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Answer > reports : At reference or see flight planning manual sep 1 figure 2 5 .given .fl 75.lean ?

Question 91-1 : 05 hours 12 minutes 05 hours 20 minutes 04 hours 42 minutes 05 hours 23 minutes

. /com en/com033 100 jpg..5 2h ==> 5h + 0 2h 0 2 x 60 = 12 minutes .5 h 12 minutes exemple 191 05 hours 12 minutes.05 hours 12 minutes.

From which of the following would you expect to find facilitation information ?

Question 91-2 : Aip nav/rad charts atcc notam

exemple 195 aip.aip.

An aircraft flies at a tas of 380 kt it flies from a to b and back to a ?

Question 91-3 : 2 h 35 min 3 h 00 min 2 h 10 min 2 h 32 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 199 2 h 35 min.2 h 35 min.

Given .distance from departure to destination 350 nm.true track 320.w/v ?

Question 91-4 : Distance 210 nm time 121 min distance 139 nm time 54 min distance 123 nm time 74 min distance 139 nm time 81 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 exemple 203 distance: 210 nm, time: 121 min.distance: 210 nm, time: 121 min.

Given .distance from departure to destination 250 nm.gs out 130 kt.gs home 100 ?

Question 91-5 : 109 nm 141 nm 125 nm 192 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 207 109 nm.109 nm.

Given .distance from departure to destination 550 nm.endurance 3 6 h.true track ?

Question 91-6 : 231 nm 305 nm 116 nm 319 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 211 231 nm.231 nm.

Given .distance from departure to destination 180 nm.endurance 2 h.tas 120 ?

Question 91-7 : Distance 118 nm time 53 min distance 79 nm time 45 min distance 59 nm time 30 min distance 62 nm time 28 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 215 distance: 118 nm, time: 53 min.distance: 118 nm, time: 53 min.

Given .distance from departure to destination 2500 nm.gs out 540 kt.gs home 470 ?

Question 91-8 : 129 min 171 min 28 min 149 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 219 129 min129 min

Given .distance from departure to destination 875 nm.true track 240 .wind ?

Question 91-9 : Distance 394 nm time 43 min distance 481 nm time 64 min distance 716 nm time 78 min distance 438 nm time 53 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 223 distance: 394 nm, time: 43 min.distance: 394 nm, time: 43 min.

The forecast period covered by the paris/charles de gaulle tafs totals hours . ?

Question 91-10 : 27h 9h 18h 20h

. /com en/com033 122 jpg..total tafs duration 27h exemple 227 27h.27h.

If cas is 190 kts.altitude 9000 ft.temp isa 10°c.true course tc 350°.w/v ?

Question 91-11 : 1213 utc 1221 utc 1233 utc 1203 utc

Convert cas to tas on your computer. /com en/com033 1156 jpg..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 231 1213 utc1213 utc

At reference or see flight planning manual mrjt 1 figure 4 4.planning a flight ?

Question 91-12 : 48 125 kg 2 250 kg 48 675 kg 49 250 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 235 48 125 kg.48 125 kg.

Given .distance from departure to destination 1385 nm .gs out 480 kt .gs home ?

Question 91-13 : 74 min 128 min 96 min 50 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 239 74 min.74 min.

Given .distance from departure to destination 256 nm .gs out 160 kt .gs home ?

Question 91-14 : 104 nm 152 nm 128 nm 176 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

Given .distance from departure to destination 480 nm.safe endurance 5 h.true ?

Question 91-15 : 280 nm 205 nm 141 nm 199 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 247 280 nm.280 nm.

Given .distance from departure to destination 150 nm .true track 142° .wind ?

Question 91-16 : 79 nm 71 nm 75 nm 134 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 251 79 nm79 nm

A metar reads .1430z 35002kt 7000 skc 21/03 q1024 =.which of the following ?

Question 91-17 : Temperature/dewpoint runway in use day/month period of validity

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 255 temperature/dewpoint.temperature/dewpoint.

The wind °/kt at 40°n 020°w is. err a 033 157 ?

Question 91-18 : 310/40 334/40 135/40 155/40

. /com en/com033 157 jpg..10 + 10 + 10 + 10 = 40 kt exemple 259 310/40.310/40.

Given .maximum allowable take off mass 64 400 kg.maximum landing mass 56200 ?

Question 91-19 : 3 000 kg 7 000 kg 5 600 kg 4 000 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 263 3 000 kg.3 000 kg.

What mean temperature °c is likely on a course of 360° t from 40°n to 50°n ?

Question 91-20 : Mean temperature 47°c mean temperature 46°c mean temperature 49°c mean temperature 50°c

. /com en/com033 167 jpg..temperatures are negative unless prefixed by ps . 46 + 47 + 47 + 48 + 49 /5 = 47 4°c exemple 267 mean temperature : -47°c.mean temperature : -47°c.

For flight planning purposes the landing mass at alternate is taken as ?

