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Question 1

5 HR 20 MIN 20 SEC corresponds to a longitude difference of ;

Answer 1

80°05’

Question 2

265 US-GAL equals ; ( Specific gravity 0.80 )

Answer 2

803 kg.

Question 3

730 ft/min equals ;

Answer 3

3.7 m/sec

Question 4

1215 UTC LAJES VORTAC ( 38°46’N 027°05’W ) RMI reads 178° , range 135 NM. Calculate the aircraft position at 1215 UTC ?

Answer 4

40°55’N 027°55’W

Question 5

1300 UTC DR position 37°30’N 021°30’W alter heading PORT SANTO NDB ( 33°03’N 016°23’W ) , TAS 450 kt , Forecast W/V 360°/30 kt. Calculate the ETA at PORT SANTO NDB ?

Answer 5

1348

Question 6

A chart has the scale 1 : 1 000 000. From A to B on the chart measures 1.5 inches ( one inch equals 2.54 centimetres ) , the
distance from A to B in NM is ;

Answer 6

20.6

Question 7

A chart has the scale 1 : 1 000 000. From A to B on the chart measures 3.8 cm , the distance from A to B in NM is ;

Answer 7

20.5

Question 8

A course of 120°( T ) is drawn between “X” ( 61°30’N ) and “Y” ( 58°30’N ) on a Lambert Conformal conic chart with a scale of 1 : 1 000 000 at 60°N. The chart distance between “X” and “Y” is ;

Answer 8

66.7 cm

Question 9

A definition of a Magnetic Track angle is ;

Answer 9

The direction of a line referenced to Magnetic North.

Question 10

A direct Mercator graticule is based on a projection that is ;

Answer 10

cylindrical.

Question 11

A direct reading compass should be swung when ;

Answer 11

there is a large , and permanent , change in magnetic latitude.

Question 12

A flight is to be made from “A” 49°S 180’E/W to “B” 58°S 180’E/W. The distance in kilometres from “A” to “B” is approximately ;

Answer 12

1000.

Question 13

A great circle intersects the equator in 030°W with a great circle direction of 035°( T ). An aircraft tracking the great circle will reach the maximum Northern/Southern latitude in position ;

Answer 13

( 55°N , 060°E ).

Question 14

A great circle on the Earth running from the North Pole to the South Pole is called ;

Answer 14

a meridian.

Question 15

A great circle track crosses the equator at 30°W has an initial track of 035°T. It’s highest or lowest North/South point is ;

Answer 15

55ºN 060ºE.

Question 16

A great circle track joins position A ( 59°S 141°W ) and B ( 61°S 148°W ). What is the difference between the great circle
track at A and B ?

Answer 16

It increases by 6°.

Question 17

A ground feature appears 30° to the left of the centre line of the CRT of an airborne weather radar. If the heading of the aircraft is 355° ( M ) and the magnetic variation is 15° East , the true bearing of the aircraft from the feature is ;

Answer 17

160°

Question 18

A ground feature was observed on a relative bearing of 315° and 3 MIN later on a relative bearing of 270°. The W/V is calm , aircraft GS 180 kt. What is the minimum distance between the aircraft and the ground feature ?

Answer 18

9 NM

Question 19

A ground feature was observed on a relative bearing of 325° and five minutes later on a relative bearing of 280°. The aircraft
heading was 165° ( M ) , variation 25°W , drift 10°Right and GS 360 kt. When the relative bearing was 280° , the distance and true
bearing of the aircraft from the feature was ;

Answer 19

30 NM and 240°.

Question 20

A Lambert conformal conic chart has a constant of the cone of 0.75. The initial course of a straight line track drawn on this chart from A ( 40°N 050°W ) to B is 043°( T ) at A , course at B is 055°( T ). What is the longitude of B ?

Answer 20

34°W

Question 21

A Lambert conformal conic chart has a constant of the cone of 0.80. A straight line course drawn on this chart from A ( 53°N
004°W ) to B is 080° at A , course at B is 092° ( T ). What is the longitude of B ?

Answer 21

011°E

Question 22

A Lambert conformal conic projection , with two standard parallels ;

Answer 22

the scale is only correct along the standard parallels.

Question 23

A line drawn on a chart which joins all points where the value of magnetic variation is zero is called an ;

Answer 23

agonic line.

Question 24

A Mercator chart has a scale at the equator = 1 : 3 704 000. What is the scale at latitude 60° S ?

Answer 24

1 : 1 852 000

Question 25

A nautical mile is ;

Answer 25

1852 metres.

Question 26

A negative ( westerly ) magnetic variation signifies that ;

Answer 26

True North is East of Magnetic North.

Question 27

A pilot accidently turning OFF the IRS in flight , and then turns it back ON a few moments later. Following this incident

Answer 27

it can only be used for attitude reference.

Question 28

A rhumb line from a position 86°N 30°W has an initial track of 085°T is it ;

Answer 28

a spiral to the North pole

Question 29

A Rhumb line is ;

Answer 29

a line on the surface of the earth cutting all meridians at the same angle.

Question 30

A rhumb line on a Direct Mercator chart appears as a ;

Answer 30

straight line.

Question 31

A ring laser gyro is ;

Answer 31

a device which measures angular movements.

Question 32

A route is flown from ( 80°S , 100°W ) to ( 80°S , 140°E ). At 160°W the Grid Track ( GT ) and True Track ( TT ) on a
Polar Stereographic chart with a grid orientated on the 180º meridian are respectively ;

Answer 32

290º(G) and 270º(T).

Question 33

A route is flown from ( 80°S , 100°W ) to ( 80°S , 140°E ). At 180°E/W the Grid Track ( GT ) and True Track ( TT ) on a Polar Stereographic chart , whose grid is aligned with the Greenwich meridian , are respectively ;

Answer 33

110º(G) and 290º(T).

Question 34

A route is flown from ( 85°S , 100°E ) to ( 85°S , 140°W ). At 160°E the Grid Track ( GT ) and True track ( TT ) on a Polar Stereographic chart with a grid orientated on the 180º meridian are respectively ;

Answer 34

070º(G) and 090º(T).

Question 35

A route is flown from ( 85°S , 100°E ) to ( 85°S , 140°W ). At 180°E/W the Grid Track ( GT ) and True Track ( TT ) on a Polar Stereographic chart , whose grid is aligned with the Greenwich meridian , are respectively ;

Answer 35

250º(G) and 070º(T).

