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Question 107-1 : 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 ? [ Exam pilot ]

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 207 633 nm.

Question 107-2 : 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 211 Fl380.

Question 107-3 : 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 215 163 min

Question 107-4 : 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 219 203 min

Question 107-5 : 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 223 (a) 17100kg (b) 6h 07 min

Question 107-6 : 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 227 142 nm.

Question 107-7 : 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 231 (a) 14200 kg (b) 5 h 30 min.

Question 107-8 : 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 235 Nil.

Question 107-9 : 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 ?

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 239 752 nm.

Question 107-10 : 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


Question 107-11 : 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 247 284 nm

Question 107-12 : 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 251 Light rain associated with thunderstorms.

Question 107-13 : 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 255 1386 nm.

Question 107-14 : 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 259 20 nm.

Question 107-15 : 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 263 Distance: 1680 nm time: 252 min

Question 107-16 : 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 267 116°.

Question 107-17 : 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 271 2 h 51 min

Question 107-18 : 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 275 34500 ft

Question 107-19 : 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 279 2255 nm

Question 107-20 : Which best describes be maximum intensity of icing if any at fl150 in the vicinity of bucharest 45°n 026°e . err a 033 430 ?

Moderate

. 629. /com en/com033 430 jpg.from below the chart base to fl 200 icing is moderate over bucarest exemple 283 Moderate.

Question 107-21 : Given .distance from departure to destination 210 nm.safe endurance 3 5 h .true track 310.w/v 270/30.tas 120 kt.what is the distance of the psr from the departure point ?

200 nm

exemple 287 200 nm

Question 107-22 : What maximum surface windspeed kt is forecast for bordeaux/merignac at 1600 utc . err a 033 434 ?

30 kt

. fc1100r 121100z 121221 amended r short taf prepared on the twelfth day of the month at 1100z valid from 12h to 21h utc .there is a tempo from 12h to 18h indicating wind from 280° for 20 kt with gust up to 30 kt tempo 1218 28020g30kt exemple 291 30 kt.

Question 107-23 : Which best describes the significant cloud forecast over toulouse 44°n001°e . err a 033 435 ?

Broken ac/cu base below fl100 tops fl150 embedded isolated cb base below fl100 tops fl270

. /com en/com033 435 png. exemple 295 Broken ac/cu base below fl100 tops fl150, embedded isolated cb base below fl100 tops fl270

Question 107-24 : At reference or see flight planning manual mrjt 1 figure 4 4 .holding planning.ngm= nam x tas+ wind/ tas .the fuel required for 45 minutes holding in a racetrack pattern at 5000 ft pressure altitude and a weight of 47000 kg is . err a 033 437 ?

1635 kg

.fuel flow for 47000kg at 5000 ft is 2180 kg/h interpolating between 2220 and 2140 kg/h .the fuel required for 45 minutes holding is . 2180 x 45/60 = 1635 kg exemple 299 1635 kg.

Question 107-25 : The wind °/kt at 50°n 015°w is . err a 033 438 ?

290/75

. /com en/com033 438 png.wind is coming from 290° .number or pennants and/or feathers correspond to speed .pennants correspond to 50 kt .feathers correspond to 10 kt .half feathers correspond to 5 kt .50+10+10+5 = 75 kt exemple 303 290/75.

Question 107-26 : At reference or see flight planning manual mrjt 1 figure 4 3 6. in order to get alternate fuel and time the twin jet aeroplane operations manual graph shall be entered with . err a 033 443 ?

Distance nm wind component landing mass at alternate

exemple 307 Distance (nm), wind component, landing mass at alternate.

Question 107-27 : Given .distance from departure to destination 320 nm.safe endurance 4 3 h.true track 120°.wind 180°/40 kt.tas 130 kt.what is the distance of the psr from the departure point ?

263 nm

..under index set true track 120° under the center bore set tas 130 kt and with the rotative scale set wind 180°/40 kt . /com en/com033 444a jpg.drift is always measured from heading to track so turn to set true heading 120° + 17° = 137° . /com en/com033 444b jpg.gs out is 105 kt .proceed the same way to find ground speed home .under index set true track 300°.13° right drift true heading is 300° 13° = 287°.gs home is 146 kt .psr = time x gs out x gs home / gs out + gs home .psr = 4 3 x 105 x 146 / 105 + 146 .psr = 263 nm exemple 311 263 nm.

Question 107-28 : The flight crew of a turbojet aeroplane prepares a flight using the following data . flight leg distance 3 500 nm. flight level fl 310 true airspeed 450 kt. headwind component at this level 5 kt. initially planned take off mass without extra fuel on board 180 000 kg. fuel price 0 35 us dollars/l ?

The fuel transport operation is not recommended in this case

.it's very simple fuel is cheaper at destination so fuel transport operation is not recommended in this case exemple 315 The fuel transport operation is not recommended in this case.

Question 107-29 : At reference or see flight planning manual mrjt 1 figure 4 3 5 .the following apply . tail wind component 10 kt.temperature isa +10°c.brake release mass 63000 kg.trip fuel available 20000 kg.what is the maximum possible trip distance . err a 033 447 ?

