Carlitos gennav

·265 US-GAL equals? (Specific gravity 0.80): 803 KG·5 HR 20 MIN 20 SEC corresponds to a longitude difference of: 80° 05·730 FT/MIN equals: 3.7M/SEC·A chart has the scale 1: 1 000 000. From A to B on the chart measures 3.8 cm, the distance from A toB in NM is: 20.5·A course of 120°(T) is drawn between 'X' (61°30'N) and 'Y' (58°30'N) on a Lambert Conformal conicchart with a scale of 1: 1 000 000 at 60°N. The chart distance between 'X' and 'Y' is: 66.7CM·A definition of a Magnetic Track angle is: THE DIRECTION OF A LINE REFERENCED TO MAGNETICNORTH·A direct Mercator graticule is based on a projection that is: CYLINDRICAL·A direct reading compass should be swung when: THERE IS A LARGE, AND PERMANENT,CHANGE IN MAGNETIC LATITUDE·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: 1000·A great circle intersects the equator in 030°W with a great circle direction of 035°(T). An aircrafttracking the great circle will reach the maximum Northern/Southern latitude in position: 55°N, 060°E·A great circle on the Earth running from the North Pole to the South Pole is called: A MERIDIAN·A great circle track crosses the equator at 30°W has an initial track of 035°T. It''s highest or lowestNorth/South point is: 55°N 060°E·A great circle track joins position A (59°S 141°W) and B (61°S 148°W). What is the differencebetween the great circle track at A and B? IT INCREASES BY 6°·A ground feature appears 30° to the left of the centre line of the CRT of an airborne weather radar. Ifthe heading of the aircraft is 355° (M) and the magnetic variation is 15° East, the true bearing of theaircraft from the feature is: 160°·A ground feature was observed on a relative bearing of 315° and 3 MIN later on a relative bearing of270°. The W/V is calm; aircraft GS 180 kt. What is the minimum distance between the aircraft and theground feature? 9NM·A ground feature was observed on a relative bearing of 325° and five minutes later on a relativebearing 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 featurewas: 30NM AND 240°·A Lambert conformal conic chart has a constant of the cone of 0.75. The initial course of a straightline 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 isthe longitude of B? 34°W·A Lambert conformal conic chart has a constant of the cone of 0.80. A straight line course drawn onthis 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?011°E·A Lambert conformal conic projection, with two standard parallels: THE SCALE IS ONLY CORRECTALONG THE STANDARD PARALLELS·A line drawn on a chart which joins all points where the value of magnetic variation is zero is calledan: AGONIC LINE·A Mercator chart has a scale at the equator = 1: 3 704 000. What is the scale at latitude 60° S?1:1.852.000·A nautical mile is equivalent to: 1852 M·A nautical mile is: 1852 METRES·A negative (westerly) magnetic variation signifies that: TRUE NORTH IS EAST OF MAGNETICNORTH

·A pilot accidently turning OFF the IRS in flight, and then turns it back ON a few moments later.Following this incident: IT CAN ONLY BE USED FOR ATTITUDE REFERENCE·A rhumb line from a position 86°N 30°W has an initial track of 085°T is it: A SPIRAL TO THE NORTHPOLE·A Rhumb line is: A LINE ON THE SURFACE OF THE EARTH CUTTING ALL MERIDIANS AT THESAME ANGLE·A rhumb line on a Direct Mercator chart appears as a: STRAIGHT LINE·A ring laser gyro is:; A DEVICE WHCH MEASURES ANGULAR MOVEMENTS·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: 290°(G) AND 270°(T)·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, arerespectively: 110° (G) AND 290°(T)·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:070°(G) AND 090° (T)·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, arerespectively: 250°(G) AND 070°(T)·A