Question 36

A straight line drawn on a chart measures 4.63 cm and represents 150 NM. The chart scale is ;

Answer 36

1 : 6 000 000

Question 37

A straight line from A ( 53ºN , 155ºW ) to B ( 53ºN , 170ºE ) is drawn on a Lambert Conformal conical chart with standard parallels at 50ºN and 56ºN. When passing the meridian 175ºE , the True Track is ;

Answer 37

260.0º

Question 38

A straight line from A ( 53ºS , 155ºE ) to B ( 53ºS , 170ºW ) is drawn on a Lambert Conformal conical chart with standard parallels at 50ºS and 56ºS. When passing 175ºW , the True Track is ;

Answer 38

078.0º

Question 39

A straight line from A ( 75ºN , 120ºE ) to B ( 75ºN , 160ºE ) is drawn on a Polar Stereographic chart. When passing the
meridian 155ºE , the True Track is ;

Answer 39

105º

Question 40

A straight line from A ( 75ºS , 120ºE ) to B ( 75ºS , 160ºE ) is drawn on a Polar Stereographic chart. When passing the meridian 155ºE , the True Track is ;

Answer 40

075º

Question 41

A straight line is drawn on a Lamberts conformal conic chart between two positions of different longitude. The angular difference between the initial true track and the final true track of the line is equal to ;

Answer 41

chart convergency

Question 42

A straight line on a chart 4.89 cm long represents 185 NM. The scale of this chart is approximately ;

Answer 42

1 : 7 000 000

Question 43

A straight line on a Lambert Conformal Projection chart for normal flight planning purposes ;

Answer 43

is approximately a Great Circle

Question 44

A useful method of a pilot resolving , during a visual flight , any uncertainty in the aircraft’s position is to maintain visual contact with the ground and ;

Answer 44

set heading towards a line feature such as a coastline , motorway , river or railway.

Question 45

A VOR is situated at position ( N55°26’ , W005°42’ ). The variation at the VOR is 9°W. The position of the aircraft is
( N60°00’N , W010°00’ ). The variation at the aircraft-position is 11°W. The initial TT-angle of the great circle from the aircraft position to the VOR is 101.5°. Which radial is the aircraft on ?

Answer 45

294

Question 46

A VOR is situated at position ( 74ºN , 094ºW ) , local variation is 50ºW. A Polar Stereographic chart supplied with a
Greenwich grid is used for navigation. To proceed along ( magnetic ) radial 238 inbound an aircraft has to follow a Grid Track of ;

Answer 46

103º

Question 47

After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM South of the intended track and 2.5 NM ahead of the dead reckoning position. To reach destination B from this position , the TH should be ;

Answer 47

292°

Question 48

After alignment of the stable platform of an Inertial Navigation System , the output data from the platform is ;

Answer 48

acceleration north/south and east/west , attitude and true heading.

Question 49

After alignment of the stable platform of the Inertial Navigation System , the output data from the INS computer to the platform is

Answer 49

rate corrections to the gyros.

Question 50

An aeronautical chart is conformal when ;

Answer 50

at any point the scale over a short distance in the direction of the parallel is equal to the scale in the direction of the meridian and the meridians are perpendicular to the parallels.

Question 51

An aeroplane flies from A ( 59°S 142°W ) to B ( 61°S 148°W ) with a TAS of 480 kt. The autopilot is engaged and coupled with an Inertial Navigation System in which AB track is active. On route AB , the true track ;

Answer 51

increases by 5°.

Question 52

An aeroplane is flying at TAS 180 kt on a track of 090°. The W/V is 045° / 50kt. How far can the aeroplane fly out from its base and return in one hour ?

Answer 52

85 NM

Question 53

An Agonic line is a line that connects ;

Answer 53

positions that have 0° variation.

Question 54

An aircraft at FL120 , IAS 200 kt , OAT –5° and wind component +30 kt , is required to reduce speed in order to cross a reporting point 5 MIN later than planned. Assuming flight conditions do not change , when 100 NM from the reporting point IAS
should be reduced to ;

Answer 54

159 kt.

Question 55

An aircraft at FL140 , IAS 210 kt , OAT –5°C and wind component minus 35 kt , is required to reduce speed in order to cross a reporting point 5 MIN later than planned. Assuming that flight conditions do not change , when 150 NM from the reporting point the IAS should be reduced by ;

Answer 55

20 kt.

Question 56

An aircraft at FL290 is required to commence descent when 50 NM from a VOR and to cross that VOR at FL80. Mean GS during descent is 271 kt. What is the minimum rate of descent required ?

Answer 56

1900 FT/MIN

Question 57

An aircraft at FL310 , M0.83 , temperature –30°C , is required to reduce speed in order to cross a reporting point five minutes later than planned. Assuming that a zero wind component remains unchanged , when 360 NM from the reporting point Mach Number should be reduced to ;

Answer 57

M0.74

Question 58

An aircraft at FL330 is required to commence descent when 65 NM from a VOR and to cross the VOR at FL100. The mean GS during the descent is 330 kt. What is the minimum rate of descent required ?

Answer 58

1950 FT/MIN

Question 59

An aircraft at FL350 is required to commence descent when 85 NM from a VOR and to cross the VOR at FL80. The mean
GS for the descent is 340 kt. What is the minimum rate of descent required ?

Answer 59

1800 FT/MIN

Question 60

An aircraft at FL350 is required to cross a VOR/DME facility at FL110 and to commence descent when 100 NM from the facility. If the mean GS for the descent is 335 kt , the minimum rate of descent required is ;

Answer 60

1340 FT/MIN

Question 61

An aircraft at FL350 is required to descend to cross a DME facility at FL80. Maximum rate of descent is 1800 FT/MIN and mean GS for descent is 276 kt. The minimum range from the DME at which descent should start is ;

Answer 61

69 NM

Question 62

An aircraft at FL360 is required to descent to FL120. The aircraft should reach FL120 at 40 NM from the next waypoint. The rate of descent is 2000 ft/min. The average GS is 420 kt. The minimum distance from the next waypoint at which descent should start

Answer 62

124 NM

Question 63

An aircraft at FL370 is required to commence descent at 120 NM from a VOR and to cross the facility at FL130. If the mean GS for the descent is 288 kt , the minimum rate of descent required is ;

Answer 63

960 FT/MIN

Question 64

An aircraft at FL370 is required to commence descent when 100 NM from a DME facility and to cross the station at FL120.
If the mean GS during the descent is 396 kt , the minimum rate of descent required is approximately ;

Answer 64

1650 FT/MIN

Question 65

An aircraft at FL370 , M0.86 , OAT −44°C , headwind component 110 kt , is required to reduce speed in order to cross a reporting point 5 MIN later than planned. If the speed reduction were to be made 420 NM from the reporting point , what Mach Number is required

Answer 65

M0.81

Question 66

An aircraft at FL390 is required to descend to cross a DME facility at FL70. Maximum rate of descent is 2500 FT/MIN ,
mean GS during descent is 248 kt. What is the minimum range from the DME at which descent should commence ?

Answer 66

53 NM

Question 67

An aircraft at latitude 02°20’N tracks 180°( T ) for 685 km. On completion of the flight the latitude will be ;

Answer 67

03°50’S

Question 68

An aircraft at latitude 10°North flies south at a groundspeed of 445 km/hr. What will be its latitude after 3 HR ?

Answer 68

02°00’S

Question 69

An aircraft at latitude 10°South flies north at a GS of 890 km/hr. What will its latitude be after 1.5 HR ?

