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Colombia Aerocivil commercial aeroplane theory foundations study materials
Colombia Aerocivil commercial aeroplane theory foundations study resources

Colombia Aerocivil commercial aeroplane theory foundations practice questions

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Question 1Aircraft systems and instruments

During a magneto check on a light airplane before a commercial VFR departure, RPM drops 175 on the left magneto and 80 on the right, both within the AFM limit, and the engine runs smoothly. What does this primarily confirm?

  1. Each magneto can ignite the mixture independently, and the larger (still allowable) drop is typically the magneto being tested when that ignition source is isolated Correct answer.
  2. The left magneto is failed and the flight must be cancelled
  3. The battery is discharged because magnetos are battery-powered
  4. Both magnetos must produce identical RPM drops or the engine is unairworthy

Explanation

Dual magnetos provide independent ignition. Switching to one magneto isolates the other; a drop within AFM limits with smooth running shows that magneto can fire the engine. Magnetos are engine-driven and do not rely on the aircraft battery.

Question 2Aircraft systems and instruments

A carburetted training airplane in a humid 12 °C descent at low power develops roughness that clears when carburettor heat is applied. A fuel-injected airplane in the same conditions would most likely:

  1. Be far less prone to carburettor ice, though impact icing of the air intake or alternate-air requirements can still apply Correct answer.
  2. Suffer identical carburettor ice because injection uses a venturi in the same way
  3. Always ice first at the injectors at 12 °C
  4. Require carburettor heat as a normal descent item

Explanation

Carburettor ice forms in the venturi and on the throttle plate as fuel vaporises and pressure drops. Fuel injection meters fuel after the throttle body, so classic carb ice is uncommon; intake icing and alternate-air procedures still matter. Commercial PIC still uses AFM anti-ice procedures for the system installed.

Question 3Aircraft systems and instruments

If both static ports ice over in climb, the airspeed indicator will typically:

  1. Under-read as altitude increases (indicated airspeed falls below true dynamic pressure) because trapped static pressure is too high relative to the outside Correct answer.
  2. Read correctly because it uses only pitot pressure
  3. Over-read in climb because trapped static pressure is lower than ambient
  4. Drop to zero immediately

Explanation

ASI measures pitot minus static. In a climb, ambient static pressure falls, but a blocked static system keeps a higher trapped pressure, so the differential is too small and IAS under-reads. Alternate static, if installed, is the AFM corrective.

Question 4Aircraft systems and instruments

On a turbocharged airplane, manifold pressure at full throttle at 8,000 ft pressure altitude is the same as at sea level. The PIC should understand that:

  1. The turbocharger is restoring deck pressure so sea-level MP can be maintained up to the critical altitude; exceeding AFM MP limits (overboost) remains a real hazard Correct answer.
  2. The turbocharger is undersized and the engine is unairworthy
  3. Turbocharging eliminates all density-altitude performance loss
  4. Manifold pressure should always fall 1 inHg per 1,000 ft, even with a turbocharger

Explanation

A turbocharger compresses induction air so rated manifold pressure can be held up to critical altitude. Density altitude still hurts excess power, climb, and true airspeed performance once the turbo is at its limit. Overboosting beyond AFM MP is a serious engine-limit exceedance.

Question 5Aircraft systems and instruments

Detonation during a high-power climb is most likely when the PIC:

  1. Uses the AFM takeoff mixture and the correct fuel grade
  2. Applies carburettor heat in icing conditions
  3. Uses a lower-than-specified octane fuel, an over-lean mixture, and high manifold pressure with low RPM Correct answer.
  4. Reduces power after takeoff using MP then RPM

Explanation

Detonation is explosive combustion from low octane, excessive temperature, over-lean mixture, or high MP with low RPM. It damages pistons rapidly. Pre-ignition is ignition before the spark from a hot spot; both are AFM emergencies, not a leaning technique.

