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Question 124-1 : 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 ? [ Preparation civilian ]

35 kt.

. /com en/com033 164.jpg.. exemple 224 35 kt.

Question 124-2 : 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 228 2300 kg/h

Question 124-3 : 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 232 By -5%

Question 124-4 : 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 236 Distance in nautical miles (nm), wind component, landing mass at alternate

Question 124-5 : 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 240 30.3 us gallons/hour

Question 124-6 : 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 244 By 7.6%.

Question 124-7 : 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 248 2500 kg.

Question 124-8 : 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 252 Yes, the pilot in command must inform atc using the phrase 'cancelling my ifr flight'.

Question 124-9 : 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 256 3700 kg.

Question 124-10 : 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 260 150 kg.

Question 124-11 : 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 264 (a) 860 nm (b) 3h 20 min.

Question 124-12 : 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 268 25 kg trip fuel and 8 kg reserve fuel.

Question 124-13 : 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 272 0.8 m

Question 124-14 : 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 276 Fl90.

Question 124-15 : 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 280 Fl 140.

Question 124-16 : 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 284 42195 kg.

Question 124-17 : 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 descent ?

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 288 19 min.

Question 124-18 : 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 292 1180 kg.

Question 124-19 : 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 296 1300 kg.

Question 124-20 : 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 ?

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 300 1720 kg.

Question 124-21 : 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 304 434 kt.

Question 124-22 : The still air distance in the climb is 189 nautical air miles nam and time 30 minutes..what ground distance would be covered in a 30 kt head wind ?

174 nm.

.ground distance nm = air distance +/ time x effective wind/60.ground distance nm = 189 nam 30 minutes x 30kt/60.ground distance nm = 189 15 = 174 nm... stanley.is it correct counting.189/30 x 60 = 378 tas.378 30 kt = 348.348/2 = 174 nm....yes, it works. exemple 308 174 nm.

Question 124-23 : At reference or see flight planning manual mrjt 1 figure 4.5.3.1.given long range cruise. outside air temperature oat 45°c in fl350.mass at the beginning of the leg 40000 kg.mass at the end of the leg 39000 kg..find the true airspeed tas at the end of the leg and the distance nam .. err a 033 345 ?

Tas 431 kt. 227 nam

.mass at the end of the leg 39000 kg, we read 422 kt tas on the table..temperature is isa+9, we must increase tas by 1 kt per degree c above isa.422 + 9 = 431 kt..we have burnt 1000 kg on the leg 40000 39000 , on the table, we read the following values..40000 >1163 nam..39000 > 936 nam....1163 936 = 227 nam. exemple 312 Tas 431 ktxsx 227 nam

Question 124-24 : At reference or see flight planning manual mrjt 1 figure 4.3.6.given.estimated dry operation mass 35 500 kg.estimated load 14 500 kg.final reserve fuel 1200 kg.distance to alternate 95 nm.average true track 219°.head wind component 10 kt.find fuel and time to alternate... err a 033 353 ?

1 100 kg.25 min

. /com en/com033 353.jpg.the time to alternate is given as decimal 0.42 h ==> 25 minutes. exemple 316 1 100 kgxsx25 min

Question 124-25 : At reference or see flight planning manual mrjt 1 figure 4.5.1. find time, fuel, still air distance and tas for an enroute climb 280/.74 to fl 350..given.brake release mass 64000 kg.isa +10°c.airport elevation 3000 ft.. err a 033 361 ?

26 min, 1975 kg, 157 nautical air miles nam , 399 kt

..on the table, you can read 26 min 2050 kg 157 nam 399 kt...but we can also read 'fuel adjustment for high elevation airport'..at 3000 ft, we must interpolate between 2000 and 4000 ft and remove 75 kg. exemple 320 26 min, 1975 kg, 157 nautical air miles (nam), 399 kt

Question 124-26 : 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 42 minutes, assuming other flight data remains constant, what fuel will be expected on arrival overhead mike.. ?

