General Navigation: 7 Recent ATPL Questions Explained

General Navigation is one of those ATPL subjects where small calculation mistakes can quickly turn into costly exam errors. Most questions demand precise units, the right formula, and a clear understanding of the navigation logic behind the numbers.

In this walkthrough, we break down seven recently reported General Navigation questions seen across multiple EASA authorities within the last 90 days. You’ll practise VOR geometry, departure and ground speed, air distance, heading conversions, TAS changes during descent, great-circle convergence, and 3° climb-path calculations.

The aim is not simply to reach the correct answer, but to recognise the quickest and safest method to use when the same question type appears in a different form.

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7 Latest ATPL GNAV Exam Questions Covered in This Blog

• AIR-248748: Calculating Range from a VOR Using Coordinates and Radials

• AIR-248330: Calculating Ground Speed Between Two Fixes

• AIR-248637: Converting Ground Distance into Air Distance

• AIR-276452: Converting Compass Heading, Magnetic Heading, True Heading and Track

• AIR-247980: How TAS Changes During Descent at Constant CAS

• AIR-248883: Great-Circle Convergence and Track Change

• AIR-248134: 3° Climb Path — Time and Ground Distance

Question 1: Calculating Range from a VOR Using Coordinates and Radials

AIR-248748: The BUN VOR/DME is located at 51°07'N 004°50'E where the variation is 0°. An aircraft is flying a true course of 090°, starting at 53°04'N 003°37'E with the BUN turned on the VOR. When the VOR shows a radial of 360°, what is the range of the aircraft from the BUN?

• 117 NM

• 46 NM

• 126 NM

• 73 NM

Correct Answer: 117 NM

Explanation

This question looks more complicated than it actually is. The quickest way to understand it is to draw the geometry.

The aircraft is flying a true course of 090°, so it is travelling eastwards. For this exam calculation, its latitude therefore remains 53°04'N.

The BUN VOR is at 51°07'N 004°50'E, and variation at the VOR is 0°. This means the magnetic and true meridians coincide.

A 360° VOR radial extends directly north from the station. When the aircraft crosses that radial, it is therefore on the same meridian as BUN.

The range can now be found simply from the difference in latitude: 53°04'N − 51°07'N = 1°57'.

Convert the degree into minutes: 1°57' = 60' + 57' = 117'.

Along a meridian: 1 minute of latitude = 1 NM. Therefore: Range = 117 NM.

The trick is recognising that once the aircraft crosses the 360° radial, the problem becomes a simple change-of-latitude calculation.

Exam Tip: The geometry often makes these questions much easier. Remember an important GNAV rule:

• VOR question → use variation at the VOR station

• NDB question → use variation at the aircraft position

That distinction comes from where the magnetic reference is established.

Question 2: Calculating Ground Speed Between Two Fixes

AIR-248330: Please refer to the annex. Calculate the GS of the aircraft on a flight between the two observed fixes stated below.

10:22 passing over Regensburg Oberhub EDNR (N49°08', E012°05'). 

10:38 passing over Arnbruck EDNB (N49°08', E012°59')

• 135 kt

• 202 kt

• 83 kt

• 172 kt

Correct Answer: 135 kt

Explanation

The aircraft takes: 10:38 − 10:22 = 16 minutes to travel between the two fixes.

You could measure the distance directly from the chart using a ruler. In the exam, however, we recommend using the coordinates where possible because it removes some of the uncertainty involved in measuring from a map.

Both positions are at the same latitude: 49°08'N.

The only difference is longitude: 012°59'E − 012°05'E = 54'.

At a latitude other than the Equator, one minute of longitude is not equal to one nautical mile. We therefore calculate the departure:

Departure = Change of Longitude × cos Latitude

So: Departure = 54 × cos 49°08'. Departure ≈ 35.3 NM

Now calculate ground speed. Sixteen minutes is: 16 / 60 = 0.267 hours.

Therefore: GS = Distance / Time. GS = 35.3 / 0.267 ≈ 132 kt

The closest available answer is: 135 kt.

Exam Tip: If the two positions have the same latitude, immediately look at the difference in longitude and think:

Departure = ΔLong × cos Latitude

We also recommend a useful speed shortcut: Distance ÷ minutes × 60 = speed in knots.

