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Key Facts: GASM Deck OOW Exam
STCW II/1
IMO Standard
STCW Code Operational Level
7 Modules
Exam Subject Papers
GASM Tablo-3 Structure
60-70%
Module Passing Score
GASM Examination Regulations
5 Years
Certificate Validity
STCW Revalidation Standard
12 Months
Required Sea Service
Deck Cadet Training Requirement
CBT
Delivery Format
GASM Exam Centres
The Turkish GASM Uzakyol Vardiya Zabiti (Deck OOW Unlimited) examination is the statutory certification assessment for operational-level watchkeeping officers on unlimited-tonnage merchant vessels, conducted under STCW Regulation II/1 and Turkish national regulations (Tablo-3). Administered by GASM via modular computer-based tests, candidates must pass seven modules: Seyir, Vardiya Standartları, Gemi İnşa ve Yük, Denizde Haberleşme, Gemicilik, Deniz Hukuku, and Denizcilik İngilizcesi. This question bank provides 100 independent English-language practice questions with comprehensive rationales and step-by-step calculations.
Sample GASM Deck OOW Practice Questions
Try these sample questions to review concepts for the GASM Deck OOW exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1When planning an ocean passage across high latitudes, what is the principal navigational advantage of a Great Circle track compared to a Rhumb Line track, and what operational limitation must the Officer of the Watch (OOW) observe?
A.A Great Circle track represents the shortest distance between two points on a sphere, but the true course continuously changes at every meridian and may lead into dangerous high-latitude ice zones.
B.A Great Circle track maintains a constant compass heading throughout the voyage, but requires significantly higher engine power due to Coriolis force.
C.A Great Circle track guarantees that the vessel remains outside all seasonal tropical storm belts, but increases the overall voyage distance by 8 to 12 percent.
D.A Great Circle track allows direct Mercator chart plotting with straight lines, but requires a celestial sight at every change of watch.
Explanation: A great circle track is the shortest geodesic arc between two points on the Earth's surface. However, because it crosses meridians of longitude at oblique angles, the true heading must be adjusted continuously (or steered as a series of rhumb line waypoints), and the vertex of the track may take the vessel into hazardous polar weather or sea ice, requiring composite sailing with a limiting latitude.
2A vessel is steering a gyro course of 078° with a gyro error of 1° Low (1°E). The standard magnetic compass reads 082°, and the charted magnetic variation for the local position is 4° West. What is the magnetic compass deviation on this heading?
A.1° East
B.1° West
C.3° East
D.5° West
Explanation: True course = Gyro course + Gyro error = 078° + (+1°) = 079°(T). Magnetic course = True course - Variation (West is minus) = 079° - (-4°) = 079° + 4° = 083°(M) [or True = Mag + Var => 079° = Mag + (-4°) => Mag = 083°]. Deviation = Magnetic course - Compass course = 083° - 082° = +1°, which is 1° East (Compass Best, Error West; Compass Least, Error East: 082° is less than 083°, so Deviation is 1°E).
3A vessel steers 140°(T) at an engine speed of 12.0 knots. A steady tidal stream is setting 230°(T) at 2.5 knots. For a 2-hour run, what are the approximate course made good (CMG) and speed made good (SMG)?
A.150°(T) and 12.6 knots
B.130°(T) and 10.2 knots
C.165°(T) and 14.2 knots
D.125°(T) and 14.5 knots
Explanation: Using vector addition for 1 hour: Ship vector is 12.0 NM on 140° (dLat = -9.19, dLong dep = +7.71). Current vector is 2.5 NM on 230° (dLat = -1.61, dLong dep = -1.92). Resultant vector dLat = -10.80 NM, dep = +5.79 NM. Course made good = 180° - arctan(5.79 / 10.80) = 180° - 28.2° = 151.8° ~ 150° to 151°(T). Speed made good = sqrt(10.80² + 5.79²) = sqrt(116.64 + 33.52) = sqrt(150.16) ~ 12.25 to 12.6 knots.
4An OOW observes a prominent lighthouse bearing 045°(T) at 0800 hours. The vessel maintains a course of 090°(T) at a constant speed of 15.0 knots. At 0840 hours, the same lighthouse bears 315°(T). Neglecting current and leeway, what was the distance of the vessel from the lighthouse at the second observation?
A.10.0 nautical miles
B.7.1 nautical miles
C.14.1 nautical miles
D.5.0 nautical miles
Explanation: At 0800 hrs, the relative bearing is 045° on the port bow (090° - 045° = 45° port). At 0840 hrs, the relative bearing is 315° - 090° = 225° or 135° from the bow, making the interior angles of the triangle formed by the lighthouse, initial position, and second position 45°, 90°, and 45° (an isosceles right triangle). The run between 0800 and 0840 is 40 minutes at 15.0 knots = (40/60) * 15 = 10.0 NM. In a 45°-90°-45° triangle, the distance run equals the distance off the object when it is 4-points (45°) abaft the beam, so distance off at 0840 is 10.0 NM.
5A vessel must transit a distance of 336 nautical miles to arrive at the pilot boarding station by 0600 hours on October 24. The vessel departs at 1400 hours on October 23. An adverse head current of 1.2 knots is expected throughout the passage. What minimum engine speed must the OOW set to ensure arrival on time?
