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Question 108-1 : 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 ? [ Exam pilot ]

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 208 1 hour 42 minutes

Question 108-2 : 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 212 259°.

Question 108-3 : 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 216 112 nm.

Question 108-4 : 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 220 (a) 131° (b) 1095 nm

Question 108-5 : 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 224 302 nm.

Question 108-6 : 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 228 275°.

Question 108-7 : 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 232 190°/95 kt.

Question 108-8 : 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 236 5 hours 30 minutes.

Question 108-9 : 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 240 40 kt headwind component.

Question 108-10 : 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 244 151 nm.

Question 108-11 : 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 248 256 nm

Question 108-12 : From which of the following would you expect to find details of the search and rescue organisation and procedures sar ?

Aip

exemple 252 Aip

Question 108-13 : 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 256 188 minutes.

Question 108-14 : 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 260 Distance: 120 nm time: 46 min

Question 108-15 : 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 264 124 min.

Question 108-16 : 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 268 Fl230.

Question 108-17 : 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 272 5 to 7 at 500 ft.

Question 108-18 : 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 276 2222 nm.

Question 108-19 : What is the earliest time utc if any that thunderstorms are forecast for doha otbd . err a 033 603 ?

1000

exemple 280 1000

Question 108-20 : Given .maximum landing mass 51300 kg.maximum allowable take off mass 56300 kg.dry operating mass 29100 kg.load 11700 kg.trip fuel 3000 kg.contingency fuel 215 kg.final reserve fuel 1250 kg.alternate fuel 1300 kg.taxi fuel 200 kg.determine the maximum possible extra fuel that can be uplifted for ?

7735 kg

Ecqb03 october 2016 ...   29100.+ 11700.+  3000.+   215.+  1250.+  1300. .46565 kg today's take off mass..maximum possible take off mass = maximum landing mass + trip fuel.maximum possible take off mass = 51300 + 3000 = 54300 kg...extra fuel = 54300 46565 = 7735 kg exemple 284 7735 kg.

Question 108-21 : Which best describes the weather if any forecast for johannesburg/jan smuts at 0400 utc ?

Patches of fog

Prob30 0305 3000 bcfg .bc bancs patches .fg brouillard fog exemple 288 Patches of fog.

Question 108-22 : What minimum visibility is forecast taf for damascus osdi . 1517 ?

200 m

Becmg 0203 200 fg exemple 292 200 m.

Question 108-23 : Metar egly 301220 24015kt 200v280 8000 ra sct010 bkn025 ovc080 18/15 q0983 tempo 3000 ra bkn008 ovc020= ?

Mean surface wind 240° true 15 kt varying between 200° and 280°

Egly issued at 1220z on 30th .surface wind mean 240 deg true 15 kt varying between 200 and 280 deg .prevailing vis 8 km .weather light rain cloud 1 2 .oktas base 1000 ft 5 7 oktas 2500 ft .temperature +18 °c dew point +15°c .qnh 983 mb .trend temporarily 3000 m in moderate rain with 5 7 oktas 800 ft exemple 296 Mean surface wind 240° (true) 15 kt varying between 200° and 280°.

Question 108-24 : For a long distance flight at fl370 long range regime divided into four flight legs with the following specifications .leg 1 ground distance 2 000 nm headwind component 50 kt.leg 2 ground distance 1 000 nm headwind component 30 kt.leg 3 ground distance 500 nm tailwind component 70 kt.leg 4 ?

4800 nm

exemple 300 4800 nm

Question 108-25 : Which best describes the maximum intensity of turbulence at fl 290 above london . 1519 ?

Moderate

exemple 304 Moderate.

Question 108-26 : What minimum visibility is forecast for paris/charles de gaulle at 2100 utc . 1521 ?

6000 m

exemple 308 6000 m.

Question 108-27 : A 'current flight plan' is a ?

Filed flight plan with amendments and clearance included

exemple 312 Filed flight plan with amendments and clearance included.

