Correct: D)
Explanation: The total distance is the sum of the individual legs: Grenchen to Kagiswil, Kagiswil to Buttwil, and Buttwil to Langenthal (since the pilot diverted instead of returning to Grenchen). Measuring these legs on the 1:500,000 ICAO chart using the given radial/distance references from Bern-Belp and Zurich-Kloten yields a total of approximately 178 km. Option A (257 km) is too long and likely adds an extra leg. Option B (154 km) and option C (145 km) are too short, probably omitting one leg of the route.
Correct: A)
Explanation: The prefix "D" in LS-D7 designates a Danger zone under the Swiss airspace classification system. The upper limit of this zone is 9000 ft AMSL (above mean sea level). Option B incorrectly calls it a prohibited zone (that would be LS-P). Options C and D refer to a "lower limit" of 9000 ft, which would mean the zone starts at 9000 ft rather than ending there — and both also either misclassify the zone type or use the wrong altitude reference (AGL vs. AMSL).
Correct: D)
Explanation: To find the map scale, convert both measurements to the same unit: 10 km = 10,000 m = 1,000,000 cm. The ratio is 4 cm on the map to 1,000,000 cm in reality, so 1 cm represents 250,000 cm, giving a scale of 1:250,000. Option A (1:25,000) would mean 4 cm = 1 km. Option B (1:100,000) would mean 4 cm = 4 km. Option C (1:400,000) would mean 4 cm = 16 km. Only 1:250,000 yields the correct 4 cm = 10 km relationship.
Correct: D)
Explanation: The Locarno CTR (Control Zone) extends from the surface up to 3,950 ft AMSL (above mean sea level), as published on the Swiss aeronautical charts. Option A confuses feet with metres — 3,950 m would be approximately 12,960 ft, far too high for a CTR. Option B uses AGL (above ground level), which is not how this CTR's upper limit is defined. Option C (FL 125) refers to a flight level reference that is unrelated to this particular CTR boundary.
Correct: C)
Explanation: At Fraubrunnen (north of Bern-Belp) at 4500 ft AMSL, the aircraft is below the BERN 2 TMA, which begins at 5500 ft AMSL in this area, and above the Bern CTR, which only extends to a lower altitude. This places the aircraft in Class E airspace. Option A is wrong because the TMA floor is above the aircraft. Option D is incorrect because the Bern CTR does not extend this far north or this high. Option B (Class G) applies to uncontrolled airspace below the Class E floor, which the aircraft is above.
Correct: C)
Explanation: Modern aviation GPS units allow the pilot to change distance display units (NM to km or vice versa) through the device's SETTING MODE menu. This is a simple user preference and requires no technical workshop intervention. Option A is incorrect because unit changes are user-accessible. Option B incorrectly suggests certification locks prevent the change. Option D confuses the aviation database (which contains waypoints and airspace data) with the display settings menu.
Correct: B)
Explanation: On 5 June, Switzerland observes Central European Summer Time (CEST), which is UTC+2. Departure is at 0945 UTC, and the flight lasts 45 minutes, so landing occurs at 0945 + 0045 = 1030 UTC. Converting to local time: 1030 UTC + 2 hours = 1230 CEST. However, the correct answer given is B (1130 LT), which would correspond to UTC+1 conversion. This suggests the question intends standard CET (UTC+1) or uses a different convention. Options A and C yield times before departure, which are impossible, and option D overshoots.
Correct: C)
Explanation: The conversion factor is 1 NM = 1.852 km. Therefore 54 NM x 1.852 km/NM = 100.008 km, which rounds to 100.00 km. Option A (27 km) appears to divide by 2 instead of multiplying by 1.852. Option B (29.16 km) uses an incorrect conversion factor. Option D (92.60 km) is close to the correct value but uses an inaccurate conversion ratio. Knowing the NM-to-km conversion factor of 1.852 is essential for cross-country flight planning.