Answer 69

02°00’N

Question 70

An aircraft at position 60°N 005°W tracks 090°( T ) for 315 km. On completion of the flight the longitude will be ;

Answer 70

000°40’E

Question 71

An aircraft departing A ( N40º00’ E080º00’ ) flies a constant true track of 270º at a ground speed of 120 kt. What are the coordinates of the position reached in 6 HR ?

Answer 71

N40º00’ E064º20’

Question 72

An aircraft departs from position A ( 04°10’S 178°22’W ) and flies northward following the meridian for 2950 NM. It then flies westward along the parallel of latitude for 382 NM to position B. The coordinates of position B are ?

Answer 72

45°00’N 172°38’E

Question 73

An aircraft departs from Schiphol airport and flies to Santa Cruz in Bolivia ( South America ) via Miami in Florida. The 
departure time ( off blocks )  is 07:45 ST at the 10th of November , taxi time before take off at Schiphol is 25 minutes. The flight time to Miami over the Atlantic Ocean is 09 h 20 m. The total taxi time in Miami to and from the gate is 25 minutes. The time spend at the gate is 02 h 40 m. From Miami to Santa Cruz the airborne time is 06 h 30 m. Calculate the time and date of touch down in Santa Cruz in ST Bolivia if the difference between ST and UTC is 5 hours ;

Answer 73

21:05 10th Nov.

Question 74

An aircraft equipped with an Inertial Navigation System ( INS ) flies with INS 1 coupled with autopilot 1. Both inertial navigation systems are navigating from way-point A to B. The inertial systems” Central Display Units ( CDU ) shows ;
− XTK on INS 1 = 0
− XTK on INS 2 = 8L ( XTK = cross track )
From this information it can be deduced that ;

Answer 74

at least one of the inertial navigation systems is drifting.

Question 75

An aircraft flies a great circle track from 56°N 070°W to 62°N 110°E. The total distance travelled is ?

Answer 75

3720 NM

Question 76

An aircraft flies at FL 250. OAT = –45°C. The QNH , given by a meteorological station with an elevation of 2830 ft , is 1033 hPa. Calculate the clearance above a mountain ridge with an elevation of 20410 ft ;

Answer 76

4200 ft

Question 77

An aircraft flies from waypoint 7 ( 63°00’N , 073°00’W ) to waypoint 8 ( 62°00’N , 073°00’W ). The aircraft position is
( 62°00’N , 073°10’W ). The cross track distance in relation to the planned track is ;

Answer 77

4.7 NM R

Question 78

An aircraft flies the following rhumb line tracks and distances from position 04°00’N 030°00’W 600 NM South , then 600 NM East , then 600 NM North , then 600 NM West. The final position of the aircraft is ;

Answer 78

04°00’N 029°58’W

Question 79

An aircraft follows a great circle in the Northern Hemisphere. At a certain moment the aircraft is in the position on the great
circle where the great circle direction is 270°(T). Continuing on the great circle the ;

Answer 79

track angle will decrease and the latitude will decrease.

Question 80

An aircraft follows a radial to a VOR/DME station. At 10:00 the DME reads 120 NM. At 10:03 the DME reads 105 NM. The estimated time overhead the VOR/DME station is ;

Answer 80

10:24

Question 81

An aircraft has to fly over a mountain ridge. The highest obstacle , indicated in the navigation chart , has an elevation of 9800 ft. The QNH , given by a meteorological station at an elevation of 6200 ft , is 1022 hPa. The OAT = ISA+5ºC. Calculate the approximate indicated altitude to obtain a clearance of 2000 ft.

Answer 81

Between 11500 ft and 11700 ft.

Question 82

An aircraft in the northern hemisphere is making an accurate rate one turn to the right. If the initial heading was 135° , after
30 seconds the direct reading magnetic compass should read ;

Answer 82

more than 225°

Question 83

An aircraft in the northern hemisphere makes an accurate rate one turn to the right/starboard. If the initial heading was 330° , after 30 seconds of the turn the direct reading magnetic compass should read ;

Answer 83

less than 060°

Question 84

An aircraft is at position ( 53ºN , 006ºW ) and has a landmark at position ( 52º47’N , 004º45’W ) , with a relative bearing of 060º. Given ;
Compass Heading = 051º
Variation = 16ºW
Deviation = 2ºE
What is the true bearing of the position line to be plotted from the landmark to the aircraft on a Lambert chart with standard parallels at 37ºN and 65ºN ?

Answer 84

278º

Question 85

An aircraft is departing from an airport which has an elevation of 2000 ft and the QNH is 1003 hPa. The TAS is 100 kt , the
head wind component is 20 kt and the rate of climb is 500 ft/min. Top of climb is FL 050. At what distance from the airport will this be achived

Answer 85

7.2 NM

Question 86

An aircraft is departing from an airport which has an elevation of 2000 ft and the QNH is 1003 hPa. The TAS is 100 kt , the head wind component is 20 kt and the rate of climb is 1000 ft/min. Top of climb is FL 050. At what distance from the airport will this be achived

Answer 86

3.6 NM

Question 87

An aircraft is departing from an airport which has an elevation of 2000 ft and the QNH is 1003 hPa. The TAS is 100 kt , the head wind component is 20 kt and the rate of climb is 1000 ft/min. Top of climb is FL 100. At what distance from the airport will this be achived ?

Answer 87

10.3 NM

Question 88

An aircraft is departing from an airport which has an elevation of 2000 ft and the QNH is 1023 hPa. The TAS is 100 kt , the
head wind component is 20 kt and the rate of climb is 1000 ft/min. Top of climb is FL 100. At what distance from the airport will this
be achieved ?

Answer 88

11.1 NM

Question 89

An aircraft is descending down a 6% slope whilst maintaining a G/S of 300 kt. The rate of descent of the aircraft is approximately ;

Answer 89

1800 FT/MIN

Question 90

An aircraft is flying according the flight log at the Annex. After 15 minutes of flying with the planned TAS and TH the
aircraft is 3 NM North of the intended track and 2.5 NM ahead of the dead reckoning position. To reach destination B from this
position the TH should be ;
(refer to image 5)

Answer 90

112º

Question 91

An aircraft is flying according the flight log at the Annex. After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM North of the intended track and 2.5 NM ahead of the dead reckoning position. To reach destination B from this position the TH should be ; (refer to image 9)

Answer 91

258º

Question 92

An aircraft is flying at FL100. The OAT = ISA – 15°C. The QNH given by a station at an elevation 3000 ft is 1035 hPa. Calculate the approximate True Altitude ;

Answer 92

10 200 ft

Question 93

An aircraft is flying at FL150 , with an outside air temperature of –30° , above an airport where the elevation is 1660 ft and the QNH is 993 hPa. Calculate the true altitude ; ( Assume 30 ft = 1 hPa )

Answer 93

13 660 ft

Question 94

An aircraft is flying at FL180 and the outside air temperature is –30°C. If the CAS is 150 kt , what is the TAS ?

