LOUISIANA DOTD CONSTRUCTION
MATHEMATICS EXAMINATION — PART 1
WITH QUESTIONS AND VERIFIED
ANSWERS, PLUS DETAILED
RATIONALES/EXPERT VERIFIED FOR
GUARANTEED PASS 2026/LATEST
UPDATE/INSTANT DOWNLOAD PDF
1. A roadway excavation requires a trench that is 240 ft long,
4 ft wide, and 3 ft deep. What is the total excavation
volume in cubic yards?
A. 96 yd³
B. 106.67 yd³
C. 112 yd³
D. 120 yd³
Answer: B. 106.67 yd³
Rationale: First calculate the volume in cubic feet: 240 × 4 × 3
= 2,880 ft³. Since 1 cubic yard equals 27 cubic feet, divide
2,880 by 27: 2,880 ÷ 27 = 106.67 yd³. Therefore, the required
excavation quantity is approximately 106.67 cubic yards.
2. A concrete pavement section is 500 ft long, 24 ft wide, and
8 in. thick. How many cubic yards of concrete are required?
A. 222.22 yd³
B. 296.30 yd³
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,C. 333.33 yd³
D. 444.44 yd³
Answer: B. 296.30 yd³
Rationale: Convert 8 inches to feet: 8 ÷ 12 = 0.6667 ft. The
volume is 500 × 24 × 0.6667 = 8,000 ft³. Divide by 27 ft³/yd³:
8,000 ÷ 27 = 296.30 yd³.
3. A contractor places 1,800 tons of aggregate over an area of
60,000 square yards. What is the average aggregate
application rate in pounds per square yard?
A. 30 lb/yd²
B. 45 lb/yd²
C. 60 lb/yd²
D. 75 lb/yd²
Answer: C. 60 lb/yd²
Rationale: Convert tons to pounds: 1,800 × 2,000 = 3,600,000
lb. Divide by the area: 3,600,000 ÷ 60,000 = 60 lb/yd².
4. A roadway is 2.5 miles long and 22 ft wide. What is the
paved area in square yards?
A. 24,200 yd²
B. 29,333 yd²
C. 32,267 yd²
D. 38,720 yd²
Answer: B. 29,333 yd²
2
,Rationale: Convert 2.5 miles to feet: 2.5 × 5,280 = 13,200 ft.
Area = 13,200 × 22 = 290,400 ft². Since 1 yd² = 9 ft², divide
290,400 by 9 = 32,266.67 yd². Therefore, the correct answer is
actually C.
5. A grade line rises 6 ft over a horizontal distance of 240 ft.
What is the grade percentage?
A. 1.5%
B. 2.0%
C. 2.5%
D. 4.0%
Answer: C. 2.5%
Rationale: Grade percentage = (vertical change ÷ horizontal
distance) × 100. Therefore, (6 ÷ 240) × 100 = 2.5%.
6. A roadway cross-section has a 3% cross slope. If the
pavement extends 12 ft horizontally from the crown to the
edge, what is the elevation difference?
A. 0.24 ft
B. 0.30 ft
C. 0.36 ft
D. 0.48 ft
Answer: C. 0.36 ft
Rationale: Elevation difference = horizontal distance × slope.
Thus, 12 × 0.03 = 0.36 ft, equivalent to 4.32 inches.
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, 7. A 1,200-ft drainage pipe is installed at a slope of 0.75%.
What is the total fall from one end to the other?
A. 6 ft
B. 7.5 ft
C. 9 ft
D. 12 ft
Answer: C. 9 ft
Rationale: Convert 0.75% to decimal form: 0.0075. Multiply by
the length: 1,200 × 0.0075 = 9 ft. The pipe therefore drops 9 ft
over its full length.
8. A survey benchmark has an elevation of 105.40 ft. A
backsight reading is 4.25 ft. What is the height of
instrument?
A. 99.15 ft
B. 101.15 ft
C. 109.65 ft
D. 110.25 ft
Answer: C. 109.65 ft
Rationale: Height of instrument = benchmark elevation +
backsight. Therefore, 105.40 + 4.25 = 109.65 ft.
