Free PE Civil Construction Practice Test

Answer: C. 20 ft.

Start with classification, not the slope table. A strength of 2.0 tsf meets the Type A threshold (1.5 tsf or more), but Appendix A says no soil is Type A if it is subject to vibration from heavy traffic. Soil that meets the Type A strength but is subject to vibration is Type B. Type B's maximum allowable slope is 1H:1V for excavations 20 ft deep or less, so each wall lays back 8 ft. Top width = 4 + 2(8) = 20 ft.

Why the other choices miss: A: Type A's short-term ½H:1V slope (4 + 2 × 4). It doesn't apply: this soil isn't Type A. B: Type A's ¾H:1V slope (4 + 2 × 6). The strength qualifies, but the traffic vibration disqualifies Type A. D: Type C's 1½H:1V slope (4 + 2 × 12). Nothing in the stem makes this Type C.

This is the width corresponding to the tabulated slope. Appendix B(c)(3) also requires site conditions, including traffic surcharge and signs of distress, to be addressed when determining the actual slope. If factors affecting the soil classification change, Appendix A(c)(5) requires the competent person to evaluate them and reclassify the deposit as necessary.

Edition check: The classification and slope provisions used here match in the July 2020 CFR (Appendix A, Appendix B) and the 2024 CFR listed by NCEES.

Source: 29 CFR 1926 Subpart P, Appendix A (soil classification), Definitions of Type A (exception ii, vibration) and Type B (item iv); (c)(1)–(2), classification; (c)(5), reclassification; 29 CFR 1926 Subpart P, Appendix B (sloping and benching), (b), maximum allowable slope; (c)(3)(ii)–(iii), distress and surcharge; Table B-1; Figure B-1.2 (Type B simple slope).

Answer: C. Day 13.

List the paths. A–B–D–F = 2 + 6 + 3 + 2 = 13 days. A–C–E–F = 2 + 4 + 5 + 2 = 13 days. Both are critical. Shortening B drops the first path to 12, but A–C–E–F still takes 13, so the project still finishes on day 13. To finish earlier, you'd have to shorten both paths or an activity they share (A or F).

Why the other choices miss: A: Subtracts the one-day change twice. B: Assumes there's only one critical path. D: Moves the finish the wrong way; a shorter activity can't lengthen this network.

Source: Hendrickson, Project Management for Construction, Chapter 10: Fundamental Scheduling Procedures, §10.3, critical path and forward pass.

Answer: 23.4 kPa. Any entry that rounds to 23.4 at the stated precision counts as correct.

Effective stress = total stress − pore-water pressure, with water pressure measured from the water table. Before: total = 1(18) + 7(20) = 158 kPa; u = 7(9.81) = 68.67 kPa; σ′ = 89.33 kPa. After: total = 4(18) + 4(20) = 152 kPa; u = 4(9.81) = 39.24 kPa; σ′ = 112.76 kPa. Increase = 112.76 − 89.33 = 23.43 kPa, or 23.4 kPa. Total stress actually drops a little, but pore pressure drops more, so the soil skeleton carries more load. That added effective stress is why dewatering can settle nearby structures.

Common wrong answers: 29.4 kPa: Keeps the 20 kN/m³ saturated weight in the drained 1–4 m zone, so it only counts the 3 m drop in water head (3 × 9.81).

Source: USACE EM 1110-1-1904, Settlement Analysis (Sept. 1990), hosted copy, Eq. 1-1, para. 1-5a(3)(a), p. 1-7: effective stress and hydrostatic pore-water pressure; Table 3-5, step 2, p. 3-29.

Answer: B. 22.0 CY.

Footing = 120 × 2 × 1 = 240 ft³. Wall = 120 × (8/12) × 4 = 320 ft³. Total = 560 ft³ ÷ 27 = 20.74 CY. With 5% waste: 20.74 × 1.05 = 21.78 CY. Round up to the next 0.5 CY: 22.0 CY.

Why the other choices miss: A: Skips the waste allowance (20.74 rounded up). C: Treats 8 in. as 0.8 ft instead of 0.667 ft. D: Divides cubic feet by 9 instead of 27.

