Given Distance from departure to destination 300 NMSafe Endurance 4 hTAS 110 ktGround Speed Out 120 ktGround Speed Home 100 ktWhat is the distance of the PSR from the ?
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What is the mean temperature deviation °C from the ISA over 50°N 010°W ERR _a_033 83 ?
Given Distance from departure to destination 4630 NMSafe Endurance 12 4 hTrue Track 240W/V 060/80TAS 530 ktWhat is the distance of the PSR from the departure point ?
Given distance from departure to destination 1100 nmtrue track 280°w/v 100/80tas 440 ktwhat the distance time of pet from departure point Distance 45 nm time 52 min. track 280° wind from 100° it's a tailwind of 80 kt ground speed out (gso) = 440 + 80 = 520 kt track 100° wind from 100° it's a headwind of 80 kt ground speed home (gsh) = 440 80 = 360 kt distance to pet = distance x gsh / (gso + gsh) distance to pet = 1100 x 360 / (520 + 360) distance to pet = 396000 / 880 = 450 nm time of pet from departure point 450 nm / 520 = 0 865 h 0 865 x 60 min = 52 minutes. 
Given distance from departure to destination 150 nmsafe endurance 3 2 htas 90 ktground speed out 100 ktground speed home 80 ktwhat the distance and time of psr from departure point Distance 42 nm time 85 min. point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) ground speed out = 100 kt ground speed home = 80 kt point of safe return (psr) = 3 2 x 80 / (100 + 80) point of safe return (psr) = 256 / 180 point of safe return (psr) = 1 42 h 1 42 x 60 = 85 minutes distance of psr from departure point at a speed of 100 kt 85 min x (100/60) = 142 nm. 
Given distance from departure to destination 270 nmtrue track 030w/v 120/35tas 125 ktwhat the distance and time of pet from departure point Distance 35 nm time 68 min. under index set true track 030° centre dot on tas 125 kt with rotative scale set wind 120°/35 kt you find a left drift of 15° img /com_en/com033 95a jpg now drift always measured from heading to track turn to set true heading 045° (030° + 15° left drift) under index you now read your ground speed out of 120 kt img /com_en/com033 95b jpg proceed in same way to find ground speed home of 120 kt (right drift of 15° true heading of 195°) ground speed out (gso) = 120 kt ground speed home (gsh) = 120 kt distance to pet = distance x gsh / (gso + gsh) distance to pet = 270 x 120 / (120 + 120) distance to pet = 32400 / 240 = 135 nm 135 nm at a ground speed out of 120 kt = 135 x (60/120) = 67 5 minutes. 
At reference or see flight planning manual sep 1 figure 2 5 given fl 75lean mixture and full throttle 2300 rpmtake off fuel 444 lbstake off from msl find endurance in hours and minutes err _a_033 100 Distance 35 nm time 68 min. img /com_en/com033 100 jpg 5 2h ==> 5h + 0 2h (0 2 x 60 = 12 minutes) 5 h 12 minutes. Question 89-8
From which of following would you expect to find facilitation information regarding customs and health formalities Distance 35 nm time 68 min. img /com_en/com033 100 jpg 5 2h ==> 5h + 0 2h (0 2 x 60 = 12 minutes) 5 h 12 minutes. Question 89-9
An aircraft flies at a tas of 380 kt it flies from a to b and back to a distance ab = 480 nm when going from a to b it experiences a headwind component = 60 kt the wind remains constant the duration of flight will be Distance 35 nm time 68 min. take care the wind remains constant headwind from a to b becomes tailwind from b to a tas 380 kt from a to b ground speed 380 60 = 320 kt 480 nm / 320 kt = 1 5 h (1h30) from b to a ground speed 380+60 = 440 kt 480 nm / 440 kt = 1 09 h (1h05) 1h30 + 1h05 = 2 h 35 min. Question 89-10
