry the flow?
Show answer and explanation
Answer: B — no; full-flow capacity is about 16.0 cfs
Full circular pipe, D = 2.0 ft: A = π(2.0)²/4 = 3.142 ft²; P = π(2.0) = 6.283 ft; R = A/P = D/4 = 0.50 ft.
Q = (1.49/n) A R^(2/3) S^(1/2) = (1.49/0.013)(3.142)(0.50)^(2/3)(0.005)^(1/2) = 16.0 cfs.
16.0 cfs < 17.1 cfs, so no — the pipe is undersized.
Why the other answers miss: A (8.0 cfs) uses half the area. C (23.2 cfs) uses n = 0.009. D (40.4 cfs) uses the diameter (2.0 ft) as the hydraulic radius; for a full circle, R = D/4.
Source: MoDOT Engineering Policy Guide 750.1, §§750.1.4.1–750.1.4.1.2
Question 8 of 40
Traffic Engineering · Multiple choice · TRN-08
A signalized lane group has a saturation flow rate of 1,980 veh/h for the whole group. Effective green is 50 s in a 110-s cycle. Demand is 945 veh/h. What is the volume-to-capacity (v/c) ratio?
Show answer and explanation
Answer: C — 1.05
Capacity c = s(g/C) = 1,980 × (50/110) = 900 veh/h.
v/c = 945 ÷ 900 = 1.05. Demand exceeds capacity.
Why the other answers miss: A (0.48) divides by saturation flow without allocating green time. B (0.95) flips capacity and demand. D (2.20) is C/g. The saturation flow is already for the whole lane group, so don't multiply by a lane count.
Source: FHWA Traffic Signal Timing Manual (FHWA-HOP-08-024), §§3.3.3–3.3.5, Eqs. 3-1 to 3-3
Question 9 of 40
Traffic Signals · Fill in the blank · TRN-09
An approach has a speed of 45 mph and a 2% downgrade. Use the ITE yellow change interval equation with a perception-reaction time of 1.0 s and a deceleration rate of 10 ft/s². What is the yellow change interval? Enter to the nearest 0.1 s.
Your answer: ______ s
Show answer and explanation
Answer: 4.5 s
y = t + 1.47V ÷ (2a + 64.4g), with g negative on a downgrade.
1.47 × 45 = 66.15 ft/s.
Denominator = 2(10) + 64.4(−0.02) = 20 − 1.288 = 18.712.
y = 1.0 + 66.15 ÷ 18.712 = 1.0 + 3.535 = 4.535, so 4.5 s.
Common mistakes: Ignoring the grade gives 1.0 + 66.15/20 = 4.3 s. A downgrade makes stopping harder, so the yellow gets longer, not shorter. Using +0.02 instead of −0.02 would wrongly give about 4.1 s.
Edition note: It comes from the ITE equation, not from an MUTCD table.
Source: ADOT Traffic Engineering Guidelines, TGP 621 (ITE-based change and clearance intervals)
Question 10 of 40
Roadside and Cross-Section Design · Matching · TRN-10
Match each smooth, unobstructed foreslope to its usual classification.
| Slope | Classification |
|---|---|
| 1V:6H | ______ |
| 1V:3.5H | ______ |
| 1V:2H | ______ |
Choices: Recoverable · Traversable but non-recoverable · Not traversable (not counted as clear zone)
Show answer and explanation
Answer: 1V:6H → recoverable; 1V:3.5H → traversable but non-recoverable; 1V:2H → not traversable
FHWA describes slopes of 1V:4H or flatter as recoverable, slopes between 1V:3H and 1V:4H as traversable but non-recoverable, and slopes steeper than 1V:3H as not traversable.
- 1V:6H is flatter than 1V:4H → recoverable.
- 1V:3.5H sits between 1V:3H and 1V:4H → traversable but non-recoverable. A driver can usually ride it down but can't stop or turn back on it, so a clear runout area at the bottom matters.
- 1V:2H is steeper than 1V:3H → not traversable.
Common mistakes: The usual mix-up is reading the ratio backward: a bigger horizontal number means a flatter slope.
Source: FHWA, Clear Zone and Horizontal Clearance; FHWA, Roadside Design Improvements at Curves (FHWA-SA-21-029)
Question 11 of 40
Project Management · Select all that apply · TRN-11
All links are finish-to-start with no lag. Select every activity on the critical path.
| Activity | Duration (days) | Immediate predecessor(s) |
|---|---|---|
| A | 4 | — |
| B | 6 | A |
| C | 3 | A |
| D | 5 | B |
| E | 7 | C |
| F | 2 | D, E |
Show answer and explanation
Answer: A, B, D, F — critical path A–B–D–F, 17 days
Path A–B–D–F: 4 + 6 + 5 + 2 = 17 days.
Path A–C–E–F: 4 + 3 + 7 + 2 = 16 days.
The longest path sets the completion time, so the critical path is A–B–D–F (17 days). C and E can each slip 1 day without delaying the project.
Why the other answers miss: E is the single longest activity, which tempts people to put it on the critical path. What counts is the total along each path, not the biggest activity. A and F sit on every path, so they're always critical here.
