Iowa Fundamentals of Surveying
Examination Advanced 100-Question
Practice Exam 2026 | Questions &
Answers with Detailed Rationales |
Complete Exam Prep & Study Guide
1. A total station measures a slope distance of 1,250.000 ft and a zenith angle
of 88°20′00″. What is the approximate horizontal distance?
A. 1,213.7 ft
B. 1,249.5 ft
C. 1,286.9 ft
D. 1,301.2 ft
Answer: 1,249.5 ft
Rationale: Horizontal distance is H=SsinZ. Thus, H=1,250sin(88°20′)≈1,249.5 ft.
2. A benchmark has an elevation of 842.36 ft. A backsight of 5.28 ft is
observed, followed by a foresight of 7.14 ft on a turning point. What is the
elevation of the turning point?
,A. 840.50 ft
B. 840.22 ft
C. 844.22 ft
D. 854.78 ft
Answer: 840.50 ft
Rationale: HI = 842.36 + 5.28 = 847.64 ft. The turning-point elevation is 847.64 −
7.14 = 840.50 ft.
3. In differential leveling, the primary purpose of balancing backsights and
foresights is to minimize the effects of:
A. Prism constant errors
B. Atmospheric refraction only
C. Instrument collimation error
D. Magnetic declination
Answer: Instrument collimation error
Rationale: Keeping backsight and foresight distances approximately equal
causes systematic collimation errors to largely cancel.
4. A closed leveling loop begins and ends on the same benchmark. The
measured elevations produce a closure error of −0.024 ft over 16
instrument setups. Using an equal-distribution adjustment, what correction
should be applied at the final station?
A. −0.0015 ft
B. +0.0015 ft
C. +0.024 ft
D. −0.024 ft
Answer: +0.024 ft
Rationale: The total closure correction must equal the negative of the
misclosure. Therefore, +0.024 ft is required at the endpoint, with corrections
distributed proportionally along the loop.
, 5. A line has a whole-circle azimuth of 237°18′40″. What is its quadrant
bearing?
A. S 57°18′40″ W
B. S 32°41′20″ W
C. N 57°18′40″ W
D. S 57°18′40″ E
Answer: S 57°18′40″ W
Rationale: An azimuth between 180° and 270° lies in the southwest quadrant.
Subtracting 180° gives 57°18′40″, measured south toward west.
6. A traverse has the following coordinate increments: ΔN = +420.35 ft and ΔE
= −315.20 ft. What is the approximate length and bearing of the course?
A. 525.3 ft, N 36°53′ W
B. 525.3 ft, N 53°07′ W
C. 735.6 ft, N 36°53′ E
D. 525.3 ft, S 36°53′ W
Answer: 525.3 ft, N 36°53′ W
Rationale: Distance = √(420.35² + 315.20²) ≈ 525.3 ft. The bearing angle is
atan(315.20/420.35) ≈ 36°53′, with north positive and east negative.
7. Which adjustment method distributes traverse closure errors according to
the lengths of individual courses?
A. Transit rule
B. Compass rule
C. Crandall method
D. Least-squares adjustment
Answer: Compass rule
Rationale: The Bowditch or compass rule distributes latitude and departure
corrections in proportion to course lengths.
, 8. A traverse has a total latitude error of +0.18 ft and total departure error of
−0.24 ft. The linear misclosure is approximately:
A. 0.06 ft
B. 0.30 ft
C. 0.42 ft
D. 0.72 ft
Answer: 0.30 ft
Rationale: Linear misclosure is √(0.18² + 0.24²) = 0.30 ft.
9. A 1,000-ft steel tape is calibrated at 68°F but used at 98°F. If the coefficient
of thermal expansion is 0.00000645/°F, the approximate temperature
correction is:
A. +0.194 ft
B. +0.065 ft
C. −0.194 ft
D. −0.065 ft
Answer: +0.194 ft
Rationale: Temperature correction = LαΔT = 1,000(0.00000645)(30) = 0.1935 ft.
The tape is longer when hot, so the measured distance requires a positive
correction.