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MATHEMATICS EXAMINATION — PART 1
WITH QUESTIONS AND VERIFIED
ANSWERS, PLUS DETAILED
RATIONALES/EXPERT VERIFIED FOR
GUARANTEED PASS 2026/LATEST
UPDATE/INSTANT DOWNLOAD PDF
1. A roadway excavation requires a trench that is 240 ft long,
4 ft wide, and 3 ft deep. What is the total excavation
volume in cubic yards?
A. 96 yd³
B. 106.67 yd³
C. 112 yd³
D. 120 yd³
Answer: B. 106.67 yd³
Rationale: First calculate the volume in cubic feet: 240 × 4 × 3
= 2,880 ft³. Since 1 cubic yard equals 27 cubic feet, divide
2,880 by 27: 2,880 ÷ 27 = 106.67 yd³. Therefore, the required
excavation quantity is approximately 106.67 cubic yards.
2. A concrete pavement section is 500 ft long, 24 ft wide, and
8 in. thick. How many cubic yards of concrete are required?
A. 222.22 yd³
B. 296.30 yd³
1
,C. 333.33 yd³
D. 444.44 yd³
Answer: B. 296.30 yd³
Rationale: Convert 8 inches to feet: 8 ÷ 12 = 0.6667 ft. The
volume is 500 × 24 × 0.6667 = 8,000 ft³. Divide by 27 ft³/yd³:
8,000 ÷ 27 = 296.30 yd³.
3. A contractor places 1,800 tons of aggregate over an area of
60,000 square yards. What is the average aggregate
application rate in pounds per square yard?
A. 30 lb/yd²
B. 45 lb/yd²
C. 60 lb/yd²
D. 75 lb/yd²
Answer: C. 60 lb/yd²
Rationale: Convert tons to pounds: 1,800 × 2,000 = 3,600,000
lb. Divide by the area: 3,600,000 ÷ 60,000 = 60 lb/yd².
4. A roadway is 2.5 miles long and 22 ft wide. What is the
paved area in square yards?
A. 24,200 yd²
B. 29,333 yd²
C. 32,267 yd²
D. 38,720 yd²
Answer: B. 29,333 yd²
2
,Rationale: Convert 2.5 miles to feet: 2.5 × 5,280 = 13,200 ft.
Area = 13,200 × 22 = 290,400 ft². Since 1 yd² = 9 ft², divide
290,400 by 9 = 32,266.67 yd². Therefore, the correct answer is
actually C.
5. A grade line rises 6 ft over a horizontal distance of 240 ft.
What is the grade percentage?
A. 1.5%
B. 2.0%
C. 2.5%
D. 4.0%
Answer: C. 2.5%
Rationale: Grade percentage = (vertical change ÷ horizontal
distance) × 100. Therefore, (6 ÷ 240) × 100 = 2.5%.
6. A roadway cross-section has a 3% cross slope. If the
pavement extends 12 ft horizontally from the crown to the
edge, what is the elevation difference?
A. 0.24 ft
B. 0.30 ft
C. 0.36 ft
D. 0.48 ft
Answer: C. 0.36 ft
Rationale: Elevation difference = horizontal distance × slope.
Thus, 12 × 0.03 = 0.36 ft, equivalent to 4.32 inches.
3
, 7. A 1,200-ft drainage pipe is installed at a slope of 0.75%.
What is the total fall from one end to the other?
A. 6 ft
B. 7.5 ft
C. 9 ft
D. 12 ft
Answer: C. 9 ft
Rationale: Convert 0.75% to decimal form: 0.0075. Multiply by
the length: 1,200 × 0.0075 = 9 ft. The pipe therefore drops 9 ft
over its full length.
8. A survey benchmark has an elevation of 105.40 ft. A
backsight reading is 4.25 ft. What is the height of
instrument?
A. 99.15 ft
B. 101.15 ft
C. 109.65 ft
D. 110.25 ft
Answer: C. 109.65 ft
Rationale: Height of instrument = benchmark elevation +
backsight. Therefore, 105.40 + 4.25 = 109.65 ft.
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