Answer 94

195 kt

Question 95

An aircraft is flying at FL 200. OAT = 0°C. When the actual air pressure on an airfield at MSL is placed in the subscale of the altimeter the indicated altitude is 19300 ft. Calculate the aircraft’s true altitude ;

Answer 95

21.200 ft

Question 96

An aircraft is flying at FL200. The QNH , given by a meteorological station at an elevation of 1300 ft is 998.2 hPa.
OAT = –40ºC. The elevation of the highest obstacle along the route is 8 000 ft. Calculate the aircraft’s approximate clearance above the highest obstacle on this route ;

Answer 96

10.500 ft

Question 97

An aircraft is flying at FL250 , OAT = –45°C. The QNH , given by a station at MSL , is 993.2 hPa. Calculate the
approximate True Altitude

Answer 97

23400 ft

Question 98

An aircraft is flying from A to B a distance of 50 NM. The True course in the flight log is 090º , the forecast wind is 225º(T)/15 kt and the TAS is 120 kt. After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM South of the intended track and 2.5 NM ahead of the dead reckoning position. To reach destination B from this position , the correction angle on the heading should be ;

Answer 98

17º

Question 99

An aircraft is flying from A to B a distance of 50 NM. The True Course in the flight log is 270º , the forecast wind is 045º(T)/15 kt and the TAS is 120 kt. After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM South of the intended track and 2.5 NM ahead of the dead reckoning position. To reach destination B from this position , the correction angle on the heading should be

Answer 99

17º

Question 100

An aircraft is flying from A to B. The true course according to the flight log is 090º , the estimated wind is 225º(T)/15 kt and the TAS is 120 kt. After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM South of the intended track and 2.5 NM ahead of the dead reckoning position. The Track angle error ( TKE ) is ;

Answer 100

5ºR

Question 101

An aircraft is flying from Inverness VORDME ( N57°32.6’’ , W004°02.5W ) to Aberdeen VORDME ( N57°18.6’’ , W002° 16.0’‘W ). At 1000 UTC the fix of the aircraft is determined by VORDME Inverness ; radial = 114 , DME-distance = 20.5 NM. At 1006 UTC the fix of the aircraft is determined by VORDME Aberdeen ; radial = 294 , DME-distance = 10.5 NM. What is the average GS of the aircraft between 1000 UTC and 1006 UTC ?
( For this question see Europe Low Altitude Enroute Chart E(LO) 1A )

Answer 101

280 kt

Question 102

An aircraft is flying from SALCO to BERRY HEAD on Magnetic Track 007º , TAS 445 kt. The wind is 050º(T)/40 kt. Variation 5ºW , deviation +2º. At 1000 UTC the RB of locator PY is 311º. At 1003 UTC the RB of locator PY is 266º. Calculate the distance of the aircraft from locator PY at 1003 UTC ;

Answer 102

21 NM

Question 103

An aircraft is flying from SALCO to BERRY HEAD on Magnetic Track 007º , TAS 445 kt. The wind is 050º(T)/40 kt.
Variation 5ºW , deviation +2º. At 1000 UTC the RB of locator PY is 311º. At 1003 UTC the RB of locator PY is 266º. Calculate the
True bearing of locator PY at 1003 UTC from the aircraft ;

Answer 103

272º(T)

Question 104

An aircraft is flying with the aid of an inertial navigation system ( INS ) connected to the autopilot. The following two points have been entered in the INS computer ;
WPT 1 : 60°N 030°W
WPT 2 : 60°N 020°W
When 025°W is passed the latitude shown on the display unit of the inertial navigation system will be ;

Answer 104

60°05.7’N

Question 105

An aircraft is following a true track of 048° at a constant TAS of 210 kt. The wind velocity is 350° / 30 kt. The GS and drift
angle are ;

Answer 105

192 kt , 7° right

Question 106

An aircraft is following the 45°N parallel of latitude. The track followed is a ;

Answer 106

rhumb line.

Question 107

An aircraft is in the position ( 86°N , 020°E ). When following a rhumb line track of 085°( T ) it will ;

Answer 107

fly via a spiral to the North Pole.

Question 108

An aircraft is maintaining a 5.2% gradient is at 7 NM from the runway , on a flat terrain , its height is approximately ;

Answer 108

2210 ft.

Question 109

An aircraft is over position HO ( 55°30’N 060°15’W ) , where YYR VOR ( 53°30’N 060°15’W ) can be received. The magnetic variation is 31°W at HO and 28°W at YYR. What is the radial from YYR ?

Answer 109

028°

Question 110

An aircraft is planned to fly from position “A” to position “B” , distance 250 NM at an average GS of 115 kt. It departs “A”
at 0900 UTC. After flying 75 NM along track from “A” , the aircraft is 1.5 MIN behind planned time. Using the actual GS experienced , what is the revised ETA at “B” ?

Answer 110

1115 UTC

Question 111

An aircraft is planned to fly from position “A” to position “B” , distance 320 NM , at an average GS of 180 kt. It departs “A” at 1200 UTC. After flying 70 NM along track from “A” , the aircraft is 3 MIN ahead of planned time. Using the actual GS experienced , what is the revised ETA at “B” ?

Answer 111

1333 UTC

Question 112

An aircraft is planned to fly from position “A” to position “B” , distance 480 NM at an average GS of 240 kt. It departs “A” at 1000 UTC. After flying 150 NM along track from “A” , the aircraft is 2 MIN behind planned time.Using the actual GS experienced , what is the revised ETA at “B” ?

Answer 112

1206

Question 113

An aircraft must fly 2000 ft above an obstacle of which the elevation is 13 600 ft. The QHN at the nearest airfield is 991 hPa , the elevation is 1500 ft and the temperature is –20°C. Calculate the minimum altitude required.

Answer 113

17 400 ft

Question 114

An aircraft obtains a relative bearing of 315° from an NDB at 0830. At 0840 the relative bearing from the same NDB is 270°.
Assuming no drift and a GS of 240 kt , what is the approximate range from the NDB at 0840 ?

Answer 114

40 NM

Question 115

An aircraft passes position A ( 60°00’N 120°00’W ) on route to position B ( 60°00’N 140°30’W ). What is the great circle track on departure from A ?

Answer 115

279°

Question 116

An aircraft takes-off from an airport 2 hours before sunset. The pilot flies a track of 090° ( T ) , W/V 130° / 20 kt , TAS 100
kt. In order to return to the point of departure before sunset , the furthest distance which may be travelled is ;

Answer 116

97 NM.

Question 117

An aircraft takes off from the aerodrome of BRIOUDE ( altitude 1483 ft , QFE = 963 hPa , temperature = 32°C ). Five minutes later , passing 5 000 ft on QFE , the second altimeter set on 1013 hPa will indicate approximately ;

Answer 117

6 500 ft.

