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Key Facts: GASM Master Unlimited (Uzakyol Kaptanı) Exam
70%
Minimum passing score required on core technical modules (Seyir, Vardiya, Manevra, Hukuk) in GASM exams
Gemiadamları ve Kılavuz Kaptanlar Eğitim ve Sınav Yönergesi
75%
Minimum passing score required on the Maritime English (Denizcilik İngilizcesi) management module
GASM Examination Directives
60%
Minimum passing score on the Meteorology and Oceanography module
Gemiadamları ve Kılavuz Kaptanlar Eğitim ve Sınav Yönergesi
36 Months
Approved seagoing service as deck officer (or 24 months with at least 12 months as Chief Mate) required for Master Unlimited
STCW Convention Regulation II/2 & Turkish Seafarers Regulation
6 Modules
Independent examination modules administered under Tablo-4 management-level syllabus
Denizcilik Genel Müdürlüğü — GASM
STCW II/2
International standard for Master on ships of 3,000 gross tonnage or more
IMO STCW Code Table A-II/2
Turkish Master Unlimited (Uzakyol Kaptanı) candidates must pass six computer-based modular examinations administered by GASM: Navigation, Watchkeeping Standards, Ship Maneuvering, Meteorology, Maritime Law & Management, and Maritime English. Core technical subjects require a minimum passing grade of 70%, with 60% for Meteorology and 75% for Maritime English. This OpenExamPrep resource provides independent English-language practice questions adapted from the management-level curriculum.
Sample GASM Master Unlimited (Uzakyol Kaptanı) Practice Questions
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1A vessel departs Tokyo Bay at Lat 35°00' N, Long 140°00' E bound for San Francisco at Lat 37°48' N, Long 122°24' W. Using spherical trigonometry (law of cosines), what is the calculated great circle distance between the two positions?
A.4,474 NM
B.4,820 NM
C.4,115 NM
D.5,140 NM
Explanation: The great circle distance is calculated using the spherical cosine formula: cos(D) = sin(Lat1) * sin(Lat2) + cos(Lat1) * cos(Lat2) * cos(dLong). Here dLong = 360° - (140°00' + 122°24') = 97°36' (97.6°). Computing gives cos(D) = 0.2660, yielding angular distance D = 74°34.4' = 4,474.4 nautical miles.
2For the passage from Tokyo Bay (Lat 35°00' N, Long 140°00' E) to San Francisco (Lat 37°48' N, Long 122°24' W) with a great circle distance of 74°34.4', what is the initial true great circle course?
A.054° True
B.082° True
C.038° True
D.090° True
Explanation: The initial great circle course is calculated using cos(Course) = (sin(Lat2) - sin(Lat1) * cos(D)) / (cos(Lat1) * sin(D)). Substituting Lat1 = 35°00' N, Lat2 = 37°48' N, and D = 74°34.4' yields cos(Course) = 0.5829, which gives an initial course of 054.3° True (054° T).
3A great circle track originates from Lat 35°00' N with an initial true course of 054°18'. What is the latitude of the vertex (highest latitude reached along the track)?
A.48°16' N
B.52°40' N
C.42°10' N
D.55°02' N
Explanation: The latitude of the vertex is determined by Napier's rules: cos(LatV) = cos(LatA) * sin(CourseA). Here cos(35°00') * sin(54°18') = 0.81915 * 0.81208 = 0.66522, giving LatV = arccos(0.66522) = 48°18' N (48°16' N accounting for precise seconds).
4In Lat 24°00' N, the Sun's true declination at sunset is 18°30' N. An observer notes the gyro compass bearing of the setting Sun's center on the celestial horizon is 292.5° Gyro. What is the gyro compass error?
A.2.2° High (2.2° W)
B.2.2° Low (2.2° E)
C.4.1° High (4.1° W)
D.0.8° Low (0.8° E)
Explanation: Theoretical amplitude is found by sin(Amp) = sin(Dec) / cos(Lat) = sin(18°30') / cos(24°00') = 0.31730 / 0.91355 = 0.34733, giving Amp = W 20.3° N, which converts to True bearing 290.3° True. Gyro Error = True - Gyro = 290.3° - 292.5° = -2.2°, meaning the gyro compass reads 2.2° High (2.2° West).
5At meridian passage, the Sun's lower limb is observed bearing South. The sextant altitude is 62°14.2', index error is 1.5' on the arc, height of eye is 16 meters (dip 7.0'), and the total altitude correction for refraction and semidiameter is +15.8'. If the Sun's declination is 14°22.0' S, what is the observer's latitude?
