Question 118-1 : Given distance from departure to destination 500 nmtrue track 090°wind 090°20kttas 150 ktwhat is the distance and time of the pet from the departure point ? [ Training professional ]

Question 118-2 : Use route manual chart e hi 1an aircraft has to fly from glasgow 55°52'n 004°27'w to benbecula 57°29'n 007°22'w cruising at 320 kt tasassuming a headwind of 40 kt and cruise fuel consumption of 2300 kgh what is the forecast fuel used for this flight err a 033 6 ?

1117 kg.

Ground speed = 320 40 = 280 kt136 nm 28060 = 2914 minutes230060 = 38333 kgmin2914 x 38333 = 1117 kg

Question 118-3 : Given distance from departure to destination 500 nmsafe endurance 4 htas 140 ktground speed out 150 ktground speed home 130 ktwhat is the distance and time of the psr from the departure point ?

Distance 279 nm time 111 min.

Point of safe return psr = endurance x homeward gs outbound gs + homeward gs ground speed out = 150 ktground speed home = 130 ktpoint of safe return psr = 4 x 130 150 + 130 point of safe return psr = 65000 280point of safe return psr = 186 h186 x 60 = 1116 mindistance of the psr from the departure point at a speed of 150 kt 1116 min x 15060 = 279 nm

Question 118-4 : On a given path it is possible to chose between four flight levels fl each associated with a mandatory flight mach number m the flight conditions static air temperature sat and headwind component hwc are given below fl 370 m = 080 ts = 60°c hwc = 15 ktfl 330 m = 078 ts = 60°c hwc= 5 ktfl 290 m ?

Fl 270.

Set corresponding mark m kt against outside temperature at flight altitude read in front of mach number on the outer scale the true air speedexample 60° and mach 08 => tas is 451 kt 1137fl 370 => tas = 451 kt => gs = 439 ktfl 330 => tas = 440 kt => gs = 435 ktfl 290 => tas = 456 kt => gs = 441 ktfl 270 => tas = 445 kt => gs = 445 kt the flight level allowing the highest ground speed is fl 270

Question 118-5 : Given distance from departure to destination 410 nm safe endurance 36 h true track 055° wv 18035 tas 120 ktwhat is the distance of the psr from the departure point ?

203 nm.

First step find the gs out given true track true air speed wind direction and velocityrequired wind correction angle and ground speed computer solution a set true track to true indexb turn the indicator to the wind direction in this case using the black azimuth graduation the angle being upwind counting anti clockwise c shift the speed arc corresponding to the true air speed so as to coincide with the wind speed on the indicatord read the wind correction angle at the same place read the ground speed under the center bore from the scale on the axis of the slidesetting 055° to true indexset the indicator to 180° on the black azimuth circle being upwind adjust the speed arc labelled 120 kt of the diagram slide to the wind speed 35 kt of the indicator scalereading under the plotted point read the wind correction angle 14° under the center bore read the ground speed 136 kt 1141step two proceed on the same way for gs home you will find 97 kt step three now apply the psr formula psr = time x gs out x gs home gs out + gs home psr = 36 x 136 x 97 136 + 97 psr = 2038 nmthis is a 4 points question at the exammathematical calculation on this kind of exercise is valid only for one right angled triangle which is not the case here only the computer enables you to find the good answer

Question 118-6 : The surface weather system over england 53°n 002°w is err a 033 22 ?

An occluded front moving east.

com encom022 33apngover england the surface weather system is an an occluded front com encom033 22jpgthe arrow attached to the occluded front indicates direction and speed 20 kmh

Question 118-7 : Which describes the intensity of icing if any at fl 150 in the vicinity of toulouse 44°n 01°e err a 033 25 ?

Question 118-8 : Given distance a to b 2050 nmmean groundspeed 'on' 440 ktmean groundspeed 'back' 540 ktthe distance to the point of equal time pet between a and b is ?

1130 nm.

Ground speed out gso = 440 ktground speed home gsh = 540 ktdistance to pet = distance x gsh gso + gsh distance to pet = 2050 x 540 440 + 540 distance to pet = 1107000 980 = 1130 nm

Question 118-9 : Which is the heaviest type of precipitation if any forecast for bordeauxmerignac at 1000 utc err a 033 27 ?

Light rain.

com encom033 27jpgat 1000 utc if any the heaviest type of precipitation forcast is light rain

Question 118-10 : Given distance from departure to destination 315 nmtrue track 343°wv 01515tas 100 ktwhat is the distance of the pet from the departure point ?

176 nm.

