Free FE Civil Practice Problems with Solutions [2026]

Correct answer: C. 23.7 kips

Reference: FE Reference Handbook 10.6, Statics — Equilibrium Requirements; Systems of Forces

Replace each distributed load by its resultant: the beam load gives W = 2.0(16) = 32 kips acting 8 ft from A, and the lateral load on the column gives H = 1.5(12) = 18 kips acting 12/2 = 6 ft above A, directed toward D.

With the roller at D on level ground, its reaction is vertical. Both resultants rotate the frame the same way about A, so summing moments about the pin picks up both: ΣM_A = 32(8) + 18(6) − D_y(22) = 0 → D_y = 364/22 = 16.55 kips

Vertical equilibrium: A_y = 32 − 16.55 = 15.45 kips

Horizontal equilibrium: only the pin can resist the lateral resultant, so A_x = 18 kips.

The pin reaction is the vector combination of both components: R_A = √(18² + 15.45²) = 23.7 kips

The lateral load enters the solution twice — once through its moment about A and once through the horizontal force balance — and the reaction magnitude is complete only when both components are combined.

Correct answer: B. 106 kips

Reference: FE Reference Handbook 10.6, Civil Engineering — Tension Members (Limit States and Available Strengths)

For net-area computation each deduction is the nominal hole diameter plus 1/16 in., and the nominal hole is itself the bolt diameter plus 1/16 in., so each hole removes 3/4 + 1/8 = 7/8 in. of width.

Rupture is checked on the weakest chain that runs across the plate from one edge to the other. The net-width rule starts from the gross width and credits each diagonal ligament with s²/4g, so it prices only a chain whose transverse steps add up to that width.

Straight transverse section (one hole): wn = 8 − 0.875 = 7.125 in.

Two-hole staggered chain (one diagonal): wn = 8 − 2(0.875) + (2)²/[4(4)] = 6.50 in. → governs

The third bolt cannot join the chain. A path that steps from the first gage line to the second and back again re-enters the edge it started from and leaves the plate in one piece; carried on to the far edge it crosses 16 in. of width for its three holes and two credits, nowhere near governing.

An = 6.50(0.375) = 2.4375 in². Every element of a flat plate is connected at the splice, so the effective net area carries no shear-lag reduction: Ae = An = 2.4375 in².

φPn = 0.75 Fu Ae = 0.75(58)(2.4375) = 106.0 kips, most nearly 106 kips

Pricing the reversing three-hole path as if it crossed the 8-in. width (wn = 5.875 in.) understates the strength at 96 kips; the straight one-hole section alone overstates it at 116 kips; ignoring the holes altogether prices the gross area at 130 kips.

Correct answer: A. 1.7

Reference: FE Reference Handbook 10.6, Civil Engineering — Singly-Reinforced Beams (Load Combinations using Strength Design (LRFD))

Stress block: a = As fy/(0.85 f′c b) = 2.4(60)/[0.85(4)(12)] = 3.53 in.

Design strength: φMn = 0.90(2.4)(60)(20 − 3.53/2)/12 = 197 kip·ft

The factored uniform load the span can carry: w_u = 8φMn/L² = 8(197)/20² = 3.94 kips/ft

Strength combination 1.2D + 1.6L: 1.2(1.0) + 1.6 w_L = 3.94 → w_L = 1.71 kips/ft, most nearly 1.7 kips/ft

The capacity is a factored quantity, so the service live load comes back out through its own load factor after the factored dead load is removed.

Correct answer: B. 1.4

Reference: FE Reference Handbook 10.6, Statics — Equilibrium Requirements

Moments about A: only the load's component perpendicular to the beam has a lever arm along it, so N_B(20) = 2.5 cos 30°(14) → N_B = 1.52 kips, acting perpendicular to the beam

Resolve the roller force into x and y: N_B = (1.52 cos 120°, 1.52 sin 120°) = (−0.76, 1.31) kips

Equilibrium at the pin: A_x = 0.76 kips and A_y = 2.5 − 1.31 = 1.19 kips → |A| = √(0.76² + 1.19²) = 1.41 kips, most nearly 1.4 kips

A roller perpendicular to an inclined member pushes partly sideways, so the pin must hold a horizontal component too — the reaction is the vector, not its vertical part.

