At reference or see Flight Planning Manual MEP 1 Figure 3 2A flight is to be made in a multi engine piston aeroplane MEP The cruising level will be ?
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Given Distance from departure to destination 180 NM Safe Endurance 2 8 h True Track 065 W/V 245/25 TAS 100 kt What is the distance of the PSR from the departure point ?
Given Distance from departure to destination 500 NM GS Out 95 ktGS Home 125 kt What is the distance of the PET from the departure point ?
Which best describes weather if any at lyon/st exupery at 1330 utc err _a_033 391 Light rain associated with thunderstorms, frequent rain showers, fog, nil. img /com_en/com033 391.jpg ts = thunderstorm ra = rain indicator ' ' means light, to indicate intensity of certain phenomena. example from annex 3 +shra = heavy shower of rain +tssngr = thunderstorm with heavy snow hail. 
Given distance x to y 2700 nm mach number 0 75temperature 45°cmean wind component 'on' 10 kt tailwind mean wind component 'back' 35 kt tailwind the distance from x to point of equal time pet between x and y 386 nm,35nm,3 3 nm,425 nm. set corresponding mark m(kt) against outside temperature at flight altitude. read in front of mach number, on outer scale, true air speed. example 45° mach 0.75 => tas 437 kt. img /com_en/com033 48.jpg ground speed out (gso) = 437 + 10 kt = 447 kt ground speed home (gsh) = 437 + 35 kt = 472 kt distance to pet = distance x gsh / (gso + gsh) distance to pet = 2700 x 472 / (447 + 472) distance to pet = 1274400 / 919 = 1386 nm. 
At reference or see flight planning manual mep1 figure 3 6a flight to be made to an airport pressure altitude 3000 ft in a multi engine piston aireroplane mep1 the forecast oat the airport 1° c the cruising level will be fl 110 where oat 10° c calculate still air descent distance 145 kias rate of descent 1000 ft/min gears and flaps up err _a_033 398 2nm, 29 nm, 36 nm, 25 nm. img /com_en/com033 398.jpg distance to descend = 29 8 = 21 nm (close to answer). 
Given distance from departure to destination 2800 nm true track 140 w/v 140/100 tas 500 kt what the distance and time of pet from departure point Distance 68nm time252 min, distance2nm time2 min, distance 4 nm time 68 min, distance2nm time 34 min We start on a true track of 140°, we have a 100 kt headwind, thus ground speed out = 500 kt 100 kt = 400 kt ground speed home = 500 kt + 100 kt = 600 kt pet = distance x gsh / (gso+ gsh) pet = 2800 x 600 / (400 + 600) pet = 1680000 / 1000 = 1680 nm. time of pet from departure point 1680 / 400 = 4.2 h 4.2 x 60 = 252 minutes. Question Pre-Flight 93 Answer 8
Route manual chart napthe initial magnetic course from c 62°n020°w to b 58°n004°e err _a_033 410 img /com_en/com033 410.jpg true course 098° + 18°w magnetic variation = 116°. Question Pre-Flight 93 Answer 9
Find time to point of safe return psr given maximum useable fuel 15000 kgminimum reserve fuel 3500 kgtas out 425 kthead wind component out 30 kttas return 430 kttailwind component return 20 ktaverage fuel flow 2150 kg/h 2 h 5min, 3 h 43 min, 2 h 59 min, 2 h 43 min point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) outbound gs = 425 30 = 395 kt homeward gs = 430 + 20 = 450 kt endurance = (15000 3500) / 2150 = 5.34 h point of safe return (psr) = 5.34 x 450 / (395 + 450) point of safe return (psr) = 5.34 x 450 / (845) point of safe return (psr) = 2.84 h 2.84 h = 2 h 51 min we are talking about a point of safe return, a decision in case of our destination finally unreachable (weather issue example) legally we need to return to our departure airport with our minimum reserve fuel. Question Pre-Flight 93 Answer 10
At reference or see flight planning manual mrjt 1 figure 4 1 find optimum altitude the twin jet aeroplane given cruise mass=54000 kglong range cruise or 74 mach err _a_033 416 345 ft, 338 ft, 353 ft, maximum operating altitude Img /com_en/com033 416.jpg . Question Pre-Flight 93 Answer 11
Given distance from departure to destination 3750 nm safe endurance 9 5 h true track 360 w/v 360/50 tas 480 ktwhat the distance of psr from departure point 2255 nm, 2 7nm, 28 nm,495 nm Question Pre-Flight 93 Answer 12
Which best describes be maximum intensity of icing if any at fl150 in vicinity of bucharest 45°n 026°e err _a_033 430 Moderate, light, severe, nil. img /com_en/com033 430.jpg from below chart base to fl 200, icing moderate over bucarest. Question Pre-Flight 93 Answer 13
