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

Question 96-1 : 76 nm 97 nm 65 nm 87 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 exemple 196 76 nm.76 nm.

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

Question 96-2 : 25000 ft 20000 ft 16000 ft 14500 ft

. /com en/com033 76 jpg. exemple 200 25000 ft.25000 ft.

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

Question 96-3 : 193 nm 128 nm 157 nm 228 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 204 193 nm.193 nm.

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

Question 96-4 : 2800 kg 2550 kg 2900 kg 2650 kg

Answer.. /com en/com033 1012 jpg exemple 208 2800 kg.2800 kg.

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

Question 96-5 : 1000 kg 24 min 800 kg 0 4 hr 800 kg 24 min 1000 kg 40 min

/com en/com033 1015 jpg exemple 212 1000 kg, 24 min.1000 kg, 24 min.

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

Question 96-6 : 5987 kg 5510 kg 6470 kg 7427 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 216 5987 kg.5987 kg.

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

Question 96-7 : 53 nm 61 nm 79 nm 85 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 220 53 nm.53 nm.

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

Question 96-8 : 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 exemple 224 420 kt.420 kt.

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

Question 96-9 : Tas 432 kt 227 nam tas 422 kt 936 nam tas 422 kt 227 nam tas 432 kt 1163 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 228 tas 432 kt, 227 nam.tas 432 kt, 227 nam.

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

Question 96-10 : 12 minutes 4 minutes 44 minutes 63 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 232 12 minutes.12 minutes.

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

Question 96-11 : 5440 kg 6070 kg 5080 kg 7240 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 236 5440 kg.5440 kg.

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

Question 96-12 : 35 kt 15 kt 70 kt 0 kt

. /com en/com033 164 jpg. exemple 240 35 kt.35 kt.

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

Question 96-13 : 2300 kg/h 1150 kg/h 2994 kg/h 1497 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 244 2300 kg/h2300 kg/h

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

Question 96-14 : By 5% by 5% by 8% by 7%

.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 248 by -5%by -5%

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

Question 96-15 : 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 252 distance in nautical miles (nm), wind component, landing mass at alternatedistance in nautical miles (nm), wind component, landing mass at alternate

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

Question 96-16 : 30 3 us gallons/hour 33 0 us gallons/hour 37 9 us gallons/hour 21 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 256 30.3 us gallons/hour30.3 us gallons/hour

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

Question 96-17 : By 7 6% by 2 3% by 3 6% by 5 3%

. /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 260 by 7.6%.by 7.6%.

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

Question 96-18 : 2500 kg 2750 kg 3050 kg 2900 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 264 2500 kg.2500 kg.

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

Question 96-19 : 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 268 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 96-20 : 3700 kg 3400 kg 3100 kg 4000 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 272 3700 kg.3700 kg.

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

Question 96-21 : 150 kg 290 kg 278 kg 140 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 276 150 kg.150 kg.

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

Question 96-22 : 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

. /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 280 (a) 860 nm (b) 3h 20 min.(a) 860 nm (b) 3h 20 min.

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

Question 96-23 : 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

.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 284 25 kg trip fuel and 8 kg reserve fuel.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 96-24 : 0 8 m 0 78 m 0 76 m 0 72 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 288 0.8 m0.8 m

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

Question 96-25 : Fl90 fl80 fl75 fl70

.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 292 fl90.fl90.

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

Question 96-26 : Fl 140 fl 130 fl 150 fl 120

.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 296 fl 140.fl 140.

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

Question 96-27 : 42195 kg 41110 kg 42210 kg 42312 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 300 42195 kg.42195 kg.

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

Question 96-28 : 19 min 17 min 8 min 10 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 exemple 304 19 min.19 min.

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

Question 96-29 : 1180 kg 2360 kg 1150 kg 2300 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 308 1180 kg.1180 kg.

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

Question 96-30 : 1300 kg 2910 kg 1510 kg 380 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 312 1300 kg.1300 kg.

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

Question 96-31 : 1720 kg 2305 kg 1450 kg 790 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 316 1720 kg.1720 kg.

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

Question 96-32 : 434 kt 424 kt 415 kt 428 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 320 434 kt.434 kt.

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

Question 96-33 : 174 nm 203 nm 188 nm 193 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 exemple 324 174 nm.174 nm.

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

Question 96-34 : Tas 431 kt. 227 nam tas 423 kt. 936 nam tas 423 kt. 227 nam tas 431 kt. 1163 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 328 tas 431 ktxsx 227 namtas 431 ktxsx 227 nam

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

Question 96-35 : 1 100 kg.25 min 800 kg.40 min 1 100 kg.44 min 800 kg.24 min

. /com en/com033 353 jpg.the time to alternate is given as decimal 0 42 h ==> 25 minutes exemple 332 1 100 kgxsx25 min1 100 kgxsx25 min

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

Question 96-36 : 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

..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 336 26 min, 1975 kg, 157 nautical air miles (nam), 399 kt26 min, 1975 kg, 157 nautical air miles (nam), 399 kt

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

Question 96-37 : 1475 kg 2245 kg 1195 kg 670 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 340 1475 kg1475 kg

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

Question 96-38 : 22000 ft 20000 ft 18000 ft 28000 ft

. /com en/com033 377 jpg. exemple 344 22000 ft.22000 ft.

Planned and actual data as shown in the flight log excerpt .actual ground speed ?

Question 96-39 : 3260 kg 3318 kg 3480 kg 3430 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 348 3260 kg.3260 kg.

How many feet you have to climb to reach fl 75 given .fl 75.departure aerodrome ?

Question 96-40 : 6300 ft 6000 ft 6600 ft 7800 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 352 6300 ft.6300 ft.

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