Given distance from departure to destination 350 nmtrue track 320w/v 350/30tas 130 ktwhat the distance and time of pet from departure point Distance 2 nm time 2 min. under index set true track 320° centre dot on tas 130 kt with rotative scale set wind 350°/30 kt you find a left drift of 7° now drift always measured from heading to track turn to set true heading 327° (320° + 7° left drift) under index you now read your ground speed out of 104 kt proceed in same way to find ground speed home of 155 kt (right drift of 6° true heading of 134°) ground speed out (gso) = 104 kt ground speed home (gsh) = 155 kt distance to pet = distance x gsh / (gso + gsh) distance to pet = 350 x 155 / (104 + 155) distance to pet = 54250 / 259 = 210 nm 210 nm at a ground speed out of 104 kt = 210 x (60/104) = 121 minutes. Question 89-11
Given distance from departure to destination 250 nmgs out 130 ktgs home 100 ktwhat the distance of pet from departure point Distance 2 nm time 2 min. Ground speed out (gso) = 130 kt ground speed home (gsh) = 100 kt distance to pet = distance x gsh / (gso + gsh) distance to pet = 250 x 100 / (130 + 100) distance to pet = 25000 / 230 = 109 nm. Question 89-12
Given distance from departure to destination 550 nmendurance 3 6 htrue track 200w/v 220/15tas 130 ktwhat the distance of psr from departure point Distance 2 nm time 2 min. under index set true track 200° centre dot on tas 130 kt with rotative scale set wind 220°/15 kt you find a left drift of 3° now drift always measured from heading to track turn to set true heading 203° (200° + 3° left drift) under index you now read your ground speed out of 115 kt proceed in same way to find ground speed home of 142 kt (true track of 020° right drift of 2° true heading of 018°) ground speed out (gso) = 115 kt ground speed home (gsh) = 142 kt point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) point of safe return (psr) = 3 6 x 142 / (115 + 142) point of safe return (psr) = 511 2 / 257 point of safe return (psr) = 1 99 h distance of psr from departure point at a speed of 115 kt 1 99 x 115 = 229 nm (closest answer 231 nm). Question 89-13
Given distance from departure to destination 180 nmendurance 2 htas 120 ktground speed out 135 ktground speed home 105 ktwhat the distance and time of psr from departure point Distance 8 nm time 53 min. point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) ground speed out = 135 kt ground speed home = 105 kt point of safe return (psr) = 2 x 105 / (135 + 105) point of safe return (psr) = 210 / 240 point of safe return (psr) = 0 875 h 0 875 x 60 = 52 5 minutes distance of psr from departure point at a speed of 135 kt 52 5 min x (135/60) = 118 125 nm. Question 89-14
Given distance from departure to destination 2500 nmgs out 540 ktgs home 470 ktwhat the time of pet from departure point Distance 8 nm time 53 min. pet = d x vsr / vsa + vsr pet = 2500 x 470 / 540 + 470 pet = 1163 nm 1163 / 540 = 2 15 h (2 x 60) + (0 15 x 60) = 129 minutes. Question 89-15
Given distance from departure to destination 875 nmtrue track 240 wind 060/50 kttas 500 ktwhat the distance and time of pet from departure point Distance 394 nm time 43 min. true track 240° wind 060/50 kt wind parallel to our course thus ground speed out 500 + 50 = 550 kt ground speed home 500 50 = 450 kt pet = d x gsh / (gso + gsh) pet = 875 x 450 /(550 + 450) = 393 75 nm 393 75 / 550 = 0 716 minutes 60 x 0 716 = 43 minutes. Question 89-16
The forecast period covered the paris/charles de gaulle tafs totals hours err _a_033 122 Distance 394 nm time 43 min. img /com_en/com033 122 jpg total tafs duration 27h. Question 89-17
If cas 190 ktsaltitude 9000 fttemp isa 10°ctrue course tc 350°w/v 320/40distance from departure to destination 350 nmendurance 3 hoursand actual time of departure 1105 utc the point of equal time pet reached at Distance 394 nm time 43 min. 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 202 67/181 = 1 12 h 1 12x0 6 = 67 minutes or 1h07 11h05+01h07 = 12h13. Question 89-18
