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Question 121-1 : 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 ? [ Preparation civilian ]

Moderate.

. 629.. /com en/com033 430.jpg..from below the chart base to fl 200, icing is moderate over bucarest. exemple 221 Moderate.

Question 121-2 : 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 225 200 nm

Question 121-3 : 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 229 30 kt.

Question 121-4 : 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 233 Broken ac/cu base below fl100 tops fl150, embedded isolated cb base below fl100 tops fl270

Question 121-5 : 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 237 1635 kg.

Question 121-6 : 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.

Question 121-7 : 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 245 Distance (nm), wind component, landing mass at alternate.

Question 121-8 : 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.

Question 121-9 : 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 at ?

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 253 The fuel transport operation is not recommended in this case.

Question 121-10 : 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 257 3740 nm.

Question 121-11 : 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 261 203 nm.

Question 121-12 : 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 265 5 to 7 oktas cu and cb base below fl100, tops fl180.

Question 121-13 : 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 269 148 nm.

Question 121-14 : 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 273 30000 ft.

Question 121-15 : 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 277 Distance: 210 nm time: 122 min

Question 121-16 : 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 281 Distance: 216 nm time: 118 min

Question 121-17 : 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 285 271°.

Question 121-18 : 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 289 92 nm

Question 121-19 : 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 293 66 nm.

Question 121-20 : 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 297 3756 nm

Question 121-21 : 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 301 720 nm.

Question 121-22 : Given. course a to b 088° t.distance 1250 nm.mean tas 330 kt.mean w/v 340°/60 kt.the time from a to the pet between a and b is ?

1 hour 42 minutes

. /com en/com033 484a.jpg.. /com en/com033 484b.jpg..ground speed out = 343 kt..proceed the same way to find ground speed home 306 kt..pet = distance x gsh / gso+ gsh.pet = 1250 x 306 / 343 + 306.pet = 382500 / 649 = 589 nm..589 nm / 343 = 1.71 h 1 hour 42 minutes. exemple 305 1 hour 42 minutes

Question 121-23 : Route manual chart nap. the average true course from a 64°n006°e to c 62°n020°w is.. err a 033 497 ?

259°.

. /com en/com033 497.jpg..we are looking for average true course .with you protractor aligned on true north, between a and c, you will find a true course of 259°. exemple 309 259°.

Question 121-24 : Given.distance from departure to destination 340 nm.true track 320°.wind 160°/40 kt.tas 110 kt.what is the distance of the pet from the departure point ?

112 nm.

...under index, set true track 320°, centre dot on tas, 110 kt, with the rotative scale, set wind 160°/40 kt, you find a right drift of 5°.. /com en/com033 500a.jpg.now, drift is always measured from heading to track.turn to set true heading 315° 320° 5° right drift under index, you now read your ground speed out of 147 kt. /com en/com033 500b.jpg.proceed in the same way to find the ground speed home of 72 kt.. left drift of 10°, true heading of 150°...ground speed out gso = 147 kt.ground speed home gsh = 72 kt..distance to pet = distance x gsh / gso + gsh.distance to pet = 340 x 72 / 147 + 72.distance to pet = 24480 / 219 = 111.78 nm. exemple 313 112 nm.

Question 121-25 : Use route manual chart nap.on a direct great circle course from reykjavik 64°10' n 022°00'w to amsterdam 52°32'n 004°50'e , the a average true course, and b distance, are.. err a 033 510 ?

A 131° b 1095 nm

.average true course.. /com en/com033 510.jpg..cos mean latitude x difference of longitude x 60 nm.cos58.31 x 26,83° x 60 nm = 846 nm..difference of latitude 11°42' = 11,70° x 60 nm = 702 nm.distance between reykjavik and amsterdam = sqrt 702² + 846² = 1099 nm. exemple 317 (a) 131° (b) 1095 nm

Question 121-26 : Given.distance from departure to destination 330 nm.safe endurance 5 h.true track 170°.wind 140/25.tas 125 kt.what is the distance of the psr from the departure point ?

302 nm.

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..on the computer.under index, set true track 170°, centre dot on tas, 125 kt, with the rotative scale, set wind 140°/25 kt we read a right drift of 7°..drift is always measured from heading to track, so turn to set true heading 163° 170° 7° right drift under index you read a ground speed out of 104 kt and a drift of 6º right...proceed on the same way for gs home you will find 140 kt...point of safe return psr = 5 x 140 / 104 + 140.point of safe return psr = 700 / 244.point of safe return psr = 2.87 h..distance of the psr from the departure point at a speed of 104 kt.2.87 x 104 = 298 nm. exemple 321 302 nm.

Question 121-27 : Route manual chart nap. the initial magnetic course from a 64°n006°e to c 62°n020°w is .. err a 033 520 ?

275°.

. /com en/com033 520.jpg..we are looking for initial magnetic course .with you protractor aligned on true north, you have a true course of 271°, add the 4°west magnetic variation = 275°. exemple 325 275°.

Question 121-28 : The maximum wind velocity °/kt immediately north of tunis 36°n010°e .. err a 033 521 ?

190°/95 kt.

