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Question 105-1 : Excluding rvsm an appropriate flight level for ifr flight in accordance with semi circular height rules on a course of 180° m is ? [ Exam pilot ]

Fl100

.the question states excluding rvsm . /com en/com033 1228 jpg.a vfr fligh level is flxx5 or flxx5 .below fl290 in accordance with semi circular rules for a magnetic heading of 180° we need a even level exemple 205 Fl100.

Question 105-2 : Given .distance from departure to destination 400 nm.safe endurance 2 5 h.tas 115 kt.ground speed out 130 kt.ground speed home 105 kt.what is the distance of the psr from the departure point ?

145 nm

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .ground speed out = 130 kt.ground speed home = 105 kt.point of safe return psr = 2 5 x 105 / 130 + 105 .point of safe return psr = 262 5 / 235.point of safe return psr = 1 12 h.distance of the psr from the departure point at a speed of 130 kt .1 12 h x 130 = 145 nm exemple 209 145 nm.

Question 105-3 : Given .distance from departure to destination 300 nm.safe endurance 4 h.tas 110 kt.ground speed out 120 kt.ground speed home 100 kt.what is the distance of the psr from the departure point ?

218 nm

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .ground speed out = 120 kt.ground speed home = 100 kt.point of safe return psr = 4 x 100 / 120 + 100 .point of safe return psr = 400 / 220.point of safe return psr = 1 81 h.distance of the psr from the departure point at a speed of 120 kt .1 81 h x 120 = 218 nm exemple 213 218 nm.

Question 105-4 : What is the mean temperature deviation °c from the isa over 50°n 010°w . err a 033 83 ?

2°c

. /com en/com033 83 jpg.temperature are negative unless prefixed by ps .at fl300 isa temperature is .15°c 2°c x 30 = 45°c .oat is 47°c deviation from isa is 2° exemple 217 -2°c.

Question 105-5 : Given .distance from departure to destination 4630 nm.safe endurance 12 4 h.true track 240.w/v 060/80.tas 530 kt.what is the distance of the psr from the departure point ?

3211 nm

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .ground speed out = 530 80 kt = 450 kt.ground speed home = 530 + 80 kt = 610 kt.point of safe return psr = 12 4 x 610 / 450 + 610 .point of safe return psr = 7564 / 1060.point of safe return psr = 7 135h.distance of the psr from the departure point at a speed of 450 kt .7 135 x 450 = 3211 nm exemple 221 3211 nm.

Question 105-6 : Given .distance from departure to destination 1100 nm.true track 280°.w/v 100/80.tas 440 kt.what is the distance time of the pet from the departure point ?

Distance 450 nm time 52 min

.track 280° wind from 100° it's a tailwind of 80 kt .ground speed out gso = 440 + 80 = 520 kt..track 100° wind from 100° it's a headwind of 80 kt .ground speed home gsh = 440 80 = 360 kt..distance to pet = distance x gsh / gso + gsh .distance to pet = 1100 x 360 / 520 + 360 .distance to pet = 396000 / 880 = 450 nm ..time of the pet from the departure point .450 nm / 520 = 0 865 h.0 865 x 60 min = 52 minutes exemple 225 Distance: 450 nm, time: 52 min.

Question 105-7 : Given .distance from departure to destination 150 nm.safe endurance 3 2 h.tas 90 kt.ground speed out 100 kt.ground speed home 80 kt.what is the distance and time of the psr from the departure point ?

Distance 142 nm time 85 min

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .ground speed out = 100 kt.ground speed home = 80 kt.point of safe return psr = 3 2 x 80 / 100 + 80 .point of safe return psr = 256 / 180.point of safe return psr = 1 42 h.1 42 x 60 = 85 minutes .distance of the psr from the departure point at a speed of 100 kt .85 min x 100/60 = 142 nm exemple 229 Distance: 142 nm, time: 85 min.

Question 105-8 : Given .distance from departure to destination 270 nm.true track 030.w/v 120/35.tas 125 kt.what is the distance and time of the pet from the departure point ?