Source: FHWA Technical Advisory T 5080.15, §9 Critical Path Method
Question 12 of 40
Intersection Geometry · Multiple choice · TRN-12
A passenger car turns left from a stop-controlled minor road onto a four-lane undivided major road with a 45-mph design speed. Approaches are level. The base time gap is 7.5 s, plus 0.5 s for each additional lane, beyond the first, that the turning car crosses. The car crosses the two lanes of traffic approaching from the left before entering its lane. What is the intersection sight distance along the major road?
Show answer and explanation
Answer: B — 529 ft
Lanes crossed from the left = 2, so one lane is beyond the first: t_g = 7.5 + 0.5 = 8.0 s.
ISD = 1.47 V t_g = 1.47 × 45 × 8.0 = 529 ft.
Why the other answers miss: A (496 ft) skips the lane adjustment. C (562 ft) adds 0.5 s for both lanes crossed instead of only the one beyond the first. D (628 ft) uses the 9.5-s single-unit-truck gap for a passenger car.
Source: TxDOT Roadway Design Manual §13.5, Tables 13-2 to 13-4
Question 13 of 40
Horizontal Design · Multiple choice · TRN-13
A horizontal curve has a radius of 1,500 ft measured to the centerline of the inside lane. The required stopping sight distance is 570 ft, and both the driver and the object are on the curve. What horizontal sightline offset (the clear distance from the inside-lane centerline to the obstruction) is needed?
Show answer and explanation
Answer: B — 27.0 ft
HSO = R[1 − cos(28.65S/R)], with the angle in degrees.
28.65 × 570 ÷ 1,500 = 10.887°.
HSO = 1,500 × (1 − cos 10.887°) = 1,500 × 0.0180 = 27.0 ft.
Why the other answers miss: A (6.8 ft) plugs in S/2 instead of S; the 28.65 already contains the halving. C (107.0 ft) uses 57.3 instead of 28.65. D (1,663 ft) enters 10.887 with the calculator in radian mode.
Source: MDT Road Design Manual, Appendix K worked calculations
Question 14 of 40
Geotechnical and Pavement · Fill in the blank · TRN-14
A field density test gives a moist unit weight of 124.0 lb/ft³ at a water content of 12.0%. The laboratory maximum dry unit weight for the specified test is 115.0 lb/ft³. What is the relative compaction? Enter to the nearest 0.1%.
Your answer: ______ %
Show answer and explanation
Answer: 96.3%
Dry unit weight: γd = γ ÷ (1 + w) = 124.0 ÷ 1.12 = 110.7 lb/ft³.
Relative compaction = γd ÷ γd,max × 100 = 110.7 ÷ 115.0 × 100 = 96.3%.
Common mistakes: Dividing the moist unit weight by the lab dry maximum (124.0 ÷ 115.0) gives 107.8%, which compares a wet number with a dry one. Convert to dry first.
Source: FHWA Geotechnical Aspects of Pavements Reference Manual (NHI-05-037), Ch. 5, Eqs. 5.1 and 5.4
Question 15 of 40
Traffic Engineering · Multiple choice · TRN-15
An intersection had 18 crashes over 3 years. The total entering volume is 22,000 vehicles per day. What is the crash rate per million entering vehicles (MEV)?
Show answer and explanation
Answer: B — 0.75 crashes/MEV
R = (C × 1,000,000) ÷ (365 × N × V).
R = 18 × 1,000,000 ÷ (365 × 3 × 22,000) = 18,000,000 ÷ 24,090,000 = 0.75 crashes per MEV.
Why the other answers miss: A (0.07) uses 100,000 instead of 1,000,000. C (2.24) leaves out the number of years. D (7.47) uses 10,000,000.
Source: FHWA Intersection Safety Manual for Local Rural Road Owners, Ch. 3
Question 16 of 40
Vertical Design · Multiple choice · TRN-16
A sag vertical curve joins a −2.0% grade to a +4.0% grade. Design stopping sight distance is 425 ft (50 mph). Use the headlight sight distance criterion and assume the sight distance is shorter than the curve. What is the minimum curve length?
Show answer and explanation
Answer: D — 574 ft
A = |+4.0 − (−2.0)| = 6.0.
For S < L: L = AS² ÷ (400 + 3.5S) = 6.0 × 425² ÷ (400 + 3.5 × 425) = 1,083,750 ÷ 1,887.5 = 574 ft.
Check: S = 425 ft < L = 574 ft, so the S < L form holds.
Why the other answers miss: A (191 ft) takes A as 4 − 2 = 2 instead of the algebraic difference, 6. B (502 ft) uses the crest constant 2,158. C (535 ft) uses the S > L form, 2S − (400 + 3.5S)/A, which doesn't apply here.
Source: MDT Road Design Manual, Appendix K worked calculations
Question 17 of 40
Traffic Signals · Multiple choice · TRN-17
A phase has a displayed green of 32 s, a yellow of 4 s, and a red clearance of 2 s. Start-up lost time is 2 s and clearance lost time is 2 s. Using g = G + Y + R − (l₁ + l₂), what is the effective green time?
Show answer and explanation
Answer: C — 34 s
g = 32 + 4 + 2 − (2 + 2) = 34 s.
