20 Landscape Math & Estimating Practice Questions & Answers
Every Landscape Math & Estimating practice question from the Landscape Technician Practice Test, with the correct answer and a short explanation.
Start practice test →1. A mulch bed measures 40 ft by 26 ft, and the specification calls for mulch 3 in deep. Mulch is sold by the cubic yard and the supplier delivers whole cubic yards only. How many cubic yards must be ordered?
- A.9.63 cubic yards
- B.28.9 cubic yards
- C.260 cubic yards
- D.10 cubic yards✓ Answer
Area is 40 x 26 = 1,040 sq ft. The depth must be converted to feet before it multiplies a square-foot area: 3 / 12 = 0.25 ft, so 1,040 x 0.25 = 260 cubic feet. A cubic yard holds 27 cubic feet, so 260 / 27 = 9.63 cubic yards, and bulk material bought in whole yards is always rounded up, giving 10. Leaving 9.63 un-rounded, dividing by 9 instead of 27, or reporting the 260 cubic feet as yards are the three standard errors.
Source: NALP Landscape Training Manual, Landscape Plan Reading & Calculations (area x depth in feet = cubic feet; cubic feet / 27 = cubic yards, rounded up)Report a problem with this question
2. A 3,000 sq ft planting bed will have 4 in of compost tilled into it. Compost is sold by the cubic yard and must be ordered in whole cubic yards. How many cubic yards are required?
- A.1,000 cubic yards
- B.111.1 cubic yards
- C.38 cubic yards✓ Answer
- D.37.04 cubic yards
Four inches is 4 / 12 = 0.3333 ft, so 3,000 x 0.3333 = 1,000 cubic feet of compost. Dividing by the 27 cubic feet in a cubic yard gives 37.04, which rounds up to 38 whole yards because a supplier will not deliver a fraction of a yard. The 111.1 figure comes from dividing by 9, which is the square-measure conversion, and 1,000 is the volume left in cubic feet.
Source: NALP Landscape Training Manual, Landscape Plan Reading & Calculations (soil amendment volume: area x depth / 27)Report a problem with this question
3. A patio pad measures 18 ft by 20 ft and calls for 6 in of compacted aggregate base. The quarry sells this base by the ton and states that it weighs 1.5 tons per cubic yard. How many tons are required?
- A.270 tons
- B.10 tons✓ Answer
- C.6.7 tons
- D.30 tons
Area is 18 x 20 = 360 sq ft and 6 in is 0.5 ft, so the base is 360 x 0.5 = 180 cubic feet, or 180 / 27 = 6.67 cubic yards. Multiplying by the supplier's stated 1.5 tons per cubic yard gives 10 tons. The tons-per-yard figure always has to come from the stem or the supplier because it changes with the material and its moisture content; 6.7 is the volume left in yards, 270 skips the division by 27, and 30 divides by 9.
Source: NALP Landscape Industry Certified Technician (Exterior), Hardscape Installation - 'Hardscape Principles and Calculations' (base course volume converted to weight using a supplier-stated tons per cubic yard)Report a problem with this question
4. A lawn measures 63 ft by 40 ft, and the sod grower quotes and sells sod by the square yard. How many square yards of sod does the lawn take?
- A.2,520 square yards
- B.280 square yards✓ Answer
- C.840 square yards
- D.31.1 square yards
The lawn is 63 x 40 = 2,520 sq ft. A yard is 3 ft on a side, so a square yard covers 3 x 3 = 9 sq ft and the conversion is a division by 9: 2,520 / 9 = 280 square yards. Dividing by 3 gives 840 and treats a linear conversion as an area conversion, 2,520 is the untouched square-foot figure, and 31.1 divides by 81.
Source: NALP Landscape Training Manual appendix 'Conversion Charts for Weights & Measures' (1 square yard = 9 square feet)Report a problem with this question
5. A seeding rate is given in pounds per 1,000 sq ft, but the plan lists the turf area as 2.5 acres. Converted to square feet, how large is that turf area?
- A.108,900 square feet✓ Answer
- B.43,560 square feet
- C.12,100 square feet
- D.17,424 square feet
One acre is a fixed 43,560 sq ft, so 2.5 x 43,560 = 108,900 sq ft, which then divides by 1,000 to give the 108.9 rate units the label calls for. The 43,560 option is a single acre, 12,100 is the same parcel expressed in square yards because it uses the 4,840 sq yd per acre figure, and 17,424 comes from dividing by 2.5 rather than multiplying.