3740 nm

.you have to use the graph backward you must go to the condition first 10 kt tailwind and then go to the ref line instead of the 'normal' way to proceed first the ref line and then the condition . /com en/com033 447 jpg. exemple 319 3740 nm.

Question 107-30 : Cas is 190 kt and altitude 9000 ft.temperature isa 10°c.true course 350°.wind 320/40.distance from departure to destination is 350 nm.endurance 3 hours and actual time of departure is 1105 utc .the distance from departure to point of equal time pet ?

203 nm

.oat at fl90 = 15°c 9 x 2°c = 3°c .we are in isa 10°c thus oat = 13°c .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 exemple 323 203 nm.

Question 107-31 : Which best describes the significant cloud if any forecast for the area southwest of bodo 67°n 014°e . err a 033 450 ?

5 to 7 oktas cu and cb base below fl100 tops fl180

. /com en/com033 450 jpg.the chart is from fl100 to fl450 exemple 327 5 to 7 oktas cu and cb base below fl100, tops fl180.

Question 107-32 : Given .distance from departure to destination 190 nm .safe endurance 2 4 h .true track 120° .wind 030°/40 kt .tas 130 kt .what is the distance of the psr from the departure point ?

148 nm

..set 120° true track on top with rotating scale mark the wind 030°/40 kt .you read 17° right drift .set on top 120° 17° = 103° .now read the drift 18° right .set on top 102° and you can read your 'ground speed out' of 123 5 kt .the wind is perpendicular to our track so gsh will also be 123 5 kt...point of safe return psr = endurance x homeward gs / outbound gs + homeward gs ..ground speed out = 123 5 kt.ground speed home = 123 5 kt..point of safe return psr = 2 4 x 123 5 / 123 5 + 123 5 .point of safe return psr = 296 4 / 247.point of safe return psr = 1 2 h..1 2 x 60 = 72 minutes..distance of the psr from the departure point at a speed of 123 5 kt .72 min x 123 5/60 = 148 nm exemple 331 148 nm.

Question 107-33 : 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 40000 kg.temperature isa + 20°c.trip distance 150 nautical air miles nam . err a 033 458 ?

30000 ft

. /com en/com033 458 jpg. exemple 335 30000 ft.

Question 107-34 : 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 122 min

exemple 339 Distance: 210 nm time: 122 min

Question 107-35 : Given .distance from departure to destination 240 nm .safe endurance 3 5 h .tas 125 kt .ground speed out 110 kt.ground speed home 140 kt.what is the distance and time of the psr from the departure point ?

Distance 216 nm time 118 min

.point of safe return psr = endurance x gs home / gs out + gs home .outbound gs = 110 kt.homeward gs = 140 kt.endurance = 3 5 h.point of safe return psr = 3 5 x 140 / 110 + 140 .point of safe return psr = 1 96 h.1 96 h = 118 min.distance of the psr from the departure 1 96 x 110 = 215 6 nm exemple 343 Distance: 216 nm time: 118 min

Question 107-36 : Route manual chart nap.the average magnetic course from a 64°n006°e to c 62°n020°w is . err a 033 464 ?

271°

.center your protractor at mid distance between a and c you will find 260° you must apply variation 11°w .average magnetic course is 260° + 11° = 271° exemple 347 271°.

Question 107-37 : Given .distance from departure to destination 165 nm .true track 055 .w/v 360/20 .tas 105 kt .what is the distance of the pet from the departure point ?

92 nm

. /com en/com033 475a jpg. /com en/com033 475b jpg.ground speed out = 92 kt..proceed the same way to find ground speed home .true track 235° drift = 8° true heading = 243° gs home = 116 kt ..pet = distance x gsh / gso+ gsh .pet = 165 x 116 / 92 + 116 .pet = 19140 / 208 = 92 nm exemple 351 92 nm

Question 107-38 : Given .distance from departure to destination 140 nm .gs out 90 kt .gs home 80 kt .what is the distance of the pet from the departure point ?

66 nm

exemple 355 66 nm.

Question 107-39 : Given .distance from departure to destination 6340 nm .safe endurance 15 h .true track 090. w/v 270/100 .tas 520 kt.what is the distance of the psr from the departure point ?

3756 nm

exemple 359 3756 nm

Question 107-40 : Route manual chart nap.the distance nm from a 64°n006°e to c 62°n020°w is . err a 033 483 ?

720 nm

.report the track distance along latitude 63°n average latitude between a and c .you will count a little bit more of 26° separation between a and c .26° x 60 nm x cos 63° = 708 nm . /com en/com033 483 jpg.you can also use meridian 1° = 60 nm . maxscail .26° change in longitude 60 x 26 x cos 63 = 708 225.between 64 and 62 2 ° ==> 2 x 60 1°=60nm = 120.now use phytagore sqrt 708 225 + sqrt 120 = 15108 225.square root of 515982 905 = 718 32 nm exemple 363 720 nm.


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