Correct answer: B. −10 in⁴

Reference: FE Reference Handbook 10.6, Statics — Product of Inertia

Each rectangle has zero product of inertia about its own centroidal axes, so the composite value is entirely parallel-axis transfer: Ixy = Σ(dx dy A), with dx and dy the signed distances from the composite centroid to each part's centroid.

Locating the composite centroid from the lower-left corner: the vertical leg (A = 6 in² at (0.5, 3.0)) and the horizontal leg (A = 3 in² at (2.5, 0.5)) give x̄ = (6 × 0.5 + 3 × 2.5)/9 = 1.167 in. and ȳ = (6 × 3.0 + 3 × 0.5)/9 = 2.167 in.

The vertical leg's centroid sits up and to the left of C (second quadrant, dx negative, dy positive): dx dy A = (−0.667)(+0.833)(6) = −3.33 in⁴

The horizontal leg's centroid sits down and to the right of C (fourth quadrant, dx positive, dy negative): dx dy A = (+1.333)(−1.667)(3) = −6.67 in⁴

Ixy = −3.33 + (−6.67) = −10.0 in⁴

Both legs occupy quadrants where the product dx dy is negative, which is the expected result for an L oriented with its legs along the negative-product diagonal; a positive answer signals dropped signs rather than a different geometry.

Correct answer: D. 136.9 mm

Reference: FE Reference Handbook 10.6, Dynamics — Potential Energy (Elastic Potential Energy)

From release to maximum compression the block is momentarily at rest at both ends, so the gravitational potential energy lost equals the elastic energy stored. The block falls through the 0.5-m gap plus the compression x itself: mg(h + x) = kx²/2

12(9.81)(0.5 + x) = 8,000x²/2 → 4,000x² − 117.72x − 58.86 = 0

x = [117.72 + √(117.72² + 4(4,000)(58.86))]/(2 × 4,000) = 0.1369 m

x = 136.9 mm

Two habits fail here. Dropping the "+ x" term — as if the block stopped falling when it touched the spring — undershoots by about 15 mm. And no static or suddenly-applied formula applies: this is a free fall through half a metre, so the spring must absorb far more than 2×(static deflection).

Correct answer: D. 4.8 in

Reference: FE Reference Handbook 10.6, Mechanics of Materials — Stresses in Beams

The nails carry the longitudinal shear flow crossing the flange-web interface. With both plank areas equal to 16 in², the composite centroid sits at ȳ = (16×4 + 16×9)/32 = 6.5 in above the bottom. The composite moment of inertia about that axis is I = [2(8)³/12 + 16(2.5)²] + [8(2)³/12 + 16(2.5)²] = 185.3 + 105.3 = 290.7 in⁴. The flange's first moment about the neutral axis is Q = 16(9 − 6.5) = 40 in³. The shear flow is q = VQ/I = 600(40)/290.7 = 82.6 lb/in, so each 400-lb nail pair may be spaced up to s = 400/82.6 = 4.8 in apart. The spacing is nothing more than capacity over demand per inch — everything difficult lives in getting Q and I about the true composite centroid.

Correct answer: B. 0.014

Reference: FE Reference Handbook 10.6, Mechanics of Materials — Uniaxial Loading and Deformation

Internal forces first: segment BC carries only the end pull, 10 kips; segment AB carries both loads, 10 + 5 = 15 kips

δ_AB = FL/(AE) = 15(20)/(1.2 × 29,000) = 0.0086 in.

δ_BC = 10(10)/(0.6 × 29,000) = 0.0057 in.

δ_C = 0.0086 + 0.0057 = 0.0144 in., most nearly 0.014 in.

The load applied at B never travels through BC — each segment stretches under the force that actually crosses it, which is why the free-body cut comes before any FL/AE arithmetic.

Correct answer: C. 0.43

Reference: FE Reference Handbook 10.6, Materials Science/Structure of Matter — Lever Rule (Binary Phase Diagrams)

At 850°C the 40% B alloy lies inside the two-phase α + L field, so its mass is shared between solid α with the tie line's solidus-end composition and liquid with its liquidus-end composition. Reading the tie line at 850°C: xα = 20% B and xL = 55% B.

mass fraction solid = (xL − x)/(xL − xα) = (55 − 40)/(55 − 20) = 15/35 = 0.43

The lever rule weighs each phase by the arm of the tie line opposite to it: the solid fraction takes the distance from the overall composition out to the liquidus end, divided by the full tie-line length between the two phase compositions.

