## PEC EPE Exam Surveying Numerical Questions (Civil / Transportation Engineering)

Click here for MCQs for Surveying for Civil and Transportation Engineering

### Other PEC EPE Exam Past Paper Questions

**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 N30ยฐE. If the magnetic declination is 3ยฐ30’W, what is the magnetic bearing of the line?**

Solution:

Magnetic bearing = True bearing – Declination correction

Magnetic bearing = N30ยฐE – 3ยฐ30’W

Convert both to the same direction (either all N or all S):

Magnetic bearing = N30ยฐE – (180ยฐ – 3ยฐ30′) = N30ยฐE – 176ยฐ30′

Answer: Magnetic bearing = N30ยฐE – 176ยฐ30′

**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 S40ยฐE. If the magnetic declination is 4ยฐ30’E, what is the magnetic bearing of the line?**

Solution:

Magnetic bearing = True bearing + Declination correction

Magnetic bearing = S40ยฐE + 4ยฐ30’E

Convert both to the same direction (either all N or all S):

Magnetic bearing = N40ยฐW + 4ยฐ30’W = N35ยฐ30’W

Answer: Magnetic bearing = N35ยฐ30’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) โ 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