Distance 135 nm time 68 min

..under index set true track 030° centre dot on tas 125 kt with the rotative scale set wind 120°/35 kt you find a left drift of 15° . /com en/com033 95a jpg.now drift is always measured from heading to track .turn to set true heading 045° 030° + 15° left drift under index you now read your ground speed out of 120 kt. /com en/com033 95b jpg.proceed in the same way to find the ground speed home of 120 kt . right drift of 15° true heading of 195° .ground speed out gso = 120 kt.ground speed home gsh = 120 kt.distance to pet = distance x gsh / gso + gsh .distance to pet = 270 x 120 / 120 + 120 .distance to pet = 32400 / 240 = 135 nm .135 nm at a ground speed out of 120 kt = 135 x 60/120 = 67 5 minutes exemple 233 Distance: 135 nm, time: 68 min.

Question 105-9 : At reference or see flight planning manual sep 1 figure 2 5 .given .fl 75.lean mixture and full throttle 2300 rpm.take off fuel 444 lbs.take off from msl .find .endurance in hours and minutes . err a 033 100 ?

05 hours 12 minutes

. /com en/com033 100 jpg.5 2h ==> 5h + 0 2h 0 2 x 60 = 12 minutes .5 h 12 minutes exemple 237 05 hours 12 minutes.

Question 105-10 : From which of the following would you expect to find facilitation information regarding customs and health formalities ?

Aip


Question 105-11 : An aircraft flies at a tas of 380 kt it flies from a to b and back to a distance ab = 480 nm .when going from a to b it experiences a headwind component = 60 kt the wind remains constant .the duration of the flight will be ?

2 h 35 min

.take care the wind remains constant .headwind from a to b becomes tailwind from b to a .tas is 380 kt from a to b ground speed is 380 60 = 320 kt .480 nm / 320 kt = 1 5 h 1h30 .from b to a ground speed is 380+60 = 440 kt .480 nm / 440 kt = 1 09 h 1h05 .1h30 + 1h05 = 2 h 35 min exemple 245 2 h 35 min.

Question 105-12 : Given .distance from departure to destination 350 nm.true track 320.w/v 350/30.tas 130 kt.what is the distance and time of the pet from the departure point ?

Distance 210 nm time 121 min

..under index set true track 320° centre dot on tas 130 kt with the rotative scale set wind 350°/30 kt you find a left drift of 7° .now drift is always measured from heading to track .turn to set true heading 327° 320° + 7° left drift under index you now read your ground speed out of 104 kt .proceed in the same way to find the ground speed home of 155 kt . right drift of 6° true heading of 134° .ground speed out gso = 104 kt.ground speed home gsh = 155 kt.distance to pet = distance x gsh / gso + gsh .distance to pet = 350 x 155 / 104 + 155 .distance to pet = 54250 / 259 = 210 nm .210 nm at a ground speed out of 104 kt = 210 x 60/104 = 121 minutes

Question 105-13 : Given .distance from departure to destination 250 nm.gs out 130 kt.gs home 100 kt.what is the distance of the pet from the departure point ?

109 nm

Ground speed out gso = 130 kt.ground speed home gsh = 100 kt.distance to pet = distance x gsh / gso + gsh .distance to pet = 250 x 100 / 130 + 100 .distance to pet = 25000 / 230 = 109 nm exemple 253 109 nm.

Question 105-14 : Given .distance from departure to destination 550 nm.endurance 3 6 h.true track 200.w/v 220/15.tas 130 kt.what is the distance of the psr from the departure point ?

231 nm

..under index set true track 200° centre dot on tas 130 kt with the rotative scale set wind 220°/15 kt you find a left drift of 3° .now drift is always measured from heading to track .turn to set true heading 203° 200° + 3° left drift under index you now read your ground speed out of 115 kt.proceed in the same way to find the ground speed home of 142 kt . true track of 020° right drift of 2° true heading of 018° .ground speed out gso = 115 kt.ground speed home gsh = 142 kt.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .point of safe return psr = 3 6 x 142 / 115 + 142 .point of safe return psr = 511 2 / 257.point of safe return psr = 1 99 h.distance of the psr from the departure point at a speed of 115 kt .1 99 x 115 = 229 nm closest answer is 231 nm exemple 257 231 nm.

Question 105-15 : Given .distance from departure to destination 180 nm.endurance 2 h.tas 120 kt.ground speed out 135 kt.ground speed home 105 kt.what is the distance and time of the psr from the departure point ?