Source: Worked from the stated inputs (unit conversion: 27 ft³ = 1 yd³).

Answer: A. 20 kN.

Take moments about A. The load acts 1.6 m from A, and B is 4.0 m from A. T_B × 4.0 = 50 × 1.6, so T_B = 20 kN. Then T_A = 50 − 20 = 30 kN. The sling farther from the center of gravity carries less.

Why the other choices miss: B: Splits the load evenly, which only works when the CG is centered. C: That's the tension at A; the lever arms are swapped. D: Puts the whole load on one sling.

Source: OpenStax, University Physics Vol. 1, Section 12.1: Conditions for Static Equilibrium, Equations 12.2–12.3 and 12.5–12.6, force and torque equilibrium.

Answer: 11.7%. Any entry that rounds to 11.7 at the stated precision counts as correct.

Water content is the mass of water divided by the mass of dry solids: w = (2,150 − 1,925) ÷ 1,925 = 225 ÷ 1,925 = 0.1169, or 11.7%. If the spec required placement within 2 percentage points of a 13.5% optimum (11.5% to 15.5%), this sample meets that moisture criterion.

Common wrong answers: 10.5%: Divides by the moist mass (225 ÷ 2,150). Against the same moisture criterion, that wrong number would reject a sample that actually meets it.

Source: FHWA NHI-05-037, Geotechnical Aspects of Pavements, Chapter 5, Table 5-11: w = Ww/Ws.

Answer: D. 1,422 psi.

For a center point load on a simple span, M = PL/4 = 250 × 96 / 4 = 6,000 lb-in. Laid flat, the bending depth is 1.5 in. and the width is 11.25 in., so S = bh²/6 = 11.25 × 1.5² / 6 = 4.219 in³. Stress = M/S = 6,000 / 4.219 = 1,422 psi. Orientation matters: on edge, the same plank would see only about 190 psi.

Why the other choices miss: A: Uses the plank on edge (h = 11.25 in.), which isn't how it's laid. B: Uses M = PL/8 by spreading the same total load uniformly over the span. C: Uses nominal dimensions (12 in. × 2 in.) instead of actual dressed size.

This is the demand stress only. Whether the plank is adequate depends on its allowable stress, which isn't given.

Source: American Wood Council, Design Aid 6: Beam Formulas with Shear and Moment Diagrams, Figure 7, simple beam with concentrated load at center, p. 7; Roylance, Stresses in Beams (MIT, 2000), Eqs. 5 and 7: I = bh³/12, σ = My/I.

Answer: A and B. Both must be selected, and nothing else, to score the point.

The rule lists two exceptions: excavations made entirely in stable rock, and excavations less than 5 ft deep where a competent person's examination of the ground gives no indication of a potential cave-in. A meets the first. B meets both parts of the second.

Why the other choices miss: C: Exactly 5.0 ft is not less than 5 ft. D: Meets the depth part only. The shallow-trench exception also needs the competent person's examination.

These exceptions cover cave-in protection only. Other applicable excavation requirements still apply.

Edition check: The provisions used in this question match in the July 2020 and 2024 CFR editions listed by NCEES.

Source: 29 CFR 1926.652, Requirements for protective systems, (a)(1)(i)–(ii).

Answer: B. 27+00.50.

Tangent length T = R tan(Δ/2) = 600 tan 20° = 218.38 ft. PC = 2,500.00 − 218.38 = 22+81.62. Curve length L = 2πRΔ/360 = 2π(600)(40)/360 = 418.88 ft. PT = PC + L = 2,281.62 + 418.88 = 27+00.50.

Why the other choices miss: A: Adds the long chord (410.42 ft) instead of the arc length. C: Adds T to the PI station instead of adding the arc length to the PC station. D: Adds L to the PI instead of the PC.

Source: Caltrans Bridge Design Details, Attachment E: Horizontal Curve Equations (June 2025), Tangent distance T = R tan(Δ/2) and curve length L = 2πRΔ/360, p. 2A.E.1.

Answer: A. SW.

