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Colombia Aerocivil airline-transport theory foundations study materials
Colombia Aerocivil airline-transport theory foundations study resources

Colombia Aerocivil airline-transport theory foundations practice questions

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Question 1Aircraft systems and instruments

Transport-category aeroplane main AC generation is typically:

  1. 28 V DC only, with no AC buses
  2. 220 V AC at 50 Hz from engine-driven automotive alternators
  3. 12 V AC at 60 Hz for galley power only
  4. 115 V AC at 400 Hz from IDGs or CSDs Correct answer.

Explanation

Most transport-category electrical systems generate 115 V, 400 Hz AC through integrated drive generators (IDGs) or constant-speed drives. 28 V DC is derived through transformer-rectifier units; 50/60 Hz utility frequencies are not the primary flight-system standard.

Question 2Aircraft systems and instruments

In a typical jet pressurization system, cabin altitude is controlled primarily by:

  1. Varying ram-air scoop area on the wing leading edge
  2. Opening the ram-air turbine to dump cabin pressure
  3. Shutting the APU bleed whenever the cabin is above 8,000 ft
  4. Modulating the outflow valve while conditioned bleed or pack air is supplied Correct answer.

Explanation

Packs or equivalent air-cycle machines supply conditioned air. Cabin altitude and differential are controlled by modulating outflow valves (with safety/dump valves as overpressure protection). RAT extension and APU-bleed scheduling are not the primary cabin-altitude control loop.

Question 3Aircraft systems and instruments

Why does a transport IDG use a constant-speed drive?

  1. So generator frequency stays near 400 Hz while N2/N3 (or engine rpm) varies Correct answer.
  2. To keep EGT constant during take-off
  3. To lock the fan to a fixed Mach number
  4. To replace the need for a battery or TRU

Explanation

AC frequency is proportional to generator rpm. The CSD/IDG holds generator speed essentially constant so frequency remains about 400 Hz across the engine operating range. It does not control EGT, fan Mach, or DC battery/TRU architecture.

Question 4Aircraft systems and instruments

On a magnetic compass in the Northern Hemisphere, acceleration on a due-east heading causes the compass to indicate a turn toward:

  1. South
  2. West, then to settle on east
  3. The nearest VOR radial
  4. North Correct answer.

Explanation

Acceleration/deceleration errors follow ANDS in the Northern Hemisphere: Accelerate North, Decelerate South on east or west headings. Most of Colombia, including Bogotá, lies north of the equator, so ANDS applies. South of the equator (for example Leticia) the sense reverses.

Question 5Aircraft systems and instruments

On a transport EFIS suite, the primary flight display (PFD) normally presents:

  1. Only engine N1, EGT, and fuel flow
  2. Only the FMS legs page and performance predictions
  3. Only hydraulic quantity and brake temperature
  4. Attitude, airspeed, altitude, vertical speed, and heading/track as the flying references Correct answer.

Explanation

The PFD concentrates the flying references: attitude, speed, altitude, vertical speed, and heading or track, often with flight-director cues. Engine parameters belong on EICAS/ECAM; FMS legs are an MCDU/ND function; hydraulics are synoptic/status pages.

Question 6Aircraft systems and instruments

Mach trim on a transport aeroplane is provided mainly to:

  1. Increase flap load-relief speed above VMO
  2. Replace the need for a yaw damper
  3. Keep cabin differential exactly 8.0 psi
  4. Counter the nose-down pitching moment as the centre of pressure moves aft when Mach rises (Mach tuck) Correct answer.

Explanation

As Mach increases, shock formation moves the centre of pressure aft and produces a nose-down moment (Mach tuck). Mach trim automatically compensates, usually by stabilizer. It does not set flap speeds, replace SAS in yaw, or control pressurization.

Question 7Aircraft systems and instruments

Cabin pressure is 800 hPa and ambient pressure is 250 hPa. Using 1 psi = 68.95 hPa, the cabin differential is closest to:

  1. 3.6 psi
  2. 11.6 psi
  3. 14.7 psi
  4. 8.0 psi Correct answer.

Explanation

Differential = (800 − 250) / 68.95 = 550 / 68.95 ≈ 8.0 psi. That is a typical cruise differential on many narrow-body jets. 14.7 psi is sea-level atmospheric pressure, not a cabin ΔP.

Question 8Aircraft systems and instruments

If one engine-driven hydraulic pump fails on a dual/triple-system transport, the power transfer unit (PTU) typically:

  1. Transfers hydraulic fluid from the failed system into the cabin humidifiers
  2. Shuts both engines down automatically
  3. Converts hydraulic pressure into 400 Hz AC as a backup IDG
  4. Uses pressure from an operative system to drive a motor-pump that pressurizes the affected system without mixing fluid Correct answer.

