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Question 119-1 : At reference or see flight planning manual mrjt 1 figure 4.4.planning a flight from paris charles de gaulle to london heathrow for a twin jet aeroplane.preplanning.dry operating mass dom 34 000 kg.traffic load 13 000 kg..the holding is planned at 1 500 ft above alternate elevation. the alternate ? [ Training professional ]
48 125 kg.
Landing mass at alternate = dry operating mass + traffic load + final reserve fuel.notice you must land at destination or alternate when pre planning with final reserve fuel in your tanks.. /com en/com033 129.jpg..34000+13000 = 47000 kg.interpolate from the table 2280 + 2220 /2 = 2250 kg/h.for 30 minutes = 1125 kg...47000 + 1125 = 48125 kg.
Question 119-2 : Given.distance from departure to destination 1385 nm.gs out 480 kt.gs home 360 kt.what is the time of the pet from the departure point ?
74 min.
.ground speed out = 480 kt..ground speed home = 360 kt....pet = distance x gsh / gso + gsh..pet = 1385 x 360 / 480 + 360..pet = 498600 / 840 = 593 nm....time of the pet from the departure point..593 / 480 = 1.23 h..1.23 x 60 = 74 minutes.
Question 119-3 : Given.distance from departure to destination 256 nm.gs out 160 kt.gs home 110 kt.what is the distance of the pet from the departure point ?
104 nm.
.ground speed out gso = 160 kt.ground speed home gsh = 110 kt..distance to pet = distance x gsh / gso + gsh.distance to pet = 256 x 110 / 160 + 110.distance to pet = 28160 / 270 = 104 nm.
Question 119-4 : Given.distance from departure to destination 480 nm.safe endurance 5 h.true track 315°.w/v 100/20.tas 115 kt.what is the distance of the psr from the departure point ?
280 nm.
...start by searching outbound ground speed on nav computer.set 115 kt under center dot, true track 315° to true index. put wind direction 100° under the red compass rose, under 20 kt you read a left drift of 4°. now drift is always measured from heading to track, turn to set true heading 319° 315° + 4° left drift under index, you now read your ground speed out of 130 kt...repeat the operation to find homeward ground speed...outbound gs 130 kt.homeward gs 99 kt..point of safe return psr = endurance x homeward gs / outbound gs + homeward gs.point of safe return psr = 5 x 99 / 130 + 99.point of safe return psr = 495 / 229.point of safe return psr = 2.16 h..0.16 x 60 = 10 minutes.10 + 120 minutes = 130 min..distance of the psr from the departure point at a speed of 132 kt.130 min x 130/60 = 281.6 nm.
Question 119-5 : Given.distance from departure to destination 150 nm.true track 142°..wind 200°/15kt.tas 132 kt.what is the distance of the pet from the departure point ?
79 nm
...under index, set true track 142°, centre dot on tas, 132 kt, with the rotative scale, set wind 200°/15 kt, you find a left drift of 5°..now, drift is always measured from heading to track.turn to set true heading 147° 142° + 5° left drift under index, you now read your ground speed out of 124 kt.. /com en/com033 142.jpg..proceed in the same way to find the ground speed home of 139 kt.. right drift of 5°, true heading of 317°...ground speed out gso = 124 kt.ground speed home gsh = 139 kt..distance to pet = distance x gsh / gso + gsh.distance to pet = 150 x 139 / 124 + 139.distance to pet = 20850 / 263 = 79.2 nm.
Question 119-6 : A metar reads.1430z 35002kt 7000 skc 21/03 q1024 =.which of the following information is contained in this metar ?
Question 119-8 : Given.maximum allowable take off mass 64 400 kg.maximum landing mass 56200 kg.maximum zero fuel mass 53 000 kg.dry operating mass 35 500 kg.estimated load 14 500 kg.estimated trip fuel 4 900 kg.minimum take off fuel 7 400 kg..find maximum additional load ?
3 000 kg.
. /com en/com033 162.jpg..we are able to add 3000 kg beore reaching our first limitation which comes from the maximum zero fuel mass 17500 14500 = 3000 kg.
Question 119-9 : What mean temperature °c is likely on a course of 360° t from 40°n to 50°n at 040°e .. err a 033 167 ?
Mean temperature 47°c.
. /com en/com033 167.jpg..temperatures are negative unless prefixed by ps. 46 + 47 + 47 + 48 + 49 /5 = 47.4°c.
Question 119-10 : For flight planning purposes the landing mass at alternate is taken as ?