Fines are under 5%, so it's a clean sand with no dual symbol. Cu = D60/D10 = 1.80/0.15 = 12. Cc = D30² / (D10 × D60) = 0.36 / 0.27 = 1.33. Cu = 12 exceeds 6, and Cc = 1.33 lies between 1 and 3. Both gradation checks pass: SW.

Why the other choices miss: B: Poorly graded sand fails at least one of these gradation checks; this sample passes both. C: Silty sand needs more than 12% fines; this has 3%. D: The sample is predominantly sand, not gravel.

Source: USDA NRCS, National Engineering Handbook Part 631, Chapter 3: Engineering Classification of Earth Materials (June 2022), Cu and Cc definitions, p. 631-3.23; Figure 3-17, p. 631-3.24 (gradation and fines criteria).

Answer: 5.0 ft. Any entry that rounds to 5.0 at the stated precision counts as correct.

Slab weight = (8/12) × 150 = 100 psf. Total = 100 + 10 + 50 = 160 psf. Maximum tributary area per shore = 4,000 / 160 = 25 ft². For a square grid, spacing = √25 = 5.0 ft each way.

Common wrong answers: 6.0 ft: Leaves out the 50 psf live load (√(4,000/110) = 6.0 ft). 25 ft: Stops at the area and skips the square root.

The live load is a stipulated value for this problem, not a value from ASCE 37. A real shoring layout also checks stringers, joists, sills, and bracing.

Source: Worked from the stated inputs (statics: force = pressure × tributary area).

Answer: B. No. Net capacity is 13,830 lb.

A 42 ft radius falls between rows, so use the 45 ft row: 16,100 lb. Net capacity = 16,100 − 650 − 420 − 1,200 = 13,830 lb. The 15,800 lb load exceeds it, so the lift can't be made in this configuration.

Why the other choices miss: A: Uses the 40 ft row, which is shorter than the actual radius. C: Uses the right row but skips the deductions. D: Uses the wrong row and skips the deductions.

OSHA prohibits operating equipment above its rated capacity. Use the manufacturer's chart and notes for the actual configuration; this chart was invented for practice.

Edition check: The provisions used in this question match in the July 2020 and 2024 CFR editions listed by NCEES.

Source: 29 CFR 1926.1417, Cranes and derricks: operation, (a), manufacturer procedures; (o)(1), compliance with rated capacity; Worked from the stated inputs (chart values and deductions are supplied in the stem).

Answer: 2.04 in. Any entry that rounds to 2.04 at the stated precision counts as correct.

S = 1,000/CN − 10 = 1,000/80 − 10 = 2.5 in. Ia = 0.2S = 0.5 in. Q = (P − 0.2S)² / (P + 0.8S) = (4.0 − 0.5)² / (4.0 + 2.0) = 12.25 / 6.0 = 2.04 in. TR-55's Table 2-1 lists the same value for CN 80 and 4.0 in. of rain.

Common wrong answers: 2.46 in: Drops the initial abstraction: P² / (P + S). 1.88 in: Uses S instead of 0.8S in the denominator.

Source: USDA NRCS (SCS), TR-55, Urban Hydrology for Small Watersheds, 2nd ed. (June 1986), university-hosted copy, Chapter 2, Eqs. 2-1 to 2-4, p. 2-1; Table 2-1, p. 2-3.

Answer: B. Day 15.

B's start must satisfy both links. Start-to-start: B can't start before day 0 + 2 = 2. Finish-to-finish: B can't finish before A finishes plus 2 = day 12, so with an 8-day duration it can't start before day 4. The later limit controls: B runs days 4–12. C runs days 12–15.

Why the other choices miss: A: Ignores the finish-to-finish lag, so B runs 2–10. C: Treats the link as plain finish-to-start with no lag: B 10–18, C 18–21. D: Treats the 2-day start-to-start lag as a finish-to-start lag: B 12–20, C 20–23.

Source: Hendrickson, Project Management for Construction, Chapter 10: Fundamental Scheduling Procedures, §10.6, Example 10-4 (start-to-start and finish-to-finish lags).

Answer: C. 70 hours.