Effective green is the time the movement actually discharges at saturation flow. It includes the usable part of the yellow and red clearance and drops the lost time.
Why the other answers miss: A (28 s) subtracts lost time from the displayed green only. B (32 s) assumes effective green equals displayed green. D (38 s) adds the yellow and red clearance and forgets the lost time.
Source: FHWA Traffic Signal Timing Manual (FHWA-HOP-08-024), §§3.3.3–3.3.5, Eqs. 3-1 to 3-3
Question 18 of 40
Horizontal Design · Multiple choice · TRN-18
Using e + f = V² ÷ (15R), what is the minimum radius for a 60-mph design speed with a maximum superelevation of 8% and a maximum side friction factor of 0.12?
Show answer and explanation
Answer: A — 1,200 ft
Use e as a decimal: e = 0.08.
R = V² ÷ [15(e + f)] = 60² ÷ [15(0.08 + 0.12)] = 3,600 ÷ 3.0 = 1,200 ft.
Why the other answers miss: B (2,000 ft) uses friction alone. C (3,000 ft) uses superelevation alone. D (18,000 ft) drops the 15.
Question 19 of 40
Project Management · Fill in the blank · TRN-19
Alternatives A and B provide the same service. A costs $140,000 now and $60,000 at the end of year 8. B costs $170,000 now and $20,000 at the end of year 8. With a 5% discount rate, no salvage, and no other costs, how much lower is A's present cost than B's? Enter the nearest whole dollar.
Your answer: ______ $
Show answer and explanation
Answer: $2,926
(1.05)⁸ = 1.47746.
PW(A) = 140,000 + 60,000 ÷ 1.47746 = 140,000 + 40,610.36 = $180,610.36.
PW(B) = 170,000 + 20,000 ÷ 1.47746 = 170,000 + 13,536.79 = $183,536.79.
Difference = 183,536.79 − 180,610.36 = $2,926. A is cheaper.
Common mistakes: Adding the nominal costs ($200,000 vs. $190,000) makes A look $10,000 more expensive. That ignores timing: A's bigger cost comes later, so it's worth less today.
Question 20 of 40
Roadside and Cross-Section Design · Multiple choice · TRN-20
Where is the width of the clear zone measured from?
Show answer and explanation
Answer: A — The edge of the traveled way
The clear zone is the total roadside border area, starting at the edge of the traveled way, that is available for errant vehicles. A shoulder is part of that area, not its starting point.
Why the other answers miss: B is the most common wrong answer: shoulders count toward the clear zone. C and D aren't reference lines for roadside recovery area.
Question 21 of 40
Traffic Engineering · Multiple choice · TRN-21
A site's safety performance function predicts 4.2 crashes per year under base conditions. Two non-base conditions apply, with crash modification factors of 1.12 and 0.85. The local calibration factor is 1.10. What is the predicted crash frequency?
Show answer and explanation
Answer: B — 4.40 crashes/yr
N_predicted = N_spf × (CMF₁ × CMF₂) × C.
N = 4.2 × 1.12 × 0.85 × 1.10 = 4.40 crashes/yr.
Why the other answers miss: A (4.00) leaves out the calibration factor. C (4.48) adds the CMF effects (+12% − 15% = −3%) instead of multiplying the factors. D (5.17) drops the 0.85.
Question 22 of 40
Horizontal Design · Fill in the blank · TRN-22
A 4° curve (arc definition) has a radius of 1,432.4 ft and a central angle of 24°. What is the curve length? Enter the nearest foot.
Your answer: ______ ft
Show answer and explanation
Answer: 600 ft
L = πRΔ ÷ 180 = π × 1,432.4 × 24 ÷ 180 = 600 ft.
With the arc definition, you get the same answer from L = 100Δ ÷ D = 100 × 24 ÷ 4 = 600 ft. That's the point of the arc definition: D is the angle subtended by a 100-ft arc.
Common mistakes: Mixing up definitions trips people up here. Under the chord definition, a 4° curve has a slightly different radius, and 100Δ/D would measure chords, not arc.
Source: MDT Road Design Manual, Appendix K worked calculations
Question 23 of 40
Traffic Control Design · Multiple choice · TRN-23
Which statement about longitudinal pavement marking colors agrees with the MUTCD?
Show answer and explanation
Answer: A — white for same-direction traffic; yellow can mark the left edge of a one-way roadway
White longitudinal lines separate same-direction traffic and mark the right edge of the roadway. Yellow lines separate opposing traffic, mark the left edge of divided highways, one-way streets, and ramps, and separate two-way left-turn and reversible lanes. So A is correct.
Why the other answers miss: B and C swap the colors. D is the tempting half-truth: yellow does separate opposing traffic, but it's also the left edge line on divided and one-way roadways.
Edition note: The color rules match; the section is 3A.05 in 2009 and 3A.03 in the 11th Edition.
Source: MUTCD 2009, §3A.05; MUTCD 11th Edition, Part 3, §3A.03
Question 24 of 40
Vertical Design · Fill in the blank · TRN-24
A sag vertical curve is 600 ft long, with g₁ = −1.8% and g₂ = +2.4%. The BVC is at station 20+00, elevation 512.40 ft. What is the elevation of the low point? Enter to the nearest 0.01 ft.