Correct: B)
Explanation: GPS is highly accurate for position determination, but satellite signals can be disrupted by terrain shading, atmospheric conditions, or intentional interference. Pilots must always cross-check GPS position against visual ground references. Option A is wrong because GPS is susceptible to interference and signal loss. Option C overstates GPS capability — it does not replace basic pilotage skills, and airspace warnings depend on database currency. Option D is incorrect because GPS does not automatically update its aviation database; this requires manual updates by the user.
Correct: C)
Explanation: An isogonic line connects all points on a chart that have the same magnetic declination (variation). These lines are printed on aeronautical charts to help pilots convert between true and magnetic bearings. Option A describes an isotherm (equal temperature). Option B describes the agonic line, which is the special case where declination equals zero — a subset, not the general definition. Option D describes an isobar (equal pressure).
Correct: C)
Explanation: Plotting both positions relative to Zurich-Kloten on the chart, the Saentis lies to the southeast (110°/65 km) and Amlikon to the east-northeast (075°/40 km). The route from Saentis to Amlikon heads northwest, yielding a true course of approximately 328°. Option D (318°) is close but inaccurate based on the chart plot. Options A (147°) and B (227°) point in roughly the opposite direction — southeast and southwest respectively — which would take the pilot away from the destination.
Correct: C)
Explanation: VDF (VHF Direction Finding) works by having a ground station take a bearing on the pilot's radio transmission. The only equipment the aircraft needs is a standard VHF radio communication system — the pilot transmits, and the ground station determines the direction. Option A (ELT) is for emergency location, not routine position finding. Option B (transponder) is for radar identification, not VDF. Option D (GPS) determines position independently and is not related to VDF bearings.
Correct: C)
Explanation: In a Mercator (normal cylindrical) projection, both meridians and parallels appear as straight lines that intersect at right angles, forming a rectangular grid. Meridians are evenly spaced vertical lines and parallels are horizontal lines (though their spacing increases toward the poles). Option A describes a conic projection where meridians converge. Option B incorrectly calls them curves. Option D reverses the convergence — in a Mercator projection, neither meridians nor parallels converge.
Correct: D)
Explanation: Above Burgdorf, the lower boundary of the Bern TMA is at 1700 m AMSL. Below this altitude, a glider may fly freely without notification or authorization in Class E or G airspace. Option A (3050 m AMSL) represents a higher TMA boundary that applies in a different area. Option B (5500 ft AGL) uses an AGL reference which is incorrect for this airspace boundary. Option C (1700 m AGL) confuses the reference — the limit is AMSL, not above ground level.
Correct: C)
Explanation: The coordinates 46°29'N / 007°15'E correspond to Saanen aerodrome, which serves the Gstaad area in the Bernese Oberland. Option B (Sion airport) is located further south and slightly east, at approximately 46°13'N / 007°20'E. Option A (Sanetsch Pass) is a mountain pass between Sion and the Bernese Oberland at a different position. Option D (Gstaad/Grund heliport) is nearby but has different precise coordinates.
Correct: D)
Explanation: Geographic longitude is the angular distance measured east or west from the Prime Meridian (0° at Greenwich) to the local meridian passing through the given location, expressed in degrees (0° to 180°E or W). Options A and B incorrectly reference the equator — distance from the equator is latitude, not longitude. Option C describes a co-latitude measurement from the north pole, which is also a form of latitude. Only option D correctly identifies longitude as the angular measure from the Greenwich meridian.
Correct: D)
Explanation: Magnetic Course (MC) is defined as the angle measured clockwise from magnetic north to the intended course line over the ground. It is the course referenced to the Earth's magnetic field rather than to true (geographic) north. Option A describes the direction of true north. Option B describes the direction to the magnetic north pole, not a course angle. Option C defines True Course (TC), which is referenced to geographic north rather than magnetic north.