Question 108-28 : The maximum permissible take off mass of an aircraft for the l wake turbulence category on an atc flight plan is ?

7 000 kg

Light aeroplane 7000 kg or less .medium aeroplane less than 136000 kg but more than 7000 kg .heavy aeroplane 136000 kg or greater .for information helicopters produce vortices when in flight and there is some evidence that per kilogram of gross mass their vortices are more intense than those of fixed wing aircraft exemple 316 7 000 kg.

Question 108-29 : How many hours in advance of eobt estimated off block time should a atc flight plan be filed in the case of flights into areas subject to air traffic flow management atfm ?

3 00 h

Atfm air traffic flow management is defined as 'a function established with the objective of contributing to a safe orderly and expeditious flow of air traffic by ensuring that atc capacity is utilised to the maximum extent possible and that the traffic volume is compatible with the capacities declared by the appropriate air traffic service providers' .in case of atfm air traffic flow management the flight plan should be filed at least three hours in advance of the eobt exemple 320 3:00 h.

Question 108-30 : Which of the following statements is are correct with regard to the advantages of computer flight plans .1 the computer can file the atc flight plan .2 wind data used by the computer is always more up to date than that available to the pilot ?

Statement 1 only

Flight planning computers can generate both atc flight plan and operational flight plan by taking in account forecast wind take off mass traffic load temperature etc but since it is based on forecast data while in flight weather info available to the pilot will always be more accurate than anything stored in a computer exemple 324 Statement 1 only.

Question 108-31 : In the ats flight plan item 10 'standard equipment' is considered to be . err a 033 112 ?

Vhf rtf adf vor and ils

. /com en/com033 112 png.icao doc 4444 exemple 328 Vhf rtf, adf, vor and ils.

Question 108-32 : In an atc flight plan item 15 in order to define a position as a bearing and distance from a vor the group of figures should consist of . 1522 ?

Vor ident magnetic bearing and distance in nautical miles

Icao doc4444 pans atm . 1193 exemple 332 Vor ident, magnetic bearing and distance in nautical miles.

Question 108-33 : When an atc flight plan has been submitted for a controlled flight the flight plan should be amended or cancelled in the event of the off block time being delayed by ?

30 minutes or more

exemple 336 30 minutes or more

Question 108-34 : When completing an atc flight plan an elapsed time item 16 of 1 hour 55 minutes should be entered ?

As 0155

exemple 340 As 0155

Question 108-35 : When a pilot fills in an atc flight plan he must indicate the wake turbulence category .this category is a function of which mass ?

Maximum certified take off mass

exemple 344 Maximum certified take-off mass

Question 108-36 : Which of the following statements is are correct with regard to the operation of flight planning computers .1 the computer can file the atc flight plan .2 in the event of in flight re routing the computer produces a new plan ?

Statement 1 only

.flight planning computers can generate both atc flight plan and operational flight plan by taking in account forecast wind take off mass traffic load temperature etc but since it is based on forecast data while in flight weather info available to the pilot will always be more accurate than anything stored in a computer the computer can not produce a new 'valid' flight plan exemple 348 Statement 1 only.

Question 108-37 : From the options given below select those flights which require flight plan notification .1 any public transport flight.2 any ifr flight.3 any flight which is to be carried out in regions which are designated to ease the provision of the alerting service or the operations of search and rescue.4 ?

2 and 4

.a flight plan is required if you are expecting to use atc services and if you are going to cross borders exemple 352 2 and 4.

Question 108-38 : In the ats flight plan for a non scheduled flight which of the following letters should be entered in item 8 type of flight . 1522 ?

N

Icao doc4444 pans atm . 1201 exemple 356 N

Question 108-39 : In the atc flight plan item 15 a cruising speed of 470 knots will be entered as ?

N0470

Icao doc4444 pans atm . 1327 exemple 360 N0470

Question 108-40 : In the event that selcal is prescribed by an appropriate authority in which section of the ats flight plan will the selcal code be entered . 1522 ?

Other information

exemple 364 Other information


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