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Colombia Aerocivil instrument-rating theory foundations study materials
Colombia Aerocivil instrument-rating theory foundations study resources

Colombia Aerocivil instrument-rating theory foundations practice questions

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Question 1Aircraft 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 2Aircraft 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 3Aircraft 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 4Aircraft 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 5Aircraft 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 6Aircraft 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 7Aircraft 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 8Aircraft 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 9Aircraft 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 10Navigation, 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 11Navigation, 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 12Navigation, 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 13Navigation, 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 14Navigation, 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 15Navigation, 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 16Navigation, 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 17Navigation, 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 18Navigation, 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 19Navigation, 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 20Navigation, 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.

Question 21Navigation, radio aids and flight planning

Passing directly over a DME station at 6,000 ft AGL, the DME readout is closest to:

  1. 0.0 NM because you are at the station
  2. 1.0 NM (slant range ≈ 6,000 ft / 6,076 ft per NM) Correct answer.
  3. 6.0 NM
  4. 10.0 NM

Explanation

DME measures slant range. Overhead at 6,000 ft AGL is about 1 NM slant (6,000/6,076). At low altitude and long distance, slant and ground range converge; near the station at high AGL they do not. Do not start a DME arc using overhead as 0.0.

Question 22Navigation, radio aids and flight planning

On a 3° ILS, using 300 ft/NM and ignoring TCH, height above threshold at 5.0 NM is about:

  1. 500 ft
  2. 1,000 ft
  3. 1,500 ft Correct answer.
  4. 3,000 ft

Explanation

5.0 × 300 = 1,500 ft above threshold elevation. Geometric 3° is about 1,590 ft (318 ft/NM). Compare with the charted GP altitude at that DME; if you are well below, you are low on the path - a CFIT threat.

Question 23Navigation, radio aids and flight planning

In a hold at 8,000 ft you want a 1:00 inbound. The last inbound took 1:20. The next outbound should be timed about:

  1. 1:20, matching the inbound
  2. 1:00, ignoring the error
  3. 0:40 (subtract the 20-second inbound excess from the previous 1:00 outbound) Correct answer.
  4. 2:00, to ‘average’ the wind

Explanation

If inbound is 20 seconds long, the outbound was too long (or headwind inbound). Shorten the next outbound by about 20 seconds (1:00 − 0:20 = 0:40) so inbound returns to 1:00. Triple-drift on the outbound is a complementary wind method. Do not double the inbound time.

Question 24Human performance and decision making

In a graveyard spiral, a pilot who is spatially disoriented typically:

  1. Sees a centred turn coordinator and a climbing altimeter
  2. Experiences only a pitch illusion with no heading change
  3. Always recovers by increasing bank to 60°
  4. Is in a descending bank, feels ‘level,’ and pulling on the controls tightens the spiral Correct answer.

Explanation

After an unnoticed bank, the aircraft spirals and loses altitude. The vestibular system adapts so the bank feels level. Pulling without rolling wings level increases load factor and descent. Recovery is wings level on the AI/TC first, then pitch to stop the descent, using ASI to avoid a secondary stall.

Question 25Human performance and decision making

Controlled flight into terrain (CFIT) on an instrument approach is most directly reduced by:

  1. Descending below DA/MDA in IMC to ‘take a look’
  2. Using only the ADF regardless of the published procedure
  3. Turning off terrain alerting so it cannot distract
  4. Respecting charted altitudes, GPWS/TAWS alerts, and a missed-approach decision at minima if the required visual references are not in sight Correct answer.

Explanation

CFIT occurs when a fully controlled aircraft hits terrain or obstacles. IFR defences are procedure altitudes, a stabilized approach, TAWS, and a mandatory missed approach if visual references are not acquired at DA/MDA. ‘Ducking under’ the glideslope or MDA in IMC is a classic CFIT path.

Question 26Human performance and decision making

An effective IFR instrument scan in IMC should:

  1. Fixate on a single instrument (for example the localizer) for the entire approach
  2. Ignore the attitude indicator once established on the ILS
  3. Look only outside from take-off until landing
  4. Cross-check attitude with supporting pitch/bank/power instruments, interpret deviations, then control - without omitting altimeter and ASI Correct answer.

Explanation

The radial/T-scan uses the AI as hub and regularly samples ASI, altimeter, VSI, HI/HSI, and TC. Fixation on the localizer while the aircraft descends through minima or into a spiral is a TEM error. Outside visual scan is added only after the required references are in sight.

Question 27Meteorology and weather information

Among common icing encounters, freezing rain is especially hazardous because it:

  1. Only produces light rime that sheds immediately
  2. Cannot exist at Colombian latitudes
  3. Occurs only above FL 300
  4. Implies large supercooled drops that can form clear ice rapidly beyond many ice-protection systems Correct answer.