Set the costs equal: 2,500 + 120h = 6,000 + 70h. So 50h = 3,500 and h = 70 hours. Below 70 hours, A is cheaper; above it, B's lower hourly cost wins.

Why the other choices miss: A: Divides the fixed-cost difference by the sum of the hourly costs instead of the difference. B: Divides B's fixed cost by A's hourly cost. D: Divides B's whole fixed cost by the hourly difference, ignoring A's $2,500.

Source: Worked from the stated inputs (equating two linear cost expressions).

Answer: D. Independent assurance.

The regulation defines an independent assurance program as an unbiased and independent evaluation of the sampling and testing procedures used in the acceptance program. The evaluator here is judging how the testing is done, not whether the material passes.

Why the other choices miss: A: Quality control is the contractor's own operational activity to meet the contract. B: Verification testing checks the quality of the product itself. C: The acceptance program covers the agency's determination of product quality; this activity evaluates the reliability of that program's testing.

Source: 23 CFR 637.203, Definitions (construction quality assurance), Definitions: independent assurance program, quality control, verification sampling and testing, acceptance program.

Answer: B. 39 kN/m.

Ka = tan²(45° − φ′/2) = tan²(30°) = 1/3. Soil thrust (triangle) = ½ Ka γ H² = ½ × (1/3) × 18 × 3.0² = 27 kN/m. Surcharge thrust (rectangle) = Ka q H = (1/3) × 12 × 3.0 = 12 kN/m. Total = 39 kN/m.

Why the other choices miss: A: Leaves out the surcharge. C: Drops the ½ on the soil triangle (54 + 12). D: Uses the passive coefficient Kp = 3 instead of Ka.

The two parts act at different heights: the triangle's resultant at H/3, the surcharge's at H/2.

Source: Caltrans Trenching and Shoring Manual, Chapter 4 (July 2025), Eq. 4-3-11, p. 4-8; §4-3.02A, p. 4-9; Eq. 4-3-17, p. 4-10; Worked from the stated inputs (the surcharge term is the stem's stated assumption).

Answer: A. 1 ksi tension.

Axial stress = P/A = 24/12 = 2 ksi compression everywhere on the section. Bending stress = M/S = 60/20 = 3 ksi, tension on one face and compression on the other. On the tension face: −2 + 3 = +1 ksi, or 1 ksi tension.

Why the other choices miss: B: That's the other face, where bending compression adds to the axial compression. C: Adds the magnitudes without signs. D: Ignores bending.

This is a stress calculation, not a capacity check; buckling and allowable stresses are outside it.

Source: Roylance, Stresses in Beams (MIT, 2000), Eq. 7, σ = My/I; OpenStax, University Physics Vol. 1, Section 12.3: Stress, Strain, and Elastic Modulus, Eq. 12.34, stress = F/A.

Answer: A and D. Both must be selected, and nothing else, to score the point.

Paragraph (a)(1) requires each scaffold to support its own weight plus at least 4 times the maximum intended load: 250 + 4(1,250) = 5,250 lb. Paragraph (f)(1) separately limits actual loading to the lesser of the maximum intended load and the rated capacity.

Why the other choices miss: B: The factor of 4 applies to the intended load, not the scaffold's own weight. C: The factor of 4 is a capacity margin, not permission to load four times higher.

Edition check: The provisions used in this question match in the July 2020 and 2024 CFR editions listed by NCEES.

Source: 29 CFR 1926.451, Scaffolds: general requirements, (a)(1) and (f)(1).

Answer: 9.09 hp. Any entry that rounds to 9.09 at the stated precision counts as correct.

Total head H = 24 + 18 = 42 ft. P = 600 × 42 × 1.0 / (3,960 × 0.70) = 25,200 / 2,772 = 9.09 hp.

Common wrong answers: 6.36 hp: The hydraulic (water) power, before dividing by efficiency. 5.19 hp: Uses only the 24 ft static lift and leaves out the losses. 4.45 hp: Multiplies by the efficiency instead of dividing.

This is shaft power. Motor or electrical input would also include motor efficiency, and pump selection would check suction conditions.

