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Question 111-1 : 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 ? [ Training professional ]

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 211 174 nm.

Question 111-2 : 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 215 Tas 431 ktxsx 227 nam

Question 111-3 : 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 219 1 100 kgxsx25 min

Question 111-4 : 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 223 26 min, 1975 kg, 157 nautical air miles (nam), 399 kt

Question 111-5 : 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 227 1475 kg

Question 111-6 : 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 231 22000 ft.

Question 111-7 : 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 235 3260 kg.

Question 111-8 : 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 239 6300 ft.

Question 111-9 : 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 mike ?

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

Question 111-10 : 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 247 (a) 4800kg (b) 2h 03min.

Question 111-11 : 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 251 The remaining fuel is not sufficient to reach the destination with reserves intact.

Question 111-12 : 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 255 35500 ft.

Question 111-13 : 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 259 280°.

Question 111-14 : 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 263 27.1 nm.

Question 111-15 : 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 267 2600 kg.

Question 111-16 : 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 271 1384 kg.

Question 111-17 : 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 275 Diversion to a nearby alternate is necessary, because the remaining fuel is not sufficient.

Question 111-18 : 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 279 The distance from take-off up to the top of climb has to be known.

Question 111-19 : 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 283 2150 kg.

Question 111-20 : Planned and actual data as shown in the flight log excerpt .actual ground speed gs on the leg beta to gamma will be 110 kt .if all other flight parameters remain unchanged what fuel remaining should be expected at waypoint gamma . err a 033 502 ?

2625 kg

.between alpha to beta actual flight time was 12 minutes and fuel consumption was 3000 2900 = 100 kg .100 kg / 12 min = 8 33 kg/min .from beta to gamma distance is 60 nm .60 nm at 110 kt = 60/110 = 0 5454 hour of flight. 8 33 x 60 x 0 5454 = 273 kg .2900 273 = 2627 kg exemple 287 2625 kg.

Question 111-21 : An aircraft is in cruising flight at fl 095 ias 155 kt the pilot intends to descend at 500 ft/min to arrive overhead the man vor at 2 000 ft qnh 1 030hpa the tas remains constant in the descent wind is negligeable temperature standard .at which distance from man should the pilot commence the ?

48 nm

.tas remains constant in the descent and wind is negligeable tas = ground speed .for every 1000 ft of pressure altitude add 2% to the ias to calculate tas .2% of 155 kt = 155 x 0 02 = 3 1 kt .3 1 x 9 5 = 29 45 kt.tas at fl95 = 155 kt + 29 kt = 184 kt .fl 95 is for 1013 hpa if we change our subscale settig we increase indicated altitue by 17 hpa x 27 ft = 459 ft .9500 + 459 2000 = 7959 ft to loose .7959 / 500 = 16 minutes . 184 /60 x 16 = 49 nm exemple 291 48 nm.

Question 111-22 : 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 1h06 min assuming other flight data remains constant what fuel will be expected on arrival overhead mike . ?

1706 kg

Flight time from alpha to gamma = 20 minutes + 35 minutes = 55 minutes.fuel used from alpha to gamma = 3400 2630 = 770 kg.fuel flow = 770/55 x 60 = 840 kg/h.gamma to mike 1 h 06 min . 840/60 x 66 = 924 kg .on arrival overhead mike fuel on board will be .2630 kg 924 kg = 1706 kg exemple 295 1706 kg.

Question 111-23 : At reference or see flight planning manual mrjt 1 figure 4 7 2. using the above table in isa conditions and at a speed of m 70/280kias in an elapsed time of 90 minutes an aircraft with mass at point of diversion 48000 kg could divert a distance of . err a 033 514 ?

584 nm

exemple 299 584 nm

Question 111-24 : Flight planning manual mrjt 1 figure 4 7 2.an aircraft on an extended range operation is required never to be more than 120 minutes from an alternate based on 1 engine inoperative lrc conditions in isa using the above table and a given mass of 40000 kg at the most critical point the maximum air ?

735 nm

. /com en/com033 518 jpg. exemple 303 735 nm.

Question 111-25 : At reference or see flight planning manual mrjt 1 figure 4 5 3 1 .given long range cruise temp 63° c at fl 330.initial gross mass enroute 54 100 kg. leg flight time 29 min.find fuel consumption for this leg . err a 033 529 ?

1 100 kg

54100 kg corresponds to 3929 nam.isa 12°c.433 1 kt per degree c below isa = 421 kt.29 minutes at speed 421 kt = 203 nm 3929 203 = 3726 corresponds to 53000 kg.54100 53000 = 1100 kg..correction isa 12°c = 0 7% ==> 0 7 x 1100 kg = 8 kg ..exact answer is 1100 8 = 1092 kg exemple 307 1 100 kg

Question 111-26 : After flying for 16 min at 100 kt tas with a 20 kt tail wind component you have to return to the airfield of departure .with a rate turn of 3°/s you will arrive after ?

