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At reference or see flight planning manual sep1 figure 2 2 table 2 2 3.a flight ? [ Answer reports ]

Question 91-1 : 325 lbs 306 lbs 349 lbs 289 lbs

.total flight time .3h12 + 3 min + 10 min = 3h25 3 42h ..fuel comsumption at fl70 isa +20°c is . 68 5+69 /2 = 68 75 pph..3 42h x 68 75 = 235 pph..235 + 10 lbs for the climb + reserve fuel 30% of 235 + 10 for taxi = 325 lbs exemple 191 325 lbs.325 lbs.

A flight has to be made with a multi engine piston aeroplane mep 1 for the ?

Question 91-2 : 47 us gallons 470 us gallons 37 us gallons 60 us gallons

exemple 195 47 us gallons.47 us gallons.

A flight has to be made with the sep for the fuel calculation .allow . 10 lbs ?

Question 91-3 : 359 lbs 336 lbs 364 lbs 348 lbs

exemple 199 359 lbs.359 lbs.

A flight has to be made with the sep1 for the fuel calculation allow . 10 lbs ?

Question 91-4 : 297 lbs 231 lbs 252 lbs 215 lbs

exemple 203 297 lbs.297 lbs.

Given .dry operating mass 33500 kg .traffic load 7600 kg .maximum allowable ?

Question 91-5 : 15900 kg 17300 kg 16300 kg 17100 kg

exemple 207 15900 kg.15900 kg.

Given .brake release mass 58 000 kg.temperature isa 15.the fuel required to ?

Question 91-6 : 1150 kg 1250 kg 1350 kg 1450 kg

exemple 211 1150 kg.1150 kg.

Given .distance from departure to destination 500 nm.true track 090°.wind ?

Question 91-7 : Distance 283 nm time 131 min distance 217 nm time 100 min distance 250 nm time 88 min distance 382 nm time 176min

Ground speed out 130 kt.ground speed home 170 kt.pet = d x gsh / gso + gsh .pet = 500 x 170 / 130 + 170 = 283 nm exemple 215 Distance: 283 nm, time: 131 min.Distance: 283 nm, time: 131 min.

Use route manual chart e hi 1.an aircraft has to fly from glasgow 55°52'n ?

Question 91-8 : 1117 kg 978 kg 869 kg 2300 kg

Ground speed = 320 40 = 280 kt .136 nm/ 280/60 = 29 14 minutes.2300/60 = 38 333 kg/min.29 14 x 38 333 = 1117 kg exemple 219 1117 kg.1117 kg.

Given .distance from departure to destination 500 nm.safe endurance 4 h.tas 140 ?

Question 91-9 : Distance 279 nm time 111 min distance 232 nm time 107 min distance 139 nm time 60 min distance 221 nm time 89 min

Point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .ground speed out = 150 kt.ground speed home = 130 kt.point of safe return psr = 4 x 130 / 150 + 130 .point of safe return psr = 65000 / 280.point of safe return psr = 1 86 h.1 86 x 60 = 111 6 min.distance of the psr from the departure point at a speed of 150 kt .111 6 min x 150/60 = 279 nm exemple 223 Distance: 279 nm, time: 111 min.Distance: 279 nm, time: 111 min.

On a given path it is possible to chose between four flight levels fl each ?

Question 91-10 : Fl 270 fl 290 fl 330 fl 370

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 . 60° and mach 0 8 => tas is 451 kt . 1137.fl 370 => tas = 451 kt => gs = 439 kt.fl 330 => tas = 440 kt => gs = 435 kt.fl 290 => tas = 456 kt => gs = 441 kt.fl 270 => tas = 445 kt => gs = 445 kt .the flight level allowing the highest ground speed is fl 270 exemple 227 Fl 270.Fl 270.

Given . distance from departure to destination 410 nm. safe endurance 3 6 h. ?

Question 91-11 : 203 nm 169 nm 102 nm 207 nm

First step find the gs out .given . true track. true air speed. wind direction and velocity.required wind correction angle and ground speed. computer solution .a set true track to true index .b turn the indicator to the wind direction in this case using the black azimuth 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 angle at the same place read the ground speed under the center bore from the scale on the axis of the slide .setting 055° to true index.set the indicator to 180° on the black azimuth circle being upwind adjust the speed arc labelled 120 kt of the diagram slide to the wind speed 35 kt of the indicator scale .reading under the plotted point read the wind correction angle 14° under the center bore read the ground speed 136 kt . 1141.step two proceed on the same way for gs home you will find 97 kt .step three now apply the psr formula .psr = time x gs out x gs home / gs out + gs home .psr = 3 6 x 136 x 97 / 136 + 97 .psr = 203 8 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 231 203 nm.203 nm.

The surface weather system over england 53°n 002°w is . err a 033 22 ?

