Q1. Which of the following statements for the included angle method is/are true?
Statement 1: Included angles can be measured either clockwise or counter-clockwise.
Statement 2: The measured clockwise angles are interior angles if the direction of progress around the survey is counter-clockwise.
Difficulty: Medium
A. Only Statement 2 is true
B. Both Statement 1 and 2 are true
C. Neither Statement 1 nor Statement 2 is true
D. Only Statement 1 is true
Correct Answer: B. Both statements are trueSolution:Both statements are verified surveying principles. Included angles can be measured in either direction. When the survey progresses counter-clockwise and angles are measured clockwise, those angles are interior angles of the traverse polygon.
Q2. Which of the following methods is used to calculate the area between irregular boundaries?
Difficulty: Easy
A. Area by geometric method
B. Departure and total latitude method
C. Double parallel distance method
D. Simpson's rule
Correct Answer: D. Simpson's ruleSolution:Simpson's rule is specifically used for areas with irregular or curved boundaries. It assumes the boundary between consecutive offsets forms a parabolic arc and requires an odd number of offsets. Geometric and latitude/departure methods apply to straight-sided traverse polygons.
Q3. Volume of a tank excavated in level ground: depth = 4 m, top area = 50 m × 40 m, side slope = 2:1 (H:V). Find the volume using the prismoidal formula.
Difficulty: Hard
A. 8866 m³
B. 6688 m³
C. 5632 m³
D. 5461 m³
Correct Answer: D. 5461 m³Solution:A₁ = 2000 m², A₂ = 34×24 = 816 m², Aₑ = 42×32 = 1344 m². V = (4/6)[2000 + 4(1344) + 816] = (2/3)(8192) = 5461 m³. Side slope 2:1 reduces each side by 2d per metre of depth.
Q4. Which of the following methods is NOT in use for earthwork estimation?
Difficulty: Easy
A. Trapezoidal method
B. Prismoidal method
C. Mid-section method
D. Crossing method
Correct Answer: D. Crossing methodSolution:Trapezoidal, prismoidal, and mid-section methods are all standard earthwork volume methods. The Crossing method is an architectural building-estimation technique used for wall-length computation at corners — it has no application in earthwork volume calculation.
Q5. The volume of an embankment with cross-section areas A₁, A₂, A₃, A₄ at interval H using the trapezoidal method is:
Difficulty: Medium
A. H × [(A₁+A₄)/2 + A₂+A₃]
B. H × [(A₁+A₄)/4 + A₂+A₃]
C. L × [(A₁+A₄)/2 + A₂+A₃]
D. (L/3) × [(A₁+A₃)/2 × (A₂+A₃)]
Correct Answer: A. H × [(A₁+A₄)/2 + A₂+A₃]Solution:General trapezoidal rule: V = H × [(A₁+Aₙ)/2 + A₂+…+Aₙ⁻₁]. For 4 sections, first = A₁, last = A₄: V = H × [(A₁+A₄)/2 + A₂+A₃]. Use H (interval), not L (total length).
Q6. Calculate the volume of earthwork by the trapezoidal method. Three sections at 20 m intervals with areas: 40 m², 50 m², 80 m².
Difficulty: Medium
A. 1067 m³
B. 1700 m³
C. 2200 m³
D. 3200 m³
Correct Answer: C. 2200 m³Solution:V = d × [(A₁+A₃)/2 + A₂] = 20 × [(40+80)/2 + 50] = 20 × [60+50] = 20 × 110 = 2200 m³.
Q7. Contour areas for a dam: 410 m = 205 ha, 420 m = 120 ha, 430 m = 145 ha, 440 m = 95 ha, 450 m = 135 ha. Find dam capacity (m³) by the trapezoidal method.
Difficulty: Hard
A. 42,000,000 m³
B. 53,000,000 m³
C. 70,000,000 m³
D. 80,000,000 m³
Correct Answer: B. 53,000,000 m³Solution:d = 10 m. V = 10 × [(205+135)/2 + 120+145+95] = 10 × [170+360] = 5300 hectare-m. Convert: 5300 × 10,000 = 53,000,000 m³. Don't forget to multiply by 10,000 to convert hectare-metres to m³.
Q8. Which of the following methods estimates the best volume of earthwork of an irregular embankment?
Difficulty: Easy
A. Average ordinate method
B. Mid-ordinate method
C. Simpson's method
D. Trapezoidal method
Correct Answer: C. Simpson's methodSolution:Simpson's method (prismoidal formula) fits a parabolic curve through groups of three sections, capturing the curvature of irregular profiles more accurately than any linear method. It gives the highest accuracy for irregular embankments.
Q9. In the mid-section formula:
Difficulty: Easy
A. The mean depth is the average of two consecutive sections
B. The area of mid-sections is calculated by using mean depth
C. The volume is calculated by multiplying mid-section area by the distance between sections
D. All of the above
Correct Answer: D. All of the aboveSolution:All three statements correctly describe the mid-section method: (1) Dₑ = (D₁+D₂)/2, (2) Aₑ = B×Dₑ + S×Dₑ², (3) V = Aₑ × L. Each statement is an accurate description of one sequential step.
Q10. The assumption on which the trapezoidal formula for volumes is based, is:
Difficulty: Medium
A. The end sections are parallel planes
B. The mid-area of a pyramid is half the average area of the ends
C. Volume of the prismoidal is over-estimated; a prismoidal correction is applied
D. All options are correct
Correct Answer: D. All options are correctSolution:All three are valid characteristics of the trapezoidal volume formula: sections must be parallel planes; the linear area assumption means the mid-area of a pyramid is treated as half the average of the ends; and the trapezoidal result overestimates for tapered solids, requiring a negative prismoidal correction.
Q11. Volume by the Trapezoidal Formula Method is determined by the formula:
Difficulty: Medium
A. D × {(A₀+Aₙ)/2 + A₂+A₄+…}
B. D × {(A₁+Aₙ)/2 + A₀+A₁+A₃+…}
C. D × {(A₀+A₁)/2 + A₁+A₃+…}
D. D × {(A₀+Aₙ)/2 + A₁+A₂+A₃+…+Aₙ⁻₁}
Correct Answer: D. D × {(A₀+Aₙ)/2 + A₁+A₂+…+Aₙ⁻₁}Solution:The trapezoidal rule sums individual trapezoidal slabs: the first and last sections contribute half each; all intermediate sections contribute fully. V = D × [(A₀+Aₙ)/2 + A₁+A₂+…+Aₙ⁻₁].
Q12. If d is the constant distance between sections, the correct prismoidal formula for volume is:
Difficulty: Medium
A. d × [first + last + ΣEven + 2ΣOdd]
B. (d/3) × [first + last + 4ΣEven + 2ΣOdd]
C. (d/3) × [first + last + 2ΣEven + 4ΣOdd]
D. (d/6) × [first + last + 2ΣEven + 4ΣOdd]
Correct Answer: B. (d/3) × [first + last + 4ΣEven + 2ΣOdd]Solution:Prismoidal (Simpson's) rule: coefficient pattern is 1, 4, 2, 4, 2, …, 4, 1. End sections get 1; odd-indexed intermediates get 4; even-indexed get 2. Multiply by d/3. Option C swaps 4 and 2; D uses wrong multiplier d/6.
