A free Premium account on the FCL.055 website! Read here
Sign up to unlock all our services and 15164 corrected and explained questions.


Question 120-1 : From which of the following would you expect to find the dates and times when temporary danger areas are active ? [ Preparation civilian ]

Notam and aip

exemple 220 Notam and aip

Question 120-2 : What is the earliest time utc , if any, that thunderstorms are forecast for tunis/carthage .. err a 033 284 ?

1800 utc.

If any, the earliest time is 1800 utc..long taf terminal area forecast released at 1020 utc, specify a tempo from 1800 to 0200 next day, with shra or tsshra... /com en/com033 284.jpg.. exemple 224 1800 utc.

Question 120-3 : A sector distance is 450 nm long..the tas is 460 kt..the wind component is 50 kt tailwind..what is the still air distance ?

406 nautical air miles nam

.with a tailwind of 50 kt, your ground speed will be 460 kt + 50 kt = 510 kt...450 nm at 510 kt = 450/510 = 0.882 h...nautical air miles = 460 x 0.882 = 406 nam...you can use the following formula.nam = ngm x tas/gs.nam = 450 x 460/510 = 406 nam. exemple 228 406 nautical air miles (nam)

Question 120-4 : Find the distance to the point of safe return psr..given.maximum useable fuel 15000 kg.minimum reserve fuel 3500 kg.outbound tas 425 kt.head wind component 30 kt.fuel flow 2150 kg/h.return tas 430 kt.tailwind component 20 kt.fuel flow 2150 kg/h ?

1125 nm.

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..outbound gs = 425 30 = 395 kt.homeward gs = 430 + 20 = 450 kt.endurance = 15000 3500 / 2150 = 5.3488 h..point of safe return psr = 5.3488 x 450 / 395 + 450.point of safe return psr = 5.3488 x 450 / 845.point of safe return psr = 2.8484 h..2.8484 h x 395 kt = 1125 nm..we are talking about a point of safe return, a decision in case of our destination is finally unreachable weather issue for example and legally we need to return to our departure airport with our minimum reserve fuel. exemple 232 1125 nm.

Question 120-5 : What is the temperature deviation °c from isa over 50° n 010°e .. err a 033 295 ?

Deviation is 10°.

.at fl300, outside air temperature is 55°c... /com en/com033 295.jpg..at fl300, standard temperature is.15°c 30 x 2°c = 45°c..deviation is 10°c isa 10°c. exemple 236 Deviation is -10°.

Question 120-6 : Given.distance from departure to destination 2000 nm.safe endurance 5 h.tas 500 kt.ground speed out 480 kt.ground speed home 520 kt.what is the distance of the psr from the departure point ?

1248 nm

exemple 240 1248 nm

Question 120-7 : At reference or see flight planning manual mrjt 1 figure 4.3.1c..for a flight of 1900 ground nautical miles the following apply. head wind component 10 kt.temperature isa 5°c.trip fuel available 15000 kg.landing mass 50000kg.what is the minimum cruise level pressure altitude which may be planned .. ?

17000 ft.

. /com en/com033 308.jpg.above 'landing weight' on the right part of the graph , you have to follow an imaginary line between the the dashed line and the continuous line continuous line is for a pressure altitude flight of 37000 ft, dashed line is for pressure altitude flight of 10000 ft. exemple 244 17000 ft.

Question 120-8 : At reference or see flight planning manual mrjt 1 figure 4.4.holding planning...the fuel required for 30 minutes holding, in a racetrack pattern, at pressure altitude 1500 ft, mean gross mass 45 000 kg, is .. err a 033 309 ?

1090 kg.

.you have to interpolate. /com en/com033 309.jpg.. 2220 + 2140 /2 = 2180 kg for one hour..2180/2 = 1090 kg for 30 minutes. exemple 248 1090 kg.

Question 120-9 : The wind °/kt at 60° n015° w is.. err a 033 315 ?

300/60.

. /com en/com033 315.jpg..50 + 10 = 60 kt. exemple 252 300/60.

Question 120-10 : Route manual chart nap..the initial true course from c 62°n020°w to b 58°n004°e is.. err a 033 319 ?

098°.

Img /com en/com033 319.jpg..true course 098°. exemple 256 098°.

Question 120-11 : Find the distance from waypoint 3 wpt 3 to the critical point..given.distance from wpt 3 to wpt 4 = 750 nm.tas out 430 kt.tas return 425 kt.tailwind component out 30 kt.head wind component return 40 kt ?

342 nm.

Ground speed out = 430 + 30 = 460 kt..ground speed home = 425 40 = 385 kt....distance to pet = d x gsh / gso + gsh..distance to pet = 750 x 385 / 460 + 385..distance to pet = 288750 / 845 = 341.7 nm. exemple 260 342 nm.