Source: Hydraulic Institute Data Tool, Pump Curves, Eqs. 1.B.3–1.B.5 (pump efficiency and input power).

Answer: C. 1.20 mm.

Strain = stress / E = 150 MPa / 200,000 MPa = 0.00075. Elongation = strain × length = 0.00075 × 1,600 mm = 1.20 mm.

Why the other choices miss: A: Mixes meters and millimeters. B: Multiplies the strain by 1,000 mm instead of the rod's 1,600 mm length. D: Treats 200 GPa as 200 MPa.

Stiffness (E) controls stretch; strength controls whether the rod yields. They're different properties.

Source: OpenStax, University Physics Vol. 1, Section 12.3: Stress, Strain, and Elastic Modulus, Eqs. 12.33–12.36 (stress, strain, Young's modulus).

Answer: 360 ft. Any entry that rounds to 360 at the stated precision counts as correct.

The curve's grade changes linearly from g1 to g2 over its length: grade at x = g1 + (g2 − g1)x/L. The high point is where that grade is zero: x = g1·L / (g1 − g2) = 0.03 × 600 / 0.05 = 360 ft, or station 15+60.

Common wrong answers: 300 ft: Assumes the high point has the PVI's station, halfway through the curve. That station is correct only when the grades are equal and opposite. 240 ft: Uses g2 in the numerator, which locates the point measured from the EVC instead.

Source: Caltrans Bridge Design Details, Attachment D: Vertical Curve Calculations (June 2025), Eq. 5 (elevation on a vertical curve, rate of grade change R/C), p. 2A.D.1; Worked from the stated inputs (high-point location found by setting the curve's slope to zero).

Answer: Advance warning area → Transition area → Activity area → Termination area.

Drivers are first warned (advance warning area), then moved out of their normal path (transition area, where merging and shifting tapers redirect traffic), then pass the work (activity area, which holds the buffer space and work space), and finally return to normal (termination area).

Common mistake: Swapping transition and activity areas is a mistake to watch for. Merging and shifting tapers redirect traffic in the transition area; downstream tapers can be in the termination area.

Edition check: The same four areas, in the same order, in the 2009 MUTCD (current exam) and the 11th edition (from April 2027). The relevant sections are 6C.03 and 6B.03.

Source: MUTCD 2009 Edition with Revisions 1 and 2 (May 2012), Part 6, §6C.03, p. 552; §6C.08, paragraph 01, p. 555; MUTCD 11th Edition (Dec. 2023), Part 6, §6B.03, p. 771; §6B.08, paragraph 01, p. 775.

Answer: C. Day 11.

Crane work totals 2 + 3 + 5 = 10 days, so with E after it the project can't finish before day 11. One schedule achieves that: A on days 0–2, D on 2–7, C on 7–10 (B already finished on day 4), and E on 10–11. The crane never sits idle.

Why the other choices miss: A: The unconstrained CPM answer, which ignores the single-crane conflict. B: Forgets activity E. D: Starts C as soon as B finishes (day 4), which leaves the crane idle on days 2–4 and pushes D to days 7–12.

Source: Hendrickson, Project Management for Construction, Chapter 10: Fundamental Scheduling Procedures, §10.9, scheduling with resource constraints; Worked from the stated inputs (lower bound from total crane-days, then a feasible schedule that meets it).

Answer: A and C. Both must be selected, and nothing else, to score the point.

RCSC §4.3 requires slip-critical joints for fatigue loading with load reversal, oversized holes, slotted holes except where the load is approximately normal (80° to 100°) to the slot, and cases where slip would be detrimental. A and C are on that list.

Why the other choices miss: B: This is the stated exception for slotted holes. D: Nothing here calls for slip resistance; a snug-tightened joint is typical.

Edition check: Same four triggers in the 2014 RCSC Specification (printed in the AISC 15th edition Manual) and the 2020 edition (16th edition Manual).

Source: RCSC, Specification for Structural Joints Using High-Strength Bolts (Aug. 2014, with April 2015 errata), §4.3, Slip-Critical Joints, pp. 16.2-29–16.2-30; RCSC, Specification for Structural Joints Using High-Strength Bolts (June 2020), §4.3, p. 16.2-35.