Source: NALP Landscape Training Manual appendix 'Conversion Charts for Weights & Measures' (1 acre = 43,560 square feet = 4,840 square yards)Report a problem with this question
6. A drainage trench will be 32 ft long, 18 in wide and 8 in deep. How many cubic feet of soil come out of it?
- A.1.2 cubic feet
- B.384 cubic feet
- C.32 cubic feet✓ Answer
- D.4,608 cubic feet
Every dimension has to be in the same unit before they are multiplied, so 18 in becomes 18 / 12 = 1.5 ft and 8 in becomes 8 / 12 = 0.667 ft, giving 32 x 1.5 x 0.667 = 32 cubic feet. Leaving the depth in inches produces 384, leaving both inch dimensions unconverted produces 4,608, and 1.2 is the correct volume already divided by 27, which makes it cubic yards rather than cubic feet.
Source: NALP Landscape Industry Certified Technician (Exterior), Common Core 'Basic math' (12 inches = 1 foot; volume = length x width x depth in consistent units)Report a problem with this question
7. A circular tree ring measures 22 ft across. Using 3.14 for pi, what is the area inside the ring?
- A.1,519.76 square feet
- B.379.94 square feet✓ Answer
- C.189.97 square feet
- D.69.08 square feet
The area of a circle uses the radius, not the diameter, so the 22 ft measured across must first be halved to r = 11 ft, giving 3.14 x 11 x 11 = 379.94 sq ft. Feeding the 22 ft diameter straight into the formula quadruples the answer to 1,519.76, the 69.08 figure is the circumference 3.14 x 22 and is a lineal measure used for edging rather than mulch, and 189.97 halves the area instead of the diameter.
Source: NALP Landscape Training Manual, Landscape Plan Reading & Calculations (area of a circle = pi x radius squared)Report a problem with this question
8. A triangular turf panel has a 46 ft base and a 30 ft height. What is its area?
- A.345 square feet
- B.690 square feet✓ Answer
- C.460 square feet
- D.1,380 square feet
A triangle fills exactly half of the rectangle that shares its base and height, so the area is (46 x 30) / 2 = 1,380 / 2 = 690 sq ft. The 1,380 option is that enclosing rectangle with the halving step forgotten, 345 halves the result a second time, and 460 divides by 3 on the mistaken idea that a three-sided figure calls for a division by three.
Source: NALP Landscape Training Manual, Landscape Plan Reading & Calculations (area of a triangle = base x height / 2)Report a problem with this question
9. A bed is a 30 ft by 14 ft rectangle with a half-circle added across one of the 14 ft ends. Using 3.14 for pi, what is the total area of the bed?
- A.573.86 square feet
- B.420 square feet
- C.1,035.44 square feet
- D.496.93 square feet✓ Answer
An irregular bed is handled by cutting it into simple shapes and adding them: the rectangle is 30 x 14 = 420 sq ft, and the half-circle sits on the 14 ft end so its diameter is 14 and its radius 7, giving (3.14 x 7 x 7) / 2 = 76.93 sq ft, for 496.93 sq ft in total. The 1,035.44 figure uses the 14 ft end as the radius and forgets to halve the circle, 573.86 adds a whole circle of the correct radius, and 420 is the rectangle alone.
Source: NALP Landscape Training Manual, Landscape Plan Reading & Calculations (irregular areas are subdivided into rectangles, triangles and circle segments and summed)Report a problem with this question
10. A rectangular lawn measures 110 ft by 90 ft. Within it sit an 18 ft by 14 ft patio and a circular bed 16 ft in diameter, neither of which is turf. Using 3.14 for pi, what is the net turf area to the nearest square foot?
- A.8,844 square feet
- B.9,447 square feet✓ Answer
- C.10,353 square feet
- D.9,648 square feet
Net area is the gross area minus every deduction, so 110 x 90 = 9,900 sq ft, the patio removes 18 x 14 = 252 sq ft, and the bed removes 3.14 x 8 x 8 = 200.96 sq ft, leaving 9,447 sq ft. The 9,648 option deducts the patio but forgets the bed, 8,844 uses the 16 ft diameter as the radius and so removes four times too much, and 10,353 adds the two features instead of subtracting them.
Source: NALP Landscape Training Manual, Landscape Plan Reading & Calculations (net area = gross area minus non-turf deductions)Report a problem with this question
11. A 20,000 sq ft turf area is to receive 1.25 lb of nitrogen per 1,000 sq ft, using a fertilizer with a 22-3-11 analysis. How many pounds of fertilizer does that take, to the nearest tenth of a pound?