Question 6Aircraft systems and instruments

During a night commercial positioning flight, the ammeter shows discharge, the low-voltage light is on, and the battery is the remaining source. The PIC should:

  1. Turn on every light and radio to test the bus
  2. Ignore the ammeter if the engine is producing oil pressure
  3. Shed non-essential electrical load, follow the AFM electrical-failure checklist, and land at a suitable aerodrome before the battery is exhausted Correct answer.
  4. Raise RPM to 2,700 to mechanically drive the radios

Explanation

A discharging ammeter means the alternator/generator is not feeding the bus. Remaining time is battery endurance. Commercial TEM means load-shed, communicate early, and land with electrical power still available for gear, flaps, and radios as required.

Question 7Navigation, radio aids and flight planning

Loading an airplane toward a forward CG (still within limits) typically:

  1. Decreases stall speed and makes the aeroplane less stable in pitch
  2. Has no effect on takeoff distance
  3. Increases stall speed, increases stick force, and lengthens the takeoff and landing distances because a more nose-down pitching moment must be overcome Correct answer.
  4. Always moves the aeroplane outside the aft envelope

Explanation

A forward CG increases the tail-down force needed for balance, so the wing must produce more lift, stall speed rises, and takeoff/landing distances increase. Pitch stability usually increases. Commercial loading must still keep CG inside the AFM envelope.

Question 8Navigation, radio aids and flight planning

If takeoff weight increases 10 percent with other factors equal, takeoff distance:

  1. Decreases about 10 percent because the wing is more efficient
  2. Is unchanged if flaps are up
  3. Increases by more than 10 percent; excess thrust is a smaller fraction of weight, so ground roll and 50-ft distance grow disproportionately Correct answer.
  4. Depends only on fuel density, not mass

Explanation

Takeoff distance varies roughly with the square of weight for a given thrust. A 10 percent weight increase produces a substantially larger percentage increase in roll. Commercial VFR loading therefore cannot treat extra payload as “a little extra roll.”

Question 9Navigation, radio aids and flight planning

BEW 1,450 lb, moment 54,375 lb-in. Occupants 340 lb at 38.0 in. Fuel 216 lb at 42.0 in. Baggage arm 95.0 in. Maximum takeoff weight 2,200 lb. Aft CG limit 40.5 in. What limits baggage?

  1. Weight only: 194 lb of baggage is available up to MTOW
  2. Fuel: baggage is unlimited if fuel is full
  3. CG: about 89 lb of baggage, because more baggage at 95 in drives CG aft of 40.5 in before MTOW is reached Correct answer.
  4. Nothing: any baggage is acceptable below 2,200 lb

Explanation

Zero-baggage weight is 2,006 lb, so 194 lb remains to MTOW. Setting CG = 40.5 gives (76,367 + 95B) / (2,006 + B) = 40.5, so 54.5B = 4,876 and B ≈ 89 lb. The aft CG limit is reached first. Commercial loading must check both mass and CG.

Question 10Navigation, radio aids and flight planning

The AFM takeoff chart at 2,400 lb, pressure altitude 3,000 ft, 20 °C gives a ground roll of 1,150 ft. It states: subtract 10 percent of the ground roll for each 9 kt of headwind. With a 12 kt headwind, the corrected ground roll is closest to:

  1. 1,150 ft (no wind correction on commercial flights)
  2. 1,304 ft
  3. 996 ft Correct answer.
  4. 575 ft

Explanation

Wind factor = 12/9 × 10 percent = 13.3 percent reduction. 1,150 × (1 − 0.133) = 996 ft. Always apply the AFM’s own wind rule; do not invent a generic 50 percent cut.

Question 11Navigation, radio aids and flight planning

Field elevation 3,200 ft, altimeter setting 29.42 inHg. Approximate pressure altitude is:

  1. 2,700 ft
  2. 3,200 ft
  3. 3,700 ft Correct answer.
  4. 4,200 ft

Explanation

Pressure altitude ≈ field elevation + (29.92 − 29.42) × 1,000 = 3,200 + 500 = 3,700 ft. Each 0.01 inHg is about 10 ft; 0.50 inHg is 500 ft. When QNH is below 29.92, pressure altitude is above field elevation.