Question 118

An aircraft travels 2.4 statute miles in 47 seconds. What is its groundspeed ?

Answer 118

160 kt

Question 119

An aircraft travels 100 statute miles in 20 MIN , how long does it take to travel 215 NM ?

Answer 119

50 MIN

Question 120

An aircraft travels from point A to point B , using the autopilot connected to the aircraft’s inertial system. The coordinates of A ( 45°S 010°W ) and B ( 45°S 030°W ) have been entered. The true course of the aircraft on its arrival at B to the nearest degree , is ;

Answer 120

277°

Question 121

An aircraft was over “A” at 1435 hours flying direct to “B”.
	Given ;
					Distance “A” to “B” 2900 NM
					True airspeed 470 kt
					Mean wind component “out” +55 kt
					Mean wind component “back” –75 kt
					Safe endurance 9 HR 30 MIN
	The distance from “A” to the Point of Safe Return ( PSR ) “A” is ;

Answer 121

2141 NM

Question 122

An aircraft was over “A” at 1435 hours flying direct to “B”. Given ;
Distance “A” to “B” 2900 NM
True airspeed 470 kt
Mean wind component “out” +55 kt
Mean wind component “back” –75 kt
The ETA for reaching the Point of Equal Time ( PET ) between “A” and “B” is ;

Answer 122

1657

Question 123

An aircraft was over “Q” at 1320 hours flying direct to “R”. Given ;
Distance “Q” to “R” 3016 NM
True airspeed 480 kt
Mean wind component “out” –90 kt
Mean wind component “back” +75 kt
Safe endurance 10:00 HR
The distance from “Q” to the Point of Safe Return ( PSR ) “Q” is ;

Answer 123

2290 NM

Question 124

An aircraft was over “Q” at 1320 hours flying direct to “R”.
Given ;
Distance “Q” to “R” 3016 NM
True airspeed 480 kt
Mean wind component “out” –90 kt
Mean wind component “back” +75 kt
The ETA for reaching the Point of Equal Time ( PET ) between “Q” and “R” is ;

Answer 124

1752

Question 125

An island appears 30° to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading ( MH ) of 020° with the magnetic variation ( VAR ) 25°W ?

Answer 125

145°

Question 126

An island appears 30° to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading of 276° with the magnetic variation
12°W ?

Answer 126

054°

Question 127

An island appears 30° to the right of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading ( MH ) of 355° with the magnetic variation ( VAR ) 15°E ?

Answer 127

220°

Question 128

An island appears 45° to the right of the centre line on an airborne weather radar display. What is the true bearing of the
aircraft from the island if at the time of observation the aircraft was on a magnetic heading ( MH ) of 215° with the magnetic variation
( VAR ) 21°W ?

Answer 128

059°

Question 129

An island appears 60° to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading ( MH ) of 276° with the magnetic variation ( VAR ) 10°E ?

Answer 129

046°

Question 130

An island is observed by weather radar to be 15° to the left. The aircraft heading is 120° ( M ) and the magnetic variation
17°W. What is the true bearing of the aircraft from the island ?

Answer 130

268°

Question 131

An island is observed to be 15° to the left. The aircraft heading is 120° ( M ) , variation 17° ( W ). The bearing ( °T ) from the aircraft to the island is ;

Answer 131

088

Question 132

An island is observed to be 30° to the right of the nose of the aircraft. The aircraft heading is 290° ( M ) , variation 10° ( E ).
The bearing ( °T ) from the aircraft to the island is ;

Answer 132

330°

Question 133

An NDB is located at position ( N55°26’ , W005°42’ ). The variation at the NDB is 9°W. The position of the aircraft is
( 56°00’N , 010°00’W ). The variation at the aircraft position is 11°W. The initial TT of the great circle from the aircraft position to the NDB position , is 101.5°. What is the Magnetic Bearing of the NDB from the aircraft ?

Answer 133

112.5º

Question 134

An Oblique Mercator projection is used specifically to produce ;

Answer 134

charts of the great circle route between two points.

Question 135

An observer is situated on the parallel of 23.5°S. Which statement about the passage of the apparent sun in relation to this
position is correct ?

Answer 135

It passes through the zenith once a year around December 22nd.

Question 136

Approximately how many nautical miles correspond to 12 cm on a map with a scale of 1 : 2 000 000 ?

Answer 136

130

Question 137

As the INS position of the departure aerodrome , coordinates 35°32.7’N 139°46.3’W are input instead of 35°32.7’N 139°46.3’E. When the aircraft subsequently passes point 52°N 180°W , the longitude value shown on the INS will be ;

Answer 137

099° 32.6’W

Question 138

Assume a Mercator chart. The distance between positions A and B , located on the same parallel and 10° longitude apart , is 6 cm. The scale at the parallel is 1 : 9 260 000. What is the latitude of A and B ?

Answer 138

60° N or S

Question 139

Assume a North polar stereographic chart whose grid is aligned with the Greenwich meridian. An aircraft flies from the geographic North pole for a distance of 480 NM along the 110°E meridian , then follows a grid track of 154° for a distance of 300 NM. Its position is now approximately ;
( For this question use annex 061-1818A )

Answer 139

80°00’N 080°E

Question 140

Assuming mid-latitudes ( 40° to 50°N/S ). At which time of year is the relationship between the length of day and night , as well as the rate of change of declination of the sun , changing at the greatest rate ?

Answer 140

Spring equinox and autumn equinox.

Question 141

Assuming zero wind , what distance will be covered by an aircraft descending 15 000 FT with a TAS of 320 kt and maintaining a rate of descent of 3000 FT/MIN ?

Answer 141

26.7 NM

Question 142

At 00:00 Local Mean Time of an observer ;

Answer 142

the mean sun is in transit with the observer’s anti-meridian.n.

Question 143

At 0020 UTC an aircraft is crossing the 310° radial at 40 NM of a VOR/DME station. At 0035 UTC the radial is 040° and DME distance is 40 NM. Magnetic variation is zero. The true track and ground speed are ;

Answer 143

085° - 226 kt

Question 144

At 0422 an aircraft at FL370 , GS 320 kt , is on the direct track to VOR “X” 185 NM distant. The aircraft is required to cross
VOR “X” at FL80. For a mean rate of descent of 1800 FT/MIN at a mean GS of 232 kt , the latest time at which to commence descent is ;

Answer 144

0445

Question 145

At 10:15 the reading from a VOR/DME station is 211°/ 90 NM , at 10:20 the reading from the same VOR/DME station is 211°/120 NM.
Compass Heading = 200º
Variation in the area = 31ºW
Deviation = +1º
TAS = 390 kt
The wind vector ( T ) is approximately ;

Answer 145

110º/70 k

Question 146

At 47° North the chart distance between meridians 10° apart is 5 inches. The scale of the chart at 47° North approximates ;

Answer 146

1 : 6 000 000

Question 147

At 47° North the chart distance between meridians 10° apart is 12.7 cm. The scale of the chart at 47° North approximates ;

Answer 147

1 : 6 000 000

Question 148

At ( 54°N , 020°W ) the sun rises on November 28th at 09:01 UTC. At ( 44°N , 020°W ) the sun will rise ;

Answer 148

earlier since the latter position lies further South.

