A free Premium account on the FCL.055 website! Read here
Sign up to unlock all our services and 15164 corrected and explained questions.

Question 116-1 : Route manual sid chart 20 3 for paris charles de gaulleplanning an ifr flight from paris charles de gaulle rwy 27 to londongiven distance from paris charles de gaulle to top of climb 50 nmdetermine the distance from the top of climb toc to abb 1166 err a 033 558 ? [ Training professional ]

245 nm

Question 116-2 : Route manual chart e hi 5 caa editionthe minimum enroute altitude available on airway ur160 from nice niz 1124 43°46'n 007°15'e to bastia bta 1162 42°32'n 009°29'e is err a 033 559 ?

Fl250.

com encom033 559jpgthe minimum enroute altitude from nice to bastia is fl250
exemple 220: Fl250
Fl200. fl210. fl260.

Question 116-3 : Route manual chart paris charles de gaulle ils rwy 27 21 2 the crossing altitude and descent instruction for a propeller aircraft at coulommiers clm are err a 033 566 ?

Cross at fl60 descend to 4000 ft.

com encom033 566jpg
exemple 224: Cross at fl60 descend to 4000 ft
Cross at fl70 descend to 4000 ft. cross at fl80 descend to fl70. cross at fl60 and maintain.

Question 116-4 : Route manual chart munich ndb dme rwy 26l approach 16 3 the frequency and identifier of the ndb for the published approachs are err a 033 568 ?

400 msw.

Key informations are displayed in big bold text 400 msw is the main beacon to be use on the munich ndb dme rwy 26l approach procedure
exemple 228: 400 msw
338 mnw 108.6 dms 112.3 mun

Question 116-5 : Route manual chart e hi 4 caa edition the radio navigation aid serving strasbourg 48°30'n 007°34'e is a err a 033 571 ?

Vortacan frequency 1156 mhz.

The distance measurement component of tacan operates with the same specifications as civil dmestherefore to reduce the number of required stations tacan stations are frequently co located with vor facilities these co located stations are known as vortacs this is a station composed of a vor for civil bearing information and a tacan for military bearing information and militarycivil distance measuring informationthe ' d ' indicates a distance measuring equipment dme since we do not have the answer 'vor dme frequency 1156 mhz' we can only choose 'vor tacan frequency 1156 mhz'
exemple 232: Vortacan frequency 1156 mhz
Dme only, channel 115.6 tacan only, frequency 115.6 mhz vor only, frequency 115.6 mhz

Question 116-6 : Use route manual chart e lo 1 what are the applicable grid moras for a flight from dean cross 1152 dcs 54°43'n 003°20'w to talla 1138 tla 55°30'n 003°21'w on airway a2 err a 033 574 ?

45 and 40.

com encom033 574jpg
exemple 236: 45 and 40
42 and 39. 42 and 40. 45 and 39.

Question 116-7 : Use route manual chart e lo 1what are the magnetic curse and distance when flying on airway b4 from the reporting point shapp n54°30' w002°38' to the reporting point eskdo n55°18' w003°12' err a 033 584 ?

343°52 nm.

com encom033 584jpgthe question asks for the magnetic course from shapp to eskdo so 163°+180° = 343°
exemple 240: 343°52 nm
343°/114 nm. 163°/39 nm. 163°/52 nm.

Question 116-8 : Use route manual chart e hi 2an aircraft has to fly from alma n55°25' e013°34' to peter n56°50' e013°35' excluding rvsm the lowest flight level above fl400 that can be flown on an ifr flight plan is err a 033 585 ?

Fl 430.