Under index set true track 343° centre dot on tas 100 kt with the rotative scale set wind 015° 15 ktnow drift is always measured from heading to track turn to set true heading 345° 343° + 2° left drift under index you now read a ground speed out of 88 ktproceed in the same way to find the ground speed home of 112 kt right drift of 4° true heading of 159° distance to pet = distance x gsh gso + gsh distance to pet = 315 x 112 88 + 112 distance to pet = 35280 200 = 176 nm

Question 118-11 : What lowest cloud conditions oktasft are forecast for johannesburgjan smuts at 0300 utc err a 033 31 ?

5 to 7 at 400 ft.

Ft0900 prob30 0305 3000 bcfg bkn00430% chance that temporarily between 0300 utc and 0500 utc there will be patches fog with 3000 m visibility and 5 to 7 oktas of cloud at 400 ft

Question 118-12 : Given distance from departure to destination 210 nmsafe endurance 25 htrue track 035wv 25020tas 105 ktwhat is the distance of the psr from the departure point ?

127 nm.

Point of safe return psr = endurance x homeward gs outbound gs + homeward gs on the computer under index set true track 035° centre dot on tas 105 kt with the rotative scale set wind 250°20 kt we read a right drift of 5°drift is always measured from heading to track so turn to set true heading 030° 035° 5° right drift under index you read a ground speed out of 122 ktproceed on the same way for gs home under index set true track 215° centre dot on tas 105 kt with the rotative scale set wind 250°20 kt we read a left drift of 4°drift is always measured from heading to track so turn to set true heading 219° 215° + 4° right drift under index you read a ground speed home of 88 ktpoint of safe return psr = 25 x 88 122 + 88 point of safe return psr = 220 210point of safe return psr = 104 hdistance of the psr from the departure point at a speed of 122 kt 104 x 122 = 127 nm

Question 118-13 : Given the following d = flight distancex = distance to point of equal timegso = groundspeed outgsr = groundspeed returnthe correct formula to find distance to point of equal time is ?

X = d x gsr gso + gsr .

Question 118-14 : The flight crew of a turbojet aeroplane prepares a flight using the following data flight leg air distance 2 700 nm flight level fl 310 true airspeed 470 kt tailwind component at this level 35 kt initially planned take off mass without extra fuel on board 195 000 kg fuel price 028 eurol at ?

The fuel transport operation is not recommended in this case.

The fuel is cheaper at destination the fuel transport operation is not recommended in this case

Question 118-15 : From which of the following would you expect to find information regarding known short unserviceability of vor tacan and ndb ?

Notam.

Question 118-16 : The wind velocity over italy is err a 033 54 ?

A maximum of 110 kt at fl380.

com encom033 54jpga jet is crossing france and italy at fl 380 with a maximum of 110 kt

Question 118-17 : Assuming the following data ground distance to be covered 2 000 nmcruise flight level fl 330cruising speed mach 082 true airspeed 470 kt head wind component 30 ktplanned destination landing mass 160 000 kgtemperature isacg 37total anti ice on pack flow hitime needed to carry out such a flight is ?

4 h 43 min.

Nam = ngm x tasgs = 2000 x 470440 = 2136 nam com encom033 56jpgbetween 2100 and 2200 we can find our answer 4 h 43 min 210883 i think that the answer is wrong because 439 + 451 2 = 445 h > 4h 27' 4h 43' > 475h so the correct answer as i think is 4h 27'no times on the annex are shown as 439 for 4h39 and 451 for 4h51it's written on the top below fuel consumed kg time hmin

Question 118-18 : When calculating the fuel required to carry out a given flight one must take into account 1 the wind2 foreseeable airborne delays3 other weather forecasts4 any foreseeable conditions which may delay landingthe combination which provides the correct statement is ?

1 2 3 4.

Question 118-19 : Which of the following flight levels if any is forecast to be clear of significant cloud icing and cat along the marked route from shannon 53°n 10°w to berlin 53°n 13°e err a 033 64 ?

Fl250.

com encom033 64jpgthere are isolated embedded cbs from below fl100 to fl220 so fl 210 is excludedyou have cat area n°2 from fl270 to fl400 so fl290 is excludedonly fl250 is forecast to be clear of significant cloud icing and cat along the marked route from shannon to berlin

Question 118-20 : Given distance from departure to destination 215 nmsafe endurance 33 htrue track 005°wv 29015tas 125 ktwhat is the distance of the psr from the departure point ?

205 nm.

Point of safe return psr = endurance x homeward gs outbound gs + homeward gs on the computer under index set true track 005° centre dot on tas 125 kt with the rotative scale set wind 290°15 kt we read a right drift of 7°drift is always measured from heading to track so turn to set true heading 358° 005° 7° right drift under index you read a ground speed out of 120 ktproceed on the same way for gs home under index set true track 185° centre dot on tas 125 kt with the rotative scale set wind 290°15 kt we read a left drift of 7°drift is always measured from heading to track so turn to set true heading 192° 185° + 7° right drift under index you read a ground speed home of 128 ktpoint of safe return psr = 33 x 128 120 + 128 point of safe return psr = 4224 248point of safe return psr = 170 hdistance of the psr from the departure point at a speed of 120 kt 17 x 120 = 204 nm

Question 118-21 : Given distance from departure to destination 200 nmsafe endurance 3 htas 130 ktground speed out 150 ktground speed home 110 ktwhat is the distance psr from the departure point ?

190 nm.

Point of safe return psr = endurance x homeward gs outbound gs + homeward gs ground speed out = 150 ktground speed home = 110 ktpoint of safe return psr = 3 x 110 150 + 110 point of safe return psr = 330 260point of safe return psr = 12692 hdistance of the psr from the departure point at a speed of 150 kt 12692 h x 150 = 190 nm

Question 118-22 : What minimum visibility m is forecast for 0600 utc at london lhr egll err a 033 71 ?

1500 m.

Question 118-23 : Excluding rvsm an appropriate flight level for ifr flight in accordance with semi circular height rules on a course of 180° m is ?

Fl100.

The question states excluding rvsm com encom033 1228jpga vfr fligh level is flxx5 or flxx5below fl290 in accordance with semi circular rules for a magnetic heading of 180° we need a even level

Question 118-24 : Given distance from departure to destination 400 nmsafe endurance 25 htas 115 ktground speed out 130 ktground speed home 105 ktwhat is the distance of the psr from the departure point ?

145 nm.

Point of safe return psr = endurance x homeward gs outbound gs + homeward gs ground speed out = 130 ktground speed home = 105 ktpoint of safe return psr = 25 x 105 130 + 105 point of safe return psr = 2625 235point of safe return psr = 112 hdistance of the psr from the departure point at a speed of 130 kt 112 h x 130 = 145 nm

Question 118-25 : 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 departure point ?

218 nm.

Point of safe return psr = endurance x homeward gs outbound gs + homeward gs ground speed out = 120 ktground speed home = 100 ktpoint of safe return psr = 4 x 100 120 + 100 point of safe return psr = 400 220point of safe return psr = 181 hdistance of the psr from the departure point at a speed of 120 kt 181 h x 120 = 218 nm

Question 118-26 : What is the mean temperature deviation °c from the isa over 50°n 010°w err a 033 83 ?

2°c.

com encom033 83jpgtemperature are negative unless prefixed by psat fl300 isa temperature is 15°c 2°c x 30 = 45°coat is 47°c deviation from isa is 2°

Question 118-27 : Given distance from departure to destination 4630 nmsafe endurance 124 htrue track 240wv 06080tas 530 ktwhat is the distance of the psr from the departure point ?

3211 nm.

Point of safe return psr = endurance x homeward gs outbound gs + homeward gs ground speed out = 530 80 kt = 450 ktground speed home = 530 + 80 kt = 610 ktpoint of safe return psr = 124 x 610 450 + 610 point of safe return psr = 7564 1060point of safe return psr = 7135hdistance of the psr from the departure point at a speed of 450 kt 7135 x 450 = 3211 nm

Question 118-28 : Given distance from departure to destination 1100 nmtrue track 280°wv 10080tas 440 ktwhat is the distance time of the pet from the departure point ?

Distance 450 nm time 52 min.

Track 280° wind from 100° it's a tailwind of 80 kt ground speed out gso = 440 + 80 = 520 kttrack 100° wind from 100° it's a headwind of 80 kt ground speed home gsh = 440 80 = 360 ktdistance to pet = distance x gsh gso + gsh distance to pet = 1100 x 360 520 + 360 distance to pet = 396000 880 = 450 nm time of the pet from the departure point 450 nm 520 = 0865 h0865 x 60 min = 52 minutes

Question 118-29 : Given distance from departure to destination 150 nmsafe endurance 32 htas 90 ktground speed out 100 ktground speed home 80 ktwhat is the distance and time of the psr from the departure point ?

Distance 142 nm time 85 min.

Point of safe return psr = endurance x homeward gs outbound gs + homeward gs ground speed out = 100 ktground speed home = 80 ktpoint of safe return psr = 32 x 80 100 + 80 point of safe return psr = 256 180point of safe return psr = 142 h142 x 60 = 85 minutesdistance of the psr from the departure point at a speed of 100 kt 85 min x 10060 = 142 nm

Question 118-30 : Given distance from departure to destination 270 nmtrue track 030wv 12035tas 125 ktwhat is the distance and time of the pet from the departure point ?

Distance 135 nm time 68 min.

