Attainment > AIM : Given .planned and actual data as shown in the flight log excerpt .arriving ?

Question 97-1 : 1900 kg 2652 kg 1852 kg 1268 kg

exemple 197 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 1900 kg

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

Question 97-2 : A 4800kg b 2h 03min a 3900kg b 1h 45min a 6200kg b 2h 10min a 4400kg b 1h 35min

exemple 201 (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 (a) 4800kg (b) 2h 03min.

The fuel plan gives a trip fuel of 65 us gallons .the alternate fuel final ?

Question 97-3 : The remaining fuel is not sufficient to reach the destination with reserves intact at the destination there will still be 30 us gallons in the tanks at departure the reserve fuel was 28 us gallons at destination the required reserves remain intact

exemple 205 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 the remaining fuel is not sufficient to reach the destination with reserves intact.

At reference or see flight planning manual mrjt 1 figure 4 1.find the optimum ?

Question 97-4 : 35500 ft 36200 ft 36700 ft maximum operating altitude

exemple 209 35500 ft. — . /com en/com033 436 jpg. 35500 ft.

You are flying a constant compass heading of 252° .variation is ?

Question 97-5 : 280° 224° 242° 262°

exemple 213 280°. — . /com en/com033 469 jpg..use this very useful table for those questions 280°.

A descent is planned from 7500 ft amsl so as to arrive at 1000 ft amsl 6 nm ?

Question 97-6 : 27 1 nm 15 0 nm 11 7 nm 30 2 nm

exemple 217 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 27.1 nm.

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

Question 97-7 : 2600 kg 2820 kg 2684 kg 2770 kg

exemple 221 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 2600 kg.

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

Question 97-8 : 1384 kg 2082 kg 1002 kg 252 kg

exemple 225 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 1384 kg.

Minimum planned take off fuel is 160 kg 30% total reserve fuel is included ?

Question 97-9 : Diversion to a nearby alternate is necessary because the remaining fuel is not sufficient diversion to a nearby alternate is not necessary because the reserve fuel has not been used completely diversion to a nearby alternate is not necessary because it is allowed to calculate without reserve fuel diversion to a nearby alternate is necessary unless the captain decides to continue on his own responsibility

exemple 229 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 diversion to a nearby alternate is necessary, because the remaining fuel is not sufficient.

Which of the following statements is relevant for forming route portions in ?

Question 97-10 : The distance from take off up to the top of climb has to be known no segment shall be more than 30 minutes of flight time each reporting point requires a new segment a small change of temperature 2°c can divide a segment

exemple 233 the distance from take-off up to the top of climb has to be known. — the distance from take-off up to the top of climb has to be known.

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

Question 97-11 : 2150 kg 2050 kg 2350 kg 2250 kg

exemple 237 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 2150 kg.

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

Question 97-12 : 2625 kg 2820 kg 2900 kg 2723 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

An aircraft is in cruising flight at fl 095 ias 155 kt the pilot intends to ?

Question 97-13 : 48 nm 40 nm 45 nm 42 nm

exemple 245 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 48 nm.

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

Question 97-14 : 1706 kg 2644 kg 1720 kg 1036 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

At reference or see flight planning manual mrjt 1 figure 4 7 2. using the above ?

Question 97-15 : 584 nm 563 nm 603 nm 608 nm

Flight planning manual mrjt 1 figure 4 7 2.an aircraft on an extended range ?

Question 97-16 : 735 nm 794 nm 810 nm 875 nm

exemple 257 735 nm. — . /com en/com033 518 jpg. 735 nm.

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

Question 97-17 : 1 100 kg 1 020 kg 1 200 kg 1 680 kg

exemple 261 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 1 100 kg

After flying for 16 min at 100 kt tas with a 20 kt tail wind component you have ?

Question 97-18 : 25 min 21 min 11 min 40 sec 16 min

exemple 265 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 25 min.

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

Question 97-19 : 1180 kg 30 minutes 2360 kg 30 minutes 2360 kg 01h00 1180 kg 45 minutes

exemple 269 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 1180 kg, 30 minutes.

In the cruise at fl 155 at 260 kt tas the pilot plans for a 500 ft/min descent ?

Question 97-20 : 120 nm 140 nm 110 nm 130 nm

exemple 273 120 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 120 nm.

For this question use annex 033 11704a .true air speed 170 kt.wind in the area ?

Question 97-21 : 1 545 litres 1 600 litres 1 326 litres 1 182 litres

exemple 277 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 1 545 litres.

At reference 033 345.a twin jet aeroplane gross mass 200000 kg begin his cruise ?

Question 97-22 : 192 500 kg 193 400 kg 193 800 kg 193 000 kg

exemple 281 192 500 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% 192 500 kg.