Your answer: ______ ft
Show answer and explanation
Answer: 510.09 ft
Low point distance from BVC: x = −g₁L ÷ (g₂ − g₁) = 1.8 × 600 ÷ 4.2 = 257.14 ft (station 22+57.14).
Elevation = BVC + g₁x + [(g₂ − g₁) ÷ (2L)]x² = 512.40 + (−0.018)(257.14) + (0.042 ÷ 1,200)(257.14)² = 512.40 − 4.629 + 2.314 = 510.09 ft.
Common mistakes: Two slips are common: putting the low point at the midpoint of the curve (300 ft from the BVC), and using grades in percent inside the elevation equation instead of decimals.
Source: MDT Road Design Manual, Appendix K worked calculations
Question 25 of 40
Traffic Engineering · Multiple choice · TRN-25
Four vehicles travel a 1,000-ft segment in 12, 14, 15, and 19 seconds. What is the space-mean speed?
Show answer and explanation
Answer: A — 45.5 mph
Space-mean speed = total distance ÷ total time = (4 × 1,000) ÷ (12 + 14 + 15 + 19) = 4,000 ÷ 60 = 66.7 ft/s.
66.7 ÷ 1.467 = 45.5 mph.
Why the other answers miss: B (46.7 mph) is the time-mean speed, the plain average of the four speeds. Space-mean speed is the harmonic mean and is always less than or equal to time-mean speed. C (66.7) is the right space-mean speed left in ft/s. D (68.5) is the time-mean speed in ft/s.
Source: Levinson et al., Fundamentals of Transportation §5.2, Traffic Flow (CC BY-SA 4.0)
Question 26 of 40
Drainage · Select all that apply · TRN-26
A culvert operates under inlet control. Select every factor that affects the headwater needed to pass a given flow.
Show answer and explanation
Answer: A, B — inlet edge configuration; barrel shape and area at the inlet
Under inlet control, the barrel can carry more than the entrance lets in. Headwater depends on the inlet geometry: edge configuration and barrel shape and area. What happens downstream of the inlet doesn't change it.
Why the other answers miss: Barrel roughness, barrel length, and tailwater all matter under outlet control. Under inlet control they don't — that's why culvert design checks inlet and outlet control separately.
Source: TxDOT Hydraulic Design Manual Ch. 8 §3, Hydraulic Operation of Culverts
Question 27 of 40
Roadside and Cross-Section Design · Multiple choice · TRN-27
A bridge pier sits only 1.5 ft behind where a new barrier face will be. The barrier must shield the pier. Which barrier type best fits this space?
Show answer and explanation
Answer: C — Concrete barrier
Barriers deflect when hit, and they need space behind them to do it. Cable is the most flexible and deflects the most. Metal-beam guardrail deflects less. Concrete barrier is rigid, with little to no deflection, so it's the one that fits when the hazard is right behind it.
Why the other answers miss: A and B are designed to deflect, and 1.5 ft leaves almost no room for that. D is wrong because deflection is mainly a property of the barrier system.
Source: FHWA, Roadside Design Improvements at Curves (FHWA-SA-21-029)
Question 28 of 40
Intersection Geometry · Multiple choice · TRN-28
A proposed single-lane roundabout has been checked only for passenger cars. Freight records show that WB-67 tractor-semitrailers regularly use the route because of a nearby distribution center. What is the most appropriate next step?
Show answer and explanation
Answer: A — check the WB-67's turning paths
The design vehicle should be the largest vehicle that regularly needs to get through. Here that's the WB-67, so check its turning paths (approach and circulating geometry) while keeping the checks for cars, pedestrians, and bikes.
Why the other answers miss: B confuses volume with accommodation: one stuck truck blocks everyone. C ignores the route's actual fleet. D measures capacity, not whether vehicles physically fit.
Source: FHWA, Accommodating Trucks: Designing Roundabouts for Truck Traffic (FHWA-SA-14-013)
Question 29 of 40
Traffic Signals · Select all that apply · TRN-29
A study shows an intersection meets one MUTCD signal warrant. Select the two correct statements.
Show answer and explanation
Answer: B, D — a study decides; a warrant alone doesn't require a signal
The MUTCD requires an engineering study of traffic conditions, pedestrian characteristics, and physical characteristics to decide whether a signal is justified (B). It also says that satisfying a warrant does not in itself require installing a signal (D).
Why the other answers miss: A reverses the rule. C drops factors the study must include. Both answers must be selected, and nothing else, to get credit.
Edition note: The wording is in §4C.01 of both editions.
Source: MUTCD 2009, §4C.01; MUTCD 11th Edition, Part 4, §4C.01
Question 30 of 40
Geotechnical and Pavement · Multiple choice · TRN-30
A flexible pavement has 4 in of asphalt (a₁ = 0.44), 8 in of aggregate base (a₂ = 0.14, m₂ = 1.0), and 10 in of subbase (a₃ = 0.11, m₃ = 0.9). Using the AASHTO 1993 structural number equation, what is the SN?