straight line drawn on a chart measures 4.63 cm and represents 150 NM. The chart scale is: 1:6.000.000·A straight line from A (53ºN, 155ºW) to B (53ºN, 170ºE) is drawn on a Lambert Conformal conicalchart with standard parallels at 50ºN and 56ºN.When passing the meridian 175ºE, the True Track is:260.0°·A straight line from A (53ºS, 155ºE) to B (53ºS, 170ºW) is drawn on a Lambert Conformal conicalchart with standard parallels at 50ºS and 56ºS. When passing 175ºW, the True Track is: 078.0°·A straight line from A (75ºN, 120ºE) to B (75ºN, 160ºE) is drawn on a Polar Stereographic chart. Whenpassing the meridian 155ºE, the True Track is: 105°·A straight line from A (75ºS, 120ºE) to B (75ºS, 160ºE) is drawn on a Polar Stereographic chart. Whenpassing the meridian 155ºE, the True Track is: 075°Near the magnetic pole: THE HORIZONALCOMPONENT OF THE EARTHS MAGNETIC FIELD IS TOO SMALL TO PERMIT THE USE OF AMAGNETIC COMPASS·A straight line is drawn on a Lamberts conformal conic chart between two positions of differentlongitude. The angular difference between the initial true track and the final true track of the line isequal to: CHART CONVERGENCY·A straight line on a chart 4.89 cm long represents 185 NM. The scale of this chart is approximately:1: 7.000.000·A straight line on a Lambert Conformal Projection chart for normal flight planning purposes: ISAPPROXIMATE A GREAT CIRCLE·A useful method of a pilot resolving, during a visual flight, any uncertainty in the aircraft's position isto maintain visual contact with the ground and: SET HEADING TOWARDS A LINE FEATURE SUCHAS A COASTLINE, MOTORWAY, RIVER O RAILWAY·A VOR is situated at position (74ºN, 094ºW); local variation is 50ºW. A Polar Stereographic chartsupplied with a Greenwich grid is used for navigation. To proceed along (magnetic) radial 238 inboundan aircraft has to follow a Grid Track of: 103°·A VOR is situated at position (N55°26', W005°42'). The variation at the VOR is 9°W. The position ofthe aircraft is (N60°00'N, W010°00'). The variation at the aircraft-position is 11°W. The initial TT-angleof the great circle from the aircraft position to the VOR is 101.5°.Which radial is the aircraft on? 294·After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM South of the intendedtrack and 2.5 NM ahead of the dead reckoning position. To reach destination B from this position, the

TH should be: 292°·After alignment of the stable platform of an Inertial Navigation System, the output data from theplatform is: ACCELERATION NORTH/SOUTH AND EAST/WEST, ATTITUDE AND TRUE HEADING·After alignment of the stable platform of the Inertial Navigation System, the output data from the INScomputer to the platform is: RATE CORRECTIONS TO THE GYROS·An aeronautical chart is conformal when: AT ANY POINT THE SCALE OVER A SHORT DISTANCEIN THE DIRECTION OF THE PARALLEL IS EQUAL TO THE SCALE IN THE DIRECTION OF THEMERIDIAN AND THE MERIDIANS ARE PERPENDICULAR TO THE PARALLELS·An aeroplane flies from A (59°S 142°W) to B (61°S 148°W) with a TAS of 480 kt. The autopilot isengaged and coupled with an Inertial Navigation System in which AB track is active. On route AB, thetrue track: INCREASES BY 5°·An aeroplane is flying at TAS 180 kt on a track of 090°. The W/V is 045° / 50kt. How far can theaeroplane fly out from its base and return in one hour? 85NM·An Agonic line is a line that connects: POSITIONS THAT HAVE 0° VARIATION·An aircraft at FL120, IAS 200kt, OAT -5° and wind component +30kt, is required to reduce speed inorder 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: 159 KT·An aircraft at FL140, IAS 210 kt, OAT -5°C and wind component minus 35 kt, is required to reducespeed in order to cross a reporting point 5 MIN later than planned. Assuming that flight conditions donot change, when 150 NM from the reporting point the aircraft should reduce IAS by: 20 KT·An aircraft at FL290 is required to commence descent when 50 NM from a VOR and to cross that VORat FL80. Mean GS during descent is 271kt. What is the minimum rate of descent required? 