ASA 105 Coastal Navigation Exam V1-V3 Bundle
(Latest 2026/2027) – Practice Tests, Chart Work
& Navigation Problems | American Sailing
Association
Section 1: Chartwork & Piloting (25 questions)
Q1. (Chart plot) You measure 3.8 nautical miles between two points on chart 18665-P.
The scale bar shows 1 nm = 1.85 cm. What distance in centimetres should you lay off
on your dividers?
A) 6.83 cm
B) 7.03 cm
C) 7.23 cm
D) 7.43 cm
Correct: B
Rationale: 3.8 nm × 1.85 cm/nm = 7.03 cm. Scale conversion is direct multiplication;
mis-scaling (A, C, D) would yield plotting errors.
Q2. At 0930 you take a compass bearing of 047° on Daybeacon “Fl G 6s” and 142° on
the radar dome. Variation is 14° E, deviation +2°. What is the magnetic bearing to the
daybeacon?
,A) 033° M
B) 035° M
C) 045° M
D) 049° M
Correct: B
Rationale: True bearing = 047°; variation 14° E → subtract; deviation +2° → add.
047° – 14° + 2° = 035° M. Sequence: T – V + D = M.
Q3. You steer 208° T at 5.2 kn. From a fixed position you drift for 18 min. Your DR
position is now how far from the fix?
A) 1.3 nm
B) 1.6 nm
C) 1.9 nm
D) 2.1 nm
Correct: B
Rationale: Distance = speed × time = 5.2 kn × 0.3 h = 1.56 nm ≈ 1.6 nm. Time must be in
hours (18 min ÷ 60).
Q4. (Multiple select) A two-bearing fix using objects 60° apart gives a good intersection.
Which three additional factors improve accuracy? (Select 3.)
A) Objects close to the vessel
, B) Bearings taken quickly and simultaneously
C) Objects at similar distances from the vessel
D) Large-scale chart used
E) Objects on opposite sides of the channel
Correct: B, C, D
Rationale: Simultaneity (B) prevents positional change; similar distance (C) minimizes
time-error magnification; large scale (D) allows precise plotting. Close objects (A)
increase angular error; opposite sides (E) is helpful but not as critical as angle and
simultaneity.
Q5. You plot a fix at 1045. You then steer 240° T at 6.0 kn. What is your DR position at
1115?
A) 1.5 nm on 240° T from the 1045 fix
B) 3.0 nm on 060° T from the 1045 fix
C) 1.5 nm on 060° T from the 1045 fix
D) 3.0 nm on 240° T from the 1045 fix
Correct: D
Rationale: Time elapsed = 30 min = 0.5 h; distance = 6.0 kn × 0.5 h = 3.0 nm along the
steered course (240° T) from the fix. DR is always along course line, not reciprocal (B,
C).
(Latest 2026/2027) – Practice Tests, Chart Work
& Navigation Problems | American Sailing
Association
Section 1: Chartwork & Piloting (25 questions)
Q1. (Chart plot) You measure 3.8 nautical miles between two points on chart 18665-P.
The scale bar shows 1 nm = 1.85 cm. What distance in centimetres should you lay off
on your dividers?
A) 6.83 cm
B) 7.03 cm
C) 7.23 cm
D) 7.43 cm
Correct: B
Rationale: 3.8 nm × 1.85 cm/nm = 7.03 cm. Scale conversion is direct multiplication;
mis-scaling (A, C, D) would yield plotting errors.
Q2. At 0930 you take a compass bearing of 047° on Daybeacon “Fl G 6s” and 142° on
the radar dome. Variation is 14° E, deviation +2°. What is the magnetic bearing to the
daybeacon?
,A) 033° M
B) 035° M
C) 045° M
D) 049° M
Correct: B
Rationale: True bearing = 047°; variation 14° E → subtract; deviation +2° → add.
047° – 14° + 2° = 035° M. Sequence: T – V + D = M.
Q3. You steer 208° T at 5.2 kn. From a fixed position you drift for 18 min. Your DR
position is now how far from the fix?
A) 1.3 nm
B) 1.6 nm
C) 1.9 nm
D) 2.1 nm
Correct: B
Rationale: Distance = speed × time = 5.2 kn × 0.3 h = 1.56 nm ≈ 1.6 nm. Time must be in
hours (18 min ÷ 60).
Q4. (Multiple select) A two-bearing fix using objects 60° apart gives a good intersection.
Which three additional factors improve accuracy? (Select 3.)
A) Objects close to the vessel
, B) Bearings taken quickly and simultaneously
C) Objects at similar distances from the vessel
D) Large-scale chart used
E) Objects on opposite sides of the channel
Correct: B, C, D
Rationale: Simultaneity (B) prevents positional change; similar distance (C) minimizes
time-error magnification; large scale (D) allows precise plotting. Close objects (A)
increase angular error; opposite sides (E) is helpful but not as critical as angle and
simultaneity.
Q5. You plot a fix at 1045. You then steer 240° T at 6.0 kn. What is your DR position at
1115?
A) 1.5 nm on 240° T from the 1045 fix
B) 3.0 nm on 060° T from the 1045 fix
C) 1.5 nm on 060° T from the 1045 fix
D) 3.0 nm on 240° T from the 1045 fix
Correct: D
Rationale: Time elapsed = 30 min = 0.5 h; distance = 6.0 kn × 0.5 h = 3.0 nm along the
steered course (240° T) from the fix. DR is always along course line, not reciprocal (B,
C).