. arrows, feathers and pennants.arrows indicate direction. number or pennants and/or feathers correspond to speed...example with a 270°/115 kt wind.. /com en/com033 346a.jpg..pennants correspond to 50 kt..feathers correspond to 10 kt..half feathers correspond to 5 kt... /com en/com033 521.jpg.. exemple 329 190°/95 kt.

Question 121-29 : Given. distance a to b 3060 nm.mean groundspeed 'out' 440 kt.mean groundspeed 'back' 540 kt.safe endurance 10 hours.the time to the point of safe return psr is ?

5 hours 30 minutes.

exemple 333 5 hours 30 minutes.

Question 121-30 : The approximate mean wind component kt at mach 0.78 along true course 270° at 50°n from 000° to 010°w is.. err a 033 541 ?

40 kt headwind component.

. /com en/com033 541.jpg.. 55 kt + 55 kt + 30 kt / 3 = 46 kt..mean wind is 240°/46kt...headwind component = wind speed x cos angle between the wind and the course..headwind component = 46 kt x cos 30° = 40 kt. exemple 337 40 kt headwind component.

Question 121-31 : Given.distance from departure to destination 340 nm.gs out 150 kt.gs home 120 kt.what is the distance of the pet from the departure point ?

151 nm.

Ground speed out gso = 150 kt.ground speed home gsh = 120 kt..distance to pet = distance x gsh / gso + gsh.distance to pet = 340 x 120 / 150 + 120.distance to pet = 40800 / 270 = 151 nm. exemple 341 151 nm.

Question 121-32 : Given.distance from departure to destination 470 nm.true track 237°.wind 300/25 kt.tas 125 kt.what is the distance of the pet from the departure point ?

256 nm

...under index, set true track 237°, centre dot on tas, 125 kt, with the rotative scale, set wind 300°/25 kt, you find a left drift of 11°..now, drift is always measured from heading to track.turn to set true heading 248° 237° + 11° left drift under index, you now read your ground speed out of 111 kt..proceed in the same way to find the ground speed home of 135 kt.. right drift of 9°, true heading of 048°...ground speed out gso = 111 kt.ground speed home gsh = 135 kt..distance to pet = distance x gsh / gso + gsh.distance to pet = 470 x 135 / 111 + 135.distance to pet = 63450 / 246 = 258 nm closest answer is 256 nm. exemple 345 256 nm

Question 121-33 : From which of the following would you expect to find details of the search and rescue organisation and procedures sar ?

Aip

exemple 349 Aip

Question 121-34 : Given.distance from departure to destination 2380 nm.gs out 420 kt.gs home 520 kt.what is the time of the pet from the departure point ?

188 minutes.

.ground speed out = 420 kt.ground speed home = 520 kt..pet = distance x gsh / gso + gsh.pet = 2380 x 520 / 420 + 520.pet = 1237600 / 940 = 1316.6 nm...time of the pet from the departure point.1316.6 / 420 = 3.13 h.3.13 x 60 = 188 minutes. exemple 353 188 minutes.

Question 121-35 : Given.distance from departure to destination 338 nm.true track 045°.wind 225°/35 kt.tas 120 kt.what is the distance and time of the pet from the departure point ?

Distance 120 nm time 46 min

.pet point of equal time.....distance to pet = d x h /o + h....d = total track distance..h = groundspeed home..o = groundspeed out....d = 338 nm..h = 120 35 kt headwind = 85 kt..o = 120 + 35 kt tailwind = 155 kt....pet distance = 338 x 85 / 155 + 85..distance = 119.7 nm = 120 nm..time = 120 / 155 ground speed out x 60 = 46 minutes. exemple 357 Distance: 120 nm time: 46 min

Question 121-36 : Given.distance from departure to destination 1500 nm.safe endurance 4.5 h.tas 450 kt.ground speed out 480 kt.ground speed home 410 kt.what is the time of the psr from the departure point ?

124 min.

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..ground speed out = 480 kt.ground speed home = 460 kt..point of safe return psr = 4.5 x 410 / 480 + 410.point of safe return psr = 1845 / 890.point of safe return psr = 2.07 h..2.07 x 60 = 124 minutes. exemple 361 124 min.

Question 121-37 : Over london 51°n 000°e/w , the lowest fl listed which is unaffected by cat is.. err a 033 567 ?

Fl230.

.london is located in the cat area n°1..this cat area extends from fl240 to fl350, thus, two levels are out of this layer, fl230 and fl360. the question asks for the lowest one. exemple 365 Fl230.

Question 121-38 : What lowest cloud conditions oktas/ft are forecast for 1900 utc at hamburg eddh .. err a 033 572 ?

5 to 7 at 500 ft.

exemple 369 5 to 7 at 500 ft.

Question 121-39 : Given.distance from departure to destination 5000 nm.safe endurance 10 h.tas 450 kt.ground speed out 500 kt.ground speed home 400 kt.what is the distance of the psr from the departure point ?

2222 nm.

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs..ground speed out = 500 kt.ground speed home = 400 kt..point of safe return psr = 10 x 400 / 500 + 400.point of safe return psr = 4000 / 900.point of safe return psr = 4.44 h..distance of the psr from the departure point at a speed of 500 kt.4.44 h x 500 = 2222 nm. exemple 373 2222 nm.

Question 121-40 : What is the earliest time utc , if any, that thunderstorms are forecast for doha otbd .. err a 033 603 ?

1000

exemple 377 1000


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