Question 91-21 : Zero fuel mass plus final reserve fuel and contingency fuel landing mass at destination plus alternate fuel zero fuel mass plus final reserve fuel and alternate fuel zero fuel mass plus final reserve 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 271 zero fuel mass plus final reserve fuel and contingency fuel.zero fuel mass plus final reserve fuel and contingency fuel.

Given .distance from departure to destination 220 nm.true track 175°.wind ?

Question 91-22 : 116 nm 103 nm 110 nm 136 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 275 116 nm116 nm

At references or see flight planning manual mrjt 1 figure 4 2 and figure 4 5 3 ?

Question 91-23 : 25000 ft 435 kt 33500 ft 430 kt 33900 ft 420 kt 24000 ft 445 kt

. /com en/com033 178 jpg..temperature isa 10°c .445 10 = 435 kt exemple 279 25000 ft, 435 kt25000 ft, 435 kt

Given .distance from departure to destination 950 nm.gs out 275 kt .gs home 225 ?

Question 91-24 : 93 min 139 min 114 min 39 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 283 93 min.93 min.

Given .distance from departure to destination 950 nm .safe endurance 3 5 h .tas ?

Question 91-25 : Distance 622 nm time 117 min distance 528 nm time 79 min distance 311 nm time 52 min distance 328 nm time 62 min

exemple 287 distance: 622 nm time: 117 mindistance: 622 nm time: 117 min

Given .distance from departure to destination 1000 nm .safe endurance 4 h .tas ?

Question 91-26 : 990 nm 450 nm 495 nm 10 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 291 990 nm.990 nm.

Given .distance from departure to destination 2200 nm.true track 150° .wind ?

Question 91-27 : Distance 980 nm time 115 min distance 1120 nm time 179 min distance 1100 nm time 179 min distance 980 nm time 144 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 295 distance: 980 nm time: 115 mindistance: 980 nm time: 115 min

Route manual chart nap..the initial true course from a 64°n006°e to c ?

Question 91-28 : 271° 259° 247° 279°

. /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 299 271°.271°.

Which best describes the maximum intensity of icing if any at fl160 in the ?

Question 91-29 : Moderate severe light nil

. 629. exemple 303 moderate.moderate.

At reference or see flight planning manual sep 1 figure 2 1. given . fl 75.oat ?

Question 91-30 : 18 nam 15 nm 14 nam 18 nm 16 nam 18 nm 18 nam 13 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 307 18 nam. 15 nm.18 nam. 15 nm.

Given .distance from departure to destination 285 nm .true track 348°.wind ?

Question 91-31 : 154 nm 131 nm 143 nm 123 nm

/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 311 154 nm.154 nm.

Given .distance from departure to destination 435 nm .gs out 110 kt .gs home ?

Question 91-32 : 236 nm 199 nm 218 nm 368 nm

exemple 315 236 nm236 nm

The approximate mean wind component kt along true course 180° from 50°n to ?

Question 91-33 : Tail wind 55 kt tail wind 40 kt tail wind 70 kt headwind 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 exemple 319 tail wind 55 kt.tail wind 55 kt.

The wind direction and velocity °/kt at 40°n 040°e is . err a 033 233 ?

Question 91-34 : 330/75 330/85 150/75 300/75

/com en/com033 233 jpg..wind is coming from 330° with a speed of 75 kt exemple 323 330/75.330/75.

In the vicinity of shannon 52° n009°w the tropopause is at about. err a 033 ?

Question 91-35 : Fl 360 fl 350 fl 300 fl 270

. /com en/com033 236 jpg. exemple 327 fl 360.fl 360.

At reference or see flight planning manual mrjt 1 figure 4 5 1..given .brake ?

Question 91-36 : 62 nam 59 nm 59 nam 62 nm 67 nam 71 nm 71 nam 67 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 exemple 331 62 nam, 59 nm.62 nam, 59 nm.

Route manual chart nap.the average magnetic course from c 62°n020°w to b ?

Question 91-37 : 119° 109° 099° 118°

.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 exemple 335 119°.119°.

Given .distance from departure to destination 360 nm .safe endurance 4 5 h ?

Question 91-38 : 308 nm 185 nm 154 nm 52 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 exemple 339 308 nm308 nm

What mean temperature °c is likely on a true course of 270° from 025°e to ?

Question 91-39 : Mean temperature is 50°c mean temperature is 48°c mean temperature is 52°c mean temperature is 54°c

. /com en/com033 246 jpg..temperatures are negative unless prefixed by ps . 53 + 51 + 50 + 47 /4 = 50 25°c exemple 343 mean temperature is -50°c.mean temperature is -50°c.

At reference or see flight planning manual mrjt 1 figure 4 3 3c. given .ground ?

Question 91-40 : 12 400 kg 03h 55 min 12 400 kg 04h 12 min 11 400 kg 03h 55 min 11 400 kg 04h 12 min

Explanation. /com en/com033 1017 jpg. exemple 347 12 400 kg, 03h 55 min.12 400 kg, 03h 55 min.

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