Explanation

A PTU is a hydraulic motor-pump pair: pressure in a healthy system mechanically drives a pump in the other system so users remain powered without sharing fluid (important if one system is leaking). It is not an electrical generator and does not dump fluid into environmental systems.

Question 9Navigation, radio aids and flight planning

Take-off distance available (TODA) is:

  1. TORA plus stopway
  2. LDA plus clearway
  3. TORA plus clearway Correct answer.
  4. ASDA minus TORA

Explanation

TODA = TORA plus clearway, the distance to reach screen height with the remaining climb in obstacle-free air. Stopway extends ASDA for a rejected take-off, not TODA.

Question 10Navigation, radio aids and flight planning

Dry operating mass (DOM) is best defined as:

  1. Maximum structural mass with zero usable fuel
  2. Take-off mass minus reserve fuel only
  3. Basic empty mass minus crew
  4. The mass of the aeroplane ready for a specified operation, excluding usable fuel and traffic load Correct answer.

Explanation

DOM includes the equipped aeroplane plus crew, catering, and pantry as specified, but excludes usable fuel and traffic load (passengers, baggage, cargo). MZFM is a structural limit, not this definition.

Question 11Navigation, radio aids and flight planning

An aeroplane masses 58,000 kg with CG 17.20 m aft of datum. 1,500 kg of cargo is added at 22.40 m. The new CG is closest to:

  1. 17.20 m
  2. 16.80 m
  3. 17.33 m Correct answer.
  4. 18.10 m

Explanation

New mass = 59,500 kg. Moment = 58,000 × 17.20 + 1,500 × 22.40 = 997,600 + 33,600 = 1,031,200 kg·m. CG = 1,031,200 / 59,500 = 17.33 m. The load is aft of the original CG, so the CG moves aft.

Question 12Navigation, radio aids and flight planning

Using TAS (kt) = Mach × 38.94 × √(273 + OAT°C), the TAS at Mach 0.82 and OAT −50°C is closest to:

  1. 390 kt
  2. 447 kt
  3. 477 kt Correct answer.
  4. 541 kt

Explanation

T = 223 K; √223 ≈ 14.93; 38.94 × 14.93 ≈ 581.4 kt (local speed of sound). TAS = 0.82 × 581.4 ≈ 477 kt. Using 661 × Mach without temperature incorrectly yields 542 kt and ignores that speed of sound falls in the cold tropopause.

Question 13Navigation, radio aids and flight planning

DOM is 41,500 kg and MZFM is 62,000 kg. Ignoring take-off and landing limits, maximum traffic load is:

  1. 20,500 kg Correct answer.
  2. 62,000 kg
  3. 103,500 kg
  4. 41,500 kg

Explanation

Maximum traffic load limited by MZFM = MZFM − DOM = 62,000 − 41,500 = 20,500 kg. Traffic load is passengers, baggage, and cargo, not the MZFM itself.

Question 14Navigation, radio aids and flight planning

Bogotá El Dorado (SKBO) sits near 8,360 ft elevation. Compared with a sea-level airport at the same reported temperature, take-off from SKBO typically requires:

  1. A shorter take-off distance because true airspeed is lower
  2. No performance change if QNH is 1013 hPa
  3. Only a landing-distance increase, never a take-off penalty
  4. A longer take-off distance and a lower WAT-limited mass, because TAS for a given IAS is higher in the thinner air Correct answer.

Explanation

High elevation reduces density. For a given IAS, TAS (and groundspeed at unstick) is higher, so the ground run and distance to screen height increase, and climb gradient/WAT limits bite earlier. QNH setting does not restore sea-level density. This is core Colombian jet performance.

Question 15Navigation, radio aids and flight planning

A balanced-field take-off is the condition in which, for the chosen V1:

  1. TODA equals LDA
  2. V1 equals VMO
  3. Stopway length equals clearway length by definition
  4. Accelerate-stop distance required equals one-engine-inoperative take-off distance required Correct answer.

Explanation

Balanced V1 is chosen so ASDR = OEI TODR. That pairing usually minimises the field length needed when ASDA and TODA are comparable. It is not a VMO condition and does not require stopway to equal clearway.

Question 16Navigation, radio aids and flight planning

Take-off mass 72,400 kg, CG 19.10 m. 6,200 kg is burned from tanks at 15.80 m. The CG after the burn is closest to:

  1. 15.80 m
  2. 19.10 m
  3. 18.55 m
  4. 19.41 m Correct answer.