Correct answer: D. 45 ksi

Reference: FE Reference Handbook 10.6, Materials Science/Structure of Matter — Yield strength (Mechanics of Materials — Uniaxial Stress-Strain)

The yield strength by the 0.2% offset method is the stress where a line drawn parallel to the elastic slope, but shifted 0.002 strain to the right, meets the stress-strain record.

The elastic slope comes from the linear range of the data: E = 10/0.001 = 10,000 ksi (the first three rows lie on this line).

The offset line is therefore σ = 10,000(ε − 0.002), and testing it against the recorded points: at ε = 0.005 the line gives 30 ksi, well below the recorded 42 ksi; at ε = 0.0065 the line gives 10,000(0.0045) = 45 ksi — exactly the recorded stress; beyond that the line passes above the record.

The offset line meets the curve at 45 ksi, the yield strength.

The 0.2% offset is a strain of 0.002 measured as permanent set, not a total strain: the construction asks where unloading along the elastic slope would leave 0.2% permanent deformation. The proportional limit — where the record first leaves the linear slope — is a different, lower landmark on the same curve.

Correct answer: B. 8,990 lb

Reference: FE Reference Handbook 10.6, Fluid Mechanics — Forces on Submerged Surfaces and the Center of Pressure

The resultant of the pressure distribution is F = γ h̄ A = 62.4(11)(24) = 16,470 lb, acting at the center of pressure, which lies below the centroid by I/(h̄A): y_cp = 11 + [4(6)³/12]/[11(24)] = 11 + 0.273 = 11.273 ft below the surface, or 3.273 ft below the hinge line. Taking moments about the hinge: R_stop(6) = 16,470(3.273), so R_stop = 8,990 lb. The quarter-foot shift of the line of action is worth 750 lb at the stop — on a deeply submerged gate the pressure distribution is nearly uniform, but never uniform enough to ignore where its resultant acts.

Correct answer: D. 369 lb

Reference: FE Reference Handbook 10.6, Fluid Mechanics — The Impulse-Momentum Principle (The Continuity Equation (Conservation of Mass); Bernoulli Equation)

Continuity sets the velocities, Bernoulli prices the outlet pressure, and momentum closes each direction:

V₁ = 2.0/0.1963 = 10.2 ft/sec

V₂ = 2.0/0.0873 = 22.9 ft/sec

p₂ = 1,440 + (1.94/2)(10.2² − 22.9²) = 1,031 psf

x: R_x = p₁A₁ + ρQV₁ = 282.7 + 39.5 = 322 lb

y: R_y = p₂A₂ + ρQV₂ = 90.0 + 88.9 = 179 lb

R = √(322² + 179²) = 369 lb

A 90° bend must absorb both the incoming stream's momentum and pressure and supply the outgoing stream's — that is why each direction carries its own pA + ρQV pair. Thrust blocks on water mains are sized by exactly this resultant, and the pressure terms usually dominate — here they carry the x-direction, while the y-direction splits nearly evenly between pressure and momentum flux.

Correct answer: A. 827 ft²

Reference: FE Reference Handbook 10.6, Mathematics — Mensuration of Areas and Volumes

The fill height is the grade difference:

h = 851.0 − 842.0 = 9.0 ft

Each side slope projects horizontally by its ratio times the height: 2(9.0) = 18 ft on the left and 3(9.0) = 27 ft on the right, so the base of the fill is 76 + 18 + 27 = 121 ft

Gross trapezoid between the finished top and the ground: A = ½(76 + 121)(9.0) = 886.5 ft²

Triangular median swale cut out of the top: A = ½(24)(5.0) = 60 ft²

Net embankment area = 886.5 − 60 = 826.5 ft² ≈ 827 ft²

Decomposing an irregular section into simple shapes is the working method for every end-area computation: build the largest simple figure the outline allows, then correct it with the pieces that are not fill. The swale is air, not embankment, so its triangle carries a minus sign — keeping track of which decomposed pieces add material and which remove it is the whole bookkeeping of the method.

Correct answer: B. 21.6 ft

Reference: FE Reference Handbook 10.6, Mathematics — Trigonometry

The vertical component of the slope distance is s·cos(z) = 285.60 × cos 85°30′ = 22.4 ft

The line runs from the instrument's trunnion axis to the prism, so the ground-to-ground difference restores both offsets: ΔElev = 22.4 + hi − ht = 22.4 + 5.2 − 6.0 = 21.6 ft

The instrument height lifts the line's origin above the station mark and the prism floats above the surveyed point — both offsets must be restored to move from line-of-sight geometry to ground-to-ground elevation.