Given distance from departure to destination 210 nmsafe endurance 3 5 h true track 310w/v 270/30tas 120 ktwhat the distance of psr from departure point Question Pre-Flight 93 Answer 14
What maximum surface windspeed kt forecast bordeaux/merignac at 1600 utc err _a_033 434 fc1100r 121100z 121221 amended (r) short taf, prepared on twelfth day of month at 1100z (valid from 12h to 21h utc). there a tempo from 12h to 18h indicating wind from 280° 20 kt, with gust up to 30 kt (tempo 1218 28020g30kt ). Question Pre-Flight 93 Answer 15
Which best describes significant cloud forecast over toulouse 44°n001°e err _a_033 435 Broken ac/cu base below fltops fl 5 , embedded isolated cb base below fltops fl27 , well separated cb base fltops to fl 27 , isolated cb embedded in layer cloud, surface to fl27 , 5 to 7 oktas cu ac base below fltops to fl27 img /com_en/com033 435.png . Question Pre-Flight 93 Answer 16
At reference or see flight planning manual mrjt 1 figure 4 4 holding planningngm= nam x tas+ wind/ tas the fuel required 45 minutes holding in a racetrack pattern at 5000 ft pressure altitude and a weight of 47000 kg err _a_033 437 635 kg, 9kg,69kg, 25 kg. fuel flow 47000kg at 5000 ft 2180 kg/h (interpolating between 2220 2140 kg/h). the fuel required 45 minutes holding is 2180 x (45/60) = 1635 kg. Question Pre-Flight 93 Answer 17
The wind °/kt at 50°n 015°w err _a_033 438 29 /75, 3/85, 3/75,/75. img /com_en/com033 438.png wind coming from 290°. number or pennants and/or feathers correspond to speed pennants correspond to 50 kt. feathers correspond to 10 kt. half feathers correspond to 5 kt. 50+10+10+5 = 75 kt. Question Pre-Flight 93 Answer 18
At reference or see flight planning manual mrjt 1 figure 4 3 6 in order to get alternate fuel and time twin jet aeroplane operations manual graph shall be entered with err _a_033 443 Distance (nm), wind component, landing mass at alternate, still air distance, wind component, zero fuel mass, flight time, wind component, landing mass at alternate, distance (nm), wind component, zero fuel mass. Question Pre-Flight 93 Answer 19
Given distance from departure to destination 320 nmsafe endurance 4 3 htrue track 120°wind 180°/40 kttas 130 ktwhat the distance of psr from departure point 263 nm,85 nm,3nm, 59 nm. under index, set true track 120°, under center bore set tas 130 kt with rotative scale, set wind 180°/40 kt img /com_en/com033 444a.jpg drift always measured from heading to track, so turn to set true heading 120° + 17° = 137°. img /com_en/com033 444b.jpg gs out 105 kt. proceed same way to find ground speed home under index, set true track 300° 13° right drift, true heading 300° 13° = 287° gs home 146 kt. psr = time x gs out x gs home / (gs out + gs home) psr = 4.3 x 105 x 146 / (105 + 146) psr = 263 nm. Question Pre-Flight 93 Answer 20
The flight crew of a turbojet aeroplane prepares a flight using following data flight leg distance 3 500 nm flight level fl 310 true airspeed 450 kt headwind component at this level 5 kt initially planned take off mass without extra fuel on board 180 000 kg fuel price 0 35 us dollars/l at departure 0 315 us dollars/l at destinationto maximize savings commander chooses to carry extra fuel in addition to that which necessary using appended annex optimum quantity of fuel which should be carried in addition to prescribed quantity err _a_033 446 The fuel transport operation not recommended in this case, 8 kg, 22 kg,5 kg. it's very simple fuel cheaper at destination, so fuel transport operation not recommended in this case. Question Pre-Flight 93 Answer 21
At reference or see flight planning manual mrjt 1 figure 4 3 5 the following apply tail wind component 10 kttemperature isa +10°cbrake release mass 63000 kgtrip fuel available 20000 kgwhat the maximum possible trip distance err _a_033 447 374nm, 364nm, 35 nm, 325nm. you have to use graph backward. you must go to condition first (10 kt tailwind) then go to ref line, instead of 'norma way to proceed (first ref line then condition). img /com_en/com033 447.jpg . Question Pre-Flight 93 Answer 22
