Free ATPL general navigation 061 questions: 10 with answers
10 general navigation questions from the ATPL bank of ForAllPilots, with no account and no sign-up. Answer in your head, then open the answer: the right letter and its explanation are under each question. The full bank has 1,003 questions for this subject.
The 10 questions
Question 1 · 061 01 Basics of navigation
Given: True track=282°, Variation=8E, Drift=10R. Calculate the Magnetic Heading.
- A249°
- B264°
- C279°
- D272°
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Answer: B. 264°
With 10° right drift, the heading is 10° left of track: true heading 272°. East variation is subtracted (variation east, magnetic least): magnetic heading 264°.
Question 2 · 061 01 Basics of navigation
An aircraft flies a true track of 240° at TAS 330 kt. Wind is 210°/45kt. Calculate the approximate groundspeed.
- A290 kt
- B246 kt
- C335 kt
- D334 kt
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Answer: A. 290 kt
The wind from 210° is 30° to the left of the 240° track: headwind component 45 x cos 30° = 39 kt, crosswind 22.5 kt, wind correction angle about 4°. GS = 330 x cos 4° - 39 = about 290 kt.
Question 3 · 061 01 Basics of navigation
Given:True Heading = 180° TAS = 500 ktW/V 225° / 100 ktCalculate the GS?
- A535 kt
- B600 kt
- C435 kt
- D450 kt
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Answer: C. 435 kt
The wind from 225° blows towards 045°, 45° off the reciprocal of the 180° heading. Resolving: southward 500 - 100 x cos 45° = 429 kt, eastward 71 kt. GS = square root of (429² + 71²) = about 435 kt.
Question 4 · 061 01 Basics of navigation
An aircraft departs from position A (04°10' S 178°22'W) and flies northward following the meridian for 2950 NM. It then flies westward along the parallel of latitude for 382 NM to position B. The coordinates of position B are?
- A45°00'N 169°22W
- B45°00'N 172°38'E
- C53°20'N 172°38'E
- D53°20'N 169°22W
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Answer: B. 45°00'N 172°38'E
Flying north 2,950 NM along a meridian is 2,950' = 49°10' of latitude: 04°10' S becomes 45°00' N. Flying west 382 NM on the 45° N parallel is 382 / cos 45° = 540' = 9°00' of longitude: 178°22' W + 9° = 187°22' W, that is 172°38' E.
Question 5 · 061 01 Basics of navigation
At latitude 60°, a change of longitude of 250' (minutes of arc) corresponds to a departure (distance) of approximately:
- A125 NM
- B162 NM
- C250 NM
- D88 NM
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Answer: A. 125 NM
Departure (NM) = change of longitude (minutes) x cos latitude = 250 x cos 60° = 125 NM.
Question 6 · 061 01 Basics of navigation
Given: GS = 435 kt.Distance from A to B = 1920 NM.What is the time from A to B?
- A3 HR 25 MIN
- B4 HR 10 MIN
- C4 HR 25 MIN
- D3 HR 26 MIN
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Answer: C. 4 HR 25 MIN
Time = distance / groundspeed = 1,920 / 435 = 4.41 h, which is 4 h 25 min.
Question 7 · 061 01 Basics of navigation
Given: FL300, Mach 0.82, OAT -34°C. Calculate the approximate TAS.
- A420 kt
- B568 kt
- C494 kt
- D602 kt
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Answer: C. 494 kt
The local speed of sound is about 38.94 x square root of the absolute temperature: OAT -34 °C = 239 K, so a = about 602 kt. TAS = Mach x a = 0.82 x 602 = about 494 kt.
Question 8 · 061 01 Basics of navigation
Isogonals are lines of equal :
- Acompass deviation.
- Bwind velocity.
- Cpressure.
- Dmagnetic variation.
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Answer: D. magnetic variation.
Isogonals are lines connecting points of equal magnetic variation.
Question 9 · 061 03 Great circles and rhumb lines
Given a change of longitude of 25° and a mean latitude of 10°, calculate the approximate conversion angle.
- A12.5°
- B3.3°
- C1.1°
- D2.2°
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Answer: D. 2.2°
Conversion angle = ½ x change of longitude x sin mean latitude = ½ x 25° x sin 10° = 12.5 x 0.174 = about 2.2°.
Question 10 · 061 05 Time
A position is located at longitude -75°. Using the standard 15°=1 hour rule, local mean time at this position is approximately how many hours behind UTC?
- A7.5 hours
- B5.0 hours
- C3.0 hours
- D7.0 hours
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Answer: B. 5.0 hours
The mean sun moves 15° of longitude per hour. At 75° W local mean time is 75 / 15 = 5 hours behind UTC.
Each answer was checked again before publication; like every explanation on the site, these may contain errors. Some explanations were rewritten for this page, sometimes with the help of an AI. Spotted a mistake? Write to us at contact@forallpilots.com.
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