Question 12Navigation, radio aids and flight planning

Pressure altitude 5,400 ft, OAT 22 °C. Using DA ≈ PA + 120 × (OAT − ISA), with ISA at 5,400 ft ≈ 4 °C, density altitude is closest to:

  1. 5,400 ft
  2. 6,500 ft
  3. 7,560 ft Correct answer.
  4. 9,000 ft

Explanation

ISA temperature at 5,400 ft is about 15 − 2 × 5.4 ≈ 4 °C. Deviation is 18 °C. DA ≈ 5,400 + 120 × 18 = 5,400 + 2,160 = 7,560 ft. Commercial takeoff data must be entered with this DA, not field elevation.

Question 13Navigation, radio aids and flight planning

Runway 21 (210° magnetic). Wind 250° magnetic at 18 kt. The crosswind component is closest to:

  1. 4 kt
  2. 18 kt
  3. 12 kt Correct answer.
  4. 25 kt

Explanation

Angle between runway and wind is 40°. Crosswind = 18 × sin(40°) ≈ 18 × 0.643 = 11.6 kt, about 12 kt. Headwind = 18 × cos(40°) ≈ 14 kt. Compare 12 kt with the AFM demonstrated crosswind before a commercial departure.

Question 14Navigation, radio aids and flight planning

CAS 120 kt at 8,000 ft pressure altitude near ISA. Using the rule of thumb TAS ≈ CAS + 2 percent per 1,000 ft, TAS is closest to:

  1. 120 kt
  2. 128 kt
  3. 139 kt Correct answer.
  4. 160 kt

Explanation

Eight thousand feet × 2 percent = 16 percent. 120 × 1.16 = 139 kt. Commercial fuel and ETA planning use TAS (and groundspeed), not CAS.

Question 15Navigation, radio aids and flight planning

Same load as the 2,538 lb / 42.6 in example (fuel 288 lb at 47.5 in). After burning 150 lb of fuel, the CG will:

  1. Move aft because fuel is always forward of the CG
  2. Jump to 47.5 in
  3. Remain exactly 42.6 in
  4. Move forward to about 42.3 in because the fuel arm (47.5 in) is aft of the original CG Correct answer.

Explanation

Removing mass aft of the CG moves CG forward. New weight 2,388 lb; moment removed 150 × 47.5 = 7,125; remaining moment 108,090 − 7,125 = 100,965; CG = 100,965 / 2,388 ≈ 42.3 in. Commercial cruise fuel burn must be checked against the envelope, especially with aft baggage.

Question 16Navigation, radio aids and flight planning

Climb rate 800 ft/min, groundspeed 90 kt. Climb gradient in feet per nautical mile is:

  1. 90 ft/NM
  2. 1,350 ft/NM
  3. 800 ft/NM
  4. 533 ft/NM Correct answer.

Explanation

Minutes per NM = 60/90 = 0.667 min/NM. Gradient = 800 × 0.667 = 533 ft/NM (or 800 / 1.5 NM per minute). Commercial obstacle clearance uses gradient, not climb rate alone; a strong headwind steepens ft/NM.

Question 17Navigation, radio aids and flight planning

AFM landing distance over 50 ft is 1,800 ft at the planned weight on a dry, level runway. The AFM says: increase the 50-ft distance 20 percent for a 10 kt tailwind and 10 percent for a 1 percent downslope (additive). With 10 kt tailwind and 1 percent downslope, landing distance required is:

  1. 1,800 ft
  2. 2,160 ft
  3. 2,340 ft Correct answer.
  4. 3,600 ft

Explanation

Additive factors 20 percent + 10 percent = 30 percent. 1,800 × 1.30 = 2,340 ft. Commercial VFR into a high Andean strip with a tailwind and downslope must use the AFM factors, then add a safety margin for a wet or contaminated surface if applicable.