Correct: C)
Explanation: True altitude accounts for non-standard temperature effects on pressure altitude. ISA temperature at approximately 6500 ft is about +2°C (15° - 2°/1000 ft x 6.5). With OAT of -9°C, the air is approximately 11°C colder than ISA. Cold air is denser, meaning pressure levels are compressed closer to the ground, so the aircraft is actually lower than the altimeter indicates. Using the correction of roughly 4 ft per 1°C per 1000 ft: 11°C x 4 x 6.5 = approximately 286 ft below QNH altitude, yielding about 6250 ft true altitude. Options A, B, and D all overestimate the true altitude.
Correct: A)
Explanation: At QNH altitude 6500 ft, ISA temperature is approximately +2°C. The OAT of +11°C is about 9-10°C warmer than ISA. In warmer-than-standard air, the atmosphere is expanded, so the aircraft sits higher than the altimeter indicates. Applying the temperature correction (approximately +10°C x 4 ft/°C/1000 ft x 6.5 = +260 ft) to the QNH altitude gives approximately 6500 + 250 = 6750 ft true altitude. Option B ignores the temperature correction entirely. Options C and D either overcorrect or correct in the wrong direction.
Correct: A)
Explanation: At QNH altitude 6500 ft, ISA temperature is approximately +2°C. The OAT of +21°C means the air is about 19-20°C warmer than standard. Warm air expands, placing the aircraft significantly higher than indicated. The correction is approximately +20°C x 4 ft/°C/1000 ft x 6.5 = +520 ft, yielding about 6500 + 500 = 7000 ft true altitude. This large warm correction brings the true altitude up to match the pressure altitude. Options B, C, and D underestimate the warm-air correction effect.
Correct: D)
Explanation: With TC 255° and wind from 200°, the wind comes from approximately 55° to the left of the course line. This crosswind pushes the aircraft to the right of track. To compensate, the pilot must crab into the wind (turn left), reducing the heading below the course value. The wind correction angle is approximately sin^-1(10 x sin55° / 100) = sin^-1(0.082) = about 5°. True heading = 255° - 5° = 250°. Option A (275°) and B (265°) incorrectly add to the heading. Option C (245°) overcorrects by 10°.
Correct: D)
Explanation: The wind from 130° on a 165° course comes from approximately 35° to the left of the nose, pushing the aircraft right of track. The pilot must crab left to compensate. WCA = sin^-1(20 x sin35° / 90) = sin^-1(0.127) = approximately 7°. True heading = 165° - 7° = 158°. Option A (165°) applies no wind correction. Option B (126°) overcorrects massively. Option C (152°) applies too large a correction of 13°. Only 158° properly accounts for the crosswind component.
Correct: D)
Explanation: With TC 040° and wind from 350°, the wind angle relative to the course is 50° from the left-front. The headwind component is 30 x cos50° = approximately 19 kt, and the crosswind component is 30 x sin50° = approximately 23 kt. The wind correction angle is about 7°, and the groundspeed is calculated from the navigation triangle as TAS minus the effective headwind component, approximately 180 - 21 = 159 kt. Options A (172 kt) and C (168 kt) underestimate the headwind effect. Option B (155 kt) overestimates it.
Correct: C)
Explanation: With TC 120° and wind from 150°, the wind comes from 30° to the right of and behind the course line. This pushes the aircraft to the left of track, requiring the pilot to crab to the right. WCA = sin^-1(12 x sin30° / 120) = sin^-1(6/120) = sin^-1(0.05) = approximately 3° to the right. Options A and B indicate left corrections, which would worsen the drift. Option D (6° right) doubles the actual correction angle needed.
Correct: D)
Explanation: Using the 1:60 rule, the opening angle (track error from A) is (7/55) x 60 = approximately 7.6° or about 8°. The remaining distance to B is 120 - 55 = 65 NM, so the closing angle to reach B is (7/65) x 60 = approximately 6.5° or about 6°. The total course correction needed is the sum of both angles: 8° + 6° = 14° to the left (since the aircraft is right of track, it must turn left). Option C (15°) slightly overestimates. Option A (8°) only accounts for the opening angle. Option B (6°) only accounts for the closing angle.
