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Question 110-1 : The navigation plan reads .trip fuel 100 kg.flight time 1h35min.taxi fuel 3 kg.block fuel 181 kg.the endurance on the icao flight plan should read ? [ Training professional ]

2h 49min

.fuel on board at take off is .181 kg 3 kg = 178 kg ..fuel flow = trip fuel / flight time = 100 kg / 1h35min = 100/95 = 1 052 kg/min.endurance = 178 / 1 052 = 169 minutes 2h 49min exemple 210 2h 49min.

Question 110-2 : You have a flight plan ifr from amsterdam to london in the flight plan it is noted that you will deviate from the ats route passing the fir boundary amsterdam/london .the airway clearance reads cleared to london via flight planned route .which of the following statements is correct ?

The route according to the flight plan is accepted

exemple 214 The route according to the flight plan is accepted.

Question 110-3 : In the atc flight plan item 13 in a flight plan submitted before departure the departure time entered is the ?

Estimated off block time

exemple 218 Estimated off-block time

Question 110-4 : Given .planned and actual data as shown in the flight log excerpt .arriving overhead gamma you are cleared for direct routeing to mike the flight time for direct flight gamma to mike will be 1 h 08 min assuming other flight data remains constant what fuel will be expected on arrival overhead mike . ?

1272 kg

Beta to gamma 25 minutes .fuel used 2760 2360 = 400 kg.fuel flow from beta to gamma 400 / 25 = 16 kg/min.fuel used from gamma to mike 68 mins x 16 kg = 1088 kg..on arrival overhead mike fuel on board will be .2360 1088 = 1272 kg exemple 222 1272 kg.

Question 110-5 : A turbojet aeroplane flies using the following data .flight level fl 330.flight regime 'long range' lr .mass 156 500 kg.tailwind component at this level 40 kt .with a remaining flight time of 1 h 10 min the ground distance that can be covered by the aeroplane at cruising speed is . err a 033 2 ?

539 nm

For a mass of 156500 kg the table shows a tas of 427 kt right column .we have 40 kt tailwind component our groundspeed is 467 kt .the ground distance that can be covered by the aeroplane at cruising speed is .467 x 70/60 = 545 nm close to 539 nm exemple 226 539 nm.

Question 110-6 : Given .true course tc 017°.w/v 340°/30 kt.true air speed tas 420 kt.find .wind correction angle wca and ground speed gs ?

Wca 2° gs 396 kt

Wind correction angle wca is another term for drift except drift is expressed as left or right the wind correction angle uses '+' or ' ' .under index set true track 017° centre dot on tas 420 kt with the rotative scale set wind 340°/30 kt . /com en/com033 38a jpg.drift is always measured from heading to track so turn to set true heading 015 5° 017° 2 5° right drift under index . /com en/com033 38b jpg.you now read a ground speed of 396 kt a drift of 2º right wca is 2º exemple 230 Wca -2°, gs 396 kt.

Question 110-7 : The true course is 042° .the variation in the area is 6° w and the wind is calm .the deviation card is reproduced in the annex .in order to follow this course the pilot must fly a compass heading of . err a 033 41 ?

052°

Use this very useful table for those questions . /com en/com033 41 jpg.the nearest compass deviation reading is taken from 045° indicating 4° less than actual magnetic heading add 4° onto the 048° magnetic heading giving the answer of 052° compass heading exemple 234 052°.

Question 110-8 : At reference or see flight planning manual mrjt 1 figure 4 3 5 .given a trip time of about 9 hours within the limits of the data given a temperature decrease of 30°c will affect the trip time by approximately . err a 033 67 ?

By 7%

. /com en/com033 67 jpg.a 9h trip becomes 8 6 for isa +20°c .a 9h trip becomes 9 2 for isa 10°c ..difference for 30°c will affect the trip time by approximately . 0 6/8 6 x 100 = 7% exemple 238 By 7%.

Question 110-9 : At reference or see flight planning manual mrjt 1 figure 4 5 4 .planning an ifr flight from paris to london for the twin jet aeroplane .given .estimated landing mass 49700 kg.fl 280.w/v 280°/40 kt.average true course 320°.procedure for descent 74 m/250 kias.determine the distance from the top of ?

76 nm

.from the table . /com en/com033 75 jpg.time 19 minutes.distance nam 86 nm.tas = distance/time = 86/19 x 60 = 272 kt .on nav computer .true course 320°.w/v 280°/40 kt.ground speed is 240 kt.distance nm from the top of descent to london = 240 x 19/60 = 76 nm

Question 110-10 : At reference or see flight planning manual mrjt 1 figure 4 7 3 .given .diversion distance 720 nm.tail wind component 25kt.mass at point of diversion 55000kg.temperature isa.diversion fuel available 4250 kg.what is the minimum pressure altitude at which the above conditions may be met . err a 033 76 ?