Zero fuel mass plus final reserve fuel and contingency fuel.
Planned landing mass at alternate = dom + traffic load + final reserve fuel + contingency..if everything goes to plan on the sector, you won't use contingency fuel, so landing mass at alternate will include final reserve fuel + contingency fuel.
Question 119-11 : Given.distance from departure to destination 220 nm.true track 175°.wind 220/10 kt.tas 135 kt.what is the distance of the pet from the departure point ?
116 nm
...under index, set true track 175°, centre dot on tas, 135 kt, with the rotative scale, set wind 220°/10 kt, you find a left drift of 7°..now, drift is always measured from heading to track.turn to set true heading 182° 175° + 7° left drift under index, you now read your ground speed out of 119 kt.. /com en/com033 169.jpg..proceed in the same way to find the ground speed home of 148 kt.. right drift of 5°, true heading of 350°...ground speed out gso = 119 kt.ground speed home gsh = 148 kt..distance to pet = distance x gsh / gso + gsh.distance to pet = 220 x 148 / 120 + 148.distance to pet = 32780 / 268 = 121 nm closest answer is 116 nm.
Question 119-12 : At references or see flight planning manual mrjt 1 figure 4.2 and figure 4.5.3.2. given.estimated take off mass 57000 kg.ground distance 150 nm.temperature isa 10°c.cruise at.74 mach.find cruise altitude and expected true air speed.. err a 033 178 ?
25000 ft, 435 kt
. /com en/com033 178.jpg..temperature isa 10°c.445 10 = 435 kt.
Question 119-13 : Given.distance from departure to destination 950 nm.gs out 275 kt.gs home 225 kt.what is the time of the pet from the departure point ?
93 min.
.distance to pet = distance x gsh / gso + gsh.distance to pet = 950 x 225 / 275 + 225.distance to pet = 213750 / 500 = 427.5 nm...time of the pet from the departure point.427.5 nm / 275 = 1.55 h.1.55 x 60 min = 93 minutes.
Question 119-14 : Given.distance from departure to destination 950 nm.safe endurance 3,5 h.tas 360 kt.ground speed out 320 kt.ground speed home 400 kt.what is the distance and time of the psr from the departure point ?
Distance 622 nm time 117 min
Question 119-15 : Given.distance from departure to destination 1000 nm.safe endurance 4 h.tas 500 kt.ground speed out 550 kt.ground speed home 450 kt.what is the distance of the psr from the departure point ?
990 nm.
.ground speed out 550 kt..ground speed home 450 kt....point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..point of safe return psr = 4 x 450 / 550 + 450..point of safe return psr = 1800 / 1000..point of safe return psr = 1.8 h....0.8 x 60 = 48 minutes..48 + 60 minutes = 108 min....distance of the psr from the departure point at a speed of 550 kt..108 min x 550/60 = 990 nm.
Question 119-16 : Given.distance from departure to destination 2200 nm.true track 150°..wind 330°/50 kt.tas 460 kt.what is the distance and time of the pet from the departure point ?
Distance 980 nm time 115 min
.track 150°, wind from 330°, it's a tailwind of 50 kt..ground speed out gso = 460 + 50 = 510 kt....return track 330°, wind from 330°, it's a headwind of 50 kt..ground speed home gsh = 460 50 = 410 kt....distance to pet = distance x gsh / gso + gsh..distance to pet = 2200 x 410 / 510 + 410..distance to pet = 902000 / 920 = 980 nm.....time of the pet from the departure point..980 nm / 510 = 1.92 h..1.92 x 60 min = 115 minutes.
Question 119-17 : Route manual chart nap..the initial true course from a 64°n006°e to c 62°n020°w is.. err a 033 202 ?
271°.
. /com en/com033 202.jpg..put your protractor on a, align it with the true north, you find an initial true course of 271°..departure is almost superposed on the 64° parallel... maercin.i couldn't found those points on the map, because there is written that a is on 006e, so its impossible to find in this map..nevertheless you may also substract 64 62 = 2 degresses. so, if we multiply it by 60nm there is 120nm in vertical range. if we use formula for distance = 60nm x cos mean longitude x g difference in latitude we have 708nm. so now, we take usual calculator, input arc tg from 120/708 = 9 deg. and we know, that track from a to b is 270 9 = 261..after that we count convertion angle = 1/2 x sin mean latitude x difference in latitude and its equal to 10..so we add 10 to 261 and here it is 271.i know that this is a little bit more difficult than reading from the map, but on the other hand if you wasn't passed gen nav and flight planning its possible to resolve all maps questions without even look on them.
