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FE Civil Surveying & Geomatics Practice
Exam 2026–2027 | Full-Length
Questions, Answers & Detailed
Rationales
1. A total station measures a slope distance of 842.60 ft and a zenith angle of
86°20′. What is the horizontal distance?
A. 789.10 ft
B. 837.98 ft
C. 842.60 ft
D. 854.40 ft
Answer: 837.98 ft
Rationale: The horizontal distance is H=SsinZH=S\sin Z. Thus,
H=842.60sin(86°20′)≈837.98H=842.60\sin(86°20′)\approx837.98 ft.
2. A differential leveling survey begins with an elevation of 1,245.36 ft. The
backsights total 18.72 ft and the foresights total 21.15 ft. What is the
ending elevation?
,A. 1,242.93 ft
B. 1,247.79 ft
C. 1,249.23 ft
D. 1,251.87 ft
Answer: 1,242.93 ft
Rationale: The elevation change is ΣBS−ΣFS=18.72−21.15=−2.43\Sigma BS-
\Sigma FS=18.72-21.15=-2.43 ft. Therefore, the ending elevation is
1,245.36−2.43=1,242.931,245.36-2.43=1,242.93 ft.
3. A benchmark has an elevation of 563.42 ft. A backsight of 5.84 ft is
observed, followed by a foresight of 7.26 ft to a turning point. What is the
elevation of the turning point?
A. 561.00 ft
B. 562.00 ft
C. 564.84 ft
D. 576.52 ft
Answer: 562.00 ft
Rationale: The height of instrument is 563.42+5.84=569.26563.42+5.84=569.26
ft. The turning-point elevation is 569.26−7.26=562.00569.26-7.26=562.00 ft.
4. A closed traverse has an angular misclosure of 24 arcseconds. If the
traverse contains six angles and all observations have equal precision, what
correction should be applied to each angle?
A. +2″
B. −2″
C. +4″
D. −4″
,Answer: −4″
Rationale: The correction per angle is the negative of the misclosure divided by
the number of angles: −24″/6=−4″-24″/6=-4″.
5. A level rod reading is 6.214 ft when the rod is held on a benchmark. If the
instrument's horizontal line of sight is at elevation 812.46 ft, what is the
benchmark elevation?
A. 806.246 ft
B. 806.246?
C. 818.674 ft
D. 812.46 ft
Answer: 806.246 ft
Rationale: Elevation equals height of instrument minus rod reading:
812.460−6.214=806.246812.460-6.214=806.246 ft.
6. A 100-ft steel tape is standardized at 68°F but used at 95°F. If the
coefficient of thermal expansion is 6.45×10−6/°F6.45\times10^{-6}/°F, what
is the approximate temperature correction for one 100-ft tape length?
A. +0.0017 ft
B. +0.0174 ft
C. −0.0174 ft
D. +0.174 ft
Answer: +0.0174 ft
Rationale: Ct=αLΔT=(6.45×10−6)(100)(27)=0.0174C_t=\alpha L\Delta
T=(6.45\times10^{-6})(100)(27)=0.0174 ft. A warmer tape is longer than
standard, so the correction is positive.
, 7. A surveyor measures a line as 1,528.40 ft using a tape that is actually
100.06 ft long when its nominal length is 100 ft. What is the corrected
distance?
A. 1,527.48 ft
B. 1,528.40 ft
C. 1,529.32 ft
D. 1,534.51 ft
Answer: 1,529.32 ft
Rationale: Correct distance
=1,528.40(100.06/100)=1,529.32=1,528.40(100.06/100)=1,529.32 ft. Because the
tape is longer than nominal, the measured distance underestimates the true
distance.
8. A 200-ft line is measured with a tape under a pull greater than standard
pull. Assuming all other conditions are unchanged, the measured distance
will generally be:
A. Too long
B. Too short
C. Unaffected
D. Randomly biased
Answer: Too long
Rationale: Excessive tension elongates the tape. A longer tape covers more
ground per nominal tape length, causing the reported distance to be too large.
9. Which coordinate system is most appropriate for representing positions on
a projected plane using northing and easting coordinates?