25 min

.180°/3 = 60 secondes for turn back .gs out 100 + 20 = 120 kt .gs home 100 20 = 80 kt .16 x 120/80 = 24 minutes .24 minutes + 1 minute = 25 minutes exemple 311 25 min.

Question 111-27 : 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.estimated mass on arrival at the destination 52525 kg.alternate elevation msl.destination elevation 1500 ft.find final reserve fuel and corresponding time . err ?

1180 kg 30 minutes

.final reserve fuel is simple calculations based on 30 minutes jet/turbo prop aeroplane hold at endurance speed calculated with the estimated mass on arrival at the alternate aerodrome or the destination aerodrome when no destination alternate aerodrome is required eu ops subpart d appendix 1 to 1 255 . /com en/com033 560 jpg.2360 / 2 = 1180 kg exemple 315 1180 kg, 30 minutes.

Question 111-28 : In the cruise at fl 155 at 260 kt tas the pilot plans for a 500 ft/min descent in order to fly overhead man vor at 2 000 feet qnh 1030 .tas will remain constant during descent wind is negligible temperature is standard .the pilot must start the descent at a distance from man of ?

120 nm

.as we are descending for reaching an altitude we have to change our subscale from 1013 to 1030 before descent .1013 hpa to 1030 hpa = 17 hpa .we start our descent at an altitude of 15500 + 17 hpa x 27 ft = 15959 ft .15959 ft 2000 ft = 13959 ft to loose .13959 ft / 500 ft/min = 28 minutes .28 min at 260 kt = 28 x 260/60 = 121 nm exemple 319 120 nm.

Question 111-29 : For this question use annex 033 11704a .true air speed 170 kt.wind in the area 270°/40 kt.according to the attached the navigation log an aircraft performs a turn overhead bulen to re route to ard via tgj the given wind conditions remaining constant .the fuel consumption during the turn is 20 ?

1 545 litres

Proceed as if there is only one leg ard to bulen 243 nm .track 123°.fuel consumption from ard to bulen = 869 432 = 437 litres ..on nav computer we find a 202 kt gso groundspeed out and 135 kt gsh groundspeed home ..time on the leg = 243 / 202 = 1 2 h.consumption = 437 / 1 2 = 364 l/h..calculate fuel for the leg bulen ard .time on this leg = 243 / 135 = 1 8 h.consumption = 364 x 1 8 = 655 l..total consumption = 869 + 20 + 655 = 1544 l exemple 323 1 545 litres.

Question 111-30 : At reference 033 345 .a twin jet aeroplane gross mass 200000 kg begin his cruise leg at the optimum level of mach 0 84 isa cg=37% total anti ice devices on head wind component 30 kt .after 500 nm to avoid a severe icing area an immediate descent must be initiated the airplane mass at the beginning ?

192 500 kg

On the reference for 200000 kg we have 6778 nm and 844 min .tas = 484 kt headwind = 30 kt.gs = 454 kt.time = 500/454 = 1 10h = 66 min.66 min + 7% = 71 min.844 min 71 min = 773 min.on the reference 773 minutes is corresponding to a value of 192 500 kg . ninorr .where did you take 7% from ..on the reference .total anti ice on .fuel= +7% exemple 327 192 500 kg.

Question 111-31 : Planned and actual data as shown in the flight log excerpt 033 145 .actual ground speed gs on the leg beta to gamma is 110 kt .if all other flight parameters remain unchanged what fuel remaining should be expected over gamma . err a 033 578 ?

2000 kg

Cqb15 january 2012 ..between alpha to beta actual consumption was 2470 2330 = 140 kg.140 kg for 20 minutes column 'ate' so 140x3 = 420 kg/heure .distance between beta and gamma is 85 nm column 'leg dist' at a ground speed of 110 kt .it will take .85 /110 = 0 773 h .0 773 x 420 = 325 kg of fuel burned .the fuel remaining over gamma should be .2330 325 = 2005 kg .closest answer 2000 kg exemple 331 2000 kg.

Question 111-32 : The trip fuel for a jet aeroplane to fly from the departure aerodrome to the destination aerodrome is 8350 kg .fuel consumption in holding mode is 2800 kg/h .the quantity of fuel which is needed to carry out one go around and land on the alternate airfield is 4380 kg .the destination aerodrome has ?