Correct answer: B. 14.2 ft³/sec

Reference: FE Reference Handbook 10.6, Civil Engineering — Weir Formulas (with Orifices)

Both openings sit in series, so the basin can only release what the tighter one passes

Riser weir: Q = CwLH^3/2 = 3.33(6.0)(2.0)^1.5 = 56.5 ft³/sec

Barrel orifice: Q = CdA√(2gh) = 0.60(1.2)√(64.4 × 6.0) = 14.2 ft³/sec

The orifice governs: 14.2 ft³/sec — water pours over the crest faster than the barrel can carry it away, so the riser simply runs full.

Correct answer: A. 15,100 gal/day

Reference: FE Reference Handbook 10.6, Civil Engineering — Darcy's Law

The hydraulic gradient comes from piezometric heads, not depths to water, so each water-table elevation is computed first from its ground surface.

Head at the first well: h₁ = 741.0 − 18.6 = 722.4 ft

Head at the second well: h₂ = 752.5 − 34.6 = 717.9 ft

Even though the ground rises toward the second well, the water table there is lower, so flow moves from the first well toward the second: i = (722.4 − 717.9)/1,500 = 0.0030

The discharge crosses the saturated section of the aquifer: A = 40 × 800 = 32,000 ft²

Darcy's law gives the volumetric discharge directly from the superficial velocity — the porosity plays no part in Q: Q = KiA = 21 × 0.0030 × 32,000 = 2,016 ft³/day

Converting, Q = 2,016 × 7.48 = 15,080 gal/day.

Correct answer: C. 21

Reference: FE Reference Handbook 10.6, Civil Engineering — Manning's Equation (with the Hydraulic-Element (Partial Flow) Graph for Circular Sewers)

At d/D = 0.70 the hydraulic-elements chart reads Q/Q_full = 0.84, so the pipe must carry Q_full = 6.5/0.84 = 7.74 cfs when flowing full

Manning's equation for a full circular pipe: Q_full = (1.486/n)(πD²/4)(D/4)^(2/3)√S → D^(8/3) = 7.74(0.013)/[1.486(0.3117)(0.0632)] → D = 1.59 ft = 19.1 in.

The smallest standard size at or above 19.1 in. is 21 in.

The depth limit forces the full-flow capacity above the design flow; sizing at full flow alone would pick a pipe that runs deeper than 0.70 D.

Correct answer: D. 8

Reference: FE Reference Handbook 10.6, Civil Engineering — Soil Classification

From the curve, the diameters at 60% and 10% passing are D60 = 0.60 mm and D10 = 0.075 mm.

Cu = D60/D10 = 0.60/0.075 = 8.0

The uniformity coefficient spans the curve between its 60% and 10% ordinates — the larger diameter over the smaller, so the ratio always exceeds one, and a value below one flags an inverted read. The 30% ordinate belongs to the other gradation coefficient, and a curve read one gridline off at the top swaps the answer for the D50 impostor.

Correct answer: C. 6,600 lb/ft

Reference: FE Reference Handbook 10.6, Civil Engineering — Retaining Walls

The driving force is the active thrust, Pa = ½KaγH² = 0.5(0.33)(120)(10)² = 1,980 lb per foot of wall. Sliding resistance comes from base friction, μW, and the safety factor is the ratio of resistance to driving force. Requiring FS = 1.5 and solving for the weight:

W = FS × Pa/μ = 1.5(1,980)/0.45 = 6,600 lb/ft.

The friction coefficient divides — a slicker base demands a heavier wall — and the factor of safety scales the requirement up front. Designing the weight from the sliding check first is standard gravity-wall practice; overturning is then verified with the geometry that delivers this mass.

Correct answer: A. 0.6

Reference: FE Reference Handbook 10.6, Civil Engineering — Coefficient of concavity (or curvature)

The exhibit's ordinate is cumulative percent RETAINED, while every defining D-size is a percent-finer diameter. Sieve results are tallied as mass caught on each screen, so the two scales are complements: percent finer = 100 − percent retained. D10 therefore reads where the curve crosses the 90% retained line, D30 at 70% retained, and D60 at 40% retained.

From those three crossings: D10 = 0.2 mm, D30 = 0.6 mm, D60 = 3.0 mm.