FE Civil Surveying & Geomatics Practice
Exam 2026–2027 | Full-Length
Questions, Answers & Detailed
Rationales
1. A total station measures a slope distance of 842.60 ft and a zenith angle of
86°20′. What is the horizontal distance?
A. 789.10 ft
B. 837.98 ft
C. 842.60 ft
D. 854.40 ft
Answer: 837.98 ft
Rationale: The horizontal distance is H=SsinZH=S\sin Z. Thus,
H=842.60sin(86°20′)≈837.98H=842.60\sin(86°20′)\approx837.98 ft.
2. A differential leveling survey begins with an elevation of 1,245.36 ft. The
backsights total 18.72 ft and the foresights total 21.15 ft. What is the
ending elevation?
,A. 1,242.93 ft
B. 1,247.79 ft
C. 1,249.23 ft
D. 1,251.87 ft
Answer: 1,242.93 ft
Rationale: The elevation change is ΣBS−ΣFS=18.72−21.15=−2.43\Sigma BS-
\Sigma FS=18.72-21.15=-2.43 ft. Therefore, the ending elevation is
1,245.36−2.43=1,242.931,245.36-2.43=1,242.93 ft.
3. A benchmark has an elevation of 563.42 ft. A backsight of 5.84 ft is
observed, followed by a foresight of 7.26 ft to a turning point. What is the
elevation of the turning point?
A. 561.00 ft
B. 562.00 ft
C. 564.84 ft
D. 576.52 ft
Answer: 562.00 ft
Rationale: The height of instrument is 563.42+5.84=569.26563.42+5.84=569.26
ft. The turning-point elevation is 569.26−7.26=562.00569.26-7.26=562.00 ft.
4. A closed traverse has an angular misclosure of 24 arcseconds. If the
traverse contains six angles and all observations have equal precision, what
correction should be applied to each angle?
A. +2″
B. −2″
C. +4″
D. −4″
,Answer: −4″
Rationale: The correction per angle is the negative of the misclosure divided by
the number of angles: −24″/6=−4″-24″/6=-4″.
5. A level rod reading is 6.214 ft when the rod is held on a benchmark. If the
instrument's horizontal line of sight is at elevation 812.46 ft, what is the
benchmark elevation?
A. 806.246 ft
B. 806.246?
C. 818.674 ft
D. 812.46 ft
Answer: 806.246 ft
Rationale: Elevation equals height of instrument minus rod reading:
812.460−6.214=806.246812.460-6.214=806.246 ft.
6. A 100-ft steel tape is standardized at 68°F but used at 95°F. If the
coefficient of thermal expansion is 6.45×10−6/°F6.45\times10^{-6}/°F, what
is the approximate temperature correction for one 100-ft tape length?
A. +0.0017 ft
B. +0.0174 ft
C. −0.0174 ft
D. +0.174 ft
Answer: +0.0174 ft
Rationale: Ct=αLΔT=(6.45×10−6)(100)(27)=0.0174C_t=\alpha L\Delta
T=(6.45\times10^{-6})(100)(27)=0.0174 ft. A warmer tape is longer than
standard, so the correction is positive.
, 7. A surveyor measures a line as 1,528.40 ft using a tape that is actually
100.06 ft long when its nominal length is 100 ft. What is the corrected
distance?
A. 1,527.48 ft
B. 1,528.40 ft
C. 1,529.32 ft
D. 1,534.51 ft
Answer: 1,529.32 ft
Rationale: Correct distance
=1,528.40(100.06/100)=1,529.32=1,528.40(100.06/100)=1,529.32 ft. Because the
tape is longer than nominal, the measured distance underestimates the true
distance.
8. A 200-ft line is measured with a tape under a pull greater than standard
pull. Assuming all other conditions are unchanged, the measured distance
will generally be:
A. Too long
B. Too short
C. Unaffected
D. Randomly biased
Answer: Too long
Rationale: Excessive tension elongates the tape. A longer tape covers more
ground per nominal tape length, causing the reported distance to be too large.
9. Which coordinate system is most appropriate for representing positions on
a projected plane using northing and easting coordinates?