Cas 190 kt and altitude 9000 fttemperature isa 10°ctrue course 350°wind 320/40distance from departure to destination 350 nmendurance 3 hours and actual time of departure 1105 utc the distance from departure to point of equal time pet 2 3 nm,47 nm,83 nm,67 nm. oat at fl90 = 15°c (9 x 2°c) = 3°c. we are in isa 10°c, thus oat = 13°c. convert cas to tas on your computer img /com_en/com033 1156.jpg calculate outbound inbound groud speed, start first with wind 40 x cos30 = 34kt. outbound ground speed 215 34 = 181 kt inbound ground speed 215+34 = 249 kt point of equal time = 350x249/(181+249) = 202.67 nm. Question Pre-Flight 93 Answer 23
Which best describes significant cloud if any forecast the area southwest of bodo 67°n 014°e err _a_033 450 5 to 7 oktas cu cb base below fl , tops fl 8 , 5 to 7 oktas cu cb base fl , tops fl 8 , 3 to 7 oktas cu cb base below fl , tops fl 8 , nil. img /com_en/com033 450.jpg the chart from fl100 to fl450. Question Pre-Flight 93 Answer 24
Given distance from departure to destination 190 nm safe endurance 2 4 h true track 120° wind 030°/40 kt tas 130 kt what the distance of psr from departure point 48 nm, 95 nm, 73 nm, 44 nm. set 120° true track on top. with rotating scale, mark wind 030°/40 kt. you read 17° right drift. set on top 120° 17° = 103°. now read drift 18° right. set on top 102°, you can read your 'ground speed out' of 123.5 kt. the wind perpendicular to our track, so gsh will also be 123.5 kt point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) ground speed out = 123.5 kt ground speed home = 123.5 kt point of safe return (psr) = 2.4 x 123.5 / (123.5 + 123.5) point of safe return (psr) = 296.4 / 247 point of safe return (psr) = 1.2 h 1.2 x 60 = 72 minutes distance of psr from departure point at a speed of 123.5 kt 72 min x (123.5/60) = 148 nm. Question Pre-Flight 93 Answer 25
At reference or see flight planning manual mrjt 1 figure 4 2find short distance cruise altitude the twin jet aeroplane given brake release mass 40000 kgtemperature isa + 20°ctrip distance 150 nautical air miles nam err _a_033 458 3 ft, 25ft, 2 ft, 275 ft. img /com_en/com033 458.jpg . Question Pre-Flight 93 Answer 26
Given distance from departure to destination 350 nm true track 320 w/v 350/30 tas 130 kt what the distance and time of pet from departure point Distance2 nm time 22 min, distance 39 nm time54 min, distance 23 nm time74 min, distance 39 nm time8min Question Pre-Flight 93 Answer 27
Given distance from departure to destination 240 nm safe endurance 3 5 h tas 125 kt ground speed out 110 ktground speed home 140 ktwhat the distance and time of psr from departure point Distance2 6 nm time8 min, distance 34 nm time58 min, distance8 nm time52 min, distance24 nm time 3 min point of safe return (psr) = endurance x gs home / (gs out + gs home) outbound gs = 110 kt homeward gs = 140 kt endurance = 3.5 h point of safe return (psr) = 3.5 x 140 / (110 + 140) point of safe return (psr) = 1.96 h 1.96 h = 118 min distance of psr from departure 1.96 x 110 = 215.6 nm. Question Pre-Flight 93 Answer 28
Route manual chart napthe average magnetic course from a 64°n006°e to c 62°n020°w err _a_033 464 27 °, 259°, 247°, 279°. center your protractor at mid distance between a c, you will find 260°, you must apply variation 11°w. average magnetic course 260° + 11° = 271°. Question Pre-Flight 93 Answer 29
Given distance from departure to destination 165 nm true track 055 w/v 360/20 tas 105 kt what the distance of pet from departure point 92 nm, 73 nm, 83 nm,32 nm img /com_en/com033 475a.jpg img /com_en/com033 475b.jpg ground speed out = 92 kt proceed same way to find ground speed home true track 235°, drift = 8°, true heading = 243°, gs home = 116 kt. pet = distance x gsh / (gso+ gsh) pet = 165 x 116 / (92 + 116) pet = 19140 / 208 = 92 nm. Question Pre-Flight 93 Answer 30
Given distance from departure to destination 140 nm gs out 90 kt gs home 80 kt what the distance of pet from departure point 66 nm, 74 nm, 7nm,24 nm. Question Pre-Flight 93 Answer 31
Given distance from departure to destination 6340 nm safe endurance 15 h true track 090 w/v 270/100 tas 520 ktwhat the distance of psr from departure point 3756 nm, 256nm,878 nm, 2584 nm Question Pre-Flight 93 Answer 32
Route manual chart napthe distance nm from a 64°n006°e to c 62°n020°w err _a_033 483 72nm, 69nm,59nm,44nm. report track distance along latitude 63°n (average latitude between a c). you will count a little bit more of 26° separation between a c. 26° x 60 nm x cos 63° = 708 nm. img /com_en/com033 483.jpg you can also use meridian 1° = 60 nm. maxscail 26° change in longitude 60 x 26 x cos(63) = 708.225 between 64 62 2 ° ==> 2 x 60(1°=60nm) = 120 now, use phytagore sqrt 708,225 + sqrt 120 = 15108,225 square root of 515982,905 = 718,32 nm. Question Pre-Flight 93 Answer 33