Question 18Navigation, radio aids and flight planning

Using the 1-in-60 rule, 6 NM off a 60 NM track is an approximate track-error of:

  1. 1°
  2. 6° Correct answer.
  3. 10°
  4. 60°

Explanation

Degrees of track error ≈ (miles off × 60) / miles along track = (6 × 60) / 60 = 6°. Commercial dead-reckoning and VOR intercepts use this mental arithmetic when GNSS is degraded.

Question 19Navigation, radio aids and flight planning

Magnetic heading 080°, ADF relative bearing 040°. Magnetic bearing to the NDB is:

  1. 040°
  2. 080°
  3. 120° Correct answer.
  4. 320°

Explanation

Magnetic bearing to the station = magnetic heading + relative bearing = 080 + 040 = 120° (modulo 360). Commercial ADF use still appears as a backup to GNSS on some Colombian routes.

Question 20Navigation, radio aids and flight planning

TAS 110 kt, wind from 270° magnetic at 20 kt, true/magnetic course 360°. Approximate heading and groundspeed are:

  1. Heading 360°, groundspeed 130 kt
  2. Heading about 350°, groundspeed about 108 kt Correct answer.
  3. Heading 090°, groundspeed 110 kt
  4. Heading 270°, groundspeed 90 kt

Explanation

Westerly wind on a northerly course is mostly crosswind from the left, so heading is left of course (into wind), about 10° (max drift = 20/110 × 60 ≈ 11°). Groundspeed stays near TAS because the wind is nearly abeam (cos 90° ≈ 0). Commercial fuel planning must use GS, not TAS, when the wind is more head-on.

Question 21Navigation, radio aids and flight planning

On a VOR with a standard CDI (full-scale 10° en-route), a two-dot fly-left deflection (about 4°) at 30 NM from the station is approximately how far off the selected course?

  1. 0.2 NM
  2. 2 NM Correct answer.
  3. 4 NM
  4. 12 NM

Explanation

1-in-60: distance off ≈ (degrees off × distance) / 60 = (4 × 30) / 60 = 2 NM. Two dots are typically 4° on a 10° full-scale en-route CDI. Commercial intercepts should start before a 2 NM error grows in the mountains.

Question 22Human performance and decision making

Selecting the wrong fuel tank despite intending the fuller tank is, in TEM terms:

  1. A threat, because fuel systems are always external
  2. Not relevant to commercial VFR
  3. An undesired aircraft state as soon as the thought occurs
  4. A crew error (procedural or handling), which if uncorrected can lead to an undesired aircraft state such as fuel starvation Correct answer.

Explanation

Wrong tank selection is a crew error. The threat might have been a poorly marked selector or time pressure. If the error is not caught, the UAS could be fuel starvation or an off-airport landing. TEM on commercial flights means traps (cross-checks) around tank selection.

Question 23Human performance and decision making

Somatogravic illusion during a commercial night takeoff into IMC-like conditions typically makes the PIC perceive:

  1. A false descent, causing an inappropriate pitch-up into a stall
  2. No illusion if the attitude indicator has failed
  3. A false turn opposite to the physical bank
  4. A false pitch-up due to linear acceleration, causing an inappropriate push that can fly the aeroplane into the ground Correct answer.

Explanation

Acceleration stimulates the otoliths similarly to a pitch-up. Pilots may push forward and descend. Commercial night departures from poorly lit Andean valleys make this a TEM threat: believe the attitude instrument, not the seat-of-the-pants.

Question 24Human performance and decision making

During a commercial VFR approach, the aeroplane is 20 kt fast, 200 ft high, and not configured by the AFM stable-approach gate. In TEM this is:

  1. Still only a threat, because the runway is in sight
  2. Acceptable if passengers are connecting
  3. An error that has not yet happened
  4. An undesired aircraft state; the recovery is to restore margins (go around) rather than to keep “fixing” an unstable approach to a commercial landing Correct answer.

Explanation

Speed, path, and configuration outside the stable-approach criteria are an undesired aircraft state. TEM recovery is to go around, not to salvage a fast, high commercial landing. Passenger connections are not a justification.

