Free PE Civil — Transportation practice problems

Ten real questions from our PE Civil: Transportation 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.

4 with figures10 knowledge areasFull worked solutions

1
medium
A. Clear zone, slopes & barrier length-of-need
On the outside of a 1,500-ft-radius curve (design speed 60 mph), the tangent clear-zone value from the RSDG table is $CZ_{base}=30$ ft. The RSDG curve-adjustment factor for this radius and speed is $K_{cz}=1.3$ . The adjusted clear-zone distance on the outside of the curve is most nearly:
1. 39 ft
2. 23.1 ft
3. 30 ft
4. 69 ft
Show the answer & worked solution
Answer: A.39 ft
$CZ = CZ_{base}\cdot K_{cz} = 30\times1.3 = 39$ ft. Distractors: applied the factor then ADDED the full base again, $CZ_{base}+CZ_{base}\cdot K_{cz}=69$ ft (over-applies the adjustment, exceeding the key); DIVIDED by $K_{cz}$ (the wrong direction — that would narrow the zone, as on the inside of the curve), 23.1 ft; and FORGOT the curve adjustment, using the tangent value $30$ ft.
2
medium
A. Signal timing, clearance intervals & warrants
A pedestrian phase at a marked crosswalk across a four-lane arterial uses a walk interval of WALK $=7$ s and a walking speed of $S_p=3.5$ ft/s across a crosswalk of length $L=60$ ft. With flashing-don't-walk FDW $=L/S_p$ , the total pedestrian crossing interval (WALK + FDW) is most nearly:
1. 24.1 s
2. 17.1 s
3. 31.1 s
4. 15.6 s
Show the answer & worked solution
Answer: A.24.1 s
FDW $=L/S_p=60/3.5=17.14$ s; PED $=$ WALK $+$ FDW $=7+17.14=24.14$ s. Distractors: reported ONLY the FDW clearance, forgetting to add the WALK interval (17.14 s); sized FDW on HALF the crosswalk length, $\tfrac{L/2}{S_p}$ (15.57 s); and DOUBLE-COUNTED the WALK interval, adding it twice, $2\,$ WALK $+$ FDW (31.14 s).
3
medium
A. Equal-tangent parabola geometry & elevations
An equal-tangent vertical curve ( $g_1=+3\%$ , $g_2=-2\%$ , $L=600$ ft) on a crest curve over a ridge has its PVC at elevation 845.00 ft. The roadway elevation $x=200$ ft beyond the PVC is most nearly:
1. 851 ft
2. 839.33 ft
3. 849.33 ft
4. 847.67 ft
Show the answer & worked solution
Answer: C.849.33 ft
$A=g_2-g_1=-2-(3)=-5\%$ ; $y=y_{PVC}+\tfrac{g_1}{100}x+\tfrac{A}{200L}x^2=845 + 6 - 1.67=849.33$ ft. Distractors: DROPPED the parabolic term, staying on the $g_1$ tangent (851 ft); used $100$ instead of $200$ , forgetting the $\tfrac12$ (847.67 ft); and used the OUTGOING grade $g_2$ in the linear term instead of $g_1$ (839.33 ft).
4
medium
A. Signs, markings & temporary traffic control (MUTCD)
A work zone on a downtown lane realignment has a lateral offset $W=12$ ft at $S=30$ mph. Compute the merging-taper length $L$ ( $L=WS^2/60$ for $S\le40$ mph, $L=WS$ for $S\ge45$ mph), then find the shifting-taper length ( $\approx 0.5L$ ). The required taper is most nearly:
1. 360 ft
2. 60 ft
3. 90 ft
4. 180 ft
Show the answer & worked solution
Answer: C.90 ft
$L=WS^2/60=12(30)^2/60=180$ ft; shifting taper $=0.5L=90$ ft. Distractors: reported the FULL merging $L=180$ ft (forgot the fraction); used the OTHER fraction (60 ft, swapping the 0.5L / ⅓L rules); and applied the WRONG speed-regime formula for $L$ , $WS=360$ ft.
5
medium
A. Uninterrupted flow, capacity & level of service
On an urban freeway in the peak direction, the peak-direction hourly volume is $V=4,800$ veh/h over $N=3$ lanes, with PHF $=0.92$ , a heavy-vehicle factor $f_{HV}=0.952$ , and a driver-population factor $f_p=1$ . The demand flow rate per lane $v_p$ is most nearly:
1. 1656 pc/h/ln
2. 1681 pc/h/ln
3. 5480 pc/h/ln
4. 1827 pc/h/ln
Show the answer & worked solution
Answer: D.1827 pc/h/ln