Planned and actual data as shown in the flight log excerpt 033 145.actual ?

Question 97-23 : 2000 kg 2062 kg 2310 kg 2160 kg

exemple 285 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 2000 kg.

The trip fuel for a jet aeroplane to fly from the departure aerodrome to the ?

Question 97-24 : 14548 kg 14363 kg 14130 kg 15948 kg

exemple 289 14548 kg. — Cqb15 january 2012..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 14548 kg.

Use reference 033 046.a turbojet aeroplane flies using the following data ?

Question 97-25 : 105 nm/1000 kg.5330 kg/h 105 nm/1000 kg.6515 kg/h 86 nm/1000 kg.6515 kg/h 71 nm/1000 kg.5330 kg/h

exemple 293 105 nm/1000 kgxsx5330 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 105 nm/1000 kgxsx5330 kg/h.

See reference 033 068.given .distance between c to d 540 nm.cruise speed 300 kt ?

Question 97-26 : 4240 kg 4620 kg 3680 kg 3350 kg

exemple 297 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 4240 kg.

Use route manual chart e hi 2..what is the meaning of the chart information for ?

Question 97-27 : Ndb only ident oe ndb ident oe and vor ident vey vor only ident vey vor/dme ident sup and ndb ident oe

exemple 301 ndb only, ident oe. — . /com fr/com033 340 jpg. ndb only, ident oe.

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

Question 97-28 : 4940 kg 5090 kg 4690 kg 5010 kg

exemple 305 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 4940 kg.

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

Question 97-29 : 5669 kg 5320 kg 6175 kg 6749 kg

exemple 309 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 5669 kg.

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

Question 97-30 : 4550 kg 4740 kg 4140 kg 4640 kg

exemple 313 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 4550 kg.

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

Question 97-31 : 4690 kg 4440 kg 4160 kg 4510 kg

exemple 317 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 4690 kg.

Vtoss is the take off safety speed for ?

Question 97-32 : Category a helicopters class 2 helicopters single engine and multi engine helicopters single engine helicopters

exemple 321 category a helicopters. — category a helicopters.

Using attached graphic according to the flight manual diagram the never exceed ?

Question 97-33 : 115 kt 110 kt 120 kt 105 kt

exemple 325 115 kt. — Vne at 10000 ft oat +10°c and 2300 kg is 115 kt ias.the graphic states at any gross mass above 2300 kg decrease vne by 10 kias.since the inflight mass is below 2300 kg we do not change the value 115 kt.

According to the flight manual diagram the never exceed speed vne at pressure ?

Question 97-34 : 105 kt 110 kt 115 kt 125 kt

exemple 329 105 kt. — You must removed 10 kias at any gross mass above 2300 kg 105 kt.

Using attached graphic which letter indicates the speed for maximum endurance. ?

Question 97-35 : B c d a

exemple 333 b — Power required curve power versus airspeed chart. 690..a power required to hover outside ground effect..b power required to hover inside ground effect..c translational lift area..d adjustment of power required to counteract sinking..e minimum power for level flight/maximum rate of climb speed/ speed for maximum endurance b

Using attached graphic which letter indicates the speed for the best rate of ?

Question 97-36 : B a c d

exemple 337 b — Power required curve power versus airspeed chart. 690..a power required to hover outside ground effect..b power required to hover inside ground effect..c translational lift area..d adjustment of power required to counteract sinking..e minimum power for level flight/ maximum rate of climb speed / speed for maximum endurance b

Using attached graphic which letter indicates the speed for best range. 688 ?

Question 97-37 : C a b d

exemple 341 c — The best range will be reach at the minimum total drag speed.power required curve power versus airspeed chart. 690..a power required to hover outside ground effect..b power required to hover inside ground effect..c translational lift area..d adjustment of power required to counteract sinking..e minimum power for level flight/maximum rate of climb speed/ speed for maximum endurance c

A head wind will ?

Question 97-38 : Increase the climb flight path angle increase the angle of climb increase the rate of climb shorten the time to a given altitude

exemple 345 increase the climb flight path angle. — increase the climb flight path angle.

A helicopter which has no guaranteed 'stay up' ability in the event of an ?

Question 97-39 : Category b category a category c category d

exemple 349 category b. — No guaranteed stay up ability means that the helicopter cannot maintain level flight in case of an engine failure.category b helicopters single engine or a multi engine helicopter that does not meet category a standards have no guaranteed ability to continue safe flight in the event of an engine failure and a forced landing is assumed category b.

A helicopter will obtain a maximum flight distance at the speed ?

Question 97-40 : For maximum range for minimum ho y fuel flow for maximum endurance the speed for minimum power required

exemple 353 for maximum range. — for maximum range.

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