Question 91-12 : An occluded front moving east a depression moving north a warm front moving southeast a cold front moving east

. /com en/com022 33a png.over england the surface weather system is an an occluded front . /com en/com033 22 jpg.the arrow attached to the occluded front indicates direction and speed 20 km/h exemple 235 An occluded front moving east.An occluded front moving east.

Which describes the intensity of icing if any at fl 150 in the vicinity of ?

Question 91-13 : Moderate or severe moderate light nil

.its due to the presence of cbs around toulouse always assume moderate or severe icing next to them exemple 239 Moderate or severe.Moderate or severe.

Given .distance a to b 2050 nm.mean groundspeed 'on' 440 kt.mean groundspeed ?

Question 91-14 : 1130 nm 1025 nm 920 nm 1153 nm

.ground speed out gso = 440 kt.ground speed home gsh = 540 kt.distance to pet = distance x gsh / gso + gsh .distance to pet = 2050 x 540 / 440 + 540 .distance to pet = 1107000 / 980 = 1130 nm

Which is the heaviest type of precipitation if any forecast for ?

Question 91-15 : Light rain rain showers heavy rain associated with thunderstorms nil

. /com en/com033 27 jpg.at 1000 utc if any the heaviest type of precipitation forcast is light rain exemple 247 Light rain.Light rain.

Given .distance from departure to destination 315 nm.true track 343°.w/v ?

Question 91-16 : 176 nm 139 nm 167 nm 148 nm

..under index set true track 343° centre dot on tas 100 kt with the rotative scale set wind 015° / 15 kt .now drift is always measured from heading to track .turn to set true heading 345° 343° + 2° left drift under index you now read a ground speed out of 88 kt .proceed in the same way to find the ground speed home of 112 kt . right drift of 4° true heading of 159° .distance to pet = distance x gsh / gso + gsh .distance to pet = 315 x 112 / 88 + 112 .distance to pet = 35280 / 200 = 176 nm exemple 251 176 nm.176 nm.

What lowest cloud conditions oktas/ft are forecast for johannesburg/jan smuts ?

Question 91-17 : 5 to 7 at 400 ft 3 to 4 at 800 ft 5 to 7 at 800 ft 3 to 4 at 400 ft

.ft0900 prob30 0305 3000 bcfg bkn004 .30% chance that temporarily between 0300 utc and 0500 utc there will be patches fog with 3000 m visibility and 5 to 7 oktas of cloud at 400 ft exemple 255 5 to 7 at 400 ft.5 to 7 at 400 ft.

Given .distance from departure to destination 210 nm.safe endurance 2 5 h.true ?

Question 91-18 : 127 nm 88 nm 64 nm 172 nm

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .on the computer .under index set true track 035° centre dot on tas 105 kt with the rotative scale set wind 250°/20 kt we read a right drift of 5° .drift is always measured from heading to track so turn to set true heading 030° 035° 5° right drift under index you read a ground speed out of 122 kt .proceed on the same way for gs home .under index set true track 215° centre dot on tas 105 kt with the rotative scale set wind 250°/20 kt we read a left drift of 4° .drift is always measured from heading to track so turn to set true heading 219° 215° + 4° right drift under index you read a ground speed home of 88 kt ..point of safe return psr = 2 5 x 88 / 122 + 88 .point of safe return psr = 220 / 210.point of safe return psr = 1 04 h.distance of the psr from the departure point at a speed of 122 kt .1 04 x 122 = 127 nm exemple 259 127 nm.127 nm.

Given the following .d = flight distance.x = distance to point of equal ?

Question 91-19 : X = d x gsr / gso + gsr x = d x gso / gso + gsr x = d/2 x gso / gso + gsr x = d/2 x gsr / gso + gsr

exemple 263 X = d x gsr / (gso + gsr).X = d x gsr / (gso + gsr).

The flight crew of a turbojet aeroplane prepares a flight using the following ?

Question 91-20 : The fuel transport operation is not recommended in this case 10 000 kg 5 000 kg 8 000 kg

.the fuel is cheaper at destination the fuel transport operation is not recommended in this case exemple 267 The fuel transport operation is not recommended in this case.The fuel transport operation is not recommended in this case.

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

Question 91-21 : Notam aip sigmet atc broadcasts

exemple 271 Notam.Notam.

The wind velocity over italy is . err a 033 54 ?

Question 91-22 : A maximum of 110 kt at fl380 130 kt at fl380 maximum velocity not shown on chart a maximum of 160 kt at fl 380 110 kt at fl380 maximum velocity not shown on chart

. /com en/com033 54 jpg.a jet is crossing france and italy at fl 380 with a maximum of 110 kt exemple 275 A maximum of 110 kt at fl380.A maximum of 110 kt at fl380.

Assuming the following data .ground distance to be covered 2 000 nm.cruise ?