$v_p=\dfrac{V}{PHF\cdot N\cdot f_{HV}\cdot f_p}=\dfrac{4800}{0.92\cdot3\cdot0.952\cdot1}=1827$ pc/h/ln. Distractors: MULTIPLIED by $f_{HV}$ instead of dividing (1656 — wrong direction, since $f_{HV}<1$ should raise $v_p$ ); FORGOT to divide by $N$ , giving a total rate not a per-lane rate (5480); and OMITTED the PHF, leaving an hourly rather than a peak-15-min rate (1681).
6
medium
A. Soil properties, classification, CBR & compaction
Field density testing on a roadway subgrade lift gives a dry unit weight of $\gamma_{d,\text{field}}=116.5$ pcf; the standard Proctor maximum is $\gamma_{d,\max}=122$ pcf and the specification requires 95% relative compaction. The relative compaction achieved is most nearly:
1. 90.7%
2. 105%
3. 95.5%
4. 94.5%
Show the answer & worked solution
Answer: C.95.5%
$RC=\dfrac{\gamma_{d,field}}{\gamma_{d,\max}}\times100=\dfrac{116.5}{122}\times100=95.5\%$ (PASSES the 95% spec). Distractors: INVERTED the ratio, $\tfrac{\gamma_{d,\max}}{\gamma_{d,field}}\times100=104.7\%$ (always >100%, a false pass); subtracted the pcf gap from 100 as if it were a percent, $100-(\gamma_{d,\max}-\gamma_{d,field})=94.5\%$ ; and multiplied the ratio by the spec 95 instead of by 100, 90.7%.
7
medium
A. Circular curve geometry & stationing
A simple circular curve has radius $R=1200$ ft and central angle $\Delta=28^\circ$ on a divided-highway curve. The length $L$ of the curve (arc PC to PT) is most nearly:
1. 293.2 ft
2. 33600 ft
3. 586.4 ft
4. 580.6 ft
Show the answer & worked solution
Answer: C.586.4 ft
$L=R\Delta\dfrac{\pi}{180}=1200\cdot28\cdot\dfrac{\pi}{180}=586.43$ ft. Distractors: FORGOT the degree→radian factor, $R\Delta=33600$ ; reported the LONG CHORD $LC=2R\sin(\Delta/2)=580.61$ ft (a straight line, shorter than the arc) instead of the arc; and HALVED $\Delta$ as if for a tangent term, $R(\Delta/2)\tfrac{\pi}{180}=293.22$ ft.
8
medium
A. Quantity/cost estimating & engineering economics
Routine corridor maintenance costs $45,000 per year for 20 years. At i = 5% per year, the present worth of this maintenance stream is most nearly:
1. $1,487,968
2. $560,799
3. $119,398
4. $900,000
Show the answer & worked solution
Answer: B.$560,799
$(P/A)=\dfrac{1-(1+i)^{-n}}{i}=\dfrac{1-(1+0.05)^{-20}}{0.05}=12.462$ ; $P=A\,(P/A)=45,000\cdot12.462=\$560,799$ . Distractors: summed $A\cdot n$ with NO discounting ($900,000); used the $(F/A)$ factor, getting the FUTURE worth of the series instead of the present ($1,487,968); and grabbed the $(F/P)$ factor, compounding a single payment forward $A(1+i)^{n}$ instead of discounting the series ($119,398).
9
medium
A. Highway hydrology & runoff
A fully-paved interchange ramp catchment has a runoff coefficient $C=0.85$ and a drainage area $A=6$ acres. The design rainfall intensity is $i=4.2$ in/hr. By the rational method ( $Q=CiA$ , $Q$ in cfs), the peak discharge is most nearly:
1. 29.6 cfs
2. 21.4 cfs
3. 18.2 cfs
4. 25.2 cfs
Show the answer & worked solution
Answer: B.21.4 cfs
$Q=CiA=0.85\cdot4.2\cdot6=21.42$ cfs. Distractors: OMITTED $C$ (used $C=1$ ), $iA=25.2$ cfs; DIVIDED by $C$ instead of multiplying, $iA/C=29.65$ cfs; and APPLIED $C$ twice (squared it), $C^2iA=18.21$ cfs.
10
medium
A. Intersection sight distance & roundabout geometry
A stopped driver making a left turn has 650 ft of clear sight distance along a major road posted at $V=45$ mph (required time gap $t_g=7.5$ s). Using $\text{ISD}=1.47\,V\,t_g$ , the time gap the available sight distance actually provides is most nearly:
1. 14.4 s
2. 8.84 s
3. 10.3 s
4. 9.83 s
Show the answer & worked solution
Answer: D.9.83 s
$t_g=\dfrac{\text{ISD}}{1.47\,V}=\dfrac{650}{1.47(45)}=9.83$ s, which is ≥ the required 7.5 s (so the gap is adequate). Distractors: DROPPED the $1.47$ factor (14.44 s); ROUNDED the $1.47$ factor down to $1.4$ (10.32 s); and mis-read the major-road speed $5$ mph too high, using $V=50$ mph (8.84 s).

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