Cc = (D30)²/(D10 × D60) = (0.6)²/[(0.2)(3.0)] = 0.36/0.60 = 0.6

The middle diameter is the one that enters squared, because the coefficient measures how the center of the gradation bends between its two anchor sizes. With Cc below 1 this soil's curve dips under the smooth well-graded shape between D10 and D60 — a gap-tending gradation despite its wide size range — and that judgment starts with reading the axis for what it actually plots before pulling a single number off the curve.

Correct answer: B. 79.7 ft

Reference: FE Reference Handbook 10.6, Civil Engineering — Horizontal Curves

The printed long-chord relation recovers the radius first: LC = 2R sin(Δ/2) → R = 860/[2 sin 21°] = 1,200 ft

The printed middle-ordinate relation then closes it: M = R[1 − cos(Δ/2)] = 1,200(1 − cos 21°) = 79.7 ft

Both formulas run on the half-angle — the chord and its ordinate belong to the half-triangles the curve's bisector cuts — and the middle ordinate hangs from the chord toward the arc, while its near-twin, the external distance, reaches from the PI. The two answer different setback questions, and only one of them is asked.

Correct answer: B. 4,310 pc/h

Reference: FE Reference Handbook 10.6, Civil Engineering — Level of Service (LOS) (Peak Hour Factor)

LOS C holds while density stays at or below the printed threshold of 26 pc/mi/ln.

The flow identity q = kv converts that ceiling to a per-lane flow rate: q = 26 × 60 = 1,560 pc/h/ln, or 4,680 pc/h across three lanes — but that is a peak 15-minute rate, not an hourly volume.

The hour can only average PHF times its worst quarter: V = 0.92 × 4,680 = 4,310 pc/h

Both table reads are load-bearing — the LOS C row (not D), and the direction of the PHF adjustment: peaking always shrinks what a full hour can serve.

Correct answer: B. 11 in.

Reference: FE Reference Handbook 10.6, Civil Engineering — AASHTO Structural Number Equation

The structural number sums the layered contributions: SN = a₁D₁ + a₂D₂m₂ + a₃D₃m₃

The surface and subbase are fixed: a₁D₁ = 0.44(4.0) = 1.76 and a₃D₃m₃ = 0.11(8.0)(1.0) = 0.88

The base must supply the remainder: a₂D₂m₂ = 4.0 − 1.76 − 0.88 = 1.36

D₂ = 1.36/(0.14 × 0.9) = 1.36/0.126 = 10.8 in.

The drainage coefficient discounts the base layer's contribution, so it lives in the divisor when the equation runs backward — a wet-side m below 1.0 always demands more thickness, and any manipulation that yields less has put m on the wrong side.

Correct answer: B. 5.4 ft

Reference: NCEES FE Civil exam specification — Construction Engineering (interpretation of engineering drawings); general knowledge item — no FE Reference Handbook coverage

Cover is measured from the finished ground surface down to the outside top — the crown — of the pipe, so three reads off the sheet chain together: the pipe grade, the ground elevation at the station, and the barrel dimensions from the callout.

A profile sheet states pipe position by invert, because grade control is set on the flowline. Station 11+40 lies 140 ft beyond MH-1 at 10+00, so the invert there comes by interpolation down the 1.5% grade: inv El. = 425.60 − 0.015(140) = 425.60 − 2.10 = 423.50 ft

The invert is the inside bottom of the pipe. From it, the outside crown sits one inside diameter plus one wall thickness higher — the callout's 24-in. bore plus a (30 − 24)/2 = 3-in. wall: crown El. = 423.50 + 2.00 + 0.25 = 425.75 ft

Cover = ground − crown = 431.10 − 425.75 = 5.35 ft, most nearly 5.4 ft

Adding the full 30-in. outside diameter to the invert counts the bottom wall too, and that wall sits below the flowline the invert already marks. The rim elevations on the same sheet describe the casting heights at the two structures, not the surface along the run — the spot elevation at 11+40 is the one that counts.

Correct answer: C. $1,000

Reference: FE Reference Handbook 10.6, Civil Engineering — CPM Precedence Relationships

Crashing shortens the project only while the crashed activity lies on every critical path - and the critical path moves as durations change.

Start: B-D = 10 days governs (A-C = 9).

Day 1: crash the cheapest B-D activity - B at $400. Duration 9; now both chains are critical at 9 days.

Day 2: both chains must shorten together. Cheapest on B-D is B again ($400); cheapest on A-C is A ($200). Day-2 cost $600.