10.A tape is 99.96 ft long when it is standardized as 100.00 ft. If a measured
distance is 1,250.00 ft, the corrected distance is:
A. 1,249.50 ft
B. 1,250.00 ft
C. 1,250.50 ft
D. 1,251.00 ft
Answer: 1,249.50 ft
Examination Advanced 100-Question
Practice Exam 2026 | Questions &
Answers with Detailed Rationales |
Complete Exam Prep & Study Guide
1. A total station measures a slope distance of 1,250.000 ft and a zenith angle
of 88°20′00″. What is the approximate horizontal distance?
A. 1,213.7 ft
B. 1,249.5 ft
C. 1,286.9 ft
D. 1,301.2 ft
Answer: 1,249.5 ft
Rationale: Horizontal distance is H=SsinZ. Thus, H=1,250sin(88°20′)≈1,249.5 ft.
2. A benchmark has an elevation of 842.36 ft. A backsight of 5.28 ft is
observed, followed by a foresight of 7.14 ft on a turning point. What is the
elevation of the turning point?
,A. 840.50 ft
B. 840.22 ft
C. 844.22 ft
D. 854.78 ft
Answer: 840.50 ft
Rationale: HI = 842.36 + 5.28 = 847.64 ft. The turning-point elevation is 847.64 −
7.14 = 840.50 ft.
3. In differential leveling, the primary purpose of balancing backsights and
foresights is to minimize the effects of:
A. Prism constant errors
B. Atmospheric refraction only
C. Instrument collimation error
D. Magnetic declination
Answer: Instrument collimation error
Rationale: Keeping backsight and foresight distances approximately equal
causes systematic collimation errors to largely cancel.
4. A closed leveling loop begins and ends on the same benchmark. The
measured elevations produce a closure error of −0.024 ft over 16
instrument setups. Using an equal-distribution adjustment, what correction
should be applied at the final station?
A. −0.0015 ft
B. +0.0015 ft
C. +0.024 ft
D. −0.024 ft
Answer: +0.024 ft
Rationale: The total closure correction must equal the negative of the
misclosure. Therefore, +0.024 ft is required at the endpoint, with corrections
distributed proportionally along the loop.
, 5. A line has a whole-circle azimuth of 237°18′40″. What is its quadrant
bearing?
A. S 57°18′40″ W
B. S 32°41′20″ W
C. N 57°18′40″ W
D. S 57°18′40″ E
Answer: S 57°18′40″ W
Rationale: An azimuth between 180° and 270° lies in the southwest quadrant.
Subtracting 180° gives 57°18′40″, measured south toward west.
6. A traverse has the following coordinate increments: ΔN = +420.35 ft and ΔE
= −315.20 ft. What is the approximate length and bearing of the course?
A. 525.3 ft, N 36°53′ W
B. 525.3 ft, N 53°07′ W
C. 735.6 ft, N 36°53′ E
D. 525.3 ft, S 36°53′ W
Answer: 525.3 ft, N 36°53′ W
Rationale: Distance = √(420.35² + 315.20²) ≈ 525.3 ft. The bearing angle is
atan(315.20/420.35) ≈ 36°53′, with north positive and east negative.
7. Which adjustment method distributes traverse closure errors according to
the lengths of individual courses?
A. Transit rule
B. Compass rule
C. Crandall method
D. Least-squares adjustment
Answer: Compass rule
Rationale: The Bowditch or compass rule distributes latitude and departure
corrections in proportion to course lengths.
, 8. A traverse has a total latitude error of +0.18 ft and total departure error of
−0.24 ft. The linear misclosure is approximately:
A. 0.06 ft
B. 0.30 ft
C. 0.42 ft
D. 0.72 ft
Answer: 0.30 ft
Rationale: Linear misclosure is √(0.18² + 0.24²) = 0.30 ft.
9. A 1,000-ft steel tape is calibrated at 68°F but used at 98°F. If the coefficient
of thermal expansion is 0.00000645/°F, the approximate temperature
correction is:
A. +0.194 ft
B. +0.065 ft
C. −0.194 ft
D. −0.065 ft
Answer: +0.194 ft
Rationale: Temperature correction = LαΔT = 1,000(0.00000645)(30) = 0.1935 ft.
The tape is longer when hot, so the measured distance requires a positive
correction.
10.A tape is 99.96 ft long when it is standardized as 100.00 ft. If a measured
distance is 1,250.00 ft, the corrected distance is:
A. 1,249.50 ft
B. 1,250.00 ft
C. 1,250.50 ft
D. 1,251.00 ft
Answer: 1,249.50 ft
