PEC EPE Exam Surveying Numerical Questions (Civil / Transportation Engineering)
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The length of a line measured on the ground is 100 meters. If it appears as 2 cm on a map, what is the representative fraction (RF) of the map?
Solution:
RF = Length on map / Actual length on the ground
RF = 2 cm / 100 meters = 1/5000
Answer: RF = 1/5000
An angle of 35 degrees is measured using a theodolite. If the declination correction is 2 degrees East, what is the corrected angle?
Solution:
Corrected angle = Measured angle + Declination correction
Corrected angle = 35 degrees + 2 degrees = 37 degrees
Answer: Corrected angle = 37 degrees
In a differential leveling survey, the staff readings on a benchmark are 1.2 m and 2.8 m. The staff readings on a point P are 1.6 m and 2.2 m. What is the elevation difference between the benchmark and point P?
Solution:
Elevation difference = (Benchmark staff reading – Point P staff reading)
Elevation difference = (2.8 m – 2.2 m) = 0.6 meters
Answer: Elevation difference = 0.6 meters
A theodolite with a vertical circle reading of 190 degrees is used to measure an angle. If the index error is -1.5 minutes, what is the corrected angle?
Solution:
Corrected angle = Vertical circle reading + Index error
Corrected angle = 190 degrees + (-1.5 minutes) = 189 degrees 58.5 minutes
Answer: Corrected angle = 189 degrees 58.5 minutes
The true bearing of a line is N30°E. If the magnetic declination is 3°30’W, what is the magnetic bearing of the line?
Solution:
Magnetic bearing = True bearing – Declination correction
Magnetic bearing = N30°E – 3°30’W
Convert both to the same direction (either all N or all S):
Magnetic bearing = N30°E – (180° – 3°30′) = N30°E – 176°30′
Answer: Magnetic bearing = N30°E – 176°30′
A topographic map with a scale of 1:10,000 is used to measure a distance of 5 cm between two points. What is the actual distance between these points on the ground?
Solution:
Actual distance = Length on map * RF
Actual distance = 5 cm * 10,000 = 50,000 cm = 500 meters
Answer: Actual distance = 500 meters
In a reciprocal leveling survey, the staff readings on a benchmark are 2.5 m and 1.3 m. The staff readings on a point P are 1.8 m and 2.0 m. What is the elevation difference between the benchmark and point P?
Solution:
Elevation difference = (Benchmark staff reading – Point P staff reading)
Elevation difference = (2.5 m – 2.0 m) = 0.5 meters
Answer: Elevation difference = 0.5 meters
A total station is used to measure the horizontal distance between two points on a level ground. The instrument height is 1.5 meters, and the staff reading on the first point is 2.7 meters, and on the second point is 1.8 meters. What is the horizontal distance between the two points?
Solution:
Horizontal distance = Staff reading on the first point – Staff reading on the second point + Instrument height
Horizontal distance = (2.7 m – 1.8 m) + 1.5 meters = 2.4 meters
Answer: Horizontal distance = 2.4 meters
The curvature correction for a 500-meter long survey line is calculated to be 0.08 meters. What is the corrected length of the survey line?
Solution:
Corrected length = Actual length + Curvature correction
Corrected length = 500 meters + 0.08 meters = 500.08 meters
Answer: Corrected length = 500.08 meters
In a leveling survey, the staff readings on a benchmark are 3.5 m and 1.8 m. The staff reading on a point P is 2.6 m. If the height of the instrument is 1.2 meters, what is the elevation of point P with respect to the benchmark?
Solution:
Elevation of point P = Benchmark staff reading – Point P staff reading + Height of instrument
Elevation of point P = (3.5 m – 2.6 m) + 1.2 meters = 1.1 meters
Answer: Elevation of point P = 1.1 meters
In a differential leveling survey, the staff readings on a benchmark are 2.5 m and 3.8 m. The staff reading on a point P is 1.2 m. If the height of the instrument is 1.5 meters, what is the elevation of point P with respect to the benchmark?
Solution:
Elevation of point P = Benchmark staff reading – Point P staff reading + Height of instrument
Elevation of point P = (3.8 m – 1.2 m) + 1.5 meters = 3.1 meters
Answer: Elevation of point P = 3.1 meters
The length of a line measured on the ground is 150 meters. If it appears as 6 cm on a map, what is the representative fraction (RF) of the map?
Solution:
RF = Length on map / Actual length on the ground
RF = 6 cm / 150 meters = 1/2500
Answer: RF = 1/2500
A theodolite with a vertical circle reading of 220 degrees is used to measure an angle. If the index error is +1.2 minutes, what is the corrected angle?
Solution:
Corrected angle = Vertical circle reading – Index error
Corrected angle = 220 degrees – 1.2 minutes = 219 degrees 58.8 minutes
Answer: Corrected angle = 219 degrees 58.8 minutes
The true bearing of a line is S40°E. If the magnetic declination is 4°30’E, what is the magnetic bearing of the line?
Solution:
Magnetic bearing = True bearing + Declination correction
Magnetic bearing = S40°E + 4°30’E
Convert both to the same direction (either all N or all S):
Magnetic bearing = N40°W + 4°30’W = N35°30’W
Answer: Magnetic bearing = N35°30’W
A topographic map with a scale of 1:25,000 is used to measure a distance of 8 cm between two points. What is the actual distance between these points on the ground?
Solution:
Actual distance = Length on map * RF
Actual distance = 8 cm * 25,000 = 200,000 cm = 2,000 meters
Answer: Actual distance = 2,000 meters
In a leveling survey, the staff readings on a benchmark are 4.5 m and 3.2 m. The staff readings on a point P are 2.8 m and 1.9 m. What is the elevation difference between the benchmark and point P?
Solution:
Elevation difference = (Benchmark staff reading – Point P staff reading)
Elevation difference = (4.5 m – 1.9 m) = 2.6 meters
Answer: Elevation difference = 2.6 meters
The curvature correction for a 600-meter long survey line is calculated to be 0.12 meters. What is the corrected length of the survey line?
Solution:
Corrected length = Actual length + Curvature correction
Corrected length = 600 meters + 0.12 meters = 600.12 meters
Answer: Corrected length = 600.12 meters
A total station is used to measure the horizontal distance between two points on a level ground. The instrument height is 1.2 meters, and the staff reading on the first point is 3.6 meters, and on the second point is 2.8 meters. What is the horizontal distance between the two points?
Solution:
Horizontal distance = Staff reading on the first point – Staff reading on the second point + Instrument height
Horizontal distance = (3.6 m – 2.8 m) + 1.2 meters = 2 meters
Answer: Horizontal distance = 2 meters
The vertical angle measured between the line of sight and the horizontal plane is 6 degrees. If the staff reading on the ground is 1.5 meters, what is the distance between the instrument and the staff?
Solution:
Distance = Staff reading / Tan(Vertical angle)
Distance = 1.5 meters / Tan(6 degrees) ≈ 15.36 meters
Answer: Distance = 15.36 meters
In a reciprocal leveling survey, the staff readings on a benchmark are 2.0 m and 1.1 m. The staff readings on a point P are 1.8 m and 1.6 m. What is the elevation difference between the benchmark and point P?
Solution:
Elevation difference = (Benchmark staff reading – Point P staff reading)
Elevation difference = (2.0 m – 1.6 m) = 0.4 meters
Answer: Elevation difference = 0.4 meters