Show answer and explanation
Answer: B — 3.87
SN = a₁D₁ + a₂D₂m₂ + a₃D₃m₃ = 0.44(4) + 0.14(8)(1.0) + 0.11(10)(0.9) = 1.76 + 1.12 + 0.99 = 3.87.
Why the other answers miss: A (2.88) drops the subbase. C (3.98) ignores the subbase drainage coefficient. D (4.10) divides by m₃ instead of multiplying.
Source: FHWA Geotechnical Aspects of Pavements Reference Manual, Appendix C, Eq. C.3
Question 31 of 40
Vertical Design · Select all that apply · TRN-31
Which two heights are used in the crest vertical curve equation for passing sight distance?
Show answer and explanation
Answer: A, B — eye 3.5 ft and object 3.5 ft
For passing sight distance, the driver needs to see an oncoming vehicle, so the object height is 3.5 ft, the same as the eye height. With h₁ = h₂ = 3.5 ft, the crest-curve constant becomes 200(√3.5 + √3.5)² ≈ 2,800 for S < L.
Why the other answers miss: C (2.0 ft) is the object height for stopping sight distance (a taillight). D (2.0-ft headlight) belongs to the sag-curve headlight criterion.
Source: MDT Road Design Manual, Appendix K worked calculations
Question 32 of 40
Traffic Engineering · Multiple choice · TRN-32
A design-year AADT is 28,000 veh/day. The K-factor is 0.11 and the D-factor is 0.58. What is the directional design-hour volume (DDHV)?
Show answer and explanation
Answer: B — 1,786 veh/h
DDHV = AADT × K × D = 28,000 × 0.11 × 0.58 = 1,786.4, so 1,786 veh/h.
Why the other answers miss: A (1,540) assumes a 50/50 split instead of the given D. C (3,080) leaves out D. D (16,240) leaves out K.
Source: FHWA, Traffic Data for Highway Engineering, §§5-1a and 5-2b
Question 33 of 40
Horizontal Design · Fill in the blank · TRN-33
A two-lane road is rotated about its centerline (one 12-ft lane on each side of the axis). Full superelevation is 6%. The maximum relative gradient is 0.50%, and the adjustment factor for one lane rotated is 1.0. How long must the superelevation runoff be (from a level section to full superelevation)? Enter the nearest foot.
Your answer: ______ ft
Show answer and explanation
Answer: 144 ft
Runoff length = (change in cross slope × width from axis to edge) ÷ maximum relative gradient × adjustment factor.
L = (6 × 12) ÷ 0.50 × 1.0 = 144 ft.
If the normal crown is 2%, the tangent runout (from normal crown to level) adds (2/6) × 144 = 48 ft.
Common mistakes: Using the full 24-ft width (both lanes) doubles the answer to 288 ft. On a centerline rotation, each edge is only 12 ft from the axis.
Source: TxDOT Roadway Design Manual §4.7.6, Tables 4-8 and 4-9
Question 34 of 40
Roadside and Cross-Section Design · Multiple choice · TRN-34
From the crown, a 3.6-m lane slopes down at 2.0%, then a 1.8-m shoulder slopes down at 4.0% on the same side. How far is the outer shoulder edge below the crown?
Show answer and explanation
Answer: C — 0.144 m
Lane drop = 3.6 × 0.020 = 0.072 m.
Shoulder drop = 1.8 × 0.040 = 0.072 m.
Total = 0.144 m.
Why the other answers miss: A (0.072 m) stops at the lane edge. B (0.108 m) applies 2% across the full 5.4 m. D (0.216 m) applies 4% across the full 5.4 m.
Source: Original calculation from the values given.
Question 35 of 40
Vertical Design · Multiple choice · TRN-35
A project requires at least 16.5 ft of vertical clearance under a structure. All elevations below share one datum and matching plan locations. Which conclusion is correct?
| Location | Roadway elevation (ft) | Structure underside elevation (ft) |
|---|---|---|
| 1 | 957.2 | 974.1 |
| 2 | 957.8 | 974.2 |
| 3 | 958.1 | 974.8 |
Show answer and explanation
Answer: C — location 2 governs and is 0.1 ft short
Clearances: 974.1 − 957.2 = 16.9 ft; 974.2 − 957.8 = 16.4 ft; 974.8 − 958.1 = 16.7 ft.
The minimum, 16.4 ft at location 2, governs and is 0.1 ft short.
Why the other answers miss: D averages the clearances (16.67 ft), but a truck only needs one low point to hit. A and B misread which location is lowest.
Source: Original calculation from the values given.
Question 36 of 40
Intersection Geometry · Multiple choice · TRN-36
On an idealized level entrance, a vehicle accelerates uniformly from 15 m/s to 25 m/s at 1.2 m/s². How far does it travel during that speed change? (Ignore tapers and merge allowances.)
Show answer and explanation
Answer: C — 166.7 m
v² = v₀² + 2ad, so d = (v² − v₀²) ÷ (2a) = (625 − 225) ÷ 2.4 = 166.7 m.
Why the other answers miss: A (8.3) is the time in seconds, not a distance. B (41.7 m) squares the speed difference, (25 − 15)², instead of subtracting the squares. D (333.3 m) forgets the 2.