25000 ft

. /com en/com033 76 jpg. exemple 246 25000 ft.

Question 110-11 : Flight planning chart for an aircraft states that the time to reach the cruising level at a given gross mass is 36 minutes and the distance travelled is 157 nm zero wind .what will be the distance travelled with an average tailwind component of 60 kt ?

193 nm

.we know that a headwind or a tailwind component does not modify the time to climb to a specific level thus we only have to calculate our speed .157 x 60 / 36 = 262 kt .adding tail wind component .262 + 60 = 322 kt ..the distance travelled in 36 minutes will be .322 x 36 / 60 = 193 nm exemple 250 193 nm.

Question 110-12 : At reference or see flight planning manual mrjt 1 figure 4 3 6 .given .distance to alternate 400 nm.landing mass at alternate 50 000kg.headwind component 25 kt.the alternate fuel required is . err a 033 79 ?

2800 kg

Answer .. /com en/com033 1012 jpg exemple 254 2800 kg.

Question 110-13 : At reference or see flight planning manual mrjt 1 figure 4 3 6 .given .twin jet aeroplane.dry operating mass 35500 kg.traffic load 14500 kg.final reserve fuel 1200 kg.distance to alternate 95 nm.tailwind component 10 kt.find .fuel required and trip time to alternate with simplified flight planning ?

1000 kg 24 min

Img /com en/com033 1015 jpg exemple 258 1000 kg, 24 min.

Question 110-14 : Given .planning data as shown in the flight log excerpt fuel planning section .after a balked landing at the destination airport you have to divert to the alternate airport with the gear extended .the re calculated flight time to the alternate due to the reduced speed is 1 h 10 min and the fuel ?

5987 kg

900 x 70/60 = 1050 kg to reach the alternate airport ..subtract 1050 kg from our estimated landing mass at destination ..7037 1050 = 5987 kg exemple 262 5987 kg.

Question 110-15 : Flight planning chart for an aeroplane states that the time to reach fl 190 at a given gross mass is 22 minutes and the distance travelled is 66 nm no wind .what will be the distance travelled with an average head wind component of 35 kt ?

53 nm

.the time to reach fl 190 is 22 minutes and the distance travelled is 66 nm no wind .with a head wind component of 35 kt our ground distance travelled will be less but the time to reach fl 190 remains unchanged .we need first to know our ground speed without wind .66/22 x 60 = 180 kt .with 35 kt head wind our ground speed becomes .180 35 = 145 kt .in 22 minutes the distance travelled will be 145/60 x 22 = 53 17 nm exemple 266 53 nm.

Question 110-16 : At reference or see flight planning manual mrjt 1 figure 4 5 3 1.given twin jet aeroplane.fl 330.long range cruise.outside air temperature 63°c.gross mass 50500 kg.true air speed tas is . err a 033 126 ?

420 kt

. /com en/com033 126 jpg.isa temperature at fl330 is 15°c 2°c x 33 = 51°c .adjustements for operation at non standard temperatures decrease tas by 1 knot per degree c below isa . difference between 63°c and 51°c is 12°c.433 12 = 421 kt . ninorr .this make no difference to the answer but on the picture you have marked wrong mass should be 50500 not 55000 as you marked exemple 270 420 kt.

Question 110-17 : At reference or see flight planning manual mrjt 1 figure 4 5 3 1. given long range cruise.oat 45°c at fl 350.gross mass at the beginning of the leg 40000 kg.gross mass at the end of the leg 39000 kg .find true air speed tas and cruise distance nam for a twin jet aeroplane . err a 033 147 ?

Tas 432 kt 227 nam

.isa at fl350 is 35 x 2° + 15° = 55°c.gross mass at the end of the leg 39000 kg from the table we can read a tas of 422 kt as we are at isa+9 we must increase tas by 1 kt per degree c above isa as indicated on the bottom of the table .422 + 10 = 432 kt ..fuel consumption on the leg was 1000 kg 40000 39000 table values are .40000 >1163 nam.39000 > 936 nam..1163 936 = 227 nam exemple 274 Tas 432 kt, 227 nam.