On the airway ug55 from point alma to point peter you have a e> indicates that an even flight level must be usedexcluding rvsm the lowest flight level above fl400 that can be flown won't be fl420 but fl 430 it's crazy isn't it you have to refer to icao semicircular cruising level diagram 710at or above fl290 only odd levels are used in order to allow a minimum separation of 2000 ft between aircrafts
exemple 244: Fl 430
Fl 410 fl 420 fl 440

Question 116-9 : Use route manual chart e hi 4 caa editionan aircraft has to fly from salzburg 48°002'n 012°536'e to klagenfurt 46°375'n 014°338'e at salzburg you can read on the chart d 1138 sbg it means err a 033 586 ?

Vordme ident sbg frequency 1138 mhz can be used.

exemple 248: Vordme ident sbg frequency 1138 mhz can be used
Salzburg ils/dme, frequency 113.8 mhz, can not be used for navigation. only the dme, ident sbg, can be used, on frequency 113.8 mhz. vor is not avalaible. vor/dme sbg will be dismantled soon, it can not be used for navigation.

Question 116-10 : Use route manual chart e lo 1the radio navigation aid at st abbs 55°55'n 002°12'w is err a 033 594 ?

A vordme frequency 1125 mhz.

exemple 252: A vordme frequency 1125 mhz
An ndb, frequency 112.5 mhz. a tacan, channel 112.5. a vor only, frequency 112.5 mhz.

Question 116-11 : For this question use route manual chart e hi 2 the radio navigation aid s at 56°17'n 010°47'e isare err a 033 597 ?

Ndb frequency 374 khz.

9996angel como he de saber si es de uso temporal o no an asterisk indicates navaid operation or service not continuousexample com encom033 597jpghere 'tu' means 'tango uniform' this is the navaid name
exemple 256: Ndb frequency 374 khz
Ndb, frequency 374 khz, for temporary use (tu). vor, frequency 374 mhz. vordme, frequency 374 khz.

Question 116-12 : Use route manual chart e hi 1an aircraft is flying towards gow 55°52'n 004°27w on airway un615 prior to an arrival in the london firaccording to the information on the chart what is the most relevant frequency to receive the broadcast weather for london fir north err a 033 598 ?

12660 mhz.

13367 mhz and 13367 mh are the frequencies for scottish control1154 mhz is the frequency of gow vor12660 mhz is the london north weather wx frequency
exemple 260: 12660 mhz
133.67 mhz 115.4 mhz. 133.67 mhz.

Question 116-13 : For this question use route manual chart e hi 2 the radio navigation aid s at 55°26' n011°38'e isare err a 033 599 ?

Vordme frequency 1128 mhz.

exemple 264: Vordme frequency 1128 mhz
Vordme, frequency 128.15 mhz. vordme, frequency 128.75 mhz. ndb, frequency 112.8 khz.

Question 116-14 : Route manual chart e hi 4 caa editionthe preferred airways routing from frankfurt ffm 1142 50°03' n008°38'e to koksy 51°06'n 002°39'e above fl245 on a wednesday is err a 033 601 ?

Ur10 ntm ub6 bub ats.

com encom033 601jpgub69 is not available on wednesday you can read in blue 'available only friday 1800 to monday 0800 lt'
exemple 268: Ur10 ntm ub6 bub ats
Ug108 spi ug1 ug1 ub69 dinki ub6 bub ats

Question 116-15 : Use route manual chart e hi 1what navigation aid s isare available to civil aircraft at the military airfield of kinloss 57°40'n 003°32'w err a 033 602 ?

The range element only of tacan on dme frequency 1098 mhz.

com encom061 453jpgfor civil aircrafts only the range element of a tacan is available
exemple 272: The range element only of tacan on dme frequency 1098 mhz
Tacan range and bearing on channel 35. the bearing element only of tacan on vor frequency 109.8 mhz. vor and dme on frequency 109.8 mhz.

Question 116-16 : On ur14 airway an aircraft overflying gibso must be at or above 1496 ?

Fl 250.

exemple 276: Fl 250
Fl 200 fl 140 fl 160

Question 116-17 : The correct track and corresponding iaf for a western arrival via ridar for runway 08lr is 1497 ?