Answer: D. $7.40/LF.

Crew cost = 62 + 4(44) + 95 = $333/hr. Unit cost = 333 / 45 = $7.40 per LF.

Why the other choices miss: A: Divides the hourly cost by a full day's output (8 × 45 = 360 LF). B: Counts only one laborer. C: Leaves out the equipment.

Source: Worked from the stated inputs (unit cost = crew cost per hour ÷ production per hour).

Answer: C. 2,560 lb.

Per foot of wall, wind = 20 × 12 = 240 lb acting at 6 ft. Moments about the pinned base: H × 9 = 240 × 6, so the brace's horizontal component H = 160 lb per ft. Per brace: 160 × 8 = 1,280 lb. The brace is at 60° from horizontal, so its axial force = 1,280 / cos 60° = 2,560 lb.

Why the other choices miss: A: Stops at the horizontal component. B: Divides by sin 60° instead of cos 60°. D: Gives the brace the whole wind load (240 × 8 = 1,920 lb) instead of sharing it with the base, then divides by cos 60°.

The brace force also has a vertical component (2,560 × sin 60° ≈ 2,217 lb) that the anchor and the form connection must carry.

Source: OpenStax, University Physics Vol. 1, Section 12.1: Conditions for Static Equilibrium, Equations 12.2–12.3 and 12.5–12.6, force and torque equilibrium.

Answer: 77.8 BCY. Any entry that rounds to 77.8 at the stated precision counts as correct.

Left square corners: 2.0, 3.0, 1.0, 2.0, average 2.0 ft, volume 400 × 2.0 = 800 ft³. Right square: 3.0, 5.0, 2.0, 3.0, average 3.25 ft, volume 400 × 3.25 = 1,300 ft³. Total 2,100 ft³ ÷ 27 = 77.8 BCY. The shortcut V = (A/4)(Σh₁ + 2Σh₂) gives the same: (400/4)(11 + 2 × 5) = 2,100 ft³, where h₁ are corners used once and h₂ corners shared by two squares.

Common wrong answers: 79.0 BCY: Averages all six depths equally (2.67 ft × 800 ft²), which over-weights the outer corners.

Source: Worked from the stated inputs (each square is treated as a prism with the average of its corner depths; consistent with the average end-area idea of averaging bounding dimensions).

Answer: D. 1/16.

Maximum deflection of a simple span under uniform load is Δ = 5wL⁴ / (384EI). With w, E and I unchanged, Δ scales with L⁴: (1/2)⁴ = 1/16.

Why the other choices miss: A: Assumes deflection is proportional to span. B: Uses the L² scaling of moment. C: Holds the total load fixed instead of the load per foot (PL³ scaling).

Halving the span by adding a support under one continuous beam is a different structural model, with a different answer.

Source: American Wood Council, Design Aid 6: Beam Formulas with Shear and Moment Diagrams, Figure 1, simple beam with uniformly distributed load, p. 4.

Answer: A. 0.95 in.

The final stress (2,000 psf) stays below the preconsolidation stress (2,500 psf), so the clay only recompresses: use Cr. Δe = Cr log₁₀(σ′f / σ′0) = 0.05 × log₁₀(2) = 0.01505. Settlement = Δe / (1 + e0) × H = 0.01505 / 1.90 × 120 in. = 0.95 in.

Why the other choices miss: B: Leaves out the 1 + e0 divisor. C: Uses the natural log instead of log base 10. D: Uses Cc, as if the clay were loaded past its preconsolidation stress.

If the fill pushed the stress past 2,500 psf, you'd add a Cc term for the part above the preconsolidation stress.

Source: USACE EM 1110-1-1904, Settlement Analysis (Sept. 1990), hosted copy, Para. 3-12; Eq. 3-20, p. 3-28; Eq. 3-23 and following instructions, p. 3-34 (when σ′f < σ′p, omit the Cc term and substitute σ′f for σ′p in the Cr term).

Answer: 3.06 m/s. Any entry that rounds to 3.06 at the stated precision counts as correct.

