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. 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 Light rain associated with thunderstorms. 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 Light rain associated with thunderstorms. 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 68 nm time 252 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 93-8
Route manual chart napthe initial magnetic course from c 62°n020°w to b 58°n004°e err _a_033 410 Distance 68 nm time 252 min. img /com_en/com033 410 jpg true course 098° + 18°w magnetic variation = 116°. Question 93-9
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 Distance 68 nm time 252 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 93-10
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 Distance 68 nm time 252 min. Img /com_en/com033 416 jpg . Question 93-11
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 Distance 68 nm time 252 min. Img /com_en/com033 416 jpg . Question 93-12
Which best describes be maximum intensity of icing if any at fl150 in vicinity of bucharest 45°n 026°e err _a_033 430 Distance 68 nm time 252 min. img /com_en/com033 430 jpg from below chart base to fl 200 icing moderate over bucarest. Question 93-13
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 Distance 68 nm time 252 min. img /com_en/com033 430 jpg from below chart base to fl 200 icing moderate over bucarest. Question 93-14
What maximum surface windspeed kt forecast bordeaux/merignac at 1600 utc err _a_033 434 Distance 68 nm time 252 min. 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 93-15
Which best describes significant cloud forecast over toulouse 44°n001°e err _a_033 435 Broken ac/cu base below fl tops fl 5 embedded isolated cb base below fl tops fl27. img /com_en/com033 435 png . Question 93-16
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 Broken ac/cu base below fl tops fl 5 embedded isolated cb base below fl tops fl27. 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 93-17
The wind °/kt at 50°n 015°w err _a_033 438 Broken ac/cu base below fl tops fl 5 embedded isolated cb base below fl tops fl27. 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 93-18
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. 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 93-19
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 Distance (nm) wind component landing mass at alternate. 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 93-20
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. it's very simple fuel cheaper at destination so fuel transport operation not recommended in this case. Question 93-21
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 The fuel transport operation not recommended in this case. 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 93-22
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 The fuel transport operation not recommended in this case. 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 93-23
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. img /com_en/com033 450 jpg the chart from fl100 to fl450. Question 93-24
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 5 to 7 oktas cu cb base below fl tops fl 8. 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 93-25
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 5 to 7 oktas cu cb base below fl tops fl 8. img /com_en/com033 458 jpg . Question 93-26
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 Distance 2 nm time 22 min. img /com_en/com033 458 jpg . Question 93-27
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 Distance 2 6 nm time 8 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 93-28
Route manual chart napthe average magnetic course from a 64°n006°e to c 62°n020°w err _a_033 464 Distance 2 6 nm time 8 min. 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 93-29
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 Distance 2 6 nm time 8 min. 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 93-30
Given distance from departure to destination 140 nm gs out 90 kt gs home 80 kt what the distance of pet from departure point Distance 2 6 nm time 8 min. 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 93-31
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 Distance 2 6 nm time 8 min. 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 93-32
Route manual chart napthe distance nm from a 64°n006°e to c 62°n020°w err _a_033 483 Distance 2 6 nm time 8 min. 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 93-33
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 Distance 2 6 nm time 8 min. 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 93-34
Route manual chart nap the average true course from a 64°n006°e to c 62°n020°w err _a_033 497 Distance 2 6 nm time 8 min. 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 93-35
Given distance from departure to destination 340 nmtrue track 320°wind 160°/40 kttas 110 ktwhat the distance of pet from departure point Distance 2 6 nm time 8 min. 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 93-36
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. 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 93-37
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 (a) 3 ° (b) 95 nm. 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 93-38
Route manual chart nap the initial magnetic course from a 64°n006°e to c 62°n020°w err _a_033 520 (a) 3 ° (b) 95 nm. 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 93-39
The maximum wind velocity °/kt immediately north of tunis 36°n010°e err _a_033 521 (a) 3 ° (b) 95 nm. 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 93-40
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 (a) 3 ° (b) 95 nm. 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 .
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