Under index set true track 030° centre dot on tas 125 kt with the rotative scale set wind 120°35 kt you find a left drift of 15° com encom033 95ajpgnow drift is 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 com encom033 95bjpgproceed in the same way to find the ground speed home of 120 kt right drift of 15° true heading of 195° ground speed out gso = 120 ktground speed home gsh = 120 ktdistance 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 60120 = 675 minutes

Question 118-31 : At reference or see flight planning manual sep 1 figure 25given fl 75lean mixture and full throttle 2300 rpmtake off fuel 444 lbstake off from mslfind endurance in hours and minutes err a 033 100 ?

05 hours 12 minutes.

com encom033 100jpg52h ==> 5h + 02h 02 x 60 = 12 minutes 5 h 12 minutes

Question 118-32 : From which of the following would you expect to find facilitation information regarding customs and health formalities ?

Aip.

Question 118-33 : An aircraft flies at a tas of 380 kt it flies from a to b and back to a distance ab = 480 nmwhen going from a to b it experiences a headwind component = 60 kt the wind remains constantthe duration of the flight will be ?

2 h 35 min.

Take care the wind remains constant headwind from a to b becomes tailwind from b to atas is 380 kt from a to b ground speed is 380 60 = 320 kt480 nm 320 kt = 15 h 1h30 from b to a ground speed is 380+60 = 440 kt480 nm 440 kt = 109 h 1h05 1h30 + 1h05 = 2 h 35 min

Question 118-34 : Given distance from departure to destination 350 nmtrue track 320wv 35030tas 130 ktwhat is the distance and time of the pet from the departure point ?

Distance 210 nm time 121 min.

Under index set true track 320° centre dot on tas 130 kt with the rotative scale set wind 350°30 kt you find a left drift of 7°now drift is 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 ktproceed in the same way to find the ground speed home of 155 kt right drift of 6° true heading of 134° ground speed out gso = 104 ktground speed home gsh = 155 ktdistance 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 60104 = 121 minutes

Question 118-35 : Given distance from departure to destination 250 nmgs out 130 ktgs home 100 ktwhat is the distance of the pet from the departure point ?

109 nm.

Ground speed out gso = 130 ktground speed home gsh = 100 ktdistance to pet = distance x gsh gso + gsh distance to pet = 250 x 100 130 + 100 distance to pet = 25000 230 = 109 nm

Question 118-36 : Given distance from departure to destination 550 nmendurance 36 htrue track 200wv 22015tas 130 ktwhat is the distance of the psr from the departure point ?

231 nm.

Under index set true track 200° centre dot on tas 130 kt with the rotative scale set wind 220°15 kt you find a left drift of 3°now drift is 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 ktproceed in the same way to find the ground speed home of 142 kt true track of 020° right drift of 2° true heading of 018° ground speed out gso = 115 ktground speed home gsh = 142 ktpoint of safe return psr = endurance x homeward gs outbound gs + homeward gs point of safe return psr = 36 x 142 115 + 142 point of safe return psr = 5112 257point of safe return psr = 199 hdistance of the psr from the departure point at a speed of 115 kt 199 x 115 = 229 nm closest answer is 231 nm

Question 118-37 : Given distance from departure to destination 180 nmendurance 2 htas 120 ktground speed out 135 ktground speed home 105 ktwhat is the distance and time of the psr from the departure point ?

Distance 118 nm time 53 min.

Point of safe return psr = endurance x homeward gs outbound gs + homeward gs ground speed out = 135 ktground speed home = 105 ktpoint of safe return psr = 2 x 105 135 + 105 point of safe return psr = 210 240point of safe return psr = 0875 h0875 x 60 = 525 minutesdistance of the psr from the departure point at a speed of 135 kt 525 min x 13560 = 118125 nm

Question 118-38 : Given distance from departure to destination 2500 nmgs out 540 ktgs home 470 ktwhat is the time of the pet from the departure point ?

129 min.

Pet = d x vsr vsa + vsrpet = 2500 x 470 540 + 470pet = 1163 nm1163 540 = 215 h 2 x 60 + 015 x 60 = 129 minutes

Question 118-39 : Given distance from departure to destination 875 nmtrue track 240 wind 06050 kttas 500 ktwhat is the distance and time of the pet from the departure point ?

Distance 394 nm time 43 min.

True track 240°wind 06050 ktwind is parallel to our course thus ground speed out 500 + 50 = 550 ktground speed home 500 50 = 450 ktpet = d x gsh gso + gsh pet = 875 x 450 550 + 450 = 39375 nm39375 550 = 0716 minutes60 x 0716 = 43 minutes

Question 118-40 : The forecast period covered by the parischarles de gaulle tafs totals hours err a 033 122 ?

27h.

com encom033 122jpgtotal tafs duration 27h