Question 25Meteorology and weather information

Which overnight conditions most favour radiation fog before a dawn departure from a Colombian valley aerodrome?

  1. A warm, moist air mass is lifted rapidly over the Andes in daytime
  2. The jet stream is directly overhead
  3. A thunderstorm outflow arrives at noon
  4. Clear night, light wind, and a moist surface so the ground cools the air to dewpoint Correct answer.

Explanation

Radiation fog needs nocturnal cooling, high humidity, and light wind. Advection fog needs a moist air mass moving over a colder surface. Upslope fog needs terrain lift. Commercial dawn departures in valleys can trap radiation fog until the sun mixes it out.

Question 26Meteorology and weather information

Flying toward a warm front at commercial VFR levels, the typical sequence (from far ahead toward the surface front) is:

  1. Cumulonimbus, then clear ice only, then unlimited ceilings
  2. Only a wind-direction change with no cloud
  3. A sudden line of dust with no cloud
  4. Cirrus, then altostratus/nimbostratus with lowering ceiling and rain, then stratus/fog near the surface front Correct answer.

Explanation

Warm-front slope produces high cloud first, then thickening middle cloud and rain, then low stratus. Embedded convection can still occur. For commercial VFR this often means a long, lowering ceiling-plan a diversion before you are forced into IMC.

Question 27Meteorology and weather information

Low-level wind shear on a commercial approach is most likely:

  1. Only above FL200
  2. Only with a direct headwind of 5 kt
  3. When QNH is 1013 hPa
  4. In a thunderstorm gust front or microburst, or with a strong low-level temperature inversion and speed change, producing sudden airspeed and flight-path changes Correct answer.

Explanation

Shear is a change of wind speed or direction with height or time. Gust fronts and microbursts are the violent convective cases. Inversions can produce a low-level jet. Commercial TEM is a go-around if airspeed and sink rate become unstabilised.

Question 28Meteorology and weather information

A microburst under a convective cell is most dangerous to a commercial VFR landing because:

  1. It only reduces visibility, never wind
  2. Flaps cannot be used in any wind
  3. It always occurs above 10,000 ft AGL
  4. The aeroplane may first meet an increasing headwind (performance increase) then a downdraught and tailwind (sudden performance loss) close to the ground Correct answer.

Explanation

A microburst is a concentrated downdraught that spreads as a vortex ring. The classic profile is increasing performance, then a sink and tailwind that can exceed light-aircraft climb capability. The commercial action is to avoid the cell and go around early; do not chase the approach.

Question 29Meteorology and weather information

An occluded front is best described as:

  1. A front that has no temperature contrast
  2. The equatorial trough with no clouds
  3. Only a sea-breeze boundary
  4. A composite front formed when a cold front catches a warm front, lifting the warm air aloft, often with a mix of stratiform and convective weather Correct answer.

Explanation

Occlusion wraps the warm sector off the ground. Weather can combine warm-front layer cloud and cold-front convection. For commercial VFR in Colombia, treat any analysed occlusion or ITCZ cluster as a wide area of reduced ceiling, rain, and embedded Cb-plan around it.

Question 30Performance, loading and principles of flight

For a PCA planning a hot-and-high takeoff or landing, where is airplane ground effect most significant?

  1. Above 1,000 ft AGL in cruise
  2. Within about one wingspan of the surface, reducing induced drag and causing a floating landing or a premature lift-off in ground effect that the aeroplane cannot sustain Correct answer.
  3. Only with a tailwind
  4. Only on jets

Explanation

Induced drag drops in ground effect. Commercial short-field technique must not “break ground” in ground effect without climb performance, especially at high density altitude. Floating on landing from excess speed is the same phenomenon.

Question 31Aircraft systems and instruments

In a typical light-aeroplane vacuum gyro layout, which instruments usually fail if the vacuum pump fails?