Distance 118 nm time 53 min

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .ground speed out = 135 kt.ground speed home = 105 kt.point of safe return psr = 2 x 105 / 135 + 105 .point of safe return psr = 210 / 240.point of safe return psr = 0 875 h.0 875 x 60 = 52 5 minutes.distance of the psr from the departure point at a speed of 135 kt .52 5 min x 135/60 = 118 125 nm exemple 261 Distance: 118 nm, time: 53 min.

Question 105-16 : Given .distance from departure to destination 2500 nm.gs out 540 kt.gs home 470 kt.what is the time of the pet from the departure point ?

129 min

.pet = d x vsr / vsa + vsr.pet = 2500 x 470 / 540 + 470.pet = 1163 nm.1163 / 540 = 2 15 h. 2 x 60 + 0 15 x 60 = 129 minutes exemple 265 129 min

Question 105-17 : Given .distance from departure to destination 875 nm.true track 240 .wind 060/50 kt.tas 500 kt.what is the distance and time of the pet from the departure point ?

Distance 394 nm time 43 min

.true track 240°.wind 060/50 kt.wind is parallel to our course thus .ground speed out 500 + 50 = 550 kt.ground speed home 500 50 = 450 kt.pet = d x gsh / gso + gsh .pet = 875 x 450 / 550 + 450 = 393 75 nm .393 75 / 550 = 0 716 minutes .60 x 0 716 = 43 minutes exemple 269 Distance: 394 nm, time: 43 min.

Question 105-18 : The forecast period covered by the paris/charles de gaulle tafs totals hours . err a 033 122 ?

27h

. /com en/com033 122 jpg.total tafs duration 27h exemple 273 27h.

Question 105-19 : If cas is 190 kts.altitude 9000 ft.temp isa 10°c.true course tc 350°.w/v 320/40.distance from departure to destination is 350 nm.endurance 3 hours.and actual time of departure is 1105 utc .the point of equal time pet is reached at ?

1213 utc

Calculate the outbound and inbound groud speed start first with the wind .40 x cos30 = 34kt .outbound ground speed 215 34 = 181 kt.inbound ground speed 215+34 = 249 kt..point of equal time = 350x249/ 181+249 = 202 67 nm .202 67/181 = 1 12 h .1 12x0 6 = 67 minutes or 1h07 ..11h05+01h07 = 12h33 exemple 277 1213 utc

Question 105-20 : 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 ?

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 exemple 281 48 125 kg.

Question 105-21 : 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 exemple 285 74 min.

Question 105-22 : 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 exemple 289 104 nm.

Question 105-23 : 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 exemple 293 280 nm.

Question 105-24 : 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 exemple 297 79 nm

Question 105-25 : A metar reads .1430z 35002kt 7000 skc 21/03 q1024 =.which of the following information is contained in this metar ?

Temperature/dewpoint

1430 = time 1430 utc .35002kt = wind 350°/02 kt.7000 = visibility 7000 meters.skc = sky clear. 21/03 = temperature/dewpoint .q1024 = qnh 1024 hpa exemple 301 Temperature/dewpoint.

Question 105-26 : The wind °/kt at 40°n 020°w is . err a 033 157 ?

310/40

. /com en/com033 157 jpg.10 + 10 + 10 + 10 = 40 kt exemple 305 310/40.

Question 105-27 : 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 exemple 309 3 000 kg.

Question 105-28 : 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 exemple 313 Mean temperature : -47°c.

Question 105-29 : 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 exemple 317 Zero fuel mass plus final reserve fuel and contingency fuel.

Question 105-30 : 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 exemple 321 116 nm

Question 105-31 : 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 exemple 325 25000 ft, 435 kt

Question 105-32 : 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 exemple 329 93 min.

Question 105-33 : 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

exemple 333 Distance: 622 nm time: 117 min

Question 105-34 : 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 exemple 337 990 nm.

Question 105-35 : 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 exemple 341 Distance: 980 nm time: 115 min

Question 105-36 : 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 exemple 345 271°.

Question 105-37 : 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. exemple 349 Moderate.

Question 105-38 : 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 exemple 353 18 nam. 15 nm.

Question 105-39 : 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 exemple 357 154 nm.

Question 105-40 : 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

exemple 361 236 nm


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