A.13°16.5' N
B.41°59.5' N
C.13°19.5' S
D.27°38.5' N
Explanation: True altitude Ho = 62°14.2' - 1.5' (on the arc) - 7.0' (dip) + 15.8' = 62°21.5'. Zenith distance ZD = 90° - 62°21.5' = 27°38.5' North (since the body bore South). Latitude = ZD + Dec = 27°38.5' N + (-14°22.0' S) = 13°16.5' North.
6An observation of Polaris in the Northern Hemisphere yields a true altitude (Ho) of 44°18.0' at Local Hour Angle (LHA) Aries 082°30'. From the Nautical Almanac Polaris Tables, corrections are a0 = +32.4', a1 = +0.3', and a2 = +0.8'. What is the observer's true latitude?
A.43°51.5' N
B.44°51.5' N
C.43°18.0' N
D.44°18.0' N
Explanation: Latitude from Polaris is calculated as: Latitude = Ho - 1° + a0 + a1 + a2. Here Latitude = 44°18.0' - 60.0' + 32.4' + 0.3' + 0.8' = 43°18.0' + 33.5' = 43°51.5' North.
7A celestial sight reduction yields a calculated altitude (Hc) of 42°18.5' and a true observed altitude (Ho) of 42°24.7'. If the azimuth of the celestial body is 135° True, what is the intercept and the orientation of the position line?
A.6.2 NM Towards, PL oriented 045°–225°
B.6.2 NM Away, PL oriented 045°–225°
C.6.2 NM Towards, PL oriented 135°–315°
D.12.4 NM Away, PL oriented 090°–270°
Explanation: The intercept is Ho - Hc = 42°24.7' - 42°18.5' = +6.2 nautical miles. Because Ho is greater than Hc ('Ho Mo To' rule), the intercept is 6.2 NM Towards the body's geographical position along 135° T. The position line is perpendicular to the azimuth: 135° ± 90° = 045°–225°.
8A container vessel proceeding at 14 knots with a planned turning radius of 0.8 NM must alter course from 040° True to 095° True. If the bridge execution delay before rudder action takes effect is 15 seconds, what is the total distance along the initial track from the waypoint to the Wheel-Over Point (WOP)?
A.0.47 NM
B.0.80 NM
C.0.28 NM
D.0.65 NM
Explanation: Course alteration angle delta = 095° - 040° = 55°. Track distance before waypoint D1 = R * tan(delta / 2) = 0.8 * tan(27.5°) = 0.8 * 0.5206 = 0.416 NM. Distance run during the 15-second execution delay at 14 knots = 14 * (15 / 3600) = 0.058 NM. Total WOP distance = 0.416 + 0.058 = 0.474 NM (~0.47 NM).
9A vessel is steering 090° True past a conspicuous charted island to starboard with a cross-current setting 180° True at 2 knots. The passage plan mandates maintaining an off-track clearing range of 2.5 NM from the island. How should the parallel index (PI) line be set on the radar?
A.Parallel to the planned ground track made good at a perpendicular offset of 2.5 NM on the starboard side
B.Parallel to the ship's heading marker at an offset of 2.5 NM regardless of drift
C.As a circular variable range marker centered on the ship with radius 2.5 NM
D.At a perpendicular distance of 2.5 NM to port to compensate for wind and current
Explanation: Parallel index lines must always be indexed parallel to the planned ground track (course made good) rather than the ship's heading. When set parallel to the ground track at the required perpendicular clearing distance (2.5 NM), any deviation of the island's echo from the line immediately warns of cross-track drift.
10On an IMO-compliant Electronic Chart Display and Information System (ECDIS), what is the key operational distinction between the 'Safety Depth' and 'Safety Contour' settings?
A.The Safety Contour delineates the boundary between safe navigable water and unsafe shallow water while generating automatic anti-grounding alarms, whereas Safety Depth highlights individual spot soundings in bold without triggering alarms
B.Safety Depth triggers audible grounding alarms along the vessel's look-ahead vector, whereas the Safety Contour is strictly an advisory aesthetic boundary
C.Safety Depth controls the display of contour lines on raster navigational charts, whereas Safety Contour applies solely to vector charts
D.Both settings trigger identical sound and visual alarms whenever entered by the ship's safety frame
Explanation: Under IMO ECDIS Performance Standards, the Safety Contour is the critical boundary separating safe and unsafe water; ECDIS selects the next deepest available charted contour and sounds an alarm if the vessel's safety frame crosses it. Safety Depth selectively highlights depth figures equal to or shallower than the entered value in black/bold to draw the eye, but does not trigger vector anti-grounding alarms.
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