At reference or see flight planning manual mrjt 1 figure 4 4planning a flight from paris charles gaulle to london heathrow a twin jet aeroplanepreplanning dry operating mass dom 34 000 kgtraffic load 13 000 kg the holding planned at 1 500 ft above alternate elevation the alternate elevation 256 ft the holding planned 30 minutes with no reductions the estimated landing mass at alternate manchester err _a_033 129 Distance 394 nm time 43 min. Landing mass at alternate = dry operating mass + traffic load + final reserve fuel notice you must land at destination or alternate (when pre planning) with final reserve fuel in your tanks! img /com_en/com033 129 jpg 34000+13000 = 47000 kg interpolate from table (2280 + 2220)/2 = 2250 kg/h for 30 minutes = 1125 kg 47000 + 1125 = 48125 kg. Question 89-19
Given distance from departure to destination 1385 nm gs out 480 kt gs home 360 kt what the time of pet from departure point Distance 394 nm time 43 min. ground speed out = 480 kt ground speed home = 360 kt pet = distance x gsh / (gso + gsh) pet = 1385 x 360 / (480 + 360) pet = 498600 / 840 = 593 nm time of pet from departure point 593 / 480 = 1 23 h 1 23 x 60 = 74 minutes. Question 89-20
Given distance from departure to destination 256 nm gs out 160 kt gs home 110 kt what the distance of pet from departure point Distance 394 nm time 43 min. ground speed out (gso) = 160 kt ground speed home (gsh) = 110 kt distance to pet = distance x gsh / (gso + gsh) distance to pet = 256 x 110 / (160 + 110) distance to pet = 28160 / 270 = 104 nm. Question 89-21
Given distance from departure to destination 480 nmsafe endurance 5 htrue track 315°w/v 100/20tas 115 ktwhat the distance of psr from departure point Distance 394 nm time 43 min. start searching outbound ground speed on nav computer set 115 kt under center dot true track 315° to true index put wind direction 100° under red compass rose under 20 kt you read a left drift of 4° now drift always measured from heading to track turn to set true heading 319° (315° + 4° left drift) under index you now read your ground speed out of 130 kt repeat operation to find homeward ground speed outbound gs 130 kt homeward gs 99 kt point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) point of safe return (psr) = 5 x 99 / (130 + 99) point of safe return (psr) = 495 / 229 point of safe return (psr) = 2 16 h 0 16 x 60 = 10 minutes 10 + 120 minutes = 130 min distance of psr from departure point at a speed of 132 kt 130 min x (130/60) = 281 6 nm. Question 89-22
Given distance from departure to destination 150 nm true track 142° wind 200°/15kt tas 132 kt what the distance of pet from departure point Distance 394 nm time 43 min. under index set true track 142° centre dot on tas 132 kt with rotative scale set wind 200°/15 kt you find a left drift of 5° now drift always measured from heading to track turn to set true heading 147° (142° + 5° left drift) under index you now read your ground speed out of 124 kt img /com_en/com033 142 jpg proceed in same way to find ground speed home of 139 kt (right drift of 5° true heading of 317°) ground speed out (gso) = 124 kt ground speed home (gsh) = 139 kt distance to pet = distance x gsh / (gso + gsh) distance to pet = 150 x 139 / (124 + 139) distance to pet = 20850 / 263 = 79 2 nm. Question 89-23
A metar reads 1430z 35002kt 7000 skc 21/03 q1024 =which of following information contained in this metar Temperature/dewpoint. 1430 = time (1430 utc) 35002kt = wind 350°/02 kt 7000 = visibility 7000 meters skc = sky clear 21/03 = temperature/dewpoint q1024 = qnh 1024 hpa. Question 89-24
The wind °/kt at 40°n 020°w err _a_033 157 Temperature/dewpoint. img /com_en/com033 157 jpg 10 + 10 + 10 + 10 = 40 kt. Question 89-25
Given maximum allowable take off mass 64 400 kgmaximum landing mass 56200 kgmaximum zero fuel mass 53 000 kgdry operating mass 35 500 kgestimated load 14 500 kgestimated trip fuel 4 900 kgminimum take off fuel 7 400 kg find maximum additional load Temperature/dewpoint. img /com_en/com033 162 jpg we are able to add 3000 kg beore reaching our first limitation which comes from maximum zero fuel mass 17500 14500 = 3000 kg. Question 89-26
What mean temperature °c likely on a course of 360° t from 40°n to 50°n at 040°e err _a_033 167 Mean temperature 47°c. img /com_en/com033 167 jpg temperatures are negative unless prefixed ps (46 + 47 + 47 + 48 + 49)/5 = 47 4°c. Question 89-27
For flight planning purposes landing mass at alternate taken as Zero fuel mass plus final reserve fuel contingency fuel. planned landing mass at alternate = dom + traffic load + final reserve fuel + contingency if everything goes to plan on sector you won't use contingency fuel so landing mass at alternate will include final reserve fuel + contingency fuel. Question 89-28
Given distance from departure to destination 220 nmtrue track 175°wind 220/10 kttas 135 ktwhat the distance of pet from departure point Zero fuel mass plus final reserve fuel contingency fuel. under index set true track 175° centre dot on tas 135 kt with rotative scale set wind 220°/10 kt you find a left drift of 7° now drift always measured from heading to track turn to set true heading 182° (175° + 7° left drift) under index you now read your ground speed out of 119 kt img /com_en/com033 169 jpg proceed in same way to find ground speed home of 148 kt (right drift of 5° true heading of 350°) ground speed out (gso) = 119 kt ground speed home (gsh) = 148 kt distance to pet = distance x gsh / (gso + gsh) distance to pet = 220 x 148 / (120 + 148) distance to pet = 32780 / 268 = 121 nm (closest answer 116 nm). Question 89-29
At references or see flight planning manual mrjt 1 figure 4 2 and figure 4 5 3 2 given estimated take off mass 57000 kgground distance 150 nmtemperature isa 10°ccruise at 74 machfind cruise altitude and expected true air speed err _a_033 178 Zero fuel mass plus final reserve fuel contingency fuel. img /com_en/com033 178 jpg temperature isa 10°c 445 10 = 435 kt. Question 89-30
Given distance from departure to destination 950 nmgs out 275 kt gs home 225 ktwhat the time of pet from departure point Zero fuel mass plus final reserve fuel contingency fuel. distance to pet = distance x gsh / (gso + gsh) distance to pet = 950 x 225 / (275 + 225) distance to pet = 213750 / 500 = 427 5 nm time of pet from departure point 427 5 nm / 275 = 1 55 h 1 55 x 60 min = 93 minutes. Question 89-31
Given distance from departure to destination 950 nm safe endurance 3 5 h tas 360 kt ground speed out 320 kt ground speed home 400 kt what the distance and time of psr from departure point Distance 622 nm time 7 min. distance to pet = distance x gsh / (gso + gsh) distance to pet = 950 x 225 / (275 + 225) distance to pet = 213750 / 500 = 427 5 nm time of pet from departure point 427 5 nm / 275 = 1 55 h 1 55 x 60 min = 93 minutes. Question 89-32
Given distance from departure to destination 1000 nm safe endurance 4 h tas 500 kt ground speed out 550 kt ground speed home 450 kt what the distance of psr from departure point Distance 622 nm time 7 min. ground speed out 550 kt ground speed home 450 kt point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) point of safe return (psr) = 4 x 450 / (550 + 450) point of safe return (psr) = 1800 / 1000 point of safe return (psr) = 1 8 h 0 8 x 60 = 48 minutes 48 + 60 minutes = 108 min distance of psr from departure point at a speed of 550 kt 108 min x (550/60) = 990 nm. Question 89-33
Given distance from departure to destination 2200 nmtrue track 150° wind 330°/50 kttas 460 ktwhat the distance and time of pet from departure point Distance 98 nm time 5 min. track 150° wind from 330° it's a tailwind of 50 kt ground speed out (gso) = 460 + 50 = 510 kt return track 330° wind from 330° it's a headwind of 50 kt ground speed home (gsh) = 460 50 = 410 kt distance to pet = distance x gsh / (gso + gsh) distance to pet = 2200 x 410 / (510 + 410) distance to pet = 902000 / 920 = 980 nm time of pet from departure point 980 nm / 510 = 1 92 h 1 92 x 60 min = 115 minutes. Question 89-34