Flow in the pipe = 1.80 − 0.45 = 1.35 m³/s. Area = π(0.75)²/4 = 0.4418 m². Velocity = Q/A = 1.35 / 0.4418 = 3.06 m/s.

Common wrong answers: 4.07 m/s: Forgets to subtract the side branch. 0.76 m/s: Uses the diameter as the radius in the area.

Source: OpenStax, University Physics Vol. 1, Section 14.5: Fluid Dynamics, Q = Av and Eq. 14.14 (continuity for incompressible flow).

Answer: A. Both have a maturity of 1,200 °C·h.

M = Σ(T − T0)Δt. Slab X: (25 + 10)(24) + (5 + 10)(24) = 840 + 360 = 1,200 °C·h. Slab Y: (15 + 10)(48) = 1,200 °C·h. Same mix, same curve, same maturity, so the same estimated strength. That's the point of the maturity method: it converts different temperature histories to a common index.

Why the other choices miss: B: The order of these temperature intervals does not change the accumulated Nurse-Saul maturity. C: Steadiness isn't a factor in the function. D: Comparing different temperature histories is exactly what the maturity method is for, as long as the curve matches the mix.

The curve has to come from this mix; a change in materials or proportions requires a new curve.

Source: FHWA-IF-06-004, Maturity Testing for Concrete Pavement Applications (TechBrief, Nov. 2005), Nurse-Saul equation, p. 2; calibration curve requirements, p. 3.

Answer: A. Cut 0.62 ft.

HI₁ = 512.40 + 3.85 = 516.25. TP = 516.25 − 7.10 = 509.15. HI₂ = 509.15 + 2.64 = 511.79. Ground = 511.79 − 5.27 = 506.52. Ground is 506.52 − 505.90 = 0.62 ft above design, so cut 0.62 ft.

Why the other choices miss: B: Right size, wrong direction: ground above grade means cut. C: Subtracts the backsight at the turning point instead of adding it. D: Compares the design elevation to the benchmark instead of to the ground at the stake.

Source: FHWA-HRT-08-056, LTPP Manual for Profile Measurements, Chapter 4 (rod and level), §4.3.5, instrument height at turning points (new HI = old HI − FS + BS).

Answer: A and C. Both must be selected, and nothing else, to score the point.

The rule gives two routes: following the removal conditions in the plans and specifications, or proper ASTM compressive-strength testing showing that the concrete has gained enough strength to support its own weight and superimposed loads.

Why the other choices miss: B: A fixed age isn't one of the routes. The rule doesn't set any default number of days. D: Appearance tells you nothing reliable about in-place strength.

Removing reshores is covered separately by 1926.703(e)(2).

Edition check: The provisions used in this question match in the July 2020 and 2024 CFR editions listed by NCEES.

Source: 29 CFR 1926.703, Requirements for cast-in-place concrete, (e)(1)(i)–(ii) and (e)(2).

Answer: B. $1,100.

Both branches are critical, so each day of time saved on the branches has to be bought on both: $200 + $350 = $550 per day. The final activity saves a day for $800 but can only give one day. Cheapest way to save two days: two days on each branch, 2($200) + 2($350) = $1,100. The next-cheapest feasible mix, one day on each branch plus the final activity, costs $550 + $800 = $1,350.

Why the other choices miss: A: Shortens only branch 1; branch 2 still controls at 22 days. C: Feasible, but $250 more than the cheapest option. D: Crashes everything available (both branches 2 days plus the final activity). That buys a third day nobody asked for.

Source: GAO-16-89G, Schedule Assessment Guide (Dec. 2015), Table 6, p. 145 (shortening a schedule by crashing); Worked from the stated inputs (enumerating the feasible whole-day options).

Answer: D. An uncoated (black) A325 bolt.

Both RCSC editions on the exam ban reuse of A490-family bolts and of galvanized bolts. Plain (black) A325 bolts are the reusable ones: the 2014 Specification allows it with the engineer of record's approval, and the 2020 edition allows it with that approval in pretensioned and slip-critical joints (and without it in snug-tightened joints). Touching up or retightening bolts loosened by nearby installation isn't reuse.