  1. Attitude indicator and heading indicator (directional gyro) Correct answer.
  2. Turn coordinator and magnetic compass
  3. Altimeter and VSI only
  4. ILS localizer and glideslope

Explanation

The attitude indicator and heading indicator are commonly vacuum-driven. The turn coordinator is usually electric and remains available for partial-panel flight, together with the pitot-static instruments and compass. Confirm the actual aircraft’s power sources in the AFM.

Question 32Aircraft systems and instruments

A common IFR rule of thumb for a 3° ILS glide path is that height above the runway threshold is about:

  1. 300 ft per nautical mile Correct answer.
  2. 100 ft per nautical mile
  3. 500 ft per nautical mile
  4. 1,000 ft per nautical mile

Explanation

A 3° path is about 318 ft per NM; pilots use 300 ft/NM. At 4 NM you should be near 1,200 ft above threshold elevation on a standard 3° slope, before adding any TCH. Use the published glide-path angle and charted altitudes when they differ from 3°.

Question 33Aircraft systems and instruments

If the vacuum-driven attitude and heading indicators fail in IMC, which instrument is normally still available to indicate rate of turn?

  1. The vertical-speed indicator
  2. The glideslope needle
  3. The electrically driven turn coordinator (or turn-and-slip indicator) Correct answer.
  4. The suction gauge itself as a turn reference

Explanation

The turn coordinator is typically electric and independent of the vacuum pump, so it remains the primary bank/rate reference on partial panel together with the compass and pitot-static instruments. Cover failed gyros to avoid fixation.

Question 34Aircraft systems and instruments

Pitot tube blocked by ice, drain hole also blocked, static port open. In a climb the airspeed indicator will tend to:

  1. Continue to indicate the airspeed that existed when the pitot blocked (like an altimeter trapped at constant pressure in the pitot line)
  2. Read zero immediately
  3. Over-read in the climb and under-read in the descent as a trapped pitot pressure is compared with falling/rising static pressure Correct answer.
  4. Unaffected, because ASI uses static pressure only

Explanation

With pitot and drain blocked, pitot pressure is trapped. As static pressure falls in the climb, the ASI over-reads; in the descent it under-reads. If only the pitot ram inlet is blocked but the drain is open, the ASI typically goes to zero. Alternate static and pitot heat are the IFR mitigations.

Question 35Aircraft systems and instruments

If the static port ices over and no alternate static source is selected, which instruments are affected?

  1. Only the heading indicator
  2. Only the magnetic compass
  3. Altimeter, VSI, and airspeed indicator Correct answer.
  4. Only the ILS glideslope

Explanation

Static pressure feeds the altimeter, VSI, and the static side of the ASI. A blocked static source freezes or reverses those indications with altitude change. Use the published alternate static source; expect small ASI/altimeter errors from cabin-static pressure in unpressurized aircraft.

Question 36Aircraft systems and instruments

A heading indicator that is not slaved must be reset to the magnetic compass periodically because of:

  1. Mach tuck
  2. Localizer reverse sensing
  3. Gyroscopic precession (real plus apparent/earth-rate and transport wander) Correct answer.
  4. QNE to QNH change at transition

Explanation

Displacement gyros precess from bearing friction (real precession) and from Earth rotation and aircraft motion (apparent precession). In straight, unaccelerated flight, reset the HI to the compass every 10–15 minutes or as the AFM states. Slaved HSI/compass systems reduce but do not eliminate the need to monitor.

Question 37Aircraft systems and instruments

Using the 300 ft/NM rule of thumb for a 3° ILS, what approximate height above threshold should you be at 4.0 NM from the threshold, ignoring threshold crossing height?

  1. 400 ft
  2. 800 ft
  3. 1,200 ft Correct answer.
  4. 2,400 ft

Explanation

4.0 NM × 300 ft/NM = 1,200 ft above threshold elevation. A 3° path is actually about 318 ft/NM, so the true geometric height is slightly higher; charts publish the exact GP altitude. Add TCH (often about 50 ft) when comparing to wheel-crossing height.