Question 37 of 40
Geotechnical and Pavement · Multiple choice · TRN-37
A two-way roadway carries 18,000 veh/day (no growth), 9% trucks, with an average of 1.3 ESALs per truck pass. The directional factor is 0.5 and the design-lane factor is 0.9. What is the 20-year design-lane ESAL total?
Show answer and explanation
Answer: B — 6.9 million
Two-way trucks per day = 18,000 × 0.09 = 1,620.
Two-way ESAL per day = 1,620 × 1.3 = 2,106.
Design lane = 2,106 × 0.5 × 0.9 = 947.7 per day.
20 years: 947.7 × 365 × 20 = 6,918,210 ≈ 6.9 million.
Why the other answers miss: A (0.35 million) stops at one year. C (7.7 million) skips the lane factor. D (13.8 million) skips the directional factor.
Source: FHWA, Traffic Data for Highway Engineering, §§5-1a and 5-2b
Question 38 of 40
Drainage · Fill in the blank · TRN-38
A rectangular channel carries steady uniform flow. Bottom width = 8.0 ft, depth = 2.0 ft, n = 0.025, slope = 0.002 ft/ft. Using Manning's equation, what is the discharge? Enter to the nearest 0.1 cfs.
Your answer: ______ cfs
Show answer and explanation
Answer: 51.7 cfs
A = 8.0 × 2.0 = 16.0 ft².
P = 8.0 + 2(2.0) = 12.0 ft (bottom and two sides; the free surface isn't wetted).
R = A/P = 1.333 ft.
Q = (1.49/0.025)(16.0)(1.333)^(2/3)(0.002)^(1/2) = 51.7 cfs.
Common mistakes: Using depth as R (2.0 ft) or counting the water surface in P are the usual slips. So is stopping at velocity (3.2 ft/s) instead of multiplying by area.
Source: MoDOT Engineering Policy Guide 750.1, §§750.1.4.1–750.1.4.1.2
Question 39 of 40
Geotechnical and Pavement · Multiple choice · TRN-39
An embankment needs 1,000 compacted yd³ at a dry unit weight of 120 lb/ft³. The borrow material has a dry unit weight of 100 lb/ft³ in place. Assuming no loss of soil solids, how many bank cubic yards are needed?
Show answer and explanation
Answer: C — 1,200 bank yd³
The dry weight of soil solids stays the same; only the volume changes.
V_bank × 100 = 1,000 × 120 → V_bank = 1,200 bank yd³.
Why the other answers miss: A (833) flips the density ratio. B (1,000) assumes volume is conserved. D (2,200) adds the volumes.
Source: FHWA Geotechnical Aspects of Pavements Reference Manual (NHI-05-037), Ch. 5, Eqs. 5.1 and 5.4
Question 40 of 40
Traffic Control Design · Multiple choice · TRN-40
A work zone shifts traffic 11 ft laterally on a 50-mph road, without closing a lane. Using MUTCD taper criteria, what is the minimum shifting taper length?
Show answer and explanation
Answer: B — 275 ft
At 45 mph or more, L = WS = 11 × 50 = 550 ft.
A shifting taper needs at least 0.5L = 275 ft.
Why the other answers miss: A (182 ft) uses the shoulder-taper fraction (0.33L). C (458 ft) uses WS²/60, the 40-mph-and-below formula. D (550 ft) is the full L for a merging taper.
Source: MUTCD 2009, §6C.08 and Tables 6C-3, 6C-4; MUTCD 11th Edition, Part 6, §6B.08 and Tables 6B-3, 6B-4
Check your answers
Give yourself 1 point for each correct answer, out of 40.
- Multiple choice: one correct letter.
- Select all that apply (Questions 11, 26, 29, and 31): the point counts only if you picked every correct choice and nothing else.
- Matching (Question 10): all three pairs must be right.
- Fill in the blank: your number must round to the key at the precision the question asks for.
- Unanswered: 0 points. Count these separately from wrong answers so you know whether time or knowledge was the problem.