Why the other choices miss: A: Galvanized A325 bolts can't be reused. B: A490 bolts can't be reused, coated or not. C: The hand-run nut check is a commentary rule of thumb for black A325 bolts only.

Edition check: The 2014 RCSC Specification (AISC 15th edition Manual) also bans reuse of twist-off tension-control assemblies. The 2020 edition (16th edition Manual) uses Group 120 (A325 family) and Group 150 (A490 family) names, bans reuse of Group 150 and of galvanized or coated bolts, and drops the EOR approval for plain Group 120 bolts in snug-tightened joints. The keyed answer is correct under both.

Source: RCSC, Specification for Structural Joints Using High-Strength Bolts (Aug. 2014, with April 2015 errata), §2.3.3, Reuse, and its commentary, p. 16.2-6; RCSC, Specification for Structural Joints Using High-Strength Bolts (June 2020), §2.11 (2.11.1–2.11.3) and commentary, pp. 16.2-19–16.2-20.

Answer: A and C. Both must be selected, and nothing else, to score the point.

EV = 0.80(200,000) + 0.40(500,000) + 0(300,000) = $360,000. AC ($450,000) is more than EV, so the work performed cost more than budgeted (CPI = 360,000 / 450,000 = 0.80). SPI = 360,000 / 400,000 = 0.90.

Why the other choices miss: B: Inverts the ratio (AC/EV). CPI is EV/AC. D: Schedule variance is EV − PV = −$40,000: behind schedule, not ahead.

Source: NSF, Earned Value Management Gold Card, EVM definitions; CPI = EV/AC, SPI = EV/PV, SV = EV − PV.

Answer: 6.0 L/s. Any entry that rounds to 6.0 at the stated precision counts as correct.

Q = K·i·A = (2.0 × 10⁻⁴)(0.25)(120) = 0.0060 m³/s = 6.0 L/s. That's about 518 m³ per day for the pumps to handle.

Common wrong answers: 24.0 L/s: Leaves out the hydraulic gradient. 0.0060 L/s: Correct in m³/s, but the question asks for L/s.

Source: FHWA/IN/JTRP-2019/04, Investigating the Need for Drainage Layers in Flexible Pavements (Feb. 2019), Eq. 2.7, p. 9 (PDF p. 20): Q = kiA and the steady, saturated, laminar-flow conditions.

Answer: C. 2.78.

Resisting moment = 2,400 × 1.25 = 3,000 lb-ft per ft. Overturning moment = 900 × 1.2 = 1,080 lb-ft per ft. FS = 3,000 / 1,080 = 2.78.

Why the other choices miss: A: Inverts the ratio (overturning ÷ resisting). B: Puts the thrust at mid-height (1.8 ft) instead of one-third of the height (1.2 ft), where a triangular pressure resultant acts. D: Uses the full 2.5 ft base width as the weight's lever arm instead of the distance to the weight's line of action.

Overturning is one check; sliding and bearing are separate.

Source: OpenStax, University Physics Vol. 1, Section 12.1: Conditions for Static Equilibrium, Eq. 12.5, torque equilibrium; Eq. 12.10, moment arm; FHWA, Rockery Design and Construction Guidelines, Chapter 4, Overturning, Figures 37–39: resisting and overturning moments about the toe; FS = resisting moment / overturning moment.

Answer: A. About 112 ft.

At 40 mph or less, L = WS²/60 = 11 × 35² / 60 = 224.6 ft. A shifting taper needs at least 0.5L = 112.3 ft, so about 112 ft.

Why the other choices miss: B: Half of WS, the formula for 45 mph and above. C: The full L, which is the merging-taper length, not a shift. D: WS, the high-speed formula, unhalved.

Edition check: Same formulas and taper fractions in the 2009 MUTCD (current exam) and the 11th edition (from April 2027); only the table numbers changed.

Source: MUTCD 2009 Edition with Revisions 1 and 2 (May 2012), Part 6, §6C.08; Tables 6C-3 and 6C-4, p. 557; MUTCD 11th Edition (Dec. 2023), Part 6, §6B.08; Tables 6B-3 and 6B-4, p. 775.