1900FT/MIN·An aircraft at FL310, M0.83, temperature -30°C, is required to reduce speed in order to cross areporting point five minutes later than planned. Assuming that a zero wind component remainsunchanged, when 360 NM from the reporting point Mach Number should be reduced to: M0.74·An aircraft at FL330 is rerquired to commence descent when 65 NM from a VOR and to cross the VORat FL100. The mean GS during the descent is 330 kt. What is the minimum rate of descent required?1950 FT/MIN·An aircraft at FL350 is required to commence descent when 85 NM from a VOR and to cross the VORat FL80. The mean GS for the descent is 340 kt. What is the minimum rate of descent required? 1800FT/MIN·An aircraft at FL350 is required to cross a VOR/DME facility at FL110 and to commence descent when100 NM from the facility. If the mean GS for the descent is 335 kt, the minimum rate of descentrequired is: 1340 FT/MIN·An aircraft at FL350 is required to descend to cross a DME facility at FL80. Maximum rate of descentis 1800 FT/MIN and mean GS for descent is 276 kt. The minimum range from the DME at whichdescent should start is: 69 NM·An aircraft at FL360 is required to descent to FL120. The aircraft should reach FL120 at 40 NM fromthe next waypoint. The rate of descent is 2000 ft/min. The average GS is 420 kt. ; The minimumdistance from the next waypoint at which descent should start is: 124NM·An aircraft at FL370 is required to commence descent at 120 NM from a VOR and to cross the facilityat FL130. If the mean GS for the descent is 288 kt, the minimum rate of descent required is: 960FT/MIN·An aircraft at FL370 is required to commence descent when 100 NM from a DME facility and to crossthe station at FL120. If the mean GS during the descent is 396 kt, the minimum rate of descentrequired is approximately: 1650 FT/MIN·An aircraft at FL370, M0.86, OAT -44°C, headwind component 110 kt, is required to reduce speed inorder to cross a reporting point 5 MIN later than planned. If the speed reduction were to be made 420NM from the reporting point, what Mach Number is required? M 0.81·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 whichdescent should commence? 53 NM·An aircraft at latitude 02°20'N tracks 180°(T) for 685 km. On completion of the flight the latitude willbe: 03°50S·An aircraft at latitude 10° South flies north at a GS of 890 km/HR. What will its latitude be after 1.5HR? 02°00N·An aircraft at latitude 10°North flies south at a groundspeed of 445 km/HR. What will be its latitudeafter 3 HR? 02°00S·An aircraft at position 60°N 005°W tracks 090°(T) for 315 km. On completion of the flight thelongitude will be: 000°40E·An aircraft departing A(N40º 00´ E080º 00´) flies a constant true track of 270º at a ground speed of120 kt. What are the coordinates of the position reached in 6 HR? N40°00 E064°20·An aircraft departs from position A (04°10' S 178°22'W) and flies northward following the meridian for2950 NM. It then flies westward along the parallel of latitude for 382 NM to position B. The coordinatesof position B are? 45°00N 182°38E·An aircraft departs from Schiphol airport and flies to Santa Cruz in Bolivia (South America) via Miamiin Florida. The departure time (off blocks) is 07:45 ST at the 10th of November, taxi time before takeoff at Schiphol is 25 minutes. The flight time to Miami over the Atlantic Ocean is 09h20m. The totaltaxi time in Miami to and from the gate is 25 minutes. The