1475 kg

Beta to gamma 20 minutes..fuel used 3025 2525 = 500 kg.fuel flow from beta to gamma 500 / 20 = 25 kg/min.fuel used from gamma to mike 42 mins x 25 kg = 1050 kg..on arrival overhead mike, fuel on board will be..2525 1050 = 1475 kg. exemple 324 1475 kg

Question 124-27 : At reference or see flight planning manual mrjt 1 figure 4.7.3.given.distance to alternate 950 nm.head wind component 20 kt.mass at point of diversion 50000 kg.diversion fuel available 5800 kg.the minimum pressure altitude at which the above conditions may be met is.. err a 033 377 ?

22000 ft.

. /com en/com033 377.jpg.. exemple 328 22000 ft.

Question 124-28 : Planned and actual data as shown in the flight log excerpt..actual ground speed gs on the leg beta to gamma will be 105 kt..if all other flight parameters remain unchanged, what fuel remaining should be expected at waypoint gamma.. err a 033 379 ?

3260 kg.

.fuel used between alpha to beta 3670 3560 = 110 kg..fuel flow was 110 kg / 12 min = 9.17 kg/min...ground speed beta to gamma is 105 kt..flight time between beta to gamma 58 / 105 x 60 = 33.1 min...fuel used between beta to gamma 9.17 x 33.1 = 304 kg...fuel on arrival at gamma 3560 304 = 3256 kg. exemple 332 3260 kg.

Question 124-29 : How many feet you have to climb to reach fl 75 given.fl 75.departure aerodrome elevation 1500 ft. qnh = 1023 hpa.temperature = isa.1 hpa = 30 ft ?

6300 ft.

.distance height between sea level and 1013 = 1023 1013 x 30ft = 300 ft...pressure at sea level = 1023 > 1013 so 1013 hpa is above seal level...thus, qnh altitude = 7500 + 300 = 7800 ft,.and height = 7800 1500 = 6300 ft. exemple 336 6300 ft.

Question 124-30 : 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 40 minutes, assuming other flight data remains constant, what fuel will be expected on arrival overhead ?

1900 kg

.from alpha to gamma, flight time is 35 minutes 1h07 to 1h42....fuel used 3400 2700 = 700 kg...700 kg / 35 minutes = 20 kg/minute...from gamma, you are cleared for direct routeing to mike, flight time will be 40 minutes..40 minutes x 20 kg/minute = 800 kg...2700 kg 800 kg = 1900 kg. exemple 340 1900 kg

Question 124-31 : At reference or see flight planning manual mrjt 1 figure 4.7.3.given.diversion distance 650 nm.diversion pressure altitude 16 000 ft.mass at point of diversion 57 000 kg.head wind component 20 kt.temperature isa + 15°c.the diversion a fuel required and b time, are approximately .. err a 033 418 ?

A 4800kg b 2h 03min.

.first, set the red lines on the graph.. /com en/com033 418.jpg..reach the ref lines first, and after join the red lines..above 'weight at point of diversion' on the right part of the graph , you have to interpolate the dashed line and the full line dashed line for high pressure alitude flights, full line for low altitude pressure flights. exemple 344 (a) 4800kg (b) 2h 03min.

Question 124-32 : The fuel plan gives a trip fuel of 65 us gallons..the alternate fuel, final reserve included, is 17 us gallons..contingency fuel is 5% of the trip fuel..the usable fuel at departure is 93 us gallons..at a certain moment the fuel consumed according to the fuel gauges is 40 us gallons and the distance ?

The remaining fuel is not sufficient to reach the destination with reserves intact.

.usable fuel at departure 93 usg..minus reserves 72.75 usg 93 17 3.25..72.75/2 distance flown is half of the total distance = 36 usg..it misses 4 usg...the remaining fuel is not sufficient to reach the destination without using a part of our fuel reserves. exemple 348 The remaining fuel is not sufficient to reach the destination with reserves intact.