Question 119-18 : Which best describes the maximum intensity of icing, if any, at fl160 in the vicinity of berlin 53° n013°e .. err a 033 204 ?
Moderate.
. 629..
Question 119-19 : At reference or see flight planning manual sep 1 figure 2.1. given. fl 75.oat +5°c.during climb average head wind component 20 kt..take off from msl with the initial mass of 3650 lbs..find still air distance nam and ground distance nm using the graph 'time, fuel, distance to climb'... err a 033 212 ?
18 nam. 15 nm.
. /com en/com033 212.jpg..time to climb 9 minutes..distance to climb 18 nam...with a headwind, ground distance to climb will be lower than air distance to climb to fl75..during 9 minutes, 20 kt of wind will reduce our ground distance by 20 kt x 9/60 = 3 nm...18 nam 3 nm = 15 nm.
Question 119-20 : Given.distance from departure to destination 285 nm.true track 348°.wind 280°/25 kt.tas 128 kt.what is the distance of the pet from the departure point ?
154 nm.
Img /com en/com033 213a.jpg.. /com en/com033 213b.jpg..ground speed out = 117 kt..proceed the same way to find ground speed home 136 kt..distance to pet = d x gsh / gso + gsh.distance to pet = 285 x 136 / 117 + 136.distance to pet = 38760 / 253 = 153.2 nm.
Question 119-21 : Given.distance from departure to destination 435 nm.gs out 110 kt.gs home 130 kt.what is the distance of the pet from the departure point ?
236 nm
Question 119-22 : The approximate mean wind component kt along true course 180° from 50°n to 40°n at 005° w is.. err a 033 225 ?
Tail wind 55 kt.
. /com en/com033 225.jpg..wind is coming from 320°, with an average speed of 70 kt..tailwind component = 70 kt x cos angle between the wind and the course..tailwind component = 70 kt x cos 40°.tailwind = 54 kt.
Question 119-23 : The wind direction and velocity °/kt at 40°n 040°e is .. err a 033 233 ?
330/75.
Img /com en/com033 233.jpg..wind is coming from 330°, with a speed of 75 kt.
Question 119-24 : In the vicinity of shannon 52° n009°w the tropopause is at about.. err a 033 236 ?
Fl 360.
. /com en/com033 236.jpg..
Question 119-25 : At reference or see flight planning manual mrjt 1 figure 4.5.1..given.brake release mass 57 500 kg, temperature isa 10°c.head wind component 16 kt.initial fl 280.find still air distance nam and ground distance nm for the climb.. err a 033 237 ?
62 nam, 59 nm.
. /com en/com033 237.jpg..we are just below 58000 kg, thus 62 nam is a good choice..with a headwind component, our ground distance will be less than our air distance, 59 nm sounds good.
Question 119-26 : Route manual chart nap.the average magnetic course from c 62°n020°w to b 58°n004°e is .. err a 033 240 ?
119°.
.to find the average magnetic course from c to b, with a protractor, between the two points, you will find an average true track of 109°. then, you must add the 10°w magnetic variation we have right in the middle of those two points..109° + 10° = 119°... gomis01.in a past question asks the true course from c to b and the answer was 098º, if right now asks about the magnetic course, i am agree about the variation, but i think that it will be 098+10 variation = 108º..the answer more near is 109º. please tell me if i wrong. thank you....the other question asks for the initial true course from c to b, not the average true course.
Question 119-27 : Given.distance from departure to destination 360 nm.safe endurance 4.5 h.true track 345°.w/v 260/30.tas 140 kt.what is the distance of the psr from the departure point ?
308 nm
...under index, set true track 345°, centre dot on tas, 140 kt, with the rotative scale, set wind 260°/30 kt, you find a right drift of 12°...now, drift is always measured from heading to track.turn to set true heading 333° 345° 12° right drift under index, you now read your ground speed out of 135 kt..proceed in the same way to find the ground speed home of 141 kt.. left drift of 9°, true heading of 174°...ground speed out gso = 135 kt.ground speed home gsh = 141 kt..apply the psr formula.psr = time x gs out x gs home / gs out + gs home.psr = 4.5 x 135 x 141 / 135 + 141.psr = 310 nm...this is a 4 points question at the exam...mathematical calculation on this kind of exercise is valid only for one right angled triangle, which is not the case here. only the computer enables you to find the good answer.