Explanation

Moment after burn = 72,400 × 19.10 − 6,200 × 15.80 = 1,382,840 − 97,960 = 1,284,880 kg·m. Mass = 66,200 kg. CG = 1,284,880 / 66,200 ≈ 19.41 m. Fuel was forward of the CG, so burning it moves the CG aft.

Question 17Navigation, radio aids and flight planning

A runway has TORA 2,900 m, stopway 250 m, clearway 350 m, and no displaced threshold. ASDA and TODA are:

  1. ASDA 2,900 m and TODA 2,900 m
  2. ASDA 3,250 m and TODA 3,150 m
  3. ASDA 3,500 m and TODA 2,900 m
  4. ASDA 3,150 m and TODA 3,250 m Correct answer.

Explanation

ASDA = TORA + stopway = 2,900 + 250 = 3,150 m. TODA = TORA + clearway = 2,900 + 350 = 3,250 m. LDA equals TORA (2,900 m) when there is no displaced threshold. Do not add stopway into TODA or clearway into ASDA.

Question 18Navigation, radio aids and flight planning

For jet cruise control, long-range cruise (LRC) compared with maximum-range cruise (MRC) is typically:

  1. A slower speed than MRC that minimises fuel flow regardless of wind
  2. Always equal to VMO
  3. The speed for holding at 1,500 ft
  4. A slightly higher speed than MRC that sacrifices about 1 percent specific range for better speed stability and schedule keeping Correct answer.

Explanation

MRC maximises nautical miles per kilogram. LRC is a manufacturer speed slightly faster than MRC, typically giving about 99 percent of MRC specific range, and is less sensitive to wind and mass changes. Holding speed is an endurance speed near minimum drag, not LRC.

Question 19Navigation, radio aids and flight planning

RNP, as distinct from RNAV, additionally specifies:

  1. That the aircraft may navigate with no monitoring of actual navigation performance
  2. On-board performance monitoring and alerting of the navigation system Correct answer.
  3. That only NDBs may be used
  4. That GPS must be ignored

Explanation

RNAV is area navigation. RNP is RNAV plus on-board monitoring and alerting (for example ANP versus RNP). That distinction is central to PBN approvals used on long-range and Andes routes.

Question 20Navigation, radio aids and flight planning

FMS position on a modern transport is usually a mixed (blended) solution. If GNSS is lost over the ocean, the crew should expect:

  1. Instant reversion to NDB tracking with no ANP change
  2. ANP to increase as the system relies more on IRS, possibly exceeding RNP and requiring a contingency procedure Correct answer.
  3. Automatic cancellation of the flight plan at the FIR boundary
  4. Cabin masks to drop

Explanation

Without GNSS, the FMS coasts on IRS (and DME-DME if available). Actual navigation performance degrades. If ANP exceeds RNP, the crew must follow the oceanic/PBN contingency (request a different clearance, use IRS tracks, or offset as published).

Question 21Navigation, radio aids and flight planning

Near the magnetic poles, long-range navigation prefers true or grid heading because:

  1. Magnetic variation and dip make the magnetic compass and magnetic heading unreliable Correct answer.
  2. True heading is always 000° at the pole by definition of the aeroplane’s nose
  3. IRS platforms cannot compute true north
  4. RVSM is suspended at all latitudes above 60°

Explanation

Magnetic dip approaches 90° and variation changes rapidly, so magnetic compasses and magnetic-heading references fail. Polar operations use true or grid heading with IRS/GNSS. IRS does compute true north; RVSM is not automatically lost above 60°.

Question 22Navigation, radio aids and flight planning

True heading is 090°, magnetic variation 12°W. Magnetic heading is:

  1. 078°
  2. 090°
  3. 102° Correct answer.
  4. 012°

Explanation

Magnetic heading = true heading + west variation (or minus east). 090° + 12° = 102°. East is least, west is best remains the memory aid. Rapidly changing variation at high latitude is why polar flight uses true/grid instead.

Question 23Navigation, radio aids and flight planning

A 2,000 NM oceanic leg, TAS 500 kt, fuel flow 2,400 kg/h. Fuel burned with a 100 kt tailwind versus a 100 kt headwind differs by:

  1. 0 kg, because TAS is the same
  2. 2,400 kg
  3. 4,000 kg Correct answer.
  4. 8,000 kg

Explanation

Tailwind GS = 600 kt, time = 2,000/600 = 3.333 h, fuel = 8,000 kg. Headwind GS = 400 kt, time = 5.0 h, fuel = 12,000 kg. Difference = 4,000 kg. Wind on a long-range leg dominates fuel more than a small TAS change. This is why jet-stream routing is planned.