Question 124-33 : At reference or see flight planning manual mrjt 1 figure 4.1.find the optimum altitude for the twin jet aeroplane, given.cruise mass 50000 kg.mach 0.78.. err a 033 436 ?

35500 ft.

. /com en/com033 436.jpg.. exemple 352 35500 ft.

Question 124-34 : You are flying a constant compass heading of 252°..variation is 22°e.deviation is 3°w and your ins is showing a drift of 9° right...true track is ?

280°.

. /com en/com033 469.jpg..use this very useful table for those questions. exemple 356 280°.

Question 124-35 : A descent is planned from 7500 ft amsl so as to arrive at 1000 ft amsl 6 nm from a vortac..with a ground speed of 156 kt and a rate of descent of 800 ft/min..the distance from the vortac when descent is started is ?

27.1 nm.

. 7500 1000 / 800 = 8.125 minutes of descent...8.125 x 156 / 60 = 21.1 nm...21.1 + 6 = 27.1 nm. exemple 360 27.1 nm.

Question 124-36 : Given.planned and actual data as shown in the flight log excerpt.actual ground speed gs on the leg beta to gamma will be 100 kt. if all other flight parameters remain unchanged, what fuel remaining should be expected at waypoint gamma.. err a 033 474 ?

2600 kg.

.actual consumption between alpha and beta was 3000 2900 = 100 kg..flight time between alpha and beta was 12 minutes...our fuel flow is 100 kg/12 minutes = 8.33 kg/minute..between beta and gamma, ground speed will be 100 kt instead of 130 kt, we have to actualize the flight log, the leg duration will be 36 minutes distance/speed 60/100...with a fuel flow of 8.33 kg/minute, 36 minutes x 8.33 = 300 kg..fuel remaining at waypoint gamma should be 2900 300 = 2600 kg. exemple 364 2600 kg.

Question 124-37 : 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 45 minutes, assuming other flight data remains constant, what fuel will be expected on arrival overhead mike.. ?

1384 kg.

.from alpha to gamma, flight time was 39 minutes 1h07 to 1h46 , fuel flown was 3400 2464 /39 = 24 kg/min...gamma to mike = 45 minutes.45 min x 24 kg/min = 1080 kg...on arrival overhead mike, fuel on board will be.2464 1080 = 1384 kg. exemple 368 1384 kg.

Question 124-38 : Minimum planned take off fuel is 160 kg 30% total reserve fuel is included..assume the groundspeed on this trip is constant..when the aircraft has done half the distance the remaining fuel is 70 kg..is diversion to a nearby alternate necessary ?

Diversion to a nearby alternate is necessary, because the remaining fuel is not sufficient.

Fuel on board at take off = 160 kg.at half the distance, it remains 70 kg, we have burned 90 kg..the remaining quantity of 70 kg is not enough for travelling the next part of the flight. exemple 372 Diversion to a nearby alternate is necessary, because the remaining fuel is not sufficient.

Question 124-39 : Which of the following statements is relevant for forming route portions in integrated range flight planning ?

The distance from take off up to the top of climb has to be known.

exemple 376 The distance from take-off up to the top of climb has to be known.

Question 124-40 : At reference or see flight planning manual mrjt 1 figure 4.5.3.1.the aeroplane gross mass at top of climb is 61500 kg..the distance to be flown is 385 nm at fl 350 and oat 54.3 °c..the wind component is 40 kt tailwind..using long range cruise procedure what fuel is required.. err a 033 499 ?

2150 kg.

.for a 61500 kg gross mass, tas is 429 kt...nam = ngm x tas/gs.nam = 385 x 429/469.nam = 352...for a starting mass of 61500 kg, range is 5313 nam..5313 nam 352 nam = 4961 nam at the end..in the table, 4961 nam is corresponding to 59350 kg...61500 kg 59350 kg = 2150 kg. exemple 380 2150 kg.


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