Question 120-12 : At reference or see flight planning manual mrjt 1 figure 4.5.1.given.brake release mass 57500 kg.initial fl 280.average temperature during climb isa 10°c.aaverage head wind component 18 kt.find climb time for enroute climb 280/.74.. err a 033 325 ?

13 minutes.

. /com en/com033 356.jpg..at 57500 kg we are closer to 58000 kg than 56000 kg, 13 minutes is our answer, no need to interpolate. exemple 264 13 minutes.

Question 120-13 : Given.distance from departure to destination 180 nm.true track 310.wind 010°/20 kt.tas 115 kt.what is the distance of the pet from the departure point ?

98 nm.

Ducksherminator.with my calculations, using crp 5w, i find..gs o = 110kts..gs h = 120kts....pet=dxh/ o+h =93,91nm, which is much closer to 92nm than 98nm......you need to be a little more accurate with the wiz wheel..you will get 104 kt for the outbound leg and 124 kt for the home leg...ground speed out gso = 104 kt..ground speed home gsh = 124 kt...distance to pet = distance x gsh / gso + gsh..distance to pet = 180 x 124 / 104 + 124..distance to pet = 22320 / 228 = 97.89 nm. exemple 268 98 nm.

Question 120-14 : The surface system over vienna 48°n016°e is a.. err a 033 330 ?

Cold front moving east.

. /com en/com033 330.jpg.cold front moving east at 10 km/h. exemple 272 Cold front moving east.

Question 120-15 : Given.dry operating mass = 33510 kg.traffic load= 7600 kg.trip fuel = 2040 kg.final reserve fuel= 983 kg.alternate fuel= 1100 kg.contingency fuel= 5% of trip fuel.which of the listed estimated masses is correct ?

Estimated landing mass at destination= 43295 kg.

exemple 276 Estimated landing mass at destination= 43295 kg.

Question 120-16 : The lowest cloud conditions oktas/ft at bordeaux/merignac at 1330 utc were.. err a 033 340 ?

1 to 2 at 3000 ft.

. /com en/com033 340.jpg..the cloud cover classification is.few 1/8 to 2/8 cloud coverage..sct scattered 3/8 to 4/8 cloud coverage..bkn broken 5/8 to 7/8 cloud coverage..ovc overcast 8/8...030 means 3000 ft... dizertie.tempo indicate a could layer of scatered at 500ft...do you see an answer 3 to 4 at 500 ft .no, and there is a reason tempo doesn't refer to the actual conditions. the question asks for the actual conditions, not the forecasted conditions. exemple 280 1 to 2 at 3000 ft.

Question 120-17 : At reference or see flight planning manual mrjt 1 figure 4.2. find the short distance cruise altitude for the twin jet aeroplane. given.brake release mass=45000 kg.temperature=isa + 20°c.trip distance=50 nautical air miles nam.. err a 033 343 ?

10000 ft

exemple 284 10000 ft

Question 120-18 : The maximum wind velocity °/kt shown in the vicinity of munich 48°n 012°e is .. err a 033 346 ?

300/140

. arrows, feathers and pennants.arrows indicate direction. number or pennants and/or feathers correspond to speed...example with a 270°/115 kt wind.. /com en/com033 346a.jpg..pennants correspond to 50 kt..feathers correspond to 10 kt..half feathers correspond to 5 kt... /com en/com033 346b.jpg..munich is below the jet axis. direction of the jet is 300°, there is two pennants and four feathers, so 140 kt..there is a decrease of speed after the two oblic lines highlighted in red. exemple 288 300/140

Question 120-19 : At references or see flight planning manual mrjt 1 paragraph 5.2 and figure 4.5.1. planning an ifr flight from paris to london for a twin jet aeroplane..given.estimated take off mass tom 52000 kg.airport elevation 387 ft.fl 280.w/v 280°/40 kt.isa deviation 10°c.average true course 340°.find ground ?

50 nm.

. /com en/com033 347.jpg..tas 353 kt and air distance 53 nm..on nav computer, set under true index the true course 340°, under the center dot, tas 353 kt. with the rotating scale, set wind 280° and under the wind speed 40 kt, you read a right drift of 6°..it means that we need to fly on a true heading of 340° 6° = 334° to stay on the course..set 334° under true index, you read a ground speed of 332 kt.....to determine the ground distance travelled in the climb, multiply the air distance by the groundspeed and divide by the tas..53 x 332 / 353 = 49.84 nm. exemple 292 50 nm.

Question 120-20 : Flight planning manual mrjt 1 figure 4.5.1. planning an ifr flight from paris charles de gaulle to london heathrow for the twin jet aeroplane..given estimated take off mass tom 52000 kg.airport elevation 387 ft.fl 280.w/v 280°/40 kt.isa deviation 10°c.average true course 340°.find time to the top of ?