It is quick, reliable, and avoids unnecessary conversions during a time-pressured exam.

Question 3: Converting Ground Distance into Air Distance

AIR-248637: Given the following information, convert the ground distance to air distance.

• Ground distance: 120 NM

• TAS: 165 kt

• IAS: 140 kt

• GS: 150 kt

Correct Answer: 132 NAM

Explanation

This is a classic air-distance question, and one of the first things to notice is that IAS is irrelevant.

You only need:

• Ground distance

• TAS

• Ground speed

The relationship is:

Air Distance = Ground Distance × TAS / GS

Insert the values:

• Air Distance = 120 × 165 / 150

• Air Distance = 132 NAM

You can also make a quick common-sense check before doing any calculation.

Here: TAS = 165 kt. But: GS = 150 kt.

The ground speed is lower than TAS, so the aircraft is experiencing a headwind.

With a headwind, the aircraft spends longer flying through the air to cover the same distance over the ground. Its air distance must therefore be greater than its ground distance.

That alone eliminates the answers below 120 NAM.

Exam Tip. Use a gross-error check:

• Headwind → Air Distance > Ground Distance

• Tailwind → Air Distance < Ground Distance

It is a simple way to eliminate wrong answers before you even touch the calculator.

Quick Note. If a question specifically gives nil wind at a very high cruising altitude, the examiner may expect you to account for the slightly greater radius of the flight path above the Earth's surface. In that specific case, air distance may be slightly greater than ground distance even without wind.

For normal questions, however, use: 

NAM = NGM × TAS / GS

Question 4: From Compass Heading to True Track

AIR-276452: The flight log gives the following data True track, Drift, True heading, Magnetic variation, Magnetic heading, Compass deviation, Compass heading. The right solution, in the same order, is:

• 119°; 3°L; 122°; 2°E; 120°; +4°; 116°

• 125°; 2°R; 123°; 2°W; 121°; -4°; 117°

• 115°; 5°R; 120°; 3°W; 123°; +2°; 121°

• 117°; 4°L; 121°; 1°E; 122°; -3°; 119°

Correct Answer: 119°; 3°L; 122°; 2°E; 120°; +4°; 116°

Explanation

These questions become much easier if you reverse the order and work from the compass heading towards the true track.

Write: C → D → M → V → T → Dr → TT.

Where:

• C = Compass Heading

• D = Deviation

• M = Magnetic Heading

• V = Variation

• T = True Heading

• Dr = Drift

• TT = True Track

Start with the first option: Compass Heading = 116°.

Apply the +4° deviation: 116° + 4° = 120° Magnetic Heading.

Next, apply 2°E variation. Use a mnemonic: Variation East → Magnetic Least.

So magnetic heading must be lower than true heading: 120° + 2° = 122° True Heading.

Finally, apply 3° left drift. Another useful mnemonic is: Left is Less.

So: 122° − 3° = 119° True Track

Everything agrees with the first option.

Exam Tip. Rather than trying to inspect all seven values at once, work through them one conversion at a time. Remember:

C → D → M → V → T → Drift → Track

And keep these two quick mnemonics:

• East → Magnetic Least

• Left → Less

These conversions are not only exam theory. You will use them repeatedly when preparing navigation logs during practical VFR training.

Question 5: How TAS Changes During Descent at Constant CAS

AIR-247980: The pilot of an aircraft during descent with constant CAS of 120 kt, observes a TAS of 135 kt at FL080. What will be the TAS, if he/she checks again the instruments minutes later?

• 119 kt

• 135 kt

• 144 kt

• 131 kt

Correct Answer: 131 kt

Explanation

This question tests the relationship between CAS and TAS during a change in altitude.

CAS represents the aerodynamic pressure experienced by the aircraft after the appropriate instrument and position corrections. If the aircraft descends while maintaining a constant CAS, air density increases. Because the same dynamic-pressure indication can now be achieved at a lower true speed, TAS decreases during the descent.

The aircraft initially has:

TAS = 135 kt

So after descending for several minutes, the TAS must be lower than 135 kt.

That immediately eliminates:

• 135 kt

• 144 kt

The remaining options are:

• 119 kt

• 131 kt

For normal ATPL conditions at positive altitude, TAS will remain above CAS. Since CAS is 120 kt, 119 kt would not make sense. The correct answer is therefore: 131 kt.

Exam Tip. For constant CAS:

• Climb → TAS increases

• Descent → TAS decreases

There is a very useful rule of thumb to remember:

TAS changes by approximately 2% per 1,000 ft relative to CAS.

It is only an approximation, but it is extremely useful for navigation planning and quick reasonableness checks. Before calculating anything, always check the direction of change. That will often eliminate half the answers immediately.

Question 6: Great-Circle Convergence and Track Change

AIR-248883: A great circle track joins position A (59°S 141°W) and B (61°S 148°W). How does the true track change when flying from position A to position B?

• It increases by 6°

• It decreases by 6°

• It increases by 3°

• It decreases by 3°

Correct Answer: It increases by 6°

Explanation

A great circle is the shortest route between two points on the Earth's surface. Because the meridians converge towards the poles, the true track along a great-circle route changes as the aircraft progresses. The difference between the great-circle tracks at two points is called convergence.

Use:

Convergence = Change of Longitude × sin Mean Latitude

First calculate the change of longitude: 148°W − 141°W = 7°.

Both positions are in the same hemisphere, so subtract.

Now calculate the mean latitude: (59° + 61°) / 2 = 60°.

Therefore:

• Convergence = 7 × sin 60°

• Convergence = 7 × 0.866

• Convergence ≈ 6°

So we now know the amount of track change. The remaining question is whether the track increases or decreases.

A quick sketch of the two positions and the converging meridians shows that the true track at B is greater than the true track at A.

Therefore: True track increases by approximately 6°.

Exam Tip. Treat these as two separate tasks:

1. Calculate how much the track changes. 2. Decide whether it increases or decreases.

If you are unsure about the direction, draw a quick diagram. That is often safer than relying entirely on a memorised mnemonic.

The essential formula is:

Convergence = ΔLong × sin Mean Latitude

Question 7: 3° Climb Path: Time and Ground Distance

AIR-248134: An aircraft flying at 3000 ft commences a climb at point A on a 3° path at an average GS of 210 kt. The aircraft will reach the altitude of 8000 ft in (1) _____ and the ground distance covered from point A will be (2) _____.

• (1) 4 minutes and 46 seconds; (2) 16.7 NM

• (1) 5 minutes and 14 seconds; (2) 18.3 NM

• (1) 7 minutes and 56 seconds; (2) 27.8 NM

• (1) 12 minutes and 42 seconds; (2) 44.5 NM

Correct Answer: (1) 4 minutes and 46 seconds; (2) 16.7 NM

Explanation

The aircraft climbs from: 3,000 ft to 8,000 ft. So the vertical height gained is: 8,000 − 3,000 = 5,000 ft. For a 3° flight path, a very useful aviation approximation is:

Vertical Speed ≈ Ground Speed × 5

The aircraft's ground speed is: 210 kt

Therefore: Vertical Speed ≈ 210 × 5 = 1,050 ft/min

Now calculate the climb time: Time = Height / Rate of Climb

• Time = 5,000 / 1,050

• Time ≈ 4.76 minutes

The decimal part is: 0.76 × 60 ≈ 46 seconds

So the climb takes approximately: 4 minutes 46 seconds

Now calculate the ground distance. A ground speed of 210 kt means: 210 / 60 = 3.5 NM per minute

Multiply by 4.76 minutes: 3.5 × 4.76 ≈ 16.7 NM

Therefore: Time = 4 min 46 sec.

Ground distance = 16.7 NM

Exam Tip: Memorise this practical shortcut: 

3° path → Vertical Speed ≈ GS × 5

It works for both climbs and descents. This is much more than an exam trick. During instrument training, you will use essentially the same mental calculation to estimate the required rate of descent on a standard 3° ILS or approach path.

For example:

• GS 100 kt → approximately 500 ft/min

• GS 140 kt → approximately 700 ft/min

• GS 180 kt → approximately 900 ft/min

Keep Practising

General Navigation gets easier when the formulas stop feeling like isolated rules and you start recognising the question type, relevant data and quickest calculation method.

Want to check whether you can apply these principles without the walkthrough beside you? Practise more General Navigation questions in the Airhead ATPL Question Bank, identify the areas that still slow you down, and revisit them until the calculation logic feels automatic.