Question 149

At 60° N the scale of a direct Mercator chart is 1 : 3 000 000. What is the scale at the equator ?

Answer 149

1 : 6 000 000

Question 150

At a specific location , the value of magnetic variation ;

Answer 150

varies slowly over time.

Question 151

At latitude 60°N the scale of a Mercator projection is 1 : 5 000 000. The length on the chart between “C” N60° E008° and “D” N60° W008° is ;

Answer 151

17.8 cm

Question 152

At reference. Magnetic heading of an aircraft is 040 degrees. On the airborne weather radar ( AWR ) display the relative bearing of the distance to the must southerly part of Lands End , ( approximate position : 50 03 N 005 40 W ) are 030 degrees R and 80 NM. What is the position of the aircraft based on these observations ? The slant range correction and the map convergency between aircraft position and Lands End may be neglected.( Refer to Image 1 )

Answer 152

( 49 25 N 007 30 W )

Question 153

At the magnetic equator , when accelerating after take off on heading West , a direct reading pivot suspended compass ;

Answer 153

indicates the correct heading.

Question 154

At what approximate date is the earth closest to the sun ( perihelion ) ?

Answer 154

Beginning of January.

Question 155

At what approximate date is the earth furthest from the sun ( aphelion ) ?

Answer 155

Beginning of July.

Question 156

ATT Mode of the Inertial Reference System ( IRS ) is a back-up mode providing ;

Answer 156

only attitude and heading information.

Question 157

Calculate the constant of the cone on a Lambert Chart given chart convergency between 010°E and 030°W as being 30° ;

Answer 157

0.75

Question 158

Calibrated Airspeed ( CAS ) is Indicated Airspeed ( IAS ) corrected for ;

Answer 158

instrument error and position error.

Question 159

Compass deviation is defined as the angle between ;

Answer 159

Magnetic North and Compass North.

Question 160

Complete line 1 of the “FLIGHT NAVIGATION LOG” , positions “A” to “B”. What is the HDG° ( M ) and ETA ?( For this question use annex 061-9437A )

Answer 160

268° - 1114 UTC

Question 161

Complete line 2 of the “FLIGHT NAVIGATION LOG” , positions “C” to “D”. What is the HDG° ( M ) and ETA ? For this question use annex 061-9437A )

Answer 161

HDG 193° - ETA 1239 UTC

Question 162

Complete line 3 of the “FLIGHT NAVIGATION LOG” , positions “E” to “F”. What is the HDG° ( M ) and ETA ? For this question use annex 061-9437A )

Answer 162

HDG 105° - ETA 1205 UTC

Question 163

Complete line 4 of the “FLIGHT NAVIGATION LOG” , positions “G” to “H”. What is the HDG° ( M ) and ETA ? For this question use annex 061-9437A )

Answer 163

HDG 344° - ETA 1336 UTC

Question 164

Complete line 5 of the “FLIGHT NAVIGATION LOG” , positions “J” to “K”. What is the HDG° ( M ) and ETA ?
For this question use annex 061-9437A )

Answer 164

HDG 337° - ETA 1422 UTC

Question 165

Complete line 6 of the “FLIGHT NAVIGATION LOG” , positions “L” to “M”. What is the HDG° ( M ) and ETA ? For this question use annex 061-9437A )

Answer 165

HDG 075° - ETA 1502 UTC

Question 166

Complete the following statement regarding magnetic variation. The charted values of magnetic variation on earth normally change annually due to

Answer 166

magnetic pole movement causing numerical values at all locations to increase or decrease.

Question 167

Concerning direct reading magnetic compasses , in the northern hemisphere , it can be said that ;

Answer 167

on an Easterly heading , a longitudinal acceleration causes an apparent turn to the North.

Question 168

Consider the following factors that determine the accuracy of a DR position ;
1. The flight time since the last position update.
2. The accuracy of the forecasted wind.
3. The accuracy of the TAS.
4. The accuracy of the steered heading.
Using the list above which of the following contains the most complete answer ?

Answer 168

1 , 2 , 3 and 4.

Question 169

Consider the positions ( 00ºN/S , 000ºE/W ) and ( 00ºN/S , 180ºE/W ) on the ellipsoid. Which statement about the distances between these positions is correct ?

Answer 169

The route via the North Pole is shorter than the route along the equator.

Question 170

Contour lines on aeronautical maps and charts connect points ;

Answer 170

having the same elevation above sea level

Question 171

Deviation applied to magnetic heading gives ;

Answer 171

compass heading.

Question 172

Deviation on the standby compass is

Answer 172

dependent on the heading of the aircraft.

Question 173

Double integration of the output from the east/west accelerometer of an inertial navigation system ( INS ) in the NAV
MODE give ;

Answer 173

distance east/west.

Question 174

During a low level flight 2 parallel roads that are crossed at right angles by an aircraft. The time between these roads can be used to check the aircraft ;

Answer 174

groundspeed.

Question 175

During approach the following data are obtained ;
					DME 12.0 NM , altitude 3000 ft
					DME 9.8 NM , altitude 2400 ft
					TAS = 160 kt , GS = 125 kt
	The rate of descent is ;

Answer 175

570 ft/min

Question 176

During the initial alignment of an inertial navigation system ( INS ) the equipment ;

Answer 176

will not accept a 10° error in initial latitude but will accept a 10° error in initial longitude.

Question 177

For a distance of 1860 NM between Q and R , a ground speed “out” of 385 kt , a ground speed “back” of 465 kt and an
endurance of 8 HR ( excluding reserves ) the distance from Q to the point of safe return ( PSR ) is ;

Answer 177

1685 NM

Question 178

For a landing on runway 23 ( 227° magnetic ) surface W/V reported by the ATIS is 180 / 30 kt. VAR is 13°E. Calculate the cross wind component ;

Answer 178

22 kt

Question 179

From Rakovnik ( 50° 05.9' N , 013° 41.5' E ) to Frankfurt FFM ( 50° 05.9' N , 008° 38.3' E ) the True Track of departure 
along the straight line is 272.0°. The constant of the cone  of this Lambert conformal projection is ;

Answer 179

0.79

Question 180

From the departure point , the distance to the point of equal time is ;

Answer 180

inversely proportional to the sum of ground speed out and ground speed back.

Question 181

Fuel flow per HR is 22 US-GAL , total fuel on board is 83 IMP GAL. What is the endurance ?

Answer 181

4 HR 32 MIN

Question 182

Geodetic latitude and geocentric latitude coincide ;

Answer 182

at the Poles and on the equator.

Question 183

Given ;
	A polar stereographic chart whose grid is aligned with the zero meridian.
					Grid track 344°
					Longitude 115°00'W
	Calculate the true course ?

