Quantity takeoff questions on contractor exams are arithmetic with a trap in the setup. The numbers are simple; the trap is a corner counted twice, an opening left in, a slab thickness in inches multiplied against dimensions in feet, or a waste factor applied to the wrong quantity. This guide shows the takeoff methods that avoid double-counting, with original worked problems for concrete, masonry, framing, drywall, and roofing, and the checks that catch the usual errors before you mark an answer.
The takeoff discipline
- Sketch the shape and label every dimension from the problem, in one unit. If the plan is in feet and a thickness is in inches, convert the thickness to feet right now (4 inches = 4/12 = 0.333 ft).
- Decide the counting unit the question wants: cubic yards, square feet, number of blocks, number of sheets, number of studs, squares of roofing.
- Measure once. For anything that runs around a perimeter, use a centerline; for anything that fills an area, deduct openings the way the question says to.
- Apply waste last, and only if the question says to. Waste is a percentage of the net quantity, added once.
- Round the way the question says (up to whole units for things you buy whole, to a stated decimal for volumes).
Original worked problems
These are original practice examples written for this guide, not official exam questions. Coverage figures (blocks per square foot, sheet sizes) are stated in the problems; on the exam, use the figures the question provides.
Problem 1: concrete slab in cubic yards
A slab is 40 feet by 30 feet and 4 inches thick. How many cubic yards of concrete does it take, with 5 percent waste, rounded up to the next whole yard?
Thickness = 4 / 12 = 0.3333 ft. Volume = 40 x 30 x 0.3333 = 400 cubic feet (keep the fraction as 4/12 or carry 0.3333; 40 x 30 x 4 / 12 = 4,800 / 12 = 400 exactly). Cubic yards = 400 / 27 = 14.815. Add waste: 14.815 x 1.05 = 15.556. Round up: 16 cubic yards. Traps: multiplying 40 x 30 x 4 = 4,800 and treating that as cubic feet (it is feet x feet x inches, a mixed unit with no physical meaning until the 4 is converted), and forgetting the 27 cubic feet in a cubic yard.
Problem 2: footing around a perimeter (the corner double-count)
A continuous rectangular footing ring measures 40 feet by 30 feet to the outside faces of the footing. The footing is 2 feet wide (measured inward from those outside faces), 12 inches deep, with square corners and no other intersecting footings. How many cubic yards of concrete, no waste?
The wrong way: outside perimeter = 2 x (40 + 30) = 140 ft; 140 x 2 x 1 = 280 cubic feet = 10.37 cubic yards. This counts each corner square (2 ft by 2 ft) twice, once from each direction.
Check by areas first, because it needs no shortcut: the outside rectangle is 40 x 30 = 1,200 square feet, the inside opening is (40 minus 4) x (30 minus 4) = 36 x 26 = 936 square feet, so the footing footprint is 1,200 minus 936 = 264 square feet, and at 1 foot deep the volume is 264 cubic feet. Cubic yards = 264 / 27 = 9.778, so 9.78 cubic yards.
The centerline shortcut gives the same result for this shape: centerline perimeter = outside perimeter minus 4 x width = 140 minus 8 = 132 ft; 132 x 2 x 1 = 264 cubic feet. The difference from the wrong way, 16 cubic feet, is exactly the four 2 x 2 x 1 corner blocks the outside-perimeter method counted twice. The shortcut is valid for a closed rectangular ring of constant width with square corners; for L-shapes, varying widths, or footings that intersect, go back to the area method.
Problem 3: concrete block wall
A wall is 100 feet long and 8 feet high with one opening 3 feet by 7 feet. The question says to use 1.125 blocks per square foot of wall (nominal 8-inch by 16-inch face), deduct the opening, and add 3 percent waste, rounding up to whole blocks.
Gross area = 100 x 8 = 800 square feet. Opening = 3 x 7 = 21 square feet. Net = 779 square feet. Blocks = 779 x 1.125 = 876.4. Waste: 876.4 x 1.03 = 902.7. Round up: 903 blocks. Where the 1.125 comes from: a nominal 8-by-16-inch face is 128 square inches, which is 0.889 square feet, and 1 / 0.889 = 1.125 blocks per square foot. Trap: applying the 3 percent to the gross area before deducting the opening (which double-counts waste on wall that is not there).
Problem 4: wall studs at 16 inches on center
A straight wall is 20 feet long, framed with studs at 16 inches on center. Using the question's rule (one stud per 16 inches of length plus one to close the end; corners and openings handled separately), how many studs?