| Q | ID | Area | Answer |
|---|---|---|---|
| 1 | TRN-01 | Vertical Design | C — 568 ft |
| 2 | TRN-02 | Traffic Engineering | B — 0.83 |
| 3 | TRN-03 | Project Management | D — $9,072 |
| 4 | TRN-04 | Horizontal Design | 48+56.11 |
| 5 | TRN-05 | Traffic Control Design | 245 ft |
| 6 | TRN-06 | Drainage | 17.1 cfs |
| 7 | TRN-07 | Drainage | B — no; full-flow capacity is about 16.0 cfs |
| 8 | TRN-08 | Traffic Engineering | C — 1.05 |
| 9 | TRN-09 | Traffic Signals | 4.5 s |
| 10 | TRN-10 | Roadside and Cross-Section Design | 1V:6H → recoverable; 1V:3.5H → traversable but non-recoverable; 1V:2H → not traversable |
| 11 | TRN-11 | Project Management | A, B, D, F — critical path A–B–D–F, 17 days |
| 12 | TRN-12 | Intersection Geometry | B — 529 ft |
| 13 | TRN-13 | Horizontal Design | B — 27.0 ft |
| 14 | TRN-14 | Geotechnical and Pavement | 96.3% |
| 15 | TRN-15 | Traffic Engineering | B — 0.75 crashes/MEV |
| 16 | TRN-16 | Vertical Design | D — 574 ft |
| 17 | TRN-17 | Traffic Signals | C — 34 s |
| 18 | TRN-18 | Horizontal Design | A — 1,200 ft |
| 19 | TRN-19 | Project Management | $2,926 |
| 20 | TRN-20 | Roadside and Cross-Section Design | A — The edge of the traveled way |
| 21 | TRN-21 | Traffic Engineering | B — 4.40 crashes/yr |
| 22 | TRN-22 | Horizontal Design | 600 ft |
| 23 | TRN-23 | Traffic Control Design | A — white for same-direction traffic; yellow can mark the left edge of a one-way roadway |
| 24 | TRN-24 | Vertical Design | 510.09 ft |
| 25 | TRN-25 | Traffic Engineering | A — 45.5 mph |
| 26 | TRN-26 | Drainage | A, B — inlet edge configuration; barrel shape and area at the inlet |
| 27 | TRN-27 | Roadside and Cross-Section Design | C — Concrete barrier |
| 28 | TRN-28 | Intersection Geometry | A — check the WB-67's turning paths |
| 29 | TRN-29 | Traffic Signals | B, D — a study decides; a warrant alone doesn't require a signal |
| 30 | TRN-30 | Geotechnical and Pavement | B — 3.87 |
| 31 | TRN-31 | Vertical Design | A, B — eye 3.5 ft and object 3.5 ft |
| 32 | TRN-32 | Traffic Engineering | B — 1,786 veh/h |
| 33 | TRN-33 | Horizontal Design | 144 ft |
| 34 | TRN-34 | Roadside and Cross-Section Design | C — 0.144 m |
| 35 | TRN-35 | Vertical Design | C — location 2 governs and is 0.1 ft short |
| 36 | TRN-36 | Intersection Geometry | C — 166.7 m |
| 37 | TRN-37 | Geotechnical and Pavement | B — 6.9 million |
| 38 | TRN-38 | Drainage | 51.7 cfs |
| 39 | TRN-39 | Geotechnical and Pavement | C — 1,200 bank yd³ |
| 40 | TRN-40 | Traffic Control Design | B — 275 ft |
Example: 26 correct, 8 wrong, and 6 blank is 26/40, or 65%, with 34 attempted.
That percentage describes your performance on these 40 questions, nothing more. NCEES converts results to a scaled score and doesn't publish a passing score (NCEES Examinee Guide, May 2026, pp. 12 and 15), so no practice percentage translates into a pass or fail. With only 3 to 6 questions per area here, area results are a pointer to what to review, not a measurement of your skill.
Turn your misses into a study list
For each miss, find the first step that went wrong. That step, not the topic name, tells you what to practice.
| What went wrong | Where it shows up here | What to do next |
|---|---|---|
| Mixed up a count, a rate, and a proportion | Q2, Q8, Q32, Q37 | Before you calculate, write each number's unit and time basis (per 15 min, per hour, per day, per lane). |
| Used the wrong branch of an equation | Q1, Q5, Q16, Q40 | Write down the condition (S < L, speed ≤ 40 mph, taper type) and check it after you get the answer. |
| Measured from the wrong reference point | Q4, Q13, Q24, Q34, Q35 | Sketch it. Mark the PI, PC, BVC, crown, or matching elevation points before you write a number. |
| Unit or denominator slip | Q3, Q7, Q14, Q15, Q19, Q25, Q36, Q38 | Carry units through every line. Compare the size of your answer with the inputs. |
| Added factors that should be multiplied or weighted | Q6, Q21, Q30, Q37, Q39 | Write the full equation with every factor before substituting numbers. |
| Misread a rule or classification | Q10, Q20, Q23, Q26, Q27, Q29 | Open the cited section and reread the exact wording. In the MUTCD, check whether it's a Standard (shall) or Guidance (should). |
| Skipped an adjustment the problem called for | Q9, Q11, Q12, Q18, Q31, Q33 | List every adjustment named in the stem (grade, lanes crossed, path totals) and tick each one off. |
A simple error log is enough. Track: question ID · the first wrong step · the corrected setup · the reference section to reread · the date you'll redo it. Redo each miss a couple of days later without looking at the solution.
What this set covers
NCEES gives a range of questions for each knowledge area on the real 80-question exam (Transportation specifications, pp. 1–2). The right-hand columns show how we spread these 40 questions. They're a sample of each area, not full coverage of every subtopic.
| NCEES knowledge area | Questions on the real exam (NCEES range) | Questions in this set | Question numbers |
|---|---|---|---|
| 1. Project Management | 6–9 | 3 | 3, 11, 19 |
| 2. Traffic Engineering (Capacity Analysis, Transportation Planning, and Safety Analysis) | 10–15 | 6 | 2, 8, 15, 21, 25, 32 |
| 3. Roadside and Cross-Section Design | 7–11 | 4 | 10, 20, 27, 34 |
| 4. Horizontal Design | 8–12 | 5 | 4, 13, 18, 22, 33 |
| 5. Vertical Design | 8–12 | 5 | 1, 16, 24, 31, 35 |
| 6. Intersection Geometry | 7–11 | 3 | 12, 28, 36 |
| 7. Traffic Signals | 5–8 | 3 | 9, 17, 29 |
| 8. Traffic Control Design | 5–8 | 3 | 5, 23, 40 |
| 9. Geotechnical and Pavement | 6–9 | 4 | 14, 30, 37, 39 |
| 10. Drainage | 8–12 | 4 | 6, 7, 26, 38 |
| Total | 80 | 40 |
The exam uses both U.S. customary and SI units. Most questions here use U.S. customary units; Questions 3, 34, and 36 use SI.