14548 kg

.minimum quantity of fuel at take off = trip fuel + alternate + contingency + 30 min final reserve jet aircraft .trip fuel = 8350 kg .alternate = 4380 kg .contingency = the greater of 5% of trip or 5 min holding at 1500 ft .5% of trip = 5% x 8350 = 418 kg.5 min holding at 1500 ft = 2800 x 5/60 = 233 kg .30 min final reserve = 2800 /2 = 1400 kg .minimum quantity of fuel at take off = 8350 + 418 + 4380 + 1400 = 14548 kg exemple 335 14548 kg.

Question 111-33 : Use reference 033 046 .a turbojet aeroplane flies using the following data .optimum flight level mach 0 80.mass 190000 kg.temperature isa.tailwind component 100 kt .fuel mileage and ho y fuel consumption are . . err a 033 580 ?

105 nm/1000 kg.5330 kg/h

From reference for 190000 kg we get 6515 nam and tas 459 kt .6515 459 = 6056 nam > 184 600 kg.190 000 184 600 = 5400 kg/h.distance accomplished in one hour with the wind is 559 nm 459 + 100 .thus fuel consumption for 1000kg is 559/5 4 = 103 5 nm/1000kg exemple 339 105 nm/1000 kgxsx5330 kg/h.

Question 111-34 : See reference 033 068 .given .distance between c to d 540 nm.cruise speed 300 kt ias at fl210.temperature deviation from isa +20°c.headwind component 50 kt.gross mass at c 60 000 kg.the fuel required from c to d is . err a 033 581 ?

4240 kg

Cqb15 january 2012 ..nam = ground distance x tas/gs .nam = 540 x 406/356 = 616 nm.from reference 033 068 at line 60000 the cruise distance nautical air miles is 3898 nautique air miles substract 616 you find 3282 nam .back to reference 033 068 what is the mass for 3282 nm we get 55800 kg.60000 55800=4200kg.last step we have to increase fuel required by 0 5 percent per 10 degrees above isa we are in isa+20°c so 1 percent more .4200x1%= 4242 kg ..see section 5 4 2 method page 24 on caa cap697 flight planning manual for that kind of questions .notice you must use the tas given for cruise 406 kt as state in the pdf document exemple 343 4240 kg.

Question 111-35 : Use route manual chart e hi 2.what is the meaning of the chart information for the beacon s at position 55°59'n 014°06'e . err a 033 582 ?

Ndb only ident oe

. /com fr/com033 340 jpg. exemple 347 Ndb only, ident oe.

Question 111-36 : Given .planned and actual data as shown in the flight log excerpt provided that flight conditions on the leg gamma to delta remain unchanged and fuel consumption remains unchanged what fuel remaining should be expected at waypoint delta . err a 033 587 ?

4940 kg

Buhoraptor .hello my answer is 4990 kg .please can you tell me why there are difference thank you and congratulations for your website ..thank you .beta to gamma 10 minutes .fuel used 5440 5340 = 100 kg.fuel flow from beta to gamma 100 / 10 = 10 kg/min.planned time from gamma to delta 1 53 to 2 33 = 40 minutes .planned fuel from gamma to delta 40 min x 10 kg = 400 kg.on arrival overhead delta fuel on board will be .5340 400 = 4940 kg exemple 351 4940 kg.

Question 111-37 : 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 1h 20min and the fuel flow ?

5669 kg

.1h40 = 1 33 h.720 x 1 33 = 960 kg.estimated mass at destination est ldg mass at dest 6629 kg .6629 960 = 5669 kg exemple 355 5669 kg.

Question 111-38 : Given .planned and actual data as shown in the flight log excerpt .provided that flight conditions on the leg gamma to delta remain unchanged and fuel consumption remains unchanged .what fuel remaining should be expected at waypoint delta . err a 033 589 ?

4550 kg

.beta to gamma 10 minutes .fuel used 5270 5150 = 120 kg.fuel flow from beta to gamma 120 / 10 = 12 kg/min.fuel used from gamma to delta 50 min x 12 kg = 600 kg.on arrival overhead delta fuel on board will be .5150 600 = 4550 kg exemple 359 4550 kg.

Question 111-39 : Given .planned and actual data as shown in the flight log excerpt.provided that flight conditions on the leg gamma to delta remain unchanged and fuel consumption remains unchanged what fuel remaining should be expected at waypoint delta . err a 033 590 ?

4690 kg

.beta to gamma 65 minutes .fuel used 5490 4970 = 520 kg.fuel flow from beta to gamma 520 / 65 = 8 kg/min.fuel used from gamma to delta 35 min x 8 kg = 280 kg..on arrival overhead delta fuel on board will be .4970 280 = 4690 kg exemple 363 4690 kg.

Question 111-40 : Vtoss is the take off safety speed for ?

Category a helicopters

exemple 367 Category a helicopters.


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