11 min

exemple 296 11 min

Question 120-21 : Which describes the maximum intensity of turbulence, if any, forecast for fl260 over toulouse 44°n001°e .. err a 033 350 ?

Severe.

.this chart goes from fl100 up to fl450, embedded cumulonimbus clouds with base below fl100 and top up to fl270 are located in the area enclosed by scalloped lines, over toulouse. /com en/com033 350.jpg..even though we are in cat n°1 aera, showing moderate turbulence symbol, isol emb cb isolated embedded cb always means moderate to severe turbulence. thus, the maximum intensity can be severe. exemple 300 Severe.

Question 120-22 : At reference or see flight planning manual sep 1 figure 2.4.given.aeroplane mass at start up 3663 lbs.aviation gasoline density 6 lbs/gal fuel load 74 gal.take off altitude sea level.headwind 40 kt.cruising altitude 8000 ft.power setting full throttle 2300 rpm 20°c lean of peak egt..calculate the ?

633 nm.

.we are looking for ground range as the answers are in nm and a wind component is given... /com en/com033 351.jpg..on the reference, we find 844 nam..range in nm = nam x gs/tas..tas is 160 kt, ground speed is tas wind = 160 40 = 120 kt..range in nm = 844 x 120/160 = 633 nm... milinoo.how we know, that true air speed is 160 kt please...it's written in the center of the graph 'true airspeed knots'. exemple 304 633 nm.

Question 120-23 : In the vicinity of paris 49°n 003°e the tropopause is at about .. err a 033 352 ?

Fl380.

.200 km west of paris, we have a box showing 400 this is the tropopause height , north of paris near amsterdam tropause height is 350..let's have a look to a jet stream cross section.. /com en/com033 352.jpg..before approaching the jet, tropopause subsides..fl380 is the correct answer at the exam. exemple 308 Fl380.

Question 120-24 : Given.distance from departure to destination 1860 nm.gs out 360 kt.gs home 400 kt.what is the time of the pet from the departure point ?

163 min

exemple 312 163 min

Question 120-25 : Given.distance from departure to destination 3000 nm.safe endurance 8 h.tas 520 kt.ground speed out 600 kt.ground speed home 440 kt.what is the time of the psr from the departure point ?

203 min

exemple 316 203 min

Question 120-26 : At reference or see flight planning manual mrjt 1 figure 4.3.1c...for a flight of 2800 ground nautical miles the following apply.tail wind component 45 kt.temperature isa 10°c.cruise altitude 29000ft.landing mass 55000kg.the a trip fuel b trip time respectively are .. err a 033 371 ?

A 17100kg b 6h 07 min

. /com en/com033 371.jpg..6.13 h = 6h 08 minutes close to the answer. exemple 320 (a) 17100kg (b) 6h 07 min

Question 120-27 : Given.distance from departure to destination 150 nm.safe endurance 2.4 h.true track 250°.w/v 280/15.tas 120 kt.what is the distance of the psr from the departure point ?

142 nm.

.....start by searching outbound ground speed on nav computer.set 120 kt under center dot, true track 250° to true index. put wind direction 280° under the red compass rose, under 15 kt you read a groudspeed of 107 kt.....repeat the opeartion to find homeward ground speed.....outbound gs 107 kt..homeward gs 134 kt....point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..point of safe return psr = 2.4 x 134 / 107 + 134..point of safe return psr = 321.6 / 241..point of safe return psr = 1.33 h....0.33 x 60 = 20 minutes..20 + 60 minutes = 80 min....distance of the psr from the departure point at a speed of 107 kt..80 min x 107/60 = 142.6 nm. exemple 324 142 nm.

Question 120-28 : At reference or see flight planning manual mrjt 1 figure 4.3.1c..for a flight of 2400 ground nautical miles the following apply.tail wind component 25 kt.temperature isa 10°c.cruise altitude 31000 ft.landing mass 52000 kg.the a trip fuel and b trip time respectively are .. err a 033 380 ?

A 14200 kg b 5 h 30 min.

. /com en/com033 380.gif.. exemple 328 (a) 14200 kg (b) 5 h 30 min.

Question 120-29 : Which best describes the maximum intensity of cat, if any, forecast for fl330 over benghazi 32°n 020°e .. err a 033 382 ?

Nil.

. /com en/com033 382.jpg..cat area n°1 extends fl350 to fl450, so no clear air turbulence at fl330. exemple 332 Nil.

Question 120-30 : At reference or see flight planning manual mep 1 figure 3.2.a flight is to be made in a multi engine piston aeroplane mep..the cruising level will be 11000 ft..the outside air temperature at fl is 15°c..the usable fuel is 123 us gallons..the power is set to economic cruise..find the range in nm with ?

752 nm.