- A.25 pounds
- B.550 pounds
- C.5.5 pounds
- D.113.6 pounds✓ Answer
The area gives 20,000 / 1,000 = 20 rate units, so the nitrogen needed is 20 x 1.25 = 25 lb. The first number of the analysis is the percent nitrogen by weight, so the pounds of product is the nitrogen divided by that percent as a decimal: 25 / 0.22 = 113.6 lb. The 25 option answers with pounds of nitrogen rather than pounds of product, 550 multiplies by 22 instead of dividing by 0.22, and 5.5 multiplies by 0.22.
Source: NALP National Collegiate Landscape Competition, Maintenance Cost Estimating Study Guide (fertilizer calculation: rate units x lb N per 1,000 sq ft, then lb N divided by the decimal percent nitrogen)Report a problem with this question
12. A 46,000 sq ft lawn is to receive 1 lb of nitrogen per 1,000 sq ft from a 25-0-10 fertilizer sold in 40 lb bags, and only whole bags can be ordered. How many bags are needed?
- A.29 bags
- B.5 bags✓ Answer
- C.4 bags
- D.4.6 bags
The lawn is 46 rate units, so it needs 46 lb of nitrogen, and 46 / 0.25 = 184 lb of product. Dividing by the 40 lb bag weight gives 4.6 bags, which must be rounded up to 5 because a partial bag cannot be ordered and rounding down would leave part of the lawn untreated. The 4.6 option is the un-rounded value, 4 rounds the wrong way, and 29 multiplies 46 by 25 instead of dividing by 0.25.
Source: NALP National Collegiate Landscape Competition, Maintenance Cost Estimating Study Guide (bags = pounds of product / bag weight, rounded up to a whole bag)Report a problem with this question
13. A 27,400 sq ft lawn is to be seeded at 5 lb of seed per 1,000 sq ft. How many pounds of seed are needed?
- A.27.4 pounds
- B.1,370 pounds
- C.137 pounds✓ Answer
- D.5,480 pounds
Seed rate is stated in pounds of seed itself, so there is no analysis percentage to divide by: 27,400 / 1,000 = 27.4 rate units, and 27.4 x 5 = 137 lb. The 27.4 option reports the rate units as if they were pounds, 1,370 divides the area by 100 instead of 1,000, and 5,480 divides the area by the rate instead of multiplying the rate units by it.
Source: NALP National Collegiate Landscape Competition, Maintenance Cost Estimating Study Guide (seed quantity = area / 1,000 x pounds of seed per 1,000 square feet)Report a problem with this question
14. A groundcover bed of 1,800 sq ft is to be planted 18 in on center on a square grid. How many plants does that take?
- A.1,200 plants
- B.800 plants✓ Answer
- C.1,800 plants
- D.924 plants
On a square grid each plant occupies a square whose side is the spacing, so 18 in becomes 1.5 ft and each plant takes 1.5 x 1.5 = 2.25 sq ft: 1,800 / 2.25 = 800 plants. The 1,200 option divides by the spacing itself rather than by its square, 1,800 treats the spacing as 12 in, and 924 applies the 0.866 triangular factor even though the plan specifies a square grid.
Source: NALP Landscape Training Manual, Landscape Plan Reading & Calculations (square spacing: plants = area / spacing squared)Report a problem with this question
15. A 2,600 sq ft bed is to be planted 24 in on center in a triangular, staggered pattern, using the 0.866 triangular factor. How many plants are needed, to the nearest whole plant?
- A.650 plants
- B.563 plants
- C.1,300 plants
- D.751 plants✓ Answer
Triangular spacing pushes each row into the gaps of the row beside it, so each plant covers spacing squared times 0.866: 2 x 2 x 0.866 = 3.464 sq ft, and 2,600 / 3.464 = 751 plants. The 650 option is the square-grid count found by dividing by 4 alone, 1,300 divides by the 2 ft spacing rather than its square, and 563 multiplies the square-grid count by 0.866 instead of dividing by it, which wrongly reduces the count instead of increasing plant density.
Source: NALP Landscape Training Manual, Landscape Plan Reading & Calculations (triangular spacing: plants = area / (spacing squared x 0.866))Report a problem with this question
16. A 420 sq ft patio will be paved with units whose face measures 4 in by 8 in, and the specification adds 10 percent for cuts and breakage. Pavers are ordered whole. How many pavers are needed?
- A.13,440 pavers
- B.1,890 pavers
- C.1,701 pavers
- D.2,079 pavers✓ Answer
One paver face is 4 x 8 = 32 sq in and a square foot holds 144 sq in, so 144 / 32 = 4.5 pavers cover a square foot and the patio takes 420 x 4.5 = 1,890 units. Applying the specified allowance gives 1,890 x 1.10 = 2,079 pavers, ordered whole. The 1,890 option leaves the allowance out, 1,701 subtracts 10 percent instead of adding it, and 13,440 multiplies the area by the paver's square inches.