Question 24Navigation, radio aids and flight planning

A typical IRS alignment before pushback must be completed:

  1. In ATT mode while taxiing through 20 kt
  2. Stationary, with a valid present position, until the system indicates aligned; moving off early can leave residual heading/position errors Correct answer.
  3. Only after passing 10,000 ft
  4. With GNSS deliberately deselected so the IRS 'learns' wander

Explanation

Alignment needs a stationary platform and a correct gate position (entered or GNSS-fed). Breaking alignment by moving too soon is a classic long-range trap. ATT mode is a degraded attitude-only mode after a navigation failure, not the alignment mode. GNSS aids, rather than being deselected for wander training.

Question 25Human performance and decision making

A steep cockpit authority gradient is most dangerous when:

  1. The monitoring pilot withholds a go-around call because the PIC is a senior inspector Correct answer.
  2. The captain invites challenges to the plan during briefing
  3. Both pilots use standard phraseology and SOPs
  4. The crew completes a sterile-cockpit climb

Explanation

Steep gradients suppress advocacy. Classic accidents show the FO failing to challenge an unsafe path. CRM flattens the operational gradient through SOPs, briefed challenge rules, and captain modelling-not by rank intimidation.

Question 26Human performance and decision making

Confirmation bias after an unstable approach is most likely to appear as:

  1. Calling go-around at the stable-approach gate
  2. Going around when visual references are lost
  3. Interpreting a high, fast path as 'it will work' because the destination is home base and the weather looks acceptable Correct answer.
  4. Asking the other pilot to compute VREF again

Explanation

Confirmation bias is seeking or interpreting cues that support a continued-approach decision already made. Home-base familiarity and 'good enough' weather are classic fuels. TEM treats the unstable approach as a UAS requiring a go-around, not a narrative that it will work.

Question 27Human performance and decision making

A crew selects flaps 3 instead of flaps 2 on a noisy taxiway. They notice before take-off and reset flaps 2. In TEM this is best classified as:

  1. An undesired aircraft state that remained untrapped through take-off
  2. A threat because ATC caused the noise
  3. An error that was trapped before it became a take-off UAS Correct answer.
  4. A successful violation because the take-off would have been legal anyway

Explanation

Wrong flap selection is a crew error. Detecting and correcting it before take-off is a trapped error. If they had begun the take-off with the wrong configuration, that would be an undesired aircraft state. Taxiway noise is a threat that contributed, but the mis-selection itself is the error.

Question 28Meteorology and weather information

In the International Standard Atmosphere, tropopause height and temperature are:

  1. 18,000 ft and −20°C
  2. FL450 and 0°C
  3. 11 km (about 36,090 ft) and −56.5°C Correct answer.
  4. 5 km and −15°C

Explanation

ISA places the tropopause at 11 km (36,089 ft) with temperature −56.5°C, then isothermal in the lower stratosphere. Real tropical tropopauses, including over Colombia, are often higher and colder than ISA.

Question 29Meteorology and weather information

Compared with the polar-front jet, the subtropical jet is typically:

  1. Lower (near 700 hPa) and easterly
  2. A surface sea-breeze
  3. Higher (near 200 hPa / FL390) and still a westerly maximum Correct answer.
  4. Identical in height and latitude all year

Explanation

The subtropical jet sits near the tropical tropopause around 200 hPa, generally poleward of the Hadley cell, and remains a westerly wind maximum. The polar-front jet is usually lower (near 300 hPa) and more mobile with mid-latitude troughs.

Question 30Meteorology and weather information

Over equatorial Colombia, the Intertropical Convergence Zone (ITCZ) is operationally important because it:

  1. Produces a permanent jet-stream core at FL250 with no convection
  2. Focuses deep thunderstorms, icing, hail, and wind shear, often in the afternoon over the Andes and plains Correct answer.
  3. Guarantees CAVOK along the entire airway
  4. Eliminates the need for alternate aerodromes in RAC 121 fuel planning

Explanation

The ITCZ is a belt of converging trade winds and deep, moist convection. Over Colombia it drives seasonal rainfall and daily CB, with severe icing, hail, turbulence, and low-level wind shear. It is not a substitute for alternates or a CAVOK guarantee.

Question 31Meteorology and weather information

At jet cruise near the Andes, which combination most strongly warns a PTL crew of mountain-wave turbulence extending well above the ridge?