Explanation

Freezing rain is supercooled liquid that freezes on contact as clear ice, often in a layer below a warm-frontal inversion. Accretion can be faster than boots, TKS, or thermal systems can shed. The IFR defence is avoidance and a prompt 180° or altitude change, not ‘riding it out.’

Question 28Meteorology and weather information

A SIGMET informs IFR crews of:

  1. Routine ATIS for a single aerodrome only
  2. Uncontrolled-airspace VFR frequencies
  3. The text of every Colombian RAC amendment
  4. Significant en-route weather (for example severe icing, severe turbulence, thunderstorms, volcanic ash) affecting a FIR or a specified area Correct answer.

Explanation

SIGMETs are issued for specified hazardous phenomena. AIRMET/GAMET products, where provided, cover less severe but still operationally important conditions. Check SIGMET before and during IFR, especially in convective Andean weather.

Question 29Meteorology and weather information

Low-level wind shear on an ILS is most likely:

  1. In the wake of a thunderstorm, near a front, or with a strong temperature inversion/low-level jet Correct answer.
  2. Only above FL 410 in RVSM airspace
  3. Impossible if the localizer is identified in Morse
  4. Cancelled by setting QNE on short final

Explanation

Shear changes indicated airspeed and glide-path tracking. Microbursts and frontal/inversion shear are classic. If airspeed and GP diverge violently, go around. Identifying the ILS does not detect wind shear; QNE on final would wreck height reference.

Question 30Meteorology and weather information

A METAR includes BKN008. The ceiling is:

  1. 8,000 ft AGL because BKN is never a ceiling
  2. Only a forecast, not an observation
  3. 8 ft AGL
  4. 800 ft AGL (broken is a ceiling layer) Correct answer.

Explanation

METAR cloud heights are hundreds of feet AGL. BKN008 is broken at 800 ft. Ceiling is the lowest BKN or OVC layer (and, in some States, vertical visibility). Compare that ceiling with IFR alternate and approach planning - without inventing a charted DA/MDA.

Question 31Meteorology and weather information

IFR across the Andes with a strong perpendicular wind, downwind of the ridge, you should expect:

  1. Smooth air and a guaranteed tailwind only
  2. Automatic 2,000 ft of extra indicated altitude from the wave
  3. That wave cannot exist because Colombia is equatorial
  4. Mountain-wave turbulence, possible rotor cloud, and large altimeter/airspeed fluctuations; avoid the rotor and respect turbulence PIREPs/SIGMETs Correct answer.

Explanation

Mountain wave and rotors are classic downwind of a ridge with stable air and strong flow. They produce severe turbulence, icing in lenticulars, and altitude excursions. Increase terrain margin, use the 2,000 ft mountainous IFR buffer as a minimum concept, and divert rather than ‘ride’ a rotor.

Question 32Meteorology and weather information

Supercooled large droplets (SLD) in IMC are a cue that:

  1. Ice protection will always keep the airframe clean
  2. The pitot heat should be turned off
  3. Only helicopters are affected
  4. Severe clear icing may form aft of protected surfaces; exit the conditions Correct answer.

Explanation

SLD (including freezing drizzle/rain) can run back and freeze behind boots or heated leading edges. Many light IFR aircraft are not certified for flight in SLD. The procedure is to leave the moisture (climb, descend, or 180°) while the aircraft is still controllable.

Question 33Meteorology and weather information

You are in IMC in icing, airspeed decaying despite adding power, and ice is visible aft of the boots. The priority action is:

  1. Continue to destination because the ILS will melt the ice
  2. Tell ATS you are in severe icing, request an immediate altitude or heading change out of the moisture, and fly the AFM ice-contaminated stall margins Correct answer.
  3. Retract the flaps to increase ice collection on the tail
  4. Descend below MSA in mountainous IMC to find warmer air without a procedure

Explanation

This is an undesired aircraft state: performance loss in icing. TEM says escape the threat (leave cloud/moisture), communicate, and use AFM speeds. An ILS does not de-ice. Blind descent below MSA in the Andes is a CFIT threat even if warmer air exists somewhere below.

Question 34Aircraft 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 35Aircraft 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 36Aircraft 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 37Aircraft 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 38Aircraft 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 39Aircraft 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 40Navigation, 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 41Navigation, 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 42Navigation, 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 43Navigation, 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 44Navigation, 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 45Navigation, 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 46Navigation, 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 47Navigation, 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 48Navigation, 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 49Navigation, 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 50Navigation, 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.