Source: NALP Landscape Industry Certified Technician (Exterior), Hardscape Installation - 'Hardscape Principles and Calculations' (paver count = area / unit face area, plus the specified waste allowance)Report a problem with this question
17. A lawn needs 4,650 sq ft of sod. Rolls measure 2 ft by 5 ft, and the specification adds 8 percent for cutting waste along the curved edges. Rolls are ordered whole. How many rolls are needed?
- A.5,022 rolls
- B.465 rolls
- C.503 rolls✓ Answer
- D.502 rolls
Adding the stated allowance first gives 4,650 x 1.08 = 5,022 sq ft of sod to buy, and each roll of the stated size covers 2 x 5 = 10 sq ft, so 5,022 / 10 = 502.2 rolls. A partial roll cannot be delivered, so the order rounds up to 503. The 465 option omits the waste allowance, 502 rounds down and would come up short at the last curve, and 5,022 is the square footage rather than a roll count.
Source: NALP Landscape Training Manual, Landscape Plan Reading & Calculations (sod quantity = area plus the specified waste allowance, divided by the coverage of one roll, rounded up)Report a problem with this question
18. A walk runs 46 ft out from a building and must be pitched away from it at 2 percent. Over that full run, how much fall is that in inches?
- A.11.04 inches✓ Answer
- B.0.92 inches
- C.92 inches
- D.5.52 inches
Percent slope means rise divided by run times 100, so fall equals the run times the slope as a decimal: 46 x 0.02 = 0.92 ft. Because the answer is asked for in inches, that figure is multiplied by 12 to give 11.04 in. The 0.92 option reports feet under an inch label, 92 treats 2 percent as 2 in of fall per foot of run, and 5.52 uses a 1 percent grade.
Source: NALP Landscape Training Manual, Landscape Plan Reading & Calculations (percent slope = rise / run x 100; fall = run x slope, converted using 12 inches per foot)Report a problem with this question
19. Finished grade drops 2 ft over a 40 ft run. Stated as a percent slope and as a rise-to-run ratio, what is that grade?
- A.0.5 percent, or a 1:200 slope
- B.2 percent, or a 1:50 slope
- C.20 percent, or a 1:5 slope
- D.5 percent, or a 1:20 slope✓ Answer
Percent slope is rise divided by run and then multiplied by 100, so 2 / 40 = 0.05, which is 5 percent, and the same fraction written as a ratio of one unit of rise to units of run is 1:20. The 20 percent option inverts the fraction to run over rise, 2 percent reads the 2 ft of rise directly as the percent, and 0.5 percent moves the decimal one place.
Source: NALP Landscape Training Manual, Landscape Plan Reading & Calculations (percent slope = rise / run x 100, also expressed as a rise-to-run ratio)Report a problem with this question
20. A three-person crew mows a 96,000 sq ft turf area at a production rate of 0.00015 labor hours per square foot. How many labor hours does one mowing take?
- A.4.8 hours
- B.144 hours
- C.1.4 hours
- D.14.4 hours✓ Answer
Labor hours for one occurrence are simply the quantity of work multiplied by the stated production rate: 96,000 x 0.00015 = 14.4 hours. The 4.8 option is the elapsed time on site once those hours are split among three people, which is a different quantity from labor hours and would understate the payroll; 144 and 1.4 both come from misplacing the decimal in the production rate.
Source: NALP National Collegiate Landscape Competition, Maintenance Cost Estimating Study Guide (hours per occurrence = quantity x production rate)Report a problem with this question
Practice questions based on the published content areas of the National Association of Landscape Professionals' Landscape Industry Certified Technician (Exterior) programme — the universal safety and hazard-communication elements every candidate takes, plus the plan reading, horticultural principles and calculation content behind the installation and maintenance specialties — together with standard horticulture, soils, turfgrass and hardscape construction references. This site is not affiliated with or endorsed by the National Association of Landscape Professionals. This bank covers written knowledge only; where a certification programme also assesses hands-on skill, that part is judged in the field and cannot be practised on a screen. Irrigation system design and troubleshooting, and arboriculture, are covered by our separate irrigation and arborist banks rather than here. Landscape work is governed by pesticide, licensing, water-use and vehicle rules that vary by state and locality — confirm what applies where you work, and confirm current certification requirements with NALP before testing. About NALP certification →