Question 91-23 : 4 h 43 min 4 h 26 min 5 h 02 min 4 h 10 min

.nam = ngm x tas/gs = 2000 x 470/440 = 2136 nam . /com en/com033 56 jpg.between 2100 and 2200 we can find our answer 4 h 43 min . 210883 .i think that the answer is wrong because . 4 39 + 4 51 /2 = 4 45 h > 4h 27' . 4h 43' > 4 75h so the correct answer as i think is 4h 27' ..no times on the annex are shown as 4 39 for 4h39 and 4 51 for 4h51 .it's written on the top below fuel consumed kg time h min exemple 279 4 h 43 min.4 h 43 min.

When calculating the fuel required to carry out a given flight one must take ?

Question 91-24 : 1 2 3 4 1 3 2 4 1 2 3

exemple 283 1, 2, 3, 4.1, 2, 3, 4.

Which of the following flight levels if any is forecast to be clear of ?

Question 91-25 : Fl250 fl210 fl290 none

. /com en/com033 64 jpg.there are isolated embedded cbs from below fl100 to fl220 so fl 210 is excluded .you have cat area n°2 from fl270 to fl400 so fl290 is excluded .only fl250 is forecast to be clear of significant cloud icing and cat along the marked route from shannon to berlin exemple 287 Fl250.Fl250.

Given .distance from departure to destination 215 nm.safe endurance 3 3 h.true ?

Question 91-26 : 205 nm 112 nm 103 nm 9 nm

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .on the computer .under index set true track 005° centre dot on tas 125 kt with the rotative scale set wind 290°/15 kt we read a right drift of 7° .drift is always measured from heading to track so turn to set true heading 358° 005° 7° right drift under index you read a ground speed out of 120 kt .proceed on the same way for gs home .under index set true track 185° centre dot on tas 125 kt with the rotative scale set wind 290°/15 kt we read a left drift of 7° .drift is always measured from heading to track so turn to set true heading 192° 185° + 7° right drift under index you read a ground speed home of 128 kt ..point of safe return psr = 3 3 x 128 / 120 + 128 .point of safe return psr = 422 4 / 248.point of safe return psr = 1 70 h.distance of the psr from the departure point at a speed of 120 kt .1 7 x 120 = 204 nm exemple 291 205 nm.205 nm.

Given .distance from departure to destination 200 nm.safe endurance 3 h.tas 130 ?

Question 91-27 : 190 nm 85 nm 95 nm 10 nm

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .ground speed out = 150 kt.ground speed home = 110 kt.point of safe return psr = 3 x 110 / 150 + 110 .point of safe return psr = 330 / 260.point of safe return psr = 1 2692 h.distance of the psr from the departure point at a speed of 150 kt .1 2692 h x 150 = 190 nm exemple 295 190 nm.190 nm.

What minimum visibility m is forecast for 0600 utc at london lhr egll . err a ?

Question 91-28 : 1500 m 2200 m 5000 m 10000 m

exemple 299 1500 m.1500 m.

Excluding rvsm an appropriate flight level for ifr flight in accordance with ?

Question 91-29 : Fl100 fl90 fl95 fl105

.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 303 Fl100.Fl100.

Given .distance from departure to destination 400 nm.safe endurance 2 5 h.tas ?

Question 91-30 : 145 nm 179 nm 73 nm 255 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 307 145 nm.145 nm.

Given .distance from departure to destination 300 nm.safe endurance 4 h.tas 110 ?

Question 91-31 : 218 nm 136 nm 109 nm 82 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 311 218 nm.218 nm.

What is the mean temperature deviation °c from the isa over 50°n 010°w . err ?

Question 91-32 : 2°c +2°c +9°c +13°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 315 -2°c.-2°c.

Given .distance from departure to destination 4630 nm.safe endurance 12 4 ?

Question 91-33 : 3211 nm 1966 nm 6106 nm 1419 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 319 3211 nm.3211 nm.

Given .distance from departure to destination 1100 nm.true track 280°.w/v ?

Question 91-34 : Distance 450 nm time 52 min distance 650 nm time 108 min distance 650 nm time 75 min distance 550 nm time 75 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 323 Distance: 450 nm, time: 52 min.Distance: 450 nm, time: 52 min.

Given .distance from departure to destination 150 nm.safe endurance 3 2 h.tas ?

Question 91-35 : Distance 142 nm time 85 min distance 67 nm time 50 min distance 71 nm time 47 min distance 8 nm time 5 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 327 Distance: 142 nm, time: 85 min.Distance: 142 nm, time: 85 min.

Given .distance from departure to destination 270 nm.true track 030.w/v ?

Question 91-36 : Distance 135 nm time 68 min distance 141 nm time 65 min distance 130 nm time 68 min distance 141 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 331 Distance: 135 nm, time: 68 min.Distance: 135 nm, time: 68 min.

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

Question 91-37 : 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 335 05 hours 12 minutes.05 hours 12 minutes.

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

Question 91-38 : Aip nav/rad charts atcc notam

exemple 339 Aip.Aip.

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

Question 91-39 : 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 343 2 h 35 min.2 h 35 min.

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

Question 91-40 : 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 347 Distance: 210 nm, time: 121 min.Distance: 210 nm, time: 121 min.

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