Question 119-28 : What mean temperature °c is likely on a true course of 270° from 025°e to 010°e at 45°n .. err a 033 246 ?
Mean temperature is 50°c.
. /com en/com033 246.jpg..temperatures are negative unless prefixed by ps. 53 + 51 + 50 + 47 /4 = 50.25°c.
Question 119-29 : At reference or see flight planning manual mrjt 1 figure 4.3.3c. given.ground distance to destination aerodrome 1600 nm. headwind component 50 kt.fl 330.cruise speed 0.78 mach. isa + 20°c. estimated landing weight 55000 kg.find simplified flight planning to determine estimated trip fuel and trip ?
12 400 kg, 03h 55 min.
Explanation.. /com en/com033 1017.jpg..
Question 119-30 : Given.distance from departure to destination 260 nm.safe endurance 4.1 h.true track 150°.w/v 100/30.tas 110 kt.what is the distance of the psr from the departure point ?
213 nm.
...under index, set true track 150°, centre dot on tas, 110 kt, with the rotative scale, set wind 100°/30 kt, you find a right drift of 9°...now, drift is always measured from heading to track.turn to set true heading 141° 150° 9° right drift under index, you now read your ground speed out of 90 kt..proceed in the same way to find the ground speed home of 127 kt.. left drift of 10°, true heading of 340°...ground speed out gso = 90 kt.ground speed home gsh = 127 kt..apply the psr formula.psr = time x gs out x gs home / gs out + gs home.psr = 4.1 x 90 x 127 / 90 + 127.psr = 215 nm...this is a 4 points question at the exam...mathematical calculation on this kind of exercise is valid only for one right angled triangle, which is not the case here. only the computer enables you to find the good answer.
Question 119-31 : Route manual chart nap. the average true course from c 62°n020°w to b 58°n004°e is.. err a 033 262 ?
109°.
. /com en/com033 262.jpg..we are looking for average true course .with you protractor aligned on true north, between c and b, you will find a true course of 109°.
Question 119-32 : Given.distance from departure to destination 2450 nm.safe endurance 7.5 h.tas 410 kt.ground speed out 360 kt.ground speed home 460 kt.what is the time of the psr from the departure point ?
252 min
.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..ground speed out = 360 kt.ground speed home = 460 kt..point of safe return psr = 7.5 x 460 / 360 + 460.point of safe return psr = 3450 / 820.point of safe return psr = 4.20 h..4.20 x 60 = 252 minutes.
Question 119-33 : Which describes the worst hazard, if any, that could be associated with the type of feature at 38°n 015°e . 2240 ?
Engine flame out and windscreen damage.
At 38°n 015°e, we have the etna volcano. 1242. read the story of british airways flight nine, you will find and understand the answer.on 24 june 1982, a 747 flew into a cloud of volcanic ash thrown up by the eruption of mount galunggung south east of jakarta, indonesia , resulting in the failure of all four engines..at approximately 13 42 utc 20 42 jakarta time , engine number four began surging and soon flamed out. the flight crew immediately performed the engine shutdown drill, quickly cutting off fuel supply and arming the fire extinguishers. less than a minute later, at 13 43 utc 20 43 jakarta time , engine two surged and flamed out. within seconds, and almost simultaneously, engines one and three flamed out prompting the flight engineer to exclaim, 'i don't believe it all four engines have failed '...the flight crew quickly determined that the aircraft was capable of gliding for 23 minutes and covering 91 nautical miles 169 km from its flight level of 37,000 feet...at 13 44 utc 20 44 jakarta time , the senior first officer declared an emergency to the local air traffic control authority, stating that all four engines had failed. however, jakarta area control misunderstood the message, interpreting the call as meaning that only engine number four had shut down. it was only after a nearby garuda indonesia flight relayed the message to air traffic control that it was understood. despite the crew 'squawking' the emergency transponder setting of 7700, the aeroplane could not be located by air traffic control on their radar screens...due to the high indonesian mountains on the south coast of the island of java, an altitude of at least 11,500 feet was required to cross the coast safely. the crew decided that if the aircraft was unable to maintain altitude by the time they reached 12,000 feet they would turn back out to sea and attempt to ditch into the indian ocean. the crew began the engine restart drills, despite being well above the recommended maximum engine in flight start envelope altitude of 28,000 feet. the attempts failed...at 13,500 feet, they were approaching the altitude at which they would have to turn over the ocean and attempt a risky ditching. although there were guidelines for the procedure, no one had ever tried it in a boeing 747, nor