time spend at the gate is 02h40m. FromMiami to Santa Cruz the airborne time is 06h30m. Calculate the time and date of touch down in SantaCruz in ST Bolivia if the difference between ST and UTC is 5 hours: 21:05 10TH NOV·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' CentralDisplay Units (CDU) shows:; - XTK on INS 1 = 0, - XTK on INS 2 = 8L, (XTK = cross track) From thisinformation it can be deduced that: AT LEAST ONE OF THE INERTIAL NAVIGATION SYSTEMS ISDRIFTING·An aircraft flies a great circle track from 56° N 070° W to 62° N 110° E. The total distance travelledis? 3720 NM·An aircraft flies from waypoint 7 (63°00' N, 073°00'W) to waypoint 8 (62°00' N, 073°00' W). Theaircraft position is (62°00' N, 073°10'W). The cross track distance in relation to the planned track is:4.7 NM R·An aircraft flies the following rhumb line tracks and distances from position 04°00'N 030°00'W: 600NM South, then 600 NM East, then 600 NM North, then 600 NM West. The final position of the aircraftis: 04°00N 029°58W·An aircraft follows a great circle in the Northern Hemisphere. At a certain moment the aircraft is inthe position on the great circle where the great circle direction is 270°(T). Continuing on the greatcircle the: TRACK ANGLE WILL DECREASE AND THE LATITUDE WILL DECREASE·An aircraft follows a radial to a VOR/DME station. At 10:00 the DME reads 120 NM. At 10:03 the DMEreads 105 NM. The estimated time overhead the VOR/DME station is; 10:24·An aircraft has to fly over a mountain ridge. The highest obstacle, indicated in the navigation chart,has an elevation of 9 800 ft. The QNH, given by a meteorological station at an elevation of 6200ft, is1022hPa. The OAT = ISA+5ºC. ; Calculate the approximate indicated altitude to obtain a clearance of2000 ft: BETWEEN 11500 FT AND 11700 FT·An aircraft in the northern hemisphere is making an accurate rate one turn to the right. If the initialheading was 135°, after 30 seconds the direct reading magnetic compass should read: MORE THAN225°·An aircraft in the northern hemisphere makes an accurate rate one turn to the right/starboard. If theinitial heading was 330°, after 30 seconds of the turn the direct reading magnetic compass shouldread: LESS THAN 060°·An aircraft is at position (53ºN, 006ºW) and has a landmark at position (52º47'N, 004º45'W), with arelative 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 Lambertchart with standard parallels at 37ºN and 65ºN? 278°·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 climbis FL 100. At what distance from the airport will this be achived? 10.3 NM·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 climbis FL 050. At what distance from the airport will this be achived? 3.6 NM·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 climbis FL 100. At what distance from the airport will this be achieved? 11.1 NM·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 climbis FL 050. At what distance from the airport will this be achived? 7.2 NM·An aircraft is descending down a 6% slope whilst maintaining a G/S of 300 kt. The rate of descent ofthe aircraft is approximately: 1800 FT/MIN·An aircraft is flying according the flight log at the Annex. After 15 minutes of flying with the plannedTAS and TH the aircraft is 3 NM North of the intended track and 2.5 NM ahead of the dead reckoningposition. To reach destination B from this position the TH should be: 112°·An aircraft is flying according the flight log at the Annex. After 15 minutes of flying with the plannedTAS and TH the aircraft is 3 NM North of the intended track and 2.5 NM ahead of the dead reckoningposition. To reach destination B from this position the TH should be: 258°·An aircraft is flying at FL 200. OAT = 0°C. When the actual air pressure on an airfield at MSL is placedin the subscale of the altimeter the indicated altitude is 19300ft. ; Calculate the aircraft''s truealtitude.; 21200 FT·An aircraft is flying at FL100. The OAT = ISA - 15°C. The QNH given by a station at an elevation 3000ft is 1035hPa. Calculate the approximate True Altitude. 