Given course a to b 088° t distance 1250 nmmean tas 330 ktmean w/v 340°/60 ktthe time from a to pet between a and b Hour 42 minutes, hour 54 minutes, hour 39 minutes, 2 hours2 minutes img /com_en/com033 484a.jpg img /com_en/com033 484b.jpg ground speed out = 343 kt proceed same way to find ground speed home (306 kt) pet = distance x gsh / (gso+ gsh) pet = 1250 x 306 / (343 + 306) pet = 382500 / 649 = 589 nm 589 nm / 343 = 1.71 h (1 hour 42 minutes). Question Pre-Flight 93 Answer 34
Route manual chart nap the average true course from a 64°n006°e to c 62°n020°w err _a_033 497 259°, 247°, 27 °,79°. img /com_en/com033 497.jpg we are looking average true course with you protractor aligned on true north, between a c, you will find a true course of 259°. Question Pre-Flight 93 Answer 35
Given distance from departure to destination 340 nmtrue track 320°wind 160°/40 kttas 110 ktwhat the distance of pet from departure point 2 nm, 228 nm,2nm, 2 9 nm. under index, set true track 320°, centre dot on tas, 110 kt, with rotative scale, set wind 160°/40 kt, you find a right drift of 5°. img /com_en/com033 500a.jpg now, drift always measured from heading to track turn to set true heading 315° (320° 5° right drift) under index, you now read your ground speed out of 147 kt img /com_en/com033 500b.jpg proceed in same way to find ground speed home of 72 kt. (left drift of 10°, true heading of 150°). ground speed out (gso) = 147 kt ground speed home (gsh) = 72 kt distance to pet = distance x gsh / (gso + gsh) distance to pet = 340 x 72 / (147 + 72) distance to pet = 24480 / 219 = 111.78 nm. Question Pre-Flight 93 Answer 36
Use route manual chart napon a direct great circle course from reykjavik 64°10' n 022°00'w to amsterdam 52°32'n 004°50'e a average true course and b distance are err _a_033 510 (a)3 ° (b) 95 nm, (a) 3° (b)824 nm, (a) 8° (b) 95 nm, (a)4 ° (b)824 nm average true course img /com_en/com033 510.jpg cos mean latitude x difference of longitude x 60 nm cos58.31 x 26,83° x 60 nm = 846 nm difference of latitude 11°42' = 11,70° x 60 nm = 702 nm distance between reykjavik amsterdam = sqrt(702² + 846²) = 1099 nm. Question Pre-Flight 93 Answer 37
Given distance from departure to destination 330 nm safe endurance 5 h true track 170°wind 140/25tas 125 ktwhat the distance of psr from departure point 3 2 nm,94 nm,5nm, 3nm. point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) on computer under index, set true track 170°, centre dot on tas, 125 kt, with rotative scale, set wind 140°/25 kt we read a right drift of 7°. drift always measured from heading to track, so turn to set true heading 163° (170° 7° right drift) under index you read a ground speed out of 104 kt (and a drift of 6º right). proceed on same way gs home (you will find 140 kt). point of safe return (psr) = 5 x 140 / (104 + 140) point of safe return (psr) = 700 / 244 point of safe return (psr) = 2.87 h distance of psr from departure point at a speed of 104 kt 2.87 x 104 = 298 nm. Question Pre-Flight 93 Answer 38
Route manual chart nap the initial magnetic course from a 64°n006°e to c 62°n020°w err _a_033 520 275°, 267°, 27 °, 262°. img /com_en/com033 520.jpg we are looking initial magnetic course with you protractor aligned on true north, you have a true course of 271°, add 4°west magnetic variation = 275°. Question Pre-Flight 93 Answer 39
The maximum wind velocity °/kt immediately north of tunis 36°n010°e err _a_033 521 9 °/95 kt, 28 °/kt, 25 °/85 kt,8 °/5 kt. arrows, feathers pennants arrows indicate direction. number or pennants and/or feathers correspond to speed. example with a 270°/115 kt wind img /com_en/com033 346a.jpg pennants correspond to 50 kt. feathers correspond to 10 kt. half feathers correspond to 5 kt. img /com_en/com033 521.jpg . Question Pre-Flight 93 Answer 40
Given distance a to b 3060 nmmean groundspeed 'out' 440 ktmean groundspeed 'back' 540 ktsafe endurance 10 hoursthe time to point of safe return psr 5 hours 3minutes, 5 hours 45 minutes, 3 hours 55 minutes, 5 hours 2minutes.
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