  1. Winds are light and variable at all levels
  2. A strong, stable airstream with a component perpendicular to the ridge exists, often marked by lenticular clouds Correct answer.
  3. A shallow surface fog layer with calm winds aloft
  4. A sea-breeze front parallel to the coast with no cross-ridge flow

Explanation

Mountain wave requires sufficiently strong cross-ridge flow and stable stratification. Lenticular or cap clouds are clues, but severe wave turbulence can also exist in clear air at transport cruise levels. Crews should use forecasts, reports, airborne indications and operator procedures rather than assuming the hazard ends at ridge height.

Question 32Meteorology and weather information

Thermal-wind reasoning says that in the Northern Hemisphere, as you climb toward a jet, the wind:

  1. Always becomes easterly regardless of temperature gradients
  2. Increases and backs onto an easterly heading if cold air is to the right of the wind
  3. Increases with height with a component parallel to the isotherms/thickness lines, colder air to the left of the thermal wind Correct answer.
  4. Becomes calm at the tropopause by definition

Explanation

Thermal wind is the vector difference of geostrophic wind between levels and is parallel to thickness contours, with cold air to the left in the Northern Hemisphere. That is why westerly jets sit above strong meridional temperature gradients. The tropopause is a wind maximum region, not a calm layer.

Question 33Meteorology and weather information

The strongest jet core is usually found:

  1. Where the tropopause is flat and temperature gradients are weak
  2. Near a tropopause break/fold on the cyclonic side of a strong thickness gradient Correct answer.
  3. Inside the equatorial easterlies at 700 hPa
  4. At the surface under a high-pressure cell

Explanation

Jet cores intensify where the tropopause folds and the horizontal temperature (thickness) gradient is tight-the same region favoured for CAT. A flat tropopause with weak gradients does not support a strong jet.

Question 34Performance, loading and principles of flight

VMO/MMO limiting is typically IAS-limited at low altitude and Mach-limited at high altitude because:

  1. MMO is always 0.50
  2. Dynamic pressure (q) limits structure at low level, while compressibility/shock effects limit Mach at high level Correct answer.
  3. The pitot system fails above FL200
  4. Flaps must be extended above FL350

Explanation

At low altitude a given IAS already means high TAS and high q, so VMO protects the structure. At high altitude TAS for a given IAS is high and the speed of sound is lower, so the aeroplane meets MMO first. The crossover altitude is where VMO and MMO coincide in TAS.

Question 35Performance, loading and principles of flight

Mach tuck is the nose-down pitching moment that occurs when:

  1. The centre of pressure moves aft as shocks form at high Mach Correct answer.
  2. The CG moves to the forward stop after a fuel transfer
  3. Flaps 40 are selected at VFE
  4. The yaw damper hard-overs

Explanation

Aft movement of the centre of pressure behind the CG produces a nose-down moment. Mach trim and a Mach-trim-compensated stabilizer are the usual protections. A forward CG would be nose-up in trim terms, the opposite of tuck.

Question 36Performance, loading and principles of flight

High-speed Mach buffet is caused by:

  1. Shock-induced flow separation on the wing Correct answer.
  2. Ice on the pitot tube only
  3. A failed cabin outflow valve
  4. Undercarriage unlocked at VLE

Explanation

The shock behind Mcrit can separate the boundary layer, producing buffet. Low-speed buffet is stall-related separation. The two meet at coffin corner. Ice, pressurization, and gear warnings are different system issues.

Question 37Performance, loading and principles of flight

Using Mach = TAS / (38.94 × √(273 + OAT°C)), TAS 490 kt at OAT −40°C corresponds to a Mach number closest to:

  1. 0.66
  2. 0.74
  3. 0.82 Correct answer.
  4. 0.90

Explanation

T = 233 K; √233 ≈ 15.26; speed of sound ≈ 38.94 × 15.26 ≈ 594 kt. Mach = 490 / 594 ≈ 0.82. This is a typical cruise Mach and shows why the same TAS is a higher Mach in colder air.

Question 38Performance, loading and principles of flight

For a jet in the clean configuration, minimum drag (maximum L/D) is the speed for:

  1. Maximum endurance; maximum range is at a slightly higher speed where TAS/drag is maximised Correct answer.
  2. VMO
  3. VMCG
  4. Always the same as VREF

Explanation

Endurance maximises time per kilogram (minimum fuel flow ≈ minimum drag for a turbojet). Range maximises distance per kilogram (specific range ≈ TAS/FF), so the jet range speed is faster than min-drag. VREF is a landing speed, VMCG a ground minimum-control speed, VMO a limit.

Question 39Aircraft 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 40Aircraft 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 41Aircraft 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 42Aircraft 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 43Aircraft 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 44Aircraft 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 45Navigation, 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 46Navigation, 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 47Navigation, 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 48Navigation, 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 49Navigation, 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 50Navigation, 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.