Question 110-18 : During an ifr flight in a beech bonanza the fuel indicators show that the remaining amount of fuel is 100 lbs after 38 minutes .fuel at take off is 160lbs .for the alternate fuel 30 lbs is necessary .the planned fuel for taxi is 13 lbs .final reserve fuel is estimated at 50 lbs .if the fuel flow ?

12 minutes

Minimum take off fuel = trip fuel + contingency + alternate + final reserve.after 38 min 60 lbs of fuel has been used .fuel flow = 60/38 = 1 579 lb/min .at landing at destination alternate fuel and final reserve must be in your tank .100 lbs 30 + 50 = 20 lbs ..we can fly 20/1 579 = 12 66 min exemple 278 12 minutes.

Question 110-19 : Given .planning data as shown in the flight log excerpt fuel planning section .after a balked landing at the destination airport you have to divert to the alternate airport with the gear extended .the re calculated flight time to the alternate due to the reduced speed is 2h 20 min and the fuel flow ?

5440 kg

Dadoki .2 333 2h 20min x 780 = 1820.1820 1100 alternate fuel 990 reserve = 270.4370 dom + 500 traffic load + 90 contingency + 210 final reserve + 270 remaining of alternate and extra fuel = 5440 kg exemple 282 5440 kg.

Question 110-20 : At reference or see flight planning manual mrjt 1 figure 4 3 1c.for a flight of 2400 ground nautical miles the following apply .temperature isa 10°c.cruise altitude 29000 ft.landing mass 45000 kg.trip fuel available 16000 kg.what is the maximum headwind component which may be accepted . err a 033 ?

35 kt

. /com en/com033 164 jpg. exemple 286 35 kt.

Question 110-21 : At reference or see flight planning manual mrjt 1 figure 4 5 3 2. find the fuel flow for the twin jet aeroplane with regard to the following data .given .mach 74 cruise.flight level 310.gross mass 50000 kg.isa conditions . err a 033 171 ?

2300 kg/h

50000 > 2994 nam maximum cruise distance tas is 434 kt .in one hour we travel 2294 434 = 2560 ..in the table the weight associated with a distance of 2560 nam is 47700 kg .50000 kg 47700 kg = 2300 kg/h .see section 5 4 2 method on caa cap697 flight planning manual for that kind of questions exemple 290 2300 kg/h

Question 110-22 : At reference or see flight planning manual mrjt 1 figure 4 3 1c. within the limits of the data given a mean temperature increase of 30°c will affect the trip time by approximately . err a 033 172 ?

By 5%

.draw a line for a ficticious flight duration of 5h for example . /com en/com033 172 jpg.at isa 10°c trip time is 5h07 min 5 12h .at isa +20°c trip time is 4h50 min 4 83h .a 30°c mean temperature increase decrease trip time by approximately 17 minutes 0 28h . 0 28/5 12 x 100 = 5 46% exemple 294 By -5%

Question 110-23 : At reference or see flight planning manual mrjt 1 figure 4 3 6. in order to find alternate fuel and time to alternate the aeroplane operating manual shall be entered with . err a 033 173 ?

Distance in nautical miles nm wind component landing mass at alternate

exemple 298 Distance in nautical miles (nm), wind component, landing mass at alternate

Question 110-24 : During a vfr flight at a navigational checkpoint the remaining usable fuel in tanks is 60 us gallons .the reserve fuel is 12 us gallons .according to the flight plan the remaining flight time is 1h35min .calculate the highest acceptable rate of consumption possible for the rest of the trip ?

30 3 us gallons/hour

.60 12 = 48 us gallons available .1 h 35 min = 95 min. 48 / 95 x 60 = 30 31 us gallons/hour .this is the highest acceptable rate of consumption possible for the rest of the trip exemple 302 30.3 us gallons/hour

Question 110-25 : At reference or see flight planning manual mrjt 1 figure 4 3 1c.for a flight of 2000 ground nautical miles cruising at 30000 ft within the limits of the data given a headwind component of 25 kt will affect the trip time by approximately . err a 033 190 ?

By 7 6%

. /com en/com033 190 jpg.with a headwind component of 25 kt we find a trip time of 307 minutes 5h07 .without headwind a trip time of 285 minutes 4h45 . 307 285 / 285 x 100 = 7 72% exemple 306 By 7.6%.

Question 110-26 : At reference or see flight planning manual mrjt 1 figure 4 3 6.given .distance to alternate 450 nm.landing mass at alternate 45 000 kg.tailwind component 50 kt.the alternate fuel required is . err a 033 191 ?

2500 kg

. /com en/com033 191 jpg.start at 450 nm go to the reference line enter condition 50 kt tailwind .go to the next condition landing mass 45 000 kg you get your answer 2500 kg exemple 310 2500 kg.