Length in inches = 20 x 12 = 240 in. 240 / 16 = 15 spaces. Studs = 15 + 1 = 16 studs. If the result of the division is not a whole number, round up before adding the end stud, because a partial space still needs a stud at its end. Corner and opening framing is added according to the problem's stated rule, not from memory of a particular framing detail.
Problem 5: drywall sheets, with the shared-wall trap
A room is 12 feet by 14 feet with 8-foot ceilings. The walls get 4-by-12-foot sheets (48 square feet each). Deduct one door at 3 by 7 feet and one window at 3 by 4 feet, per the question. The question asks for an area-based sheet count: ignore cutting layout, joint placement, sheet orientation, and waste, and round up to whole sheets. How many sheets for the walls?
Perimeter = 2 x (12 + 14) = 52 ft. Gross wall area = 52 x 8 = 416 square feet. Openings = 21 + 12 = 33 square feet. Net = 383 square feet. Sheets = 383 / 48 = 7.979, round up to 8 sheets. That is the exam answer for the stated rule; it is not a purchase quantity for a real job, where layout and waste would add sheets. The double-count trap in drywall questions appears when two rooms are estimated: the wall between them is one wall with two faces. Count faces, not walls, and count each face once.
Problem 6: roofing squares
A roof has a total surface area of 2,400 square feet. Shingles are sold by the square (100 square feet). With 10 percent waste for cuts and starter courses, how many squares must be ordered, rounded up?
2,400 / 100 = 24 squares net. 24 x 1.10 = 26.4. Round up: 27 squares (waste is applied to the net quantity once, then the result is rounded once). Trap: using the building's footprint instead of the sloped roof area when the question gives both. The roof area is larger than the footprint by the slope factor, and the question will either give the roof area directly or give the pitch and expect you to apply the factor it supplies.
Problem 7: a strip footing that meets a slab (overlap)
A 30-by-20-foot slab, 6 inches thick, sits on top of the footing from a problem like Problem 2. The question asks for the slab concrete only. What is it?
Thickness = 0.5 ft. Volume = 30 x 20 x 0.5 = 300 cubic feet = 11.11 cubic yards. The trap is adding the footing volume "to be safe" or subtracting the footing width from the slab because the two touch. The question defined the quantity; answer that quantity.
Double-count checklist
| Situation | What gets double-counted | Fix |
|---|---|---|
| A closed rectangular ring of constant width (footing, curb, wall) | The corner squares | Centerline length = outside perimeter minus 4 x width, or use outside area minus inside area for any other shape |
| Two rooms sharing a wall | The shared wall, counted from both rooms as a wall | Count wall faces; a shared wall has two faces, one per room, counted once each |
| Waste applied to gross area, then openings deducted | Waste on material that is not installed | Deduct openings, then apply waste to the net |
| Percentages stacked | Waste plus "overage" plus rounding up | Apply the one factor the question gives, then round once |
| Overlapping assemblies (footing under slab, beam through wall) | The overlap volume | Take off each assembly to the boundary the question defines |
The five-second unit check
Before computing, write the unit of every number and the unit the answer needs. Feet x feet x feet = cubic feet, and cubic feet / 27 = cubic yards (1 yard = 3 feet by definition, so 1 cubic yard = 27 cubic feet, and 1 foot = 12 inches). Feet x feet = square feet, and square feet x (units per square foot) = units. If a step produces a unit that is not on that list, an inch did not get converted. Round only at the end and in the direction stated (up for materials you buy whole).
Practice next
The Estimating topic covers unit costs, overhead, and markup versus margin, which is where takeoff quantities go next. Then take the free contractor license practice test and sketch every takeoff problem before you calculate. The contractor study guide shows where estimating sits in the overall exam preparation.
Sources and verification notes
Coverage figures (1.125 blocks per square foot, sheet sizes, stud spacing rules, waste percentages) are stated inside each problem and are original practice examples, not official exam questions or product specifications. The 27 cubic feet per cubic yard and 12 inches per foot relationships follow from the definitions of the yard and the foot; NIST SP 811 is listed as a general unit-conversion reference only, and no NIST factor is used in the problems. Problem 2 is stated for a closed rectangular ring of constant width so that the centerline shortcut and the area method agree. Arithmetic was re-checked by calculator during this revision and is covered by the repository's guide-math check script.
- NIST Special Publication 811, Appendix B.8 (general unit-conversion reference)Checked September 18, 2026
Written by the ExamsLib editorial team. Practice examples in this guide are original and are not official exam questions. Exam rules change; the candidate bulletin from your licensing authority is the final word. Found an error? Contact us.