10200 FT·An aircraft is flying at FL150, with an outside air temperature of -30°, above an airport where theelevation is 1660 ft and the QNH is 993 hPa. Calculate the true altitude. (Assume 30 ft = 1 hPa)13660 FT·An aircraft is flying at FL180 and the outside air temperature is -30°C. If the CAS is 150 kt, what is theTAS? 195 KT·An aircraft is flying at FL200. The QNH, given by a meteorological station at an elevation of 1300ft is998.2 hPa. OAT = - 40ºC. The elevation of the highest obstacle along the route is 8 000 ft. ; Calculatethe aircraft's approximate clearance above the highest obstacle on this route: 10500 FT·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: 23400 FT·An aircraft is flying from A to B a distance of 50 NM. The True Course in the flight log is 270º, theforecast wind is 045º(T)/15kt and the TAS is 120kt. ; After 15 minutes of flying with the planned TASand TH the aircraft is 3 NM South of the intended track and 2.5 NM ahead of the dead reckoningposition. ; To reach destination B from this position, the correction angle on the heading should be:17°·An aircraft is flying from A to B. The true course according to the flight log is 090º, the estimatedwind is 225º(T)/15kt and the TAS is 120 kt. ; After 15 minutes of flying with the planned TAS and THthe 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: 5°R·An aircraft is flying from SALCO to BERRY HEAD on Magnetic Track 007º, TAS 445 kt. The wind is050º(T)/40 kt. ; Variation 5ºW, deviation +2º; At 1000 UTC the RB of locator PY is 311º.; At 1003 UTCthe RB of locator PY is 266º. ; Calculate the True bearing of locator PY at 1003 UTC from the Aircraft:272° (T)·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 UTCthe RB of locator PY is 266º. ; Calculate the distance of the aircraft from locator PY at 1003 UTC: 21NM·An aircraft is flying with the aid of an inertial navigation system (INS) connected to the autopilot. Thefollowing two points have been entered in the INS computer:; WPT 1: 60°N 030°W, WPT 2: 60°N020°W. When 025°W is passed the latitude shown on the display unit of the inertial navigation systemwill be: 60° 05.7N·An aircraft is following a true track of 048° at a constant TAS of 210 kt. The wind velocity is 350° / 30kt. The GS and drift angle are: 192 KT, 7° RIGHT·An aircraft is following the 45°N parallel of latitude. The track followed is a: RHUMB LINE·An aircraft is in the position (86°N, 020°E). When following a rhumb line track of 085°(T) it will: FLYVIA A SPIRAL TO THE NORTH POLE·An aircraft is in the position (86ºN, 020ºE). When following a rhumb line track of 085º(T) it will: FLYVIA A SPIRAL TO THE NORTH POLE·An aircraft is maintaining a 5.2% gradient is at 7 NM from the runway, on a flat terrain; its height isapproximately: 2210 FT·An aircraft is over position HO (55°30'N 060°15'W), where YYR VOR (53°30'N 060°15'W) can bereceived. The magnetic variation is 31°W at HO and 28°W at YYR. What is the radial from YYR? 028°·An aircraft is planned to fly from position 'A' to position 'B', distance 250 NM at an average GS of 115kt. It departs 'A' at 0900 UTC. After flying 75 NM along track from 'A', the aircraft is 1.5 MIN behindplanned time. Using the actual GS experienced, what is the revised ETA at 'B'? 1115UTC·An aircraft is planned to fly from position 'A' to position 'B', distance 480 NM at an average GS of 240kt. It departs 'A' at 1000 UTC. After flying 150 NM along track from 'A', the aircraft is 2 MIN behindplanned time. Using the actual GS experienced, what is the revised ETA at 'B'? 