A turbojet aeroplane flies using the following data .flight level fl 330.flight ? [ Validation Marking ]

Question 98-1 : 539 nm 518 nm 471 nm 493 nm

exemple 198 539 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 539 nm.

Given .true course tc 017°.w/v 340°/30 kt.true air speed tas 420 kt.find ?

Question 98-2 : Wca 2° gs 396 kt wca +2° gs 396 kt wca 2° gs 426 kt wca +2° gs 416 kt

exemple 202 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 Wca -2°, gs 396 kt.

The true course is 042° .the variation in the area is 6° w and the wind is ?

Question 98-3 : 052° 044° 040° 058°

exemple 206 052°. — 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 052°.

At reference or see flight planning manual mrjt 1 figure 4 3 5 .given a trip ?

Question 98-4 : By 7% by 3% by 4% by 10%

exemple 210 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% By 7%.

At reference or see flight planning manual mrjt 1 figure 4 5 4 .planning an ifr ?

Question 98-5 : 76 nm 97 nm 65 nm 87 nm

exemple 214 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 76 nm.

At reference or see flight planning manual mrjt 1 figure 4 7 3 .given ?

Question 98-6 : 25000 ft 20000 ft 16000 ft 14500 ft

exemple 218 25000 ft. — . /com en/com033 76 jpg. 25000 ft.

Flight planning chart for an aircraft states that the time to reach the ?

Question 98-7 : 193 nm 128 nm 157 nm 228 nm

exemple 222 193 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 193 nm.

At reference or see flight planning manual mrjt 1 figure 4 3 6 .given .distance ?

Question 98-8 : 2800 kg 2550 kg 2900 kg 2650 kg

exemple 226 2800 kg. — Answer .. /com en/com033 1012 jpg 2800 kg.

At reference or see flight planning manual mrjt 1 figure 4 3 6 .given .twin jet ?

Question 98-9 : 1000 kg 24 min 800 kg 0 4 hr 800 kg 24 min 1000 kg 40 min

exemple 230 1000 kg, 24 min. — Img /com en/com033 1015 jpg 1000 kg, 24 min.

Given .planning data as shown in the flight log excerpt fuel planning section ?

Question 98-10 : 5987 kg 5510 kg 6470 kg 7427 kg

exemple 234 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 5987 kg.

Flight planning chart for an aeroplane states that the time to reach fl 190 at ?

Question 98-11 : 53 nm 61 nm 79 nm 85 nm

exemple 238 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 53 nm.

At reference or see flight planning manual mrjt 1 figure 4 5 3 1.given twin jet ?

Question 98-12 : 420 kt 433 kt 431 kt 418 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

At reference or see flight planning manual mrjt 1 figure 4 5 3 1. given long ?

Question 98-13 : Tas 432 kt 227 nam tas 422 kt 936 nam tas 422 kt 227 nam tas 432 kt 1163 nam

exemple 246 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 Tas 432 kt, 227 nam.

During an ifr flight in a beech bonanza the fuel indicators show that the ?

Question 98-14 : 12 minutes 4 minutes 44 minutes 63 minutes

exemple 250 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 12 minutes.

Given .planning data as shown in the flight log excerpt fuel planning section ?

Question 98-15 : 5440 kg 6070 kg 5080 kg 7240 kg

exemple 254 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 5440 kg.

At reference or see flight planning manual mrjt 1 figure 4 3 1c.for a flight of ?

Question 98-16 : 35 kt 15 kt 70 kt 0 kt

exemple 258 35 kt. — . /com en/com033 164 jpg. 35 kt.

At reference or see flight planning manual mrjt 1 figure 4 5 3 2. find the fuel ?

Question 98-17 : 2300 kg/h 1150 kg/h 2994 kg/h 1497 kg/h

exemple 262 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 2300 kg/h

At reference or see flight planning manual mrjt 1 figure 4 3 1c. within the ?

Question 98-18 : By 5% by 5% by 8% by 7%

exemple 266 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% By -5%

At reference or see flight planning manual mrjt 1 figure 4 3 6. in order to ?

Question 98-19 : Distance in nautical miles nm wind component landing mass at alternate distance in nautical air miles nam wind component landing mass at alternate distance in nautical miles nm wind component zero fuel mass distance in nautical miles nm wind component dry operating mass plus holding fuel

exemple 270 Distance in nautical miles (nm), wind component, landing mass at alternate — Distance in nautical miles (nm), wind component, landing mass at alternate

During a vfr flight at a navigational checkpoint the remaining usable fuel in ?

Question 98-20 : 30 3 us gallons/hour 33 0 us gallons/hour 37 9 us gallons/hour 21 3 us gallons/hour

exemple 274 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 30.3 us gallons/hour

At reference or see flight planning manual mrjt 1 figure 4 3 1c.for a flight of ?

Question 98-21 : By 7 6% by 2 3% by 3 6% by 5 3%

exemple 278 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% By 7.6%.

At reference or see flight planning manual mrjt 1 figure 4 3 6.given .distance ?

Question 98-22 : 2500 kg 2750 kg 3050 kg 2900 kg

exemple 282 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 2500 kg.

An aeroplane is on an ifr flight the flight is to be changed from ifr to vfr is ?

Question 98-23 : Yes the pilot in command must inform atc using the phrase 'cancelling my ifr flight' no you have to remain ifr in accordance to the filed flight plan no only atc can order you to do this but only with permission from atc

exemple 286 Yes, the pilot in command must inform atc using the phrase 'cancelling my ifr flight'. — Yes, the pilot in command must inform atc using the phrase 'cancelling my ifr flight'.

See flight planning manual mrjt 1 figure 4 5 2 and 4 5 3 1 . given .distance c ?

Question 98-24 : 3700 kg 3400 kg 3100 kg 4000 kg

exemple 290 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 3700 kg.

Flight planning manual mrjt 1 figure 4 5 4.a descent is planned at 74/250kias ?

Question 98-25 : 150 kg 290 kg 278 kg 140 kg

exemple 294 150 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 150 kg.

At reference or see flight planning manual mrjt 1 figure 4 7 3.given .diversion ?

Question 98-26 : A 860 nm b 3h 20 min a 1000 nm b 3h 40 min a 760 nm b 4h 30 min a 1130 nm b 3h 30 min

exemple 298 (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 (a) 860 nm (b) 3h 20 min.

For a planned flight the calculated fuel is as follows .flight time 2h42min ?

Question 98-27 : 25 kg trip fuel and 8 kg reserve fuel 33 kg trip fuel and 10 kg reserve fuel 23 kg trip fuel and 10 kg reserve fuel 33 kg trip fuel and no reserve fuel

exemple 302 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 25 kg trip fuel and 8 kg reserve fuel.

An aircraft is flying at mach 0 84 at fl 330 .the static air temperature is ?

Question 98-28 : 0 8 m 0 78 m 0 76 m 0 72 m

exemple 306 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 0.8 m

You must fly ifr on an airway orientated 135° magnetic with a msa at 7800 ft ?

Question 98-29 : Fl90 fl80 fl75 fl70

exemple 310 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 Fl90.

An aircraft following a 215° true track must fly over a 10 600 ft obstacle ?

Question 98-30 : Fl 140 fl 130 fl 150 fl 120

exemple 314 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 Fl 140.

Given .dry operating mass 33000 kg .traffic load 8110 kg .final reserve fuel ?

Question 98-31 : 42195 kg 41110 kg 42210 kg 42312 kg

exemple 318 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 42195 kg.

At reference or see flight planning manual mrjt 1 figure 4 5 4.planning an ifr ?

Question 98-32 : 19 min 17 min 8 min 10 min

exemple 322 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 can you explain to me please ..the wind will affect the distance but not the time 19 min.

At reference or see flight planning manual mrjt 1 figure 4 4.given .twin jet ?

Question 98-33 : 1180 kg 2360 kg 1150 kg 2300 kg

exemple 326 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 1180 kg.

Given .planned and actual data as shown in the flight log excerpt .arriving ?

Question 98-34 : 1300 kg 2910 kg 1510 kg 380 kg

exemple 330 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 1300 kg.

Given .planned and actual data as shown in the flight log excerpt .arriving ?

Question 98-35 : 1720 kg 2305 kg 1450 kg 790 kg

exemple 334 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 1720 kg.

Given .fl 370.mach 0 74.oat 47°c.the tas is ?

Question 98-36 : 434 kt 424 kt 415 kt 428 kt

exemple 338 434 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 434 kt.

The still air distance in the climb is 189 nautical air miles nam and time 30 ?

Question 98-37 : 174 nm 203 nm 188 nm 193 nm

exemple 342 174 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 .. it works 174 nm.

At reference or see flight planning manual mrjt 1 figure 4 5 3 1.given long ?

Question 98-38 : Tas 431 kt. 227 nam tas 423 kt. 936 nam tas 423 kt. 227 nam tas 431 kt. 1163 nam

exemple 346 Tas 431 ktxsx 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 Tas 431 ktxsx 227 nam

At reference or see flight planning manual mrjt 1 figure 4 3 6.given .estimated ?

Question 98-39 : 1 100 kg.25 min 800 kg.40 min 1 100 kg.44 min 800 kg.24 min

exemple 350 1 100 kgxsx25 min — . /com en/com033 353 jpg.the time to alternate is given as decimal 0 42 h ==> 25 minutes 1 100 kgxsx25 min

At reference or see flight planning manual mrjt 1 figure 4 5 1. find time fuel ?

Question 98-40 : 26 min 1975 kg 157 nautical air miles nam 399 kt 26 min 2050 kg 157 nautical air miles nam 399 kt 20 min 1750 kg 117 nautical air miles nam 288 kt 25 min 1875 kg 148 nautical air miles nam 391 kt

exemple 354 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 26 min, 1975 kg, 157 nautical air miles (nam), 399 kt

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