Question 38Aircraft systems and instruments

Earth-rate apparent drift is 15°/h × sin(latitude). Near Bogotá (about 5°N) compared with 60°N, the earth-rate component is:

  1. About the same 13°/h at both latitudes
  2. Maximum at the equator and zero at 60°N
  3. Nearly zero at 5°N (15 × sin 5° ≈ 1.3°/h) and about 13°/h at 60°N Correct answer.
  4. Independent of latitude and equal to 15° per minute

Explanation

Sin 5° ≈ 0.087, so earth-rate is only about 1.3°/h near Bogotá; sin 60° ≈ 0.87 gives about 13°/h. Heading gyros still need regular compass resets because real precession and transport wander remain. Low latitude does not make the HI a slaved compass.

Question 39Aircraft systems and instruments

Flying a constant east heading in the Northern Hemisphere, you accelerate during a missed-approach go-around. The magnetic compass will momentarily indicate:

  1. A turn toward the south
  2. No heading change, because acceleration error is zero on east headings
  3. A turn toward the north Correct answer.
  4. A 180° reversal to west

Explanation

ANDS: acceleration on east or west headings displaces the compass toward north. A go-around acceleration can look like a turn. Cross-check the turn coordinator and keep wings level on the AI rather than chasing the compass during the acceleration.

Question 40Navigation, radio aids and flight planning

Flying from a high-pressure area toward a low-pressure area without updating QNH, true altitude will be:

  1. Higher than indicated (you are higher than you think)
  2. Equal to density altitude by definition
  3. Unchanged, because QNH errors cancel in the altimeter
  4. Lower than indicated (high to low, look out below) Correct answer.

Explanation

With a too-high subscale setting, the altimeter over-reads: indicated altitude is higher than true. Flying toward lower pressure without resetting QNH therefore puts you lower than the altimeter shows. The same ‘look out below’ applies when flying toward colder air.

Question 41Navigation, radio aids and flight planning

Cruise consumption is 48 L/h. Planned time to destination is 1 hour 20 minutes, time from destination to alternate is 30 minutes, and the 45-minute final reserve is required. Ignoring taxi and contingency, the fuel needed is:

  1. 72 L
  2. 96 L
  3. 124 L Correct answer.
  4. 192 L

Explanation

Time = 1 h 20 min + 30 min + 45 min = 2 h 35 min = 2.583 h. Fuel = 2.583 × 48 ≈ 124 L. (1.333 + 0.50 + 0.75) × 48 = 2.583 × 48 = 124.0 L.

Question 42Navigation, radio aids and flight planning

You must lose 4,500 ft on a 3° descent using the 300 ft/NM rule of thumb. Approximately how many nautical miles before the point at which you want to be at the lower altitude should you start down?

  1. 7.5 NM
  2. 10 NM
  3. 15 NM Correct answer.
  4. 45 NM

Explanation

Distance (NM) ≈ altitude to lose (ft) / 300. 4,500 / 300 = 15 NM. That is a still-air geometric start-down; wind, ATC constraints, and published TOD/STARs override the rule of thumb.

Question 43Navigation, radio aids and flight planning

On a 3° ILS, a common rate-of-descent rule of thumb is groundspeed (kt) × 5, in feet per minute. At 120 kt groundspeed the target ROD is about:

  1. 240 ft/min
  2. 400 ft/min
  3. 600 ft/min Correct answer.
  4. 1,200 ft/min

Explanation

120 × 5 = 600 ft/min. The more precise factor for 3° is about × 5.2 (≈ 318 ft/NM × GS/60). Check the ILS GP and adjust for wind so the needles stay centred.

Question 44Navigation, radio aids and flight planning

In air colder than ISA, with QNH correctly set, true altitude compared with indicated altitude is:

  1. Higher than indicated (you have extra terrain clearance)
  2. Equal to flight level even below transition
  3. Identical at all temperatures by ICAO definition
  4. Lower than indicated (true altitude is less than the altimeter shows) Correct answer.

Explanation

Cold air is denser; a given pressure surface sits closer to the terrain. The altimeter, calibrated to ISA, over-reads: you are lower than indicated. Approach charts may publish cold-temperature corrections; apply them when required. This matters in the Andes and on high Colombian plateaus.