has anyone since. as they performed the engine restart procedure, engine number four started, and at 13 56 utc 20 56 jakarta time , the captain used its power to reduce the rate of descent. shortly thereafter, engine three restarted, allowing him to climb slowly. shortly after that, engines one and two successfully restarted as well..the crew subsequently requested and expedited an increase in altitude to 11,500 feet in order to clear the high mountains of indonesia... as flight 9 approached jakarta, the crew found it difficult to see anything through the windscreen , and had to make the approach almost entirely on instruments, despite reports of good visibility..although the runway lights could be made out through a small strip of the windscreen, the landing lights on the aircraft seemed to be inoperable. after landing, the flight crew found it impossible to taxi, due to glare from apron floodlights which made the already sandblasted windscreen opaque...aftermath it was found that the b747's problems had been caused by flying through a cloud of volcanic ash from the eruption of mount galunggung. because the ash cloud was dry, it did not show up on the weather radar, which is designed to detect the moisture in clouds. the cloud sandblasted the windscreen and landing light covers and clogged the engines. as the ash entered the engines, it melted in the combustion chambers and adhered to the inside of the power plant. as the engine cooled from not running and as the aircraft descended out of the ash cloud, the molten ash solidified and enough broke off to allow air to flow smoothly through the engine allowing a successful restart. the engines had enough electrical power to restart because one generator and the onboard batteries were still operating.
Question 119-34 : Route manual chart nap.the distance nm from c 62°n020°w to b 58°n004°e is .. err a 033 266 ?
760 nm.
.report the track distance along latitude 60°n average latitude between a and c...you will count a little bit more of 25° separation between a and c..25° x 60 nm x cos 60° = 750 nm... /com en/com033 266.jpg..you can also use meridian 1° = 60 nm....take care if you simply calculate latitude distance between 020°w and 004°e, you will have 24° of latitude... but points are not on the same longitude, this is the reason that we need to report the track distance along latitude 60°n, you will now count 25° of latitude on the mid latitude.
Question 119-35 : The planned flight is over a distance of 440 nm.based on the wind charts at altitude the following components are found.fl50 30kt.fl100 50kt.fl180 70kt.the operations manual in appendix details the aircraft's performances.which of the following flight levels fl gives the best range performance.. ?
Fl 180.
.you have to extrapolate data....fl50 tas= 194kt ho y fuel flow= 206 l/hr gs= 164kt flight time= 2.6h fuel burn= 552 l.....fl100 tas= 200kt ho y fuel flow= 192 l/hr gs= 150kt flight time= 2.9h fuel burn= 563.2 l.....fl180 tas= 216kt ho y fuel flow= 163 l/hr gs= 146kt flight time= 3.01h fuel burn= 491 l.....flight level 180 gives the best range performance lowest comsumption.
Question 119-36 : At reference or see flight planning manual mrjt 1 figure 4.3.5.for a flight of 2800 ground nautical miles the following apply. head wind component 20 kt.temperature isa +15°c.brake release mass 64700 kg.the a trip fuel, and b trip time respectively are .. err a 033 268 ?
A 17000 kg b 6h 45 min
. /com en/com033 268.jpg..you always need to go first to the ref line, and then, apply the condition mass, temperature, wind.
Question 119-37 : Given.distance from departure to destination 1950 nm.gs out 400 kt.gs home 300 kt.what is the time of the pet from the departure point ?
Question 119-38 : Given.distance from departure to destination 95 nm.true track 105.wind 060/15.tas 140 kt.what is the distance of the pet from the departure point ?
51 nm.
Question 119-39 : Given.distance from departure to destination 150 nm..true track 020°..wind 180/30.tas 130 kt.what is the distance of the pet from the departure point ?
59 nm.
.find the wind correction angle and the ground speed on the computer.. computer solution.a set true track to true index..b turn the indicator to the wind direction, in this case using the black azeimuth graduation the angle being upwind counting anti clockwise..c shift the speed arc corresponding to the true air speed so as to coincide with the wind speed on the indicator..d read the wind correction at the same place. read the ground speed under the center bore from the scale on the axis slide...gs out = 157 kt..gs home = 102 kt...distance to pet = d x h / o + h.distance to pet = 150 x 102 / 157 + 102 = 59 nm.
Question 119-40 : The surface wind velocity °/kt at paris/charles de gaulle at 1330 utc was.. err a 033 281 ?
270/04.
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