. /com en/com033 384.jpg..outside air temperature is 15°c, it is 8°c below isa...reserve fuel is based at 45% power, but for the flight we plan to use economic power 65%. exemple 336 752 nm.

Question 120-31 : Given.distance from departure to destination 180 nm.safe endurance 2,8 h.true track 065.w/v 245/25.tas 100 kt.what is the distance of the psr from the departure point ?

131 nm

exemple 340 131 nm

Question 120-32 : Given.distance from departure to destination 500 nm.gs out 95 kt.gs home 125 kt.what is the distance of the pet from the departure point ?

284 nm

exemple 344 284 nm

Question 120-33 : Which best describes the weather, if any, at lyon/st exupery at 1330 utc .. err a 033 391 ?

Light rain associated with thunderstorms.

. /com en/com033 391.jpg..ts = thunderstorm.ra = rain..indicator ' ' means light, to indicate the intensity of certain phenomena..example from annex 3.+shra = heavy shower of rain.+tssngr = thunderstorm with heavy snow and hail. exemple 348 Light rain associated with thunderstorms.

Question 120-34 : Given.distance x to y 2700 nm.mach number 0.75.temperature 45°c.mean wind component 'on' 10 kt tailwind.mean wind component 'back' 35 kt tailwind.the distance from x to the point of equal time pet between x and y is ?

1386 nm.

...set corresponding mark m kt against outside temperature at flight altitude. read in front of mach number, on the outer scale, the true air speed...example.. 45° and mach 0.75 => tas is 437 kt.... /com en/com033 48.jpg...ground speed out gso = 437 + 10 kt = 447 kt..ground speed home gsh = 437 + 35 kt = 472 kt....distance to pet = distance x gsh / gso + gsh..distance to pet = 2700 x 472 / 447 + 472..distance to pet = 1274400 / 919 = 1386 nm. exemple 352 1386 nm.

Question 120-35 : At reference or see flight planning manual mep1 figure 3.6..a flight is to be made to an airport, pressure altitude 3000 ft, in a multi engine piston aireroplane mep1..the forecast oat for the airport is 1° c..the cruising level will be fl 110, where oat is 10° c..calculate the still air descent ?

20 nm.

. /com en/com033 398.jpg..distance to descend = 29 8 = 21 nm close to the answer. exemple 356 20 nm.

Question 120-36 : Given.distance from departure to destination 2800 nm.true track 140.w/v 140/100.tas 500 kt.what is the distance and time of the pet from the departure point ?

Distance 1680 nm time 252 min

We start on a true track of 140°, we have a 100 kt headwind, thus....ground speed out = 500 kt 100 kt = 400 kt..ground speed home = 500 kt + 100 kt = 600 kt...pet = distance x gsh / gso+ gsh..pet = 2800 x 600 / 400 + 600..pet = 1680000 / 1000 = 1680 nm....time of the pet from the departure point..1680 / 400 = 4.2 h..4.2 x 60 = 252 minutes. exemple 360 Distance: 1680 nm time: 252 min

Question 120-37 : Route manual chart nap..the initial magnetic course from c 62°n020°w to b 58°n004°e is.. err a 033 410 ?

116°.

. /com en/com033 410.jpg..true course is 098° + 18°w magnetic variation = 116°. exemple 364 116°.

Question 120-38 : Find the time to the point of safe return psr. given.maximum useable fuel 15000 kg.minimum reserve fuel 3500 kg.tas out 425 kt.head wind component out 30 kt.tas return 430 kt.tailwind component return 20 kt.average fuel flow 2150 kg/h ?

2 h 51 min

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..outbound gs = 425 30 = 395 kt.homeward gs = 430 + 20 = 450 kt.endurance = 15000 3500 / 2150 = 5.34 h..point of safe return psr = 5.34 x 450 / 395 + 450.point of safe return psr = 5.34 x 450 / 845.point of safe return psr = 2.84 h..2.84 h = 2 h 51 min..we are talking about a point of safe return, a decision in case of our destination is finally unreachable weather issue for example and legally we need to return to our departure airport with our minimum reserve fuel. exemple 368 2 h 51 min

Question 120-39 : At reference or see flight planning manual mrjt 1 figure 4.1. find the optimum altitude for the twin jet aeroplane..given.cruise mass=54000 kg.long range cruise or.74 mach... err a 033 416 ?

34500 ft

Img /com en/com033 416.jpg.. exemple 372 34500 ft

Question 120-40 : Given.distance from departure to destination 3750 nm.safe endurance 9,5 h.true track 360.w/v 360/50.tas 480 kt.what is the distance of the psr from the departure point ?

2255 nm

exemple 376 2255 nm


~

Exclusive rights reserved. Reproduction prohibited under penalty of prosecution.

4759 Free Training Exam