1206·An aircraft is planned to fly from position 'A' to position 'B',distance 320 NM, at an average GS of 180kt. It departs 'A' at 1200 UTC. After flying 70 NM along track from 'A', the aircraft is 3 MIN ahead ofplanned time. Using the actual GS experienced, what is the revised ETA at 'B'? 1333 UTC·An aircraft must fly 2000 ft above an obstacle of which the elevation is 13 600 ft. The QNH at thenearest airfield is 991 hPa, the elevation is 1500 ft and the temperature is - 20°C. Calculate theminimum altitude required: 17400 FT·An aircraft obtains a relative bearing of 315° from an NDB at 0830. At 0840 the relative bearing fromthe same position is 270°. Assuming no drift and a GS of 240 kt, what is the approximate range fromthe NDB at 0840? 40 NM·An aircraft passes position A (60°00'N 120°00'W) on route to position B (60°00'N 140°30'W). What isthe great circle track on departure from A? 279°·An aircraft takes off from the aerodrome of BRIOUDE (altitude 1 483 FT, QFE = 963 hPa, temperature= 32°C). Five minutes later, passing 5 000 FT on QFE, the second altimeter set on 1 013 hPa willindicate approximately : 6.500 FT·An aircraft takes-off from an airport 2 hours before sunset. The pilot flies a track of 090°(T), W/V130°/ 20 kt, TAS 100 kt. In order to return to the point of departure before sunset, the furthestdistance which may be travelled is: 97 NM·An aircraft tracks radial 200 inbound to a VOR station with a Magnetic Heading (MH) of 010º. Afterbeing overhead the VOR station the aircraft tracks radial 090 outbound with a MH of 080º. The TAS is240 kt and the magnetic variation in the area is 5ºW: 320°/50KT·An aircraft travels 100 statute miles in 20 MIN, how long does it take to travel 215 NM? 50 MIN·An aircraft travels 2.4 statute miles in 47 seconds. What is its groundspeed? 160 KT·An aircraft travels from point A to point B, using the autopilot connected to the aircraft's inertialsystem. The coordinates of A (45°S 010°W) and B (45°S 030°W) have been entered. The true courseof the aircraft on its arrival at B, to the nearest degree, is: 277°·An aircraft was over 'A' at 1435 hours flying direct to 'B'. Given: Distance 'A' to 'B' 2900 NM Trueairspeed 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: 1657·An aircraft was over 'A' at 1435 hours flying direct to 'B'. Given: Distance 'A' to 'B' 2900 NM Trueairspeed 470 kt Mean wind component 'out' +55 kt Mean wind component 'back' -75 kt Safeendurance 9 HR 30 MIN The ETA at the point of equal time is: 1657·An aircraft was over 'Q' at 1320 hours flying direct to 'R'. Given: Distance 'Q' to 'R' 3016 NM Trueairspeed 480 kt Mean wind component 'out' -90 kt Mean wind component 'back' +75 kt. What is theETA at the Point of Equal Time (PET) ? 1752·An aircraft was over 'Q' at 1320 hours flying direct to 'R'. Given: Distance 'Q' to 'R' 3016 NM Trueairspeed 480 kt Mean wind component 'out' -90 kt Mean wind component 'back' +75 kt. The ETA forreaching the Point of Equal Time (PET) between 'Q' and 'R' is: 1752·An aircraft was over 'Q' at 1320 hours flying direct to 'R'. Given: Distance 'Q' to 'R' 3016 NM Trueairspeed 480 kt Mean wind component 'out' -90 kt Mean wind component 'back' +75 kt. Safeendurance 10:00 HR. The distance from 'Q' to the Point of Safe Return (PSR) 'Q' is: 2290 NM·An island appears 30° to the left of the centre line on an airborne weather radar display. What is thetrue bearing of the aircraft from the island if at the time of observation the aircraft was on a magneticheading (MH) of 020° with the magnetic variation (VAR) 25°W? 145°·An island appears 30° to the left of the centre