Ridar 2m iaf miq.

exemple 280: Ridar 2m iaf miq
Buram 1m, iaf miq ridar 2m, iaf rokil buram 2m, iaf rokil

Question 116-18 : Refer to jeppesen manual chart e hi 4 or figure hi 08 the magnetic course and distance from st prex spr 1139 n4628 e00627 to fribourg fri 1151 n4647 e00714 on airway ug60 are 1498 ?

061° 37 nm.

exemple 284: 061° 37 nm
048° / 46 nm 041° / 78 nm 061° / 28 nm

Question 116-19 : Refer to jeppesen student manual london area chart you are required to hold at bovingdon 51°436 n000°330 w however the vor is unserviceable what procedure should now follow 1499 ?

Enter the bovva alternate hold which starts at 32 dme biggin on 141° qdm.

exemple 288: Enter the bovva alternate hold which starts at 32 dme biggin on 141° qdm
Enter the bovva alternate hold which is on the 141° radial at 37 dme biggin. enter the bovva alternate hold which starts at 37 dme biggin on 141° qdm. ask atc for radar vectors instead.

Question 116-20 : Refer to jeppesen manual chart e hi 4 your aircraft registered in the uk is outbound from london towards new york via ug1 and shanwick ctafir prior to crossing w002 what should you have obtained 1500 ?

Oceanic entry clearance on 12765 mhz.

exemple 292: Oceanic entry clearance on 12765 mhz
Airways clearance from london on 133,60 mhz. oceanic entry clearance on 133,80 mhz. oceanic entry clearance on123,95 mhz.

Question 116-21 : Refer to jeppesen manual chart e hi 5 flying eastbound between bologna 44°322 n 011°175 e and ancona 43°352 n 013°283 e what is the minimum icao ifr cruising level 1501 ?

Fl 310.

exemple 296: Fl 310
Fl 250 fl 260 fl 330

Question 116-22 : Refer to jeppesen student manual trm sabadell area chart 19 1 or figure vfr 12 what is the published frequency for barcelona atis 1502 ?

11865 mhz.

exemple 300: 11865 mhz
120.80 mhz. 124.70 mhz. 119.10 mhz.

Question 116-23 : Planning an ifr flight from paris charles de gaulle to london sid is abb 8a assume variation 3° w tas 430 kts wv 28040 distance to top of climb 50 nmdetermine the magnetic course ground speed and wind correction angle from top of climb to abb 1166 1503 ?

Mc 349° gs 414 kts wca 5°.

exemple 304: Mc 349° gs 414 kts wca 5°
Mc 349°, gs 450 kts, wca +5°. mc 169°, gs 414 kts, wca +5°. mc 169°, gs 450 kts, wca +4°.

Question 116-24 : Refer to jeppesen student route manual chart e lo 1 the minimum en route altitude mea that can be maintained continuously on airway l602 from talla 1138 tla 55°30 n 003°21 w to newcastle 11425 new 55°02 n 001°24w is 1504 ?

Fl200.

Img1505
exemple 308: Fl200
Fl100. 3900 ft. 3100 ft.

Question 116-25 : The quantity of fuel which is calculated to be necessary for a jet aeroplane to fly ifr from departure aerodrome to the destination aerodrome is 5352 kgfuel consumption in holding mode is 6000 kgh alternate fuel is 4380 kgcontingency should be 5% of trip fuelwhat is the minimum required quantity of ?

13000 kg.

Warning this question indicates contingency should be 5% of trip fuel normal rules for contingency the greater of 5% of trip or 5 min holding at 1500 ftwe must assume the operator has an exemptionminimum quantity of fuel at take off = trip fuel + alternate + contingency 5% of trip fuel + 30 min final reserve jet aircraft trip fuel = 5350 kgalternate = 4380 kgcontingency = 5% of trip fuel 5% x 5350 = 267 kg30 min final reserve = 6000 2 = 3000 kgminimum quantity of fuel at take off = 5350 + 4380 + 267 + 3000 = 12997 kg
exemple 312: 13000 kg
14500 kg. 13370 kg. 13220 kg.

Question 116-26 : Using the power setting table for the single engine aeroplane determine the manifold pressure and fuel flow lbshr with full throttle and cruise lean mixture in the following conditions oat +13°cpressure altitude 8000 ftrpm 2300 2161 ?

224 inhg and 693 lbshr.

First search isa temperature at 8000 ft oat is +13°cin standard atmosphere temperature at 8000 ft is 15°c 2°c x 8 = 1°cwe are in isa +14°cwe need now to interpolate between isa and isa +20°c on the table 1139isa fuel flow 711 pph and man press 224 isa +20°c fuel flow 685 pph and man press 224 711 685 x 14 20 = 182 fuel flow = 711 182 = 693 lbshr
exemple 316: 224 inhg and 693 lbshr
23.0 in.hg and 69.0 lbs/hr. 22.4 in.hg and 71.1 lbs/hr. 22.4 in.hg and 73.8 lbs/hr.

Question 116-27 : Given the following head wind component 50 kttemperature isa +10°cbrake release mass 65000 kgtrip fuel available 18000 kgwhat is the maximum possible trip distance 2363 ?

2740 nm.

1156
exemple 320: 2740 nm
3100 nm. 3480 nm. 2540 nm.

Question 116-28 : The fuel burn off is 200 kgh with a relative fuel density of 08if the relative density is 075 the fuel burn will be ?

200 kgh.

Volume changes mass consumption will not changeif they ask for litrehour instead of kgh then fuel burn will be 235 lh
exemple 324: 200 kgh
213 kg/h. 188 kg/h. 267 kg/h.

Question 116-29 : The trip fuel for a jet aeroplane to fly from the departure aerodrome to the destination aerodrome is 5 350 kg fuel consumption in holding mode is 6 000 kghthe quantity of fuel which is needed to carry out one go around and land on the alternate airfield is 4 380 kgthe destination aerodrome has a ?

13 230 kg.

Minimum quantity of fuel at take off = trip fuel + alternate + contingency + 30 min final reserve jet aircraft trip fuel = 5350 kgalternate = 4380 kgcontingency = the greater of 5% of trip or 5 min holding at 1500 ft 5% of trip = 5% x 5350 = 267 kg5 min holding at 1500 ft = 6000 x 560 = 500 kg30 min final reserve = 6000 2 = 3000 kgminimum quantity of fuel at take off = 5350 + 4380 + 500 + 3000 = 13230 kg
exemple 328: 13 230 kg
13 000 kg. 14 730 kg. 11 730 kg.

Question 116-30 : Given fl 75 oat +10°c lean mixture 2300 rpmfind fuel flow in gallons per hour gph and tas 2461 ?

116 gph tas 160 kt.

At fl75 in standard atmosphere isa is 15°c 2°c x 75 = 0°coat is +10°c thus we are at isa +10°cno need for long interpolations fuel flow is around 114 to 12 gph and tas between 158 to 160 kt 1162
exemple 332: 116 gph tas 160 kt
11.6 gph, tas: 143 kt. 71.1 gph, tas: 143 kt. 68.5 gph, tas: 160 kt.

Question 116-31 : A public transport aeroplane with reciprocating engines the final reserve should be ?

Fuel to fly for 45 minutes.

Eu ops 1255 fuel policy c an operator shall ensure that the pre flight calculation of usable fuel required for a flight includes 1 taxi fuel and2 trip fuel and3 reserve fuel consisting of i contingency fuel see eu ops 1192 and ii alternate fuel if a destination alternate aerodrome is required this does not preclude selection of the departure aerodrome as the destination alternate aerodrome and iii final reserve fuel see appendix 1 to ops 1255 below and iv additional fuel if required by the type of operation eg etops and4 extra fuel if required by the commanderappendix 1 to eu ops 1255 final reserve fuel which shall be a for aeroplanes with reciprocating engines fuel to fly for 45 minutes or b for aeroplanes with turbine engines fuel to fly for 30 minutes at holding speed at 1 500 ft 450 m above aerodrome elevation in standard conditions calculated with the estimated mass on arrival at the destination alternate aerodrome or the destination aerodrome when no destination alternate aerodrome is required
exemple 336: Fuel to fly for 45 minutes
Fuel to fly for 2 hours. fuel to fly for 1 hour at holding speed. fuel to fly for 30 minutes at holding speed.

Question 116-32 : Assuming the following data ground distance to be covered 1 500 nmcruise flight level fl 310cruising speed mach 082 true airspeed 470 kthead wind component 40 ktplanned destination landing mass 140 000 kgtemperature isa +15°ccg 37%total anti ice on pack flow hifuel consumption for such a flight is ?

23 500 kg.

Proceed like this 1500 nm 470kt 40kt = 349 h349 x 470 = 1640 namon the table to find the corresponding value for 1640 nam you have to interpolate between line 1600 and 1700 nam 22221 20955 10 x4 = 506 kg20955 + 506 = 21461 kgyou must add +1% for pack flow highandyou have to add +7% for total anti ice onit means 8% of 21461 = 1717 kg21461 + 1717 = 23178 kgwe are on isa+15 condition for each degree above isa temperature apply fuel correction 0010 x 15 x 1640 = 246 kg23178 + 246 = 23424 kg
exemple 340: 23 500 kg
21 500 kg. 21 700 kg. 19 900 kg.

Question 116-33 : For a flight of 2400 ground nautical miles the following apply tail wind 25 kttemperature isa 10°cbrake release mass 66000 kgthe a trip fuel and b trip time respectively are 2463 ?

A 14000 kg b 5h 35 min.

1169
exemple 344: A 14000 kg b 5h 35 min
(a) 15800 kg (b)6h 20 min. (a) 14600 kg (b) 5h 45 min. (a) 15000 kg (b) 6h 00 min.

Question 116-34 : Given brake release mass 58 000 kgtemperature isa +15the fuel required to climb from an aerodrome at elevation 4000 ft to fl300 is 1515 ?

1250 kg.

1173you have to decrease the fuel by 100 kg as prescribe in the fuel adjustment table
exemple 348: 1250 kg
1400 kg. 1450 kg. 1350 kg.

Question 116-35 : The flight crew of a turbojet aeroplane prepares a flight using the following data flight level fl 370 at 'long range' lr cruise regime mass at brake release 212 800 kg flight leg ground distance 2 500 nm temperatures isa cg 37% headwind component 30 kt 'total anti ice' set on 'on' for the entire ?

34 430 kg.

Nam = ngm x tasgs nam = 2500 x 470440 nam = 2670 nam212 800 kg is corresponding to 8083 nam in the table com encom033 99jpg8083 2670 = 5413 nam5413 nam is corresponding to 180 329 kg in the tablefuel = 212 800 180 329 = 32 471 kgfuel consumption with total anti ice 0n = 32 471 x 106 = 34 419 kg renfernandes i still did not understand what do de values on top of the columns 0 2 4 in this case 8 mean i couldn't choose a column unless i knew what they meanthanks mass at brake release is 212 800 kg ==> 212 ton + 08 tonif mass at brake release was 213 600 kg ==> 212 ton + 16 ton miguel would you be so kind to explain why 5413 are exactly 180 329 kg because i'm trying to calculate it via interpolation but i cannot find a successful result thank you180400 180200 = 200 kg5419 5402 = 17 nam17 nam for 200 kg1 nam = 118 kg5419 5413 = 6 nam6 x 118 kg = 71 kg180400 71 = 180329 kg
exemple 352: 34 430 kg
30 440 kg. 32 480 kg. 28 720 kg.

Question 116-36 : The purpose of the decision point procedure is ?

To reduce the minimum required fuel and therefore be able to increase the traffic load.

Reduced contingency fuel rcf procedure decision point procedure this is a technique for increasing the traffic load by reducing the minimum fuel required you reduce the contingency figures by using it only from the decision point to the destinationif an operator's fuel policy includes pre flight planning to a destination 1 aerodrome commercial destination with a reduced contingency fuel procedure using a decision point along the route and a destination 2 aerodrome optional refuel destination the amount of usable fuel on board for departure shall be the greater of 21 or 22 below 21 the sum of a taxi fuel and b trip fuel to the destination 1 aerodrome via the decision point and c contingency fuel equal to not less than 5% of the estimated fuel consumption from the decision point to the destination 1 aerodrome and d alternate fuel or no alternate fuel if the decision point is at less than six hours from the destination 1 aerodrome are fulfilled and e final reserve fuel and f additional fuel and g extra fuel if required by the commander22 the sum of a taxi fuel and b trip fuel to the destination 2 aerodrome via the decision point and c contingency fuel from departure aerodrome to the destination 2 aerodrome and d alternate fuel if a destination 2 alternate aerodrome is required and e final reserve fuel and f additional fuel and g extra fuel if required by the commander
exemple 356: To reduce the minimum required fuel and therefore be able to increase the traffic load
To reduce the landing weight and thus reduce the structural stress on the aircraft. to increase the safety of the flight. to increase the amount of extra fuel.

Question 116-37 : At reference or see flight planning manual mrjt 1 figure 472for the purpose of planning an extended range flight it is required that with a start of diversion mass of 55000kg a diversion of 600 nautical miles should be achieved in 90 minutesusing the above table the only listed cruise technique to ?

Mkias 74330.

com encom033 120jpg
exemple 360: Mkias 74330
Lrc m/kias .74/290 m/kias .70/280

Question 116-38 : Using the power setting table for the single engine aeroplane determine the cruise tas and fuel flow lbsh with full throttle and cruise lean mixture in the following conditions oat 3°c pressure altitude 6000 ft power 21 inhg 2100 rpm 2155 ?

134 kt and 557 lbsh.

First step search for isa temperature oat is +3°c isa at 6000 ft = 15°c 2 x 6 = +3°cwe are in standard condition isa for standard day isa ktas is the same 134 kt step two search for fuel flow for standard condition isa fuel flow is 557 pph pound per hour or lbsh
exemple 364: 134 kt and 557 lbsh
131 kt and 56.9 lbs/h. 125 kt and 55.7 lbs/h. 136 kt and 56.9 lbs/h.

Question 116-39 : For turbojet engine driven aeroplane given taxi fuel 600 kgfuel flow for cruise 10 000 kghfuel flow for holding 8 000 kghalternate fuel 10 200 kgplanned flight time to destination 6 hforecast visibility at destination 2000 mthe minimum ramp fuel required is ?

77 800 kg.

Fuel flow for cruise 6h x 10000 kg = 60000 kgcontingency fuel 5% x 60000 = 3000 kgtaxi fuel 600 kgalternate fuel 10200 kg30 minutes fuel for holding at 8000kgh = 4000 kgtotal = 60000 + 3000 + 600 + 10200 + 4000total = 77800 kg
exemple 368: 77 800 kg
76 100 kg. 80 500 kg. 79 200 kg.

Question 116-40 : The following apply temperature isa +15°cbrake release mass 62000 kgtrip time 5h 20 minwhat is the trip fuel 2180 ?

13500 kg.

1194enter the graph at 533 5h20
exemple 372: 13500 kg
13000 kg. 13200 kg. 13800 kg.

~

Exclusive rights reserved. Reproduction prohibited under penalty of prosecution.

4599 Free Training Exam