A.22.2 knots
B.19.8 knots
C.18.0 knots
D.21.0 knots
Explanation: Time available from 1400 hours on October 23 to 0600 hours on October 24 is 16 hours. Required Speed Over Ground (SOG) = Distance / Time = 336 NM / 16 hours = 21.0 knots. Because the current is directly opposed (adverse head current of 1.2 knots), the required engine speed through the water = Required SOG + Adverse Current = 21.0 + 1.2 = 22.2 knots.
6Under IMO ECDIS Performance Standards, what is the critical difference between the 'Safety Depth' and the 'Safety Contour' configured by the navigator?
A.Safety Depth highlights individual soundings that are equal to or less than the set depth value, whereas Safety Contour defines the boundary between navigable water and shallow water that triggers an automatic audible crossing alarm.
B.Safety Depth triggers an audible alarm upon vessel approach, while Safety Contour only changes color without triggering an audible warning.
C.Safety Depth is calculated using only the vessel draft, while Safety Contour requires adding five meters of under keel clearance regardless of ship size.
D.Safety Depth applies exclusively to paper chart folios, while Safety Contour is the only parameter recognized by electronic charts.
Explanation: Safety Depth applies to spot soundings: soundings equal to or shallower than this value are displayed in bold black to draw the mariner's attention. In contrast, the Safety Contour is a continuous depth contour selected from the ENC database (or the next deeper available contour if the exact value does not exist) that demarcates safe water from no-go water, automatically generating visual and audible alarms if the ship's look-ahead safety frame intersects it.
7Which IMO ECDIS display mode contains the absolute minimum chart information that cannot be removed from the display under any circumstance, but is not intended to be safe for navigation on its own?
A.Display Base
B.Standard Display
C.All Other Information (Full Display)
D.Custom Mariners Navigational Layer
Explanation: IMO Resolution MSC.232(82) defines 'Display Base' as the level of S-52 display information that cannot be removed from the screen under any setting. It includes coastline, own-ship safety contour, indications of isolated underwater dangers within safe water, and traffic separation boundaries. However, it is explicitly not intended for safe passage monitoring on its own; officers must navigate in 'Standard Display' or a customized display with supplementary layers.
8At 1000 hours, an isolated light tower bears 030° relative on the starboard bow. At 1030 hours, the same light tower bears 060° relative on the starboard bow. The vessel has steered a steady course of 000°(T) at 16.0 knots. What is the predicted distance at which the vessel will pass abeam of the light tower?
A.6.9 nautical miles
B.8.0 nautical miles
C.4.0 nautical miles
D.9.2 nautical miles
Explanation: This is a classic 'doubling the angle on the bow' problem (30° doubled to 60°). The run between observations is 30 minutes at 16.0 knots = 8.0 NM. In an isosceles triangle where the second angle is double the first, the distance from the second position to the object equals the distance run between bearings, which is 8.0 NM. The distance abeam is then: Distance abeam = Distance at 2nd bearing * sin(60°) = 8.0 * 0.866 = 6.93 NM.
9On a vessel's standard magnetic compass binnacle, what components are specifically designed to correct for the coefficient B and coefficient C of the ship's permanent horizontal magnetic field?
A.Fore-and-aft and athwartships permanent corrector magnets placed in horizontal trays within the binnacle
B.Soft iron spheres (Kelvin's balls) mounted on the port and starboard arms of the binnacle
C.The vertical soft iron Flinders bar mounted on the forward or after face of the binnacle
D.The vertical heeling error magnet suspended on a chain directly below the compass bowl
Explanation: Coefficient B (permanent fore-and-aft component) and Coefficient C (permanent athwartships component) are corrected using longitudinal (fore-and-aft) and transverse (athwartships) permanent magnets placed in sliding trays inside the compass binnacle. Soft iron spheres correct for symmetrical horizontal induced magnetism (Coefficient D), the Flinders bar corrects for asymmetrical vertical induced magnetism (induced B), and the bucket magnet corrects for heeling error.
10What causes the speed and latitude error (damping error) in a marine mechanical gyrocompass, and on which courses is this error at its maximum value?
A.The combination of the ship's northerly or southerly velocity component with the Earth's easterly rotational velocity alters the direction of the resultant angular velocity vector; the error is greatest on due North or due South courses.
B.The rolling and pitching of the vessel induces mechanical friction in the vertical ring gimbal bearings; the error is greatest on due East or due West courses.
C.Variations in local magnetic anomaly accelerate the gyro wheel rotors; the error is greatest when navigating near the magnetic poles.
D.Changes in ambient seawater temperature reduce rotor electrical resistance; the error is greatest on high-latitude westerly tracks.
Explanation: A gyrocompass aligns itself with the true geographic meridian by seeking the horizontal component of the Earth's rotation vector. When a vessel travels north or south, its own linear velocity adds vectorially to the Earth's peripheral speed (V_earth = 900 * cos(lat) knots), tilting the apparent axis of rotation slightly westward on northerly headings and eastward on southerly headings. This speed-latitude error is maximum on pure North or South courses (where north-south velocity is highest) and zero on pure East or West courses.
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