Total: $400 + $600 = $1,000 for an 8-day project.

Re-identifying the critical path after every crash day is the discipline: once the chains tie, a day of project compression must be purchased on every critical chain simultaneously.

Correct answer: D. 301 ft²

Reference: FE Reference Handbook 10.6, Mathematics — Test for a Minimum

The required volume in consistent units is V = 3,000/7.48 = 401.1 ft³.

For a closed cylinder, V = πr²h fixes h = V/(πr²), so the total area of shell plus both caps becomes a function of r alone: A(r) = 2πr² + 2πrh = 2πr² + 2V/r

Setting the derivative to zero: dA/dr = 4πr − 2V/r² = 0 → r³ = V/(2π) → r = 3.997 ft

The second derivative, d²A/dr² = 4π + 4V/r³, is positive there, so this critical point is a minimum. The optimal height is h = V/(πr²) = 7.99 ft, which is exactly twice the radius — the well-known least-material proportion for a closed can.

A_min = 2π(3.997)² + 2π(3.997)(7.99) = 100.4 + 200.7 = 301.1 ft²

Correct answer: D. $112,000

Reference: FE Reference Handbook 10.6, Engineering Probability and Statistics — Least Squares; Linear Regression and Goodness of Fit

The intercept comes at the end of a slope-first chain, and both sums of squares need their correction terms before the slope exists.

Sxy = Σxy − (Σx)(Σy)/n = 2,338.0 − (42)(399.0)/6 = −455.0

Sxx = Σx² − (Σx)²/n = 364 − (42)²/6 = 70.0

b = Sxy/Sxx = −455.0/70.0 = −6.50 ($1000s per year)

The means are x̄ = 42/6 = 7.0 yr and ȳ = 399.0/6 = 66.5 ($1000s).

â = ȳ − b·x̄ = 66.5 − (−6.50)(7.0) = 112.0 ($1000s)

The prices were tallied in $1000s, so the intercept converts to $112,000 — the fitted price at age zero. The negative slope makes the subtraction in â = ȳ − b·x̄ add magnitude, which is why the intercept lands well above the mean price.

Correct answer: A. 3 ft

Reference: FE Reference Handbook 10.6, Mathematics — Quadratic Equation

The gravel fixes the border area. Converting the volume, 8 yd³ × 27 ft³/yd³ = 216 ft³, and spreading it 6 in. (0.5 ft) deep gives 216/0.5 = 432 ft² of border.

With border width x on all four sides, the bordered footprint is (40 + 2x) by (25 + 2x), so the border area is (40 + 2x)(25 + 2x) − 40(25) = 4x² + 130x

Setting that equal to 432 ft²: 4x² + 130x − 432 = 0 → 2x² + 65x − 216 = 0

The quadratic formula gives x = [−65 + √(65² + 4(2)(216))]/(2 · 2) = (−65 + 77.16)/4 = 3.04 ft

The formula's other root is negative, and a border cannot have a negative width, so that root is rejected on physical grounds. The width is most nearly 3.0 ft.

Correct answer: D. No contract exists - the conditional reply was a counteroffer that ended the original offer

Reference: NCEES FE Civil exam specification — Ethics and Professional Practice (contract-law elements); general knowledge item — no FE Reference Handbook coverage

An acceptance must mirror the offer's terms. The owner's reply changed the price, and a reply that changes the terms is not an acceptance at all — it is a counteroffer, and a counteroffer TERMINATES the offer it answers.

From that moment the $180,000 offer was dead. The owner's later message could not accept what no longer existed; it stands only as a fresh offer to the contractor, who never accepted it. Silence bound nobody in either direction — a party's failure to respond to a counteroffer forms nothing.

Correct answer: C. $13,580

Reference: FE Reference Handbook 10.6, Engineering Economics — Interest Rate Tables

Bring both deposits to time zero:

P = 50,000 + 20,000(P/F, 6%, 3) = 50,000 + 20,000(0.8396) = $66,792

The A/P factor prices an annuity whose first payment falls one period after the valuation point, so a series that begins at the end of year 5 must be valued at the end of year 4:

V₄ = 66,792(F/P, 6%, 4) = 66,792(1.2625) = $84,325

A = 84,325(A/P, 6%, 8) = 84,325(0.16104) = $13,580

Rolling the account forward to year 5 — the date the first withdrawal is actually taken — overshoots by one period of interest, because that valuation point demands a beginning-of-period series the tables do not price.