line on an airborne weather radar display. What is thetrue bearing of the aircraft from the island if at the time of observation the aircraft was on a magneticheading of 276° with the magnetic variation 12 °W? 054°·An island appears 30° to the right of the centre line on an airborne weather radar display. What is thetrue bearing of the aircraft from the island if at the time of observation the aircraft was on a magneticheading (MH) of 355° with the magnetic variation (VAR) 15°E? 220°In a navigation chart a distance of49 NM is equal to 7 cm. The scale of the chart is approximately: 1: 1.300.000·An island appears 45° to the right of the centre line on an airborne weather radar display. What is thetrue bearing of the aircraft from the island if at the time of observation the aircraft was on a magneticheading (MH) of 215° with the magnetic variation 21ºW: 059°·An island appears 60° to the left of the centre line on an airborne weather radar display. What is thetrue bearing of the aircraft from the island if at the time of observation the aircraft was on a magneticheading (MH) of 276° with the magnetic variation 10ºE: 046°·An island is observed by weather radar to be 15° to the left. The aircraft heading is 120°(M) and themagnetic variation 17°W. What is the true bearing of the aircraft from the island? 268°·An island is observed to be 15° to the left. The aircraft heading is 120°(M), variation 17°(W). Thebearing °(T) from the aircraft to the island is: 088·An island is observed to be 30° to the right of the nose of the aircraft. The aircraft heading is290°(M), variation 10°(E). The bearing °(T) from the aircraft to the island is: 330·An NDB is located at position (N55°26', W005°42'). The variation at the NDB is 9°W. The position ofthe aircraft is (56°00'N, 010°00'W). The variation at the aircraft-position is 11°W. The initial TT- of thegreat circle from the aircraft position to the NDB position, is 101.5°.; What is the Magnetic Bearing ofthe NDB from the aircraft? 112.5°·An Oblique Mercator projection is used specifically to produce: CHARTS OF THE GREAT CIRCLEROUTE BETWEEN TWO POINTS·An observer is situated on the parallel of 23.5°S. Which statement about the passage of the apparentsun in relation to this position is correct? IT PASES THROUGH THE ZENITH ONCE A YEARAROUND DECEMBER 22ND·Approximately how many nautical miles correspond to 12 cm on a map with a scale of 1: 2 000 000?130·As the INS position of the departure aerodrome, coordinates 35°32.7'N 139°46.3'W are input insteadof 35°32.7'N 139°46.3'E. When the aircraft subsequently passes point 52°N 180°W, the longitudevalue shown on the INS will be: 099°-32.6W·Assume a Mercator chart. The distance between positions A and B, located on the same parallel and10° longitude apart, is 6 cm. The scale at the parallel is 1: 9 260 000. What is the latitude of A and B?

60° N OR S·Assume a North polar stereographic chart whose grid is aligned with the Greenwich meridian. Anaircraft flies from the geographic North pole for a distance of 480 NM along the 110°E meridian, thenfollows a grid track of 154° for a distance of 300 NM. Its position is now approximately: 80°00N080°E·Assuming mid-latitudes (40° to 50°N/S). At which time of year is the relationship between the lengthof day and night, as well as the rate of change of declination of the sun, changing at the greatestrate? SPRING EQUINOX AND AUTUMN EQUINOX·Assuming zero wind, what distance will be covered by an aircraft descending 15000 FT with a TAS of320 kt and maintaining a rate of descent of 3000 FT/MIN? 26.7 NM·At (54°N, 020°W) the sun rises on November 28th at 09:01 UTC.At (44°N, 020°W) the sun will rise:EARTH SINCE THE LATTER POSITION LIES FURTHER SOUTH·At 0000 Local