Answer 183

229°

Question 184

Given ;
					ETA to cross a meridian is 2100 UTC
					GS is 441 kt
					TAS is 491 kt
	At 2010 UTC , ATC requests a speed reduction to cross the meridian at 2105 UTC. The reduction to TAS will be approximately ;

Answer 184

40 kt

Question 185

Given ;
An aircraft is flying a track of 255°( M )
2254 UTC , it crosses radial 360° from a VOR station
2300 UTC , it crosses radial 330° from the same station
At 2300 UTC , the distance between the aircraft and the station is ;

Answer 185

the same as it was at 2254 UTC.

Question 186

Given ;
Aircraft height 2500 ft
ILS GP angle 3°
At what approximate distance from THR can you expect to capture the GP ?

Answer 186

8.3 NM

Question 187

Given ;
An aircraft is flying at FL100
OAT = ISA – 15ºC
The QNH given by a meteorological station with an elevation of 100 ft below MSL is 1032 hPa
1 hPa = 27 ft
Calculate the approximate True Altitude of this aircraft ;

Answer 187

9.900 ft

Question 188

Given ;
					True course 300°
					drift 8°R
					variation 10°W
					deviation –4°
	Calculate the compass heading ;

Answer 188

306°

Question 189

Given ;
					true track 070°
					variation 30°W
					deviation +1°
					drift 10°R
	Calculate the compass heading ;

Answer 189

089°

Question 190

Given ;
					true track 352°
					variation 11° W
					deviation is –5°
					drift 10°R
	Calculate the compass heading ?

Answer 190

358°

Question 191

Given ;
					True course from A to B = 090°
					TAS = 460 kt
					W/V = 360 / 100 kt
					Average variation = 10°E
					Deviation = –2°
	Calculate the compass heading and GS ;

Answer 191

070° - 450 kt

Question 192

Given ;
Runway direction 210° ( M )
Surface W/V 230° ( M ) / 30 kt
Calculate the cross-wind component ;

Answer 192

10 kt

Question 193

Given ;
Runway direction 305° ( M )
Surface W/V 260° ( M ) / 30 kt
Calculate the cross-wind component ?

Answer 193

21 kt

Question 194

Given ;
Pressure Altitude 29000 ft
OAT –55°C
Calculate the Density Altitude ?

Answer 194

27500 ft.

Question 195

Given ;
					True Track 245°
					Drift 5° right
					Variation 3°E
					Compass Hdg 242°
	Calculate the deviation ;

Answer 195

5°W

Question 196

Given ;
					TAS = 95 kt
					HDG ( T ) = 075°
					W/V = 310 / 20 kt
	Calculate the drift and GS ;

Answer 196

9R - 108 kt

Question 197

Given ;
					TAS = 140 kt
					HDG ( T ) = 005°
					W/V = 265 / 25kt
	Calculate the drift and GS

Answer 197

10R - 146 kt

Question 198

Given ;
					TAS = 190 kt 
					HDG ( T ) = 355°
					W/V = 165 / 25kt 
	Calculate the drift and GS ;

Answer 198

1L - 215 kt

Question 199

Given ;
					TAS = 205 kt
					HDG ( T ) = 180°
					W/V = 240 / 25 kt
	Calculate the drift and GS ;

Answer 199

6L - 194 kt

Question 200

Given ;
					TAS = 230 kt
					HDG ( T ) = 250°
					W/V = 205 / 10 kt
	Calculate the drift and GS ;

Answer 200

2R - 223 kt

Question 201

Given ;
					TAS = 250 kt
					HDG ( T ) = 029°
					W/V = 035 / 45 kt
	Calculate the drift and GS ;

Answer 201

1L - 205 kt

Question 202

Given ;
					TAS = 440 kt
					HDG ( T ) = 349°
					W/V = 040 / 40 kt
	Calculate the drift and GS ;

Answer 202

4L - 415 kt

Question 203

Given ;
					TAS = 465 kt
					HDG ( T ) = 124°
					W/V = 170 / 80 kt
	Calculate the drift and GS ;

Answer 203

8L - 415 kt

Question 204

Given ;
					Magnetic heading = 255°
					VAR = 40°W
					GS = 375 kt
					W/V = 235°( T ) / 120 kt
	Calculate the drift angle ?

Answer 204

6° left

Question 205

Given ;
					TAS = 132 kt
					True HDG = 257°
					W/V = 095° ( T ) / 35 kt
	Calculate the drift angle and GS ;

Answer 205

4°R - 165 kt

Question 206

Given ;
					TAS = 140 kt
					True HDG = 302°
					W/V = 045°( T ) / 45 kt
	Calculate the drift angle and GS ;

Answer 206

16°L - 156 kt

Question 207

Given ;
					TAS = 190 kt
					True HDG = 085°
					W/V = 110°( T ) / 50 kt
	Calculate the drift angle and GS ;

Answer 207

8°L - 146 kt

Question 208

Given ;
					TAS = 235 kt
					HDG ( T ) = 076°
					W/V = 040 / 40kt
	Calculate the drift angle and GS ;

Answer 208

7R - 204 kt

Question 209

Given ;
					TAS = 270 kt
					True HDG = 145°
					Actual wind = 205° ( T ) / 30 kt
	Calculate the drift angle and GS ;

Answer 209

6°L - 256 kt

Question 210

Given ;
					TAS = 270 kt
					True HDG = 270°
					Actual wind 205°( T ) / 30 kt
	Calculate the drift angle and GS ;

Answer 210

6°R - 259 kt

Question 211

Given ;
					TAS = 290 kt
					True HDG = 171°
					W/V = 310°( T ) / 30 kt
	Calculate the drift angle and GS ;

Answer 211

4°L - 314 kt

Question 212

Given ;
					TAS = 470 kt
					True HDG = 317°
					W/V = 045°( T ) / 45 kt
	Calculate the drift angle and GS ;

Answer 212

5°L - 470 kt

Question 213

Given ;
					TAS = 472 kt 
					True HDG = 005°
					W/V = 110°( T ) / 50 kt
	Calculate the drift angle and GS ;

Answer 213

6°L - 490 kt

Question 214

Given ;
					TAS = 485 kt
					True HDG = 226°
					W/V = 110°( T ) / 95 kt
	Calculate the drift angle and GS ;

Answer 214

9°R - 533 kt

Question 215

Given ;
Runway direction 230°( T )
Surface W/V 280°( T ) / 40 kt
Calculate the effective cross-wind component ;

Answer 215

31 kt

Question 216

Given ;
Runway direction 083°( M )
Surface W/V 035 / 35 kt
Calculate the effective headwind component ;

Answer 216

24 kt

Question 217

Given ;
					TAS = 220 kt
					Magnetic course = 212 º
					W/V 160º( M ) / 50kt
	Calculate the GS ?