Question 45Navigation, radio aids and flight planning

A common altimetry rule of thumb is that 1 hPa of subscale error equals about 27 ft of altitude error. If the altimeter is set 10 hPa too high, indicated altitude is approximately:

  1. 270 ft lower than true altitude
  2. Unaffected until you pass transition level
  3. 1,000 ft higher than true altitude
  4. 270 ft higher than true altitude Correct answer.

Explanation

A too-high QNH setting makes the altimeter over-read: indicated is about 10 × 27 = 270 ft above true. You are lower than you think. Some schools use 30 ft/hPa (300 ft for 10 hPa); either rule shows the same hazard. Reset QNH from ATS or ATIS.

Question 46Navigation, radio aids and flight planning

Cruise burn is 48 L/h. Plan 1 h 20 min to destination, 35 min destination-to-alternate, 45 min final reserve, plus 15 min of expected holding. Ignoring taxi and contingency, fuel required is:

  1. 96 L
  2. 124 L
  3. 140 L Correct answer.
  4. 168 L

Explanation

Time = 80 + 35 + 45 + 15 = 175 minutes = 2.917 hours. Fuel = 2.917 × 48 = 140 L. The extra 15 min of holding is 12 L on top of dest + alternate + 45 min reserve (which alone was 124 L at this burn).

Question 47Navigation, radio aids and flight planning

You must descend from 11,000 ft to 2,000 ft (9,000 ft to lose) on a 3° path at 150 kt groundspeed. Using 300 ft/NM and ROD ≈ GS × 5, which pair is correct?

  1. Start down 15 NM out; 450 ft/min
  2. Start down 45 NM out; 300 ft/min
  3. Start down 9 NM out; 1,500 ft/min
  4. Start down 30 NM out; 750 ft/min Correct answer.

Explanation

Distance = 9,000 / 300 = 30 NM. Time = 30 NM / 150 kt = 0.2 h = 12 min. Rate = 9,000 ft / 12 min = 750 ft/min, which matches 150 × 5. Wind and ATC may require an earlier or later TOD.

Question 48Navigation, radio aids and flight planning

You are 80 NM from the filed IFR alternate at 120 kt groundspeed, burning 36 L/h. Fuel to reach the alternate plus a 45-minute final reserve at that burn is:

  1. 24 L
  2. 27 L
  3. 51 L Correct answer.
  4. 80 L

Explanation

Time to the alternate = 80 / 120 = 0.667 h = 40 min, using 0.667 × 36 = 24 L. Forty-five minutes reserve is 0.75 × 36 = 27 L. Total 24 + 27 = 51 L. This is the diversion-plus-reserve slice only; it does not include remaining trip fuel to destination.

Question 49Navigation, radio aids and flight planning

With a VOR selected and the OBS set to a course that puts the CDI centred with a TO flag, the aircraft is:

  1. On the selected radial, going away from the station if it continues that heading without wind
  2. On a bearing to the station along the selected course (TO), so that course will take it toward the VOR (wind permitting) Correct answer.
  3. Always overhead the VOR
  4. On an ILS glideslope

Explanation

TO means the selected OBS course is a course to the station. FROM means you are on that radial going away. Reverse sensing occurs if you fly the reciprocal heading with the OBS still set to the inbound course. Identify the VOR before tracking.

Question 50Navigation, radio aids and flight planning

A standard ILS provides:

  1. Azimuth from an NDB only, with no descent information
  2. Lateral guidance from the localizer and vertical guidance from the glideslope (plus markers/DME as published) Correct answer.
  3. GPS RAIM in place of the localizer
  4. A circling MDA only, never a DA

Explanation

ILS is a precision approach: localizer for azimuth, glideslope for a published descent angle. Marker beacons or DME identify ranges. Minima are a DA/DH on the chart - do not invent a Colombian DA. GNSS may overlay or replace ILS only when a published RNP/LPV procedure is used.