Mean Time of an observer: THE MEAN SUN IS IN TRANSIT WITH THE OBSERVERSANTI-MERIDIAN·At 0020 UTC an aircraft is crossing the 310° radial at 40 NM of a VOR/DME station. At 0035 UTC theradial is 040° and DME distance is 40 NM. Magnetic variation is zero. The true track and ground speedare: 085°-226 KT·At 0422 an aircraft at FL370, GS 320kt, is on the direct track to VOR 'X' 185 NM distant. ; The aircraftis required to cross VOR 'X' at FL80. For a mean rate of descent of 1800 FT/MIN at a mean GS of 232kt, the latest time at which to commence descent is: 0445·At 10:15 the reading from a VOR/DME station is 211°/ 90NM, at 10:20 the reading from the sameVOR/DME station is 211°/120NM.Compass Heading = 200ºVariation in the area = 31ºWDeviation =+1ºTAS = 390 ktThe wind vector (T) is approximately: 110°/70 KT·At 47° North the chart distance between meridians 10° apart is 12.7 cm. The scale of the chart at 47°North approximates: 1:6.000.000·At 60° N the scale of a direct Mercator chart is 1: 3 000 000. What is the scale at the equator? 1:6.000.000·At a specific location, the value of magnetic variation: VARIES SLOWLY OVER TIME·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: 17.8 CM·At reference or see Europe Low Altitude Enroute Chart E(LO) 1A. An aircraft is flying from InvernessVORDME (N57°32.6'', W 004°02.5W) to Aberdeen VORDME (N57°18.6'', W002°16.0''W). At 1000 UTCthe fix of the aircraft is determined by VORDME Inverness: radial = 114; DME-distance = 20.5 NM. At1006 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?; 280 KT·At reference. 1215 UTC LAJES VORTAC (38°46'N 027°05'W) RMI reads 178°, range 135 NM. Calculatethe aircraft position at 1215 UTC? 40°55N 027°55W·At reference. 1300 UTC DR position 37°30'N 021°30'W alter heading PORT SANTO NDB (33°03'N016°23'W) ; TAS 450 kt, Forecast W/V 360°/30kt. Calculate the ETA at PORT SANTO NDB? 1348·At the magnetic equator, when accelerating after take off on heading West, a direct reading pivotsuspended compass: INDICATES THE CORRECT HEADING·At what approximate date is the earth closest to the sun (perihelion)? BEGINNING OF JULY·ATT Mode of the Inertial Reference System (IRS) is a back-up mode providing: ONLY ATTITUDE ANDHEADING INFORMATION·Calculate the clearance above a mountain ridge with an elevation of 20410 ft: 4200 FT·Calculate the constant of the cone on a Lambert Chart given chart convergency between 010°E and030°W as being 30°: 0.75·Calibrated Airspeed (CAS) is Indicated Airspeed (IAS) corrected for: INSTRUMENT ERROR ANDPOSITION ERROR·Compass deviation is defined as the angle between: MAGNETIC NORTH AND COMPASS NORTH·Complete line 1 of the 'FLIGHT NAVIGATION LOG'; positions 'A' to 'B'. What is the HDG°(M) and ETA?

268°-1114 UTC·Complete line 2 of the 'FLIGHT NAVIGATION LOG', positions 'C' to 'D'. What is the HDG°(M) and ETA?HDG 193°- ETA 1239 UTC·Complete line 3 of the 'FLIGHT NAVIGATION LOG', positions 'E' to 'F'. What is the HDG°(M) and ETA?HDG 105°- ETA 1205 UTC·Complete line 4 of the 'FLIGHT NAVIGATION LOG', positions 'G' to 'H'. What is the HDG°(M) and ETA?HDG 344°-ETA 1336 UTC·Complete line 5 of the 'FLIGHT NAVIGATION LOG', positions 'J' to 'K'. What is the HDG°(M) and ETA?HDG 337°- ETA 1422 UTC·Complete line 6 of the 'FLIGHT NAVIGATION LOG', positions 'L' to 'M'. What is the HDG°(M) and ETA?HDG 075°- ETA 1502 UTC·Complete the following statement regarding magnetic variation. The charted values of magneticvariation on earth normally change annually due to: MAGNETIC POLE MOVEMENT CAUSINGNUMERICAL VALUES AT ALL LOCATIONS TO INCREASE OR DECREASE·Concerning direct reading magnetic compasses, in the northern hemisphere, it can be said that: ONA EASTERLY HEADING, A LONGITUDINAL ACCELERATION CAUSES AN APARENