Free PPL navigation 060 questions: 9 with answers
9 navigation questions from the PPL 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 842 questions for this subject.
The 9 questions
Question 1 · 1 The Earth and time
Two points with the same longitude are necessarily:
- Aat the same location.
- Bon the same meridian.
- Con the Equator.
- Don the same parallel.
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Answer: B. on the same meridian.
Since meridians represent a location's longitude, two points with the same longitude must be located on the same meridian.
Question 2 · 1 The Earth and time
On June 12th, the legal time in Paris is: UTC + 2 h. It is 1145. You deduce that the UTC time is:
- A1345.
- B0945.
- C1045.
- D1245.
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Answer: B. 0945.
Summer legal time in Paris is UTC + 2 h: to obtain UTC, subtract 2 h from legal time. 1145 − 2 h = 0945 UTC.
Question 3 · 3 Charts
Points A and B have the following coordinates: A: 30°N - 002°E B: 31°N - 002°E The distance between them is:
- A60 NM.
- B0 NM.
- C30.5 NM.
- D1 NM.
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Answer: A. 60 NM.
The two points are located on the same meridian (002°E) and are separated by 1° (between 30°N and 31°N). Now, one minute of angle on a meridian measures 1 NM, so 1° measures 60 NM.
Question 4 · 2 Directions, distances and magnetism
The true track is 350°. The local magnetic variation is 5° East. What is the corresponding magnetic track?
- A350°
- B345°
- C355°
- D000°
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Answer: B. 345°
Rule "variation West, add; variation East, subtract": MT = 350° − 5° = 345°.
Question 5 · 2 Directions, distances and magnetism
The magnetic heading is 235°. The local compass deviation is 4° West. What compass heading should be set?
- A243°
- B235°
- C239°
- D231°
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Answer: C. 239°
Rule "deviation West, add; deviation East, subtract": CH = 235° + 4° = 239°.
Question 6 · 4 Dead reckoning
An aircraft covers 66 NM in 22 minutes. What is its ground speed?
- A270 kt
- B90 kt
- C200 kt
- D180 kt
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Answer: D. 180 kt
Ground speed = distance / time = 66 NM / (22/60 h) = 180 kt.
Question 7 · 4 Dead reckoning
Track to be followed: 130°. TAS = 105 kt. Wind from 295° at 28 kt. What is the approximate resulting ground speed once the heading has been corrected?
- A116 kt
- B105 kt
- C132 kt
- D146 kt
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Answer: C. 132 kt
After drift correction (WCA ≈ 4.0°), GS = TAS·cos(WCA) - Ws·cos(wind/track angle) = 105·cos(4.0°) - 28·cos(165°) ≈ 132 kt.
Question 8 · 6 Radio navigation
An aircraft receives a radial (QDR) of 181° from a VOR station. What is the bearing (QDM) to fly directly to the station?
- A091°
- B271°
- C001°
- D181°
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Answer: C. 001°
QDM = QDR + 180° = 181° + 180° = 001° (reciprocal).
Question 9 · 6 Radio navigation
When a GPS is switched on, it:
- Aresynchronizes to one of the points in its database (VOR, aerodrome, etc.).
- Bneeds the user to enter the coordinates of the location.
- Ccalculates its position by itself.
- Dcannot initialize its position if the aircraft is on the ground.
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Answer: C. calculates its position by itself.
When switched on, the GPS receiver searches for the visible satellites and calculates its position by itself from the signals received (at least four satellites for a three-dimensional position), without any user input, including on the ground.
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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