Route manual chart napthe initial true course from a 64°n006°e to c 62°n020°w err _a_033 202 Distance 98 nm time 5 min. img /com_en/com033 202 jpg put your protractor on a align it with true north you find an initial true course of 271° departure almost superposed on 64° parallel maercin i couldn't found those points on map because there written that a on 006e so its impossible to find in this map nevertheless you may also substract 64 62 = 2 degresses so if we multiply it 60nm there 120nm in vertical range if we use formula distance = 60nm x cos(mean longitude) x g difference in latitude we have 708nm so now we take usual calculator input arc tg from 120/708 = 9 deg we know that track from a to b 270 9 = 261 after that we count convertion angle = 1/2 x sin (mean latitude) x difference in latitude its equal to 10 so we add 10 to 261 here it 271! i know that this a little bit more difficult than reading from map but on other hand if you wasn't passed gen nav flight planning its possible to resolve all maps questions without even look on them. Question 89-35
Which best describes maximum intensity of icing if any at fl160 in vicinity of berlin 53° n013°e err _a_033 204 Distance 98 nm time 5 min. img /com_en/com033 202 jpg put your protractor on a align it with true north you find an initial true course of 271° departure almost superposed on 64° parallel maercin i couldn't found those points on map because there written that a on 006e so its impossible to find in this map nevertheless you may also substract 64 62 = 2 degresses so if we multiply it 60nm there 120nm in vertical range if we use formula distance = 60nm x cos(mean longitude) x g difference in latitude we have 708nm so now we take usual calculator input arc tg from 120/708 = 9 deg we know that track from a to b 270 9 = 261 after that we count convertion angle = 1/2 x sin (mean latitude) x difference in latitude its equal to 10 so we add 10 to 261 here it 271! i know that this a little bit more difficult than reading from map but on other hand if you wasn't passed gen nav flight planning its possible to resolve all maps questions without even look on them. Question 89-36
At reference or see flight planning manual sep 1 figure 2 1 given fl 75oat +5°cduring climb average head wind component 20 kt take off from msl with initial mass of 3650 lbs find still air distance nam and ground distance nm using graph 'time fuel distance to climb' err _a_033 212 Distance 98 nm time 5 min. img /com_en/com033 212 jpg time to climb 9 minutes distance to climb 18 nam with a headwind ground distance to climb will be lower than air distance to climb to fl75 during 9 minutes 20 kt of wind will reduce our ground distance by 20 kt x (9/60) = 3 nm 18 nam 3 nm = 15 nm. Question 89-37
Given distance from departure to destination 285 nm true track 348°wind 280°/25 kttas 128 ktwhat the distance of pet from departure point Distance 98 nm time 5 min. Img /com_en/com033 213a jpg img /com_en/com033 213b jpg ground speed out = 117 kt proceed same way to find ground speed home (136 kt) distance to pet = d x gsh / (gso + gsh) distance to pet = 285 x 136 / (117 + 136) distance to pet = 38760 / 253 = 153 2 nm. Question 89-38
Given distance from departure to destination 435 nm gs out 110 kt gs home 130 kt what the distance of pet from departure point Distance 98 nm time 5 min. Img /com_en/com033 213a jpg img /com_en/com033 213b jpg ground speed out = 117 kt proceed same way to find ground speed home (136 kt) distance to pet = d x gsh / (gso + gsh) distance to pet = 285 x 136 / (117 + 136) distance to pet = 38760 / 253 = 153 2 nm. Question 89-39
The approximate mean wind component kt along true course 180° from 50°n to 40°n at 005° w err _a_033 225 Distance 98 nm time 5 min. img /com_en/com033 225 jpg wind coming from 320° with an average speed of 70 kt tailwind component = 70 kt x cos (angle between wind the course) tailwind component = 70 kt x cos 40° tailwind = 54 kt. Question 89-40
The wind direction and velocity °/kt at 40°n 040°e err _a_033 233 Distance 98 nm time 5 min. Img /com_en/com033 233 jpg wind coming from 330° with a speed of 75 kt.
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