Question 110-27 : An aeroplane is on an ifr flight the flight is to be changed from ifr to vfr is it possible ?

Yes the pilot in command must inform atc using the phrase 'cancelling my ifr flight'

exemple 314 Yes, the pilot in command must inform atc using the phrase 'cancelling my ifr flight'.

Question 110-28 : See flight planning manual mrjt 1 figure 4 5 2 and 4 5 3 1 . given .distance c d 680 nm.long range cruise at fl340.temperature deviation from isa 0° c.headwind component 60 kt.gross mass at c 44700 kg.the fuel required from c d is . err a 033 218 ?

3700 kg

For 44700 kg we have a tas of 431 kt no temperature correction since it is isa condition ..nam = ground distance x tas/gs .nam = 680 x 431/371 = 790 nam ..at line 44700 the cruise distance nautical air miles is 2150 nautique air miles substract 790 you find 1360 nam .what is the mass for 1360 nm . /com en/com033 218 jpg.we find 41000 kg this is our end mass ..44700 41000 = 3700 kg ..see section 5 4 2 method on caa cap697 flight planning manual for that kind of questions exemple 318 3700 kg.

Question 110-29 : Flight planning manual mrjt 1 figure 4 5 4.a descent is planned at 74/250kias from 35000ft to 5000ft .how much fuel will be consumed during this descent . err a 033 220 ?

150 kg

. /com en/com033 220 jpg.from 35000 ft to 0 ft 290 kg .from 5000 ft to 0 ft 140 kg .from 35000 ft to 5000 ft 290 140 = 150 kg exemple 322 150 kg.

Question 110-30 : At reference or see flight planning manual mrjt 1 figure 4 7 3.given .diversion fuel available 8500kg.diversion cruise altitude 10000ft.mass at point of diversion 62500kg.head wind component 50kt.temperature isa 5°c.the a maximum diversion distance and b elapsed time alternate are approximately . ?

A 860 nm b 3h 20 min

. /com en/com033 253 jpg.the reference quality is not fantastic sorry for that .for information 'dash lines' over 'weigth at point of diversion' serve as adjustment variables we have to follow a 'continous line' slope because it is more representative of our pressure altitude continous lines are for low pressure altitudes exemple 326 (a) 860 nm (b) 3h 20 min.

Question 110-31 : For a planned flight the calculated fuel is as follows .flight time 2h42min .the reserve fuel at any time should not be less than 30% of the remaining trip fuel .taxi fuel 9 kg .block fuel 136 kg .how much fuel should remain after 2 hours flight ?

25 kg trip fuel and 8 kg reserve fuel

.127 kg at take off = 130/100 x trip fuel.trip fuel = 127 x 100/130 .trip fuel is 98 kg.after after 2 hours flight it remains 42 minutes 0 7 h the fraction of trip fuel remaining is .0 7/2 7 x 98 = 25 kg .the reserve fuel at any time should not be less than 30% of the remaining trip fuel .30/100 x 25 = 8 kg exemple 330 25 kg trip fuel and 8 kg reserve fuel.

Question 110-32 : An aircraft is flying at mach 0 84 at fl 330 .the static air temperature is 48°c and the headwind component 52 kt .at 1338 utc the controller requests the pilot to cross the meridian of 030w at 1500 utc .given the distance to go is 570 nm the reduced mach required is ?

0 8 m

.13h38 to 15h00 = 1h42 1 37h .first step .ground speed = distance/time = 570/1 37 = 416 kt .second step .true air speed = gs + headwind = 416 kt + 52 kt = 468 kt .last step .now on the computer .in airspeed window 48°c under mach index.in front of 468 kt on the outer scale you read the reduced mach number 0 8 m exemple 334 0.8 m

Question 110-33 : You must fly ifr on an airway orientated 135° magnetic with a msa at 7800 ft knowing the qnh is 1025 hpa and the temperature is isa +10°c the minimum flight level you must fly at is ?

Fl90

.1025 1013 = 12 hpa.12 hpa x 27ft/hpa = 324 ft.temperature correction 7 5 x 4 x 10 = +300 ft.7800 324 300 = 7176 ft.it is 'minus' 324 ft because we decrease our pressure setting from 1025 to 1013 indicated altitude will also decrease .it is 'minus' 300 ft because air mass is warmer than isa .on a 135° heading we need an odd level the first available is fl90 exemple 338 Fl90.