Match your references to your exam date
NCEES's topic list for Transportation has been in effect since April 2024 and isn't changing in April 2027. What changes is two of the nine design standards supplied on screen during the exam.
| Design standard | Exams before April 2027 | Exams beginning April 2027 |
|---|---|---|
| Highway Capacity Manual (HCM) | 6th edition, 2016 | 7th edition, 2022 |
| Manual on Uniform Traffic Control Devices (MUTCD) | 2009, with Revisions 1 and 2 (May 2012) | 11th edition, 2023 |
| AASHTO Green Book (7th ed., 2018), Roadside Design Guide (4th ed., 2011), Highway Safety Manual (1st ed., 2010), Guide for Design of Pavement Structures (1993), MEPDG (3rd ed., 2020), Pedestrian Facilities Guide (2nd ed., 2021), FHWA HDS-5 (3rd ed., 2012) | Listed | Same editions |
Sources: standards effective before April 2027 and standards beginning April 2027, last two pages of each. NCEES scores answers against the listed editions only, so practice with the edition that matches your test date.
Three things to know:
- FHWA now calls the 11th Edition with Revision 1 (December 2025) the current MUTCD (FHWA). NCEES's April 2027 list names the 11th edition, 2023, and doesn't mention Revision 1. Go by the NCEES list.
- Section numbers move between MUTCD editions. Work-zone tapers are §6C.08 in the 2009 edition and §6B.08 in the 11th. The taper-length formulas are the same, but some details changed: the downstream taper is "100 ft per lane" in 2009 Table 6C-3 and 50 to 100 ft in 11th Edition Table 6B-3 (2009; 11th Edition, Part 6). Search by topic on exam day, not by a memorized section number.
- We picked these 40 questions so the answer doesn't depend on which list applies. None relies on a value that differs between the two lists. Where a criterion could vary, such as a time gap or relative gradient, the question gives it to you.
How this set compares with the real exam
- Length and time. The exam has 80 questions. The 9-hour appointment includes 8 hours of exam time and a 50-minute scheduled break (NCEES PE Civil). That works out to an average of 6 minutes per question. To practice at that pace, give yourself about 4 hours for these 40.
- Halfway lock. You review and submit about the first half of the questions before the scheduled break. After that, you can't go back to them (Examinee Guide, pp. 12–13).
- Question formats. The exam mixes multiple choice with alternative item types: multiple correct, point and click, drag and drop, and fill in the blank. Every question is scored right or wrong, with no partial credit (NCEES computer-based testing). This set uses multiple choice, fill-in, select-all, and matching. It doesn't reproduce point-and-click items or the testing software.
- No penalty for guessing. Wrong answers aren't deducted (Examinee Guide, p. 12), so never leave a question blank.
- References are on screen. NCEES supplies the PE Civil Reference Handbook and the listed design standards as searchable PDFs. You can't bring your own copies (NCEES PE Civil). Standards open one chapter at a time (specifications), so practice finding things by searching. The handbook is available through your MyNCEES account.
- Calculator. You can bring one NCEES-approved calculator, and a TI-30XS is available on screen (Examinee Guide, p. 9).
- These questions are original. They aren't NCEES questions and aren't drawn from any exam. The difficulty hasn't been statistically matched to the real test.
Where to go next
- Redo your misses in a couple of days, starting with the area where you missed the most.
- Practice with the references for your exam date. NCEES provides the current PE Civil Reference Handbook through MyNCEES and lists the supplied Transportation design standards on the official specifications.
- For a broader study plan and practice in other PE disciplines, see our PE exam prep guide.
- For eligibility and registration, start at the NCEES PE exam page and your state licensing board. Boards may have requirements beyond NCEES's.
Sources and independence
Exam facts last verified October 8, 2026, against NCEES's PE Civil exam page, the Transportation specifications and design standards effective before April 2027, the design standards beginning April 2027, the NCEES Examinee Guide (May 2026), the computer-based testing page.
Each explanation above links the technical source for its method, such as FHWA manuals, MUTCD editions, and state DOT design manuals. We recalculated every numeric answer and distractor from the values in its question. Those supporting documents explain the methods; they aren't the references NCEES supplies on exam day. For the exam itself, use the NCEES list for your test date.
By the Castleport Test Prep Editorial Team.
Castleport Test Prep is an independent exam prep publisher and is not affiliated with, endorsed by, or approved by the National Council of Examiners for Engineering and Surveying (NCEES). Exam and credential names identify their subjects; trademarks belong to their respective owners. These practice questions are original and unofficial.
Drafting was AI-assisted. No licensed-engineer review has been completed.