Answer 217

186 kt

Question 218

Given ;
					True Heading = 090°
					TAS = 200 kt
					W/V = 220° / 30 kt
	Calculate the GS ?

Answer 218

220 kt

Question 219

Given ;
					True Heading = 180°
					TAS = 500 kt
					W/V 225° / 100 kt
	Calculate the GS ?

Answer 219

435 kt

Question 220

Given ;
					TAS = 130 kt
					Track ( T ) = 003°
					W/V = 190 / 40 kt
	Calculate the HDG ( °T ) and GS ;

Answer 220

001 - 170 kt

Question 221

Given ;
					TAS = 155 kt
					Track ( T ) = 305°
					W/V = 160 / 18 kt
	Calculate the HDG ( °T ) and GS ;

Answer 221

301 - 169 kt

Question 222

Given ;
					TAS = 200 kt
					Track ( T ) = 073°
					W/V = 210 / 20 kt
	Calculate the HDG ( °T ) and GS ;

Answer 222

077 - 214 kt

Question 223

Given ;
					TAS = 200 kt
					Track ( T ) = 110°
					W/V = 015 / 40 kt
	Calculate the HDG ( °T ) and GS ;

Answer 223

099 - 199 kt

Question 224

Given ;
					TAS = 227 kt
					Track ( T ) = 316°
					W/V = 205 / 15 kt
	Calculate the HDG ( °T ) and GS ;

Answer 224

312 - 232 kt

Question 225

Given ;
					TAS = 270 kt
					Track ( T ) = 260°
					W/V = 275 / 30 kt
	Calculate the HDG ( °T ) and GS ;

Answer 225

262 - 241 kt

Question 226

Given ;
					TAS = 465 kt
					Track ( T ) = 007°
					W/V = 300 / 80 kt
	Calculate the HDG ( °T ) and GS ;

Answer 226

358 - 428 kt

Question 227

Given ;
					FL250
					OAT –15 ºC
					TAS 250 kt
	Calculate the Mach No. ?

Answer 227

0.40

Question 228

Given ;
					TAS = 485 kt
					OAT = ISA +10°C
					FL 410
	Calculate the Mach Number ?

Answer 228

0.825

Question 229

Given ;
TAS 487 kt , FL 330 , Temperature ISA + 15
Calculate the MACH Number ?

Answer 229

0.81

Question 230

Given ;
					True Track 245°
					Drift 5° right
					Variation 3° E
					Compass Hdg 242°
	Calculate the Magnetic Heading ;

Answer 230

237°

Question 231

Given ;
Maximum allowable crosswind component is 20 kt
Runway 06 , RWY QDM 063°( M )
Wind direction 100°( M )
Calculate the maximum allowable wind speed ;

Answer 231

33 kt

Question 232

Given ;
Maximum allowable tailwind component for landing
10 kt. Planned runway 05 ( 047° magnetic )
The direction of the surface wind reported by ATIS 210° Variation is 17°E

Calculate the maximum allowable wind speed that can be accepted without exceeding the tailwind limit ;

Answer 232

10 kt

Question 233

Given ;
For take-off an aircraft requires a headwind component of at least 10 kt and has a
cross-wind limitation of 35 kt The angle between the wind direction and the
runway is 60°

Calculate the minimum and maximum allowable wind speeds ;

Answer 233

20 kt and 40 kt.

Question 234

Given ;
					TAS = 90 kt 
					HDG ( T ) = 355°
					W/V = 120 / 20 kt
	Calculate the Track ( °T ) and GS ;

Answer 234

346 - 102 kt

Question 235

Given ;
					TAS = 132 kt
					HDG ( T ) = 053°
					W/V = 205 / 15 kt
	Calculate the Track ( °T ) and GS

Answer 235

050 - 145 kt

Question 236

Given ;
					TAS = 135 kt
					HDG ( °T ) = 278
					W/V = 140 / 20 kt
	Calculate the Track ( °T ) and GS ;

Answer 236

283 - 150 kt

Question 237

Given ;
					TAS = 155 kt
					HDG ( T ) = 216°
					W/V = 090 / 60 kt
	Calculate the Track ( °T ) and GS ;

Answer 237

231 - 196 kt

Question 238

Given ;
					TAS = 170 kt
					HDG ( T ) = 100°
					W/V = 350 / 30 kt
	Calculate the Track ( °T ) and GS ;

Answer 238

109 - 182 kt

Question 239

Given ;
					TAS = 198 kt
					HDG ( °T ) = 180
					W/V = 359 / 25
	Calculate the Track ( °T ) and GS ;

Answer 239

180 - 223 kt

Question 240

Given ;
					TAS = 225 kt
					HDG ( °T ) = 123°
					W/V = 090 / 60kt
	Calculate the Track ( °T ) and GS ;

Answer 240

134 - 178 kt

Question 241

Given ;
					TAS = 480 kt 
					HDG ( °T ) = 040°
					W/V = 090 / 60kt
	Calculate the Track ( °T ) and GS ;

Answer 241

034 - 445 kt

Question 242

Given ;
					TAS = 485 kt
					HDG ( T ) = 168°
					W/V = 130 / 75 kt
	Calculate the Track ( °T ) and GS ;

Answer 242

174 - 428 kt

Question 243

Given ;
					Course required = 085°( T )
					Forecast W/V 030 / 100 kt 
					TAS = 470 kt
					Distance = 265 NM
	Calculate the true HDG and flight time ;

Answer 243

075° , 39 MIN

Question 244

Given ;
					TAS = 125 kt
					True HDG = 355°
					W/V = 320°( T ) / 30kt
	Calculate the true track and GS ;

Answer 244

005 - 102 kt

Question 245

Given ;
					TAS = 370 kt
					True HDG = 181°
					W/V = 095°( T ) / 35 kt
	Calculate the true track and GS ;

Answer 245

186 - 370 kt

Question 246

Given ;
					TAS = 375 kt
					True HDG = 124°
					W/V = 130°( T ) / 55 kt
	Calculate the true track and GS ;

Answer 246

123 - 320 kt

Question 247

Given ;
					M 0.80
					OAT –50°C
					FL 330
					GS 490 kt
					VAR 20°W
					Magnetic heading 140°
					Drift is 11° Right
	Calculate the true W/V ?

Answer 247

020° / 95 kt

Question 248

Given ;
					Magnetic track = 075°
					HDG = 066°( M )
					VAR = 11°E
					TAS = 275 kt
					Aircraft flies 48 NM in 10 MIN
	Calculate the true W/V ?

Answer 248

340° / 45 kt

Question 249

Given ;
					Magnetic track = 210°
					Magnetic HDG = 215°
					VAR = 15°E
					TAS = 360 kt
					Aircraft flies 64 NM in 12 MIN
	Calculate the true W/V ?

Answer 249

265° / 50 kt

Question 250

Given ;
FL 350 , Mach 0.80 , OAT –55°C

Calculate the values for TAS and local speed of sound ( LSS ) ?

Answer 250

461 kt , LSS 576 kt