Question 110-34 : An aircraft following a 215° true track must fly over a 10 600 ft obstacle with a minimum obstacle clearance of 1 500 ft knowing the qnh received from an airport close by which is almost at sea level is 1035 and the temperature is isa 15°c .what is the minimum fl the aircraft should fly at ?

Fl 140

.local qnh is 1035 and we gonna fly with a 1013 hpa setting .1035 1013 = 22 hpa.22 hpa x 27 ft/hpa = 594 ft. new learning objectives state 27ft per 1 hpa .we need 10600 + 1500 594 = 11506 ft.we must correct for temperature .to determine the true altitude/height the following rule of thumb called the 4% rule shall be used .the altitude/height changes by 4% for each 10°c temperature deviation from isa .4%/15 degrees = 6%.6% x 11506 = 690 ft .air mass is colder we need to flight higher .11506 + 690 = 12196 ft .on a 215° true track we need an even flight level the first one available is fl 140 exemple 342 Fl 140.

Question 110-35 : Given .dry operating mass 33000 kg .traffic load 8110 kg .final reserve fuel 983 kg .alternate fuel 1100 kg .contingency fuel not used 102 kg .the estimated landing mass at alternate should be ?

42195 kg

.at alternate you must land with the final reserve in your tanks and contingency fuel if not used .33000 kg dom + 8110 kg traffic load + 983 kg final reserve + 102 kg contingency fuel = 42195 kg exemple 346 42195 kg.

Question 110-36 : At reference or see flight planning manual mrjt 1 figure 4 5 4.planning an ifr flight from paris to london for the twin jet aeroplane .given estimated landing mass 49700 kg.fl 280.wind 280°/40 kt.average true course 320°.procedure for descent 74 m/250 kias.determine the time from the top of ?

19 min

. /com en/com033 303 jpg.no need for calculations . cmarzocchini .whats the reason you dont make any correction about the wind coz the descend is at constant mach number if yes can you explain to me please ..the wind will affect the distance but not the time exemple 350 19 min.

Question 110-37 : At reference or see flight planning manual mrjt 1 figure 4 4.given .twin jet aeroplane.estimated mass on arrival at the alternate 50000 kg.elevation at destination aerodrome 3500 ft.elevation at alternate aerodrome 30 ft.find final reserve fuel . err a 033 304 ?

1180 kg

. /com en/com033 304 jpg.30 minutes at 1500 ft above alternate 2260/2 = 1180 kg . ninorr .why we do not take into account 3500ft of the airfield elevation we have to fly 1500ft above airfiled what makes 5000ft am i wrong with that calculations ..the question states find final reserve fuel and you do not have the mass on arrival at destination aerodrome furthermore you have to land with this final reserve in your wing at alternate exemple 354 1180 kg.

Question 110-38 : Given .planned and actual data as shown in the flight log excerpt .arriving overhead gamma you are cleared for direct routing to mike .the flight time for direct flight gamma to mike will be 1h30 min assuming other flight data remains constant what fuel will be expected on arrival overhead mike . ?

1300 kg

.from gamma you are cleared direct to mike .gamma to mike 1h30 min.from beta to gamma fuel consumption was 3200 2700 = 500 kg .flight time from beta to gamma 1h42 1h37 = 25 min .fuel flow between beta to gamma 500 kg / 25 min = 20 kg/min ..gamma to mike 1h30 x 20 kg/min = 1400 kg/.remaining fuel at gamma 2700 kg.remaining fuel at mike = 2700 1400 kg = 1300 kg exemple 358 1300 kg.

Question 110-39 : Given .planned and actual data as shown in the flight log excerpt .arriving overhead gamma you are cleared for direct routeing to mike .the flight time for direct flight gamma to mike will be 57 minutes assuming other flight data remains constant what fuel will be expected on arrival overhead mike ?

1720 kg

.beta to gamma 25 minutes 1h37 to 2h02 .fuel used 2950 2575 = 375 kg.fuel flow from beta to gamma 375 / 25 = 15 kg/min.fuel used from gamma to mike 57 min x 15 kg = 855 kg.on arrival overhead mike fuel on board will be .2575 855 = 1720 kg exemple 362 1720 kg.

Question 110-40 : Given .fl 370.mach 0 74.oat 47°c.the tas is ?

434 kt

. /com en/com033 333 jpg.by calculation .mach number = tas/ lss..lss= 39 square root t° kelvin.lss= 39 square root 273 47 °k.lss= 39 square root 226°k = 586 30..mach number = tas / 586 30 = 0 74.tas = 0 74 x 586 30 = 433 86 kt exemple 366 434 kt.


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