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Free ATPL flight planning and monitoring 033 questions: 8 with answers

8 flight planning and monitoring 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 906 questions for this subject.

The 8 questions

  1. Question 1 · 033 01 VFR flight planning

    For a VFR flight following a magnetic route of 090°, which flight level complies with the semi-circular rule?

    • AFL 55
    • BFL 45
    • CFL 65
    • DFL 105
    Show the answer+

    Answer: A. FL 55

    Under the SERA semicircular rule, VFR flights on magnetic tracks from 000° to 179° fly at odd thousands of feet plus 500 ft (FL 35, 55, 75...). FL 55 is the only such level offered for a track of 090°.

  2. Question 2 · 033 02 IFR flight planning

    A letter prefix is added to the basic designator of an ATS route. What does the prefix 'U' indicate?

    • AA route established primarily for use by helicopters
    • BA route established exclusively for use by supersonic aircraft
    • CA route where only advisory service is provided
    • DA route or portion thereof established in the upper airspace
    Show the answer+

    Answer: D. A route or portion thereof established in the upper airspace

    Under ICAO Annex 11 (Appendix 1), the prefix U on an ATS route designator indicates a route, or portion of it, established in the upper airspace (e.g. UA 12).

  3. Question 3 · 033 03 Fuel planning

    Given: DOM=54,000 kg, Load=10,400 kg, Trip fuel=5,100 kg, Contingency=255 kg, Alternate=2,200 kg, Final reserve=550 kg, Taxi=190 kg. Calculate the Estimated Landing Mass.

    • A72,505 kg
    • B67,405 kg
    • C67,215 kg
    • D67,565 kg
    Show the answer+

    Answer: B. 67,405 kg

    Take-off fuel = trip + contingency + alternate + final reserve = 5100 + 255 + 2200 + 550 = 8105 kg (taxi fuel is used before take-off). TOM = 54,000 + 10,400 + 8105 = 72,505 kg, so landing mass = 72,505 − 5100 = 67,405 kg.

  4. Question 4 · 033 03 Fuel planning

    Given: DOM=60,000 kg, Load=4,600 kg, Trip fuel=14,700 kg, Contingency=735 kg, Alternate=1,400 kg, Final reserve=1,450 kg, Taxi=260 kg. Calculate the Block (Ramp) Fuel.

    • A18,545 kg
    • B14,700 kg
    • C18,325 kg
    • D18,725 kg
    Show the answer+

    Answer: A. 18,545 kg

    Block fuel = taxi + trip + contingency + alternate + final reserve = 260 + 14,700 + 735 + 1400 + 1450 = 18,545 kg.

  5. Question 5 · 033 04 pre-flight preparation

    Unlike the PNR, the Critical Point (CP) calculation does not directly depend on:

    • Athe outbound and return groundspeeds
    • Bwind conditions along the route
    • Cthe aircraft's fuel endurance
    • Dthe total route distance
    Show the answer+

    Answer: C. the aircraft's fuel endurance

    CP is a pure time-balance calculation based on distance and speeds; PNR additionally incorporates fuel endurance as a limiting factor.

  6. Question 6 · 033 05 ICAO flight plan

    In the ATS flight plan Item 15 (Cruising speed), when not expressed as a Mach number, cruising speed is expressed as :

    • AGroundspeed
    • BIAS
    • CCAS
    • DTAS
    Show the answer+

    Answer: D. TAS

    When not given as a Mach number, cruising speed in Item 15 is expressed as True Air Speed (TAS).

  7. Question 7 · 033 05 ICAO flight plan

    When an ATS flight plan is submitted for a flight outside designated ATS routes, points included in Item 15 (route) should not normally be at intervals of more than :

    • A20 minutes flying time or 150 km
    • B30 minutes flying time or 370 km
    • C1 hour flying time or 500 km
    • D15 minutes flying time or 100 km
    Show the answer+

    Answer: B. 30 minutes flying time or 370 km

    Outside designated ATS routes, route points in Item 15 should not normally be spaced more than 30 minutes' flying time or 370 km apart.

  8. Question 8 · 033 06 flight monitoring

    An aircraft departs at 1210 UTC and flies a 760 NM route at a groundspeed of 210 kt. The ETA is:

    • A1532 UTC
    • B1617 UTC
    • C1602 UTC
    • D1547 UTC
    Show the answer+

    Answer: D. 1547 UTC

    Flight time = 760 / 210 ≈ 3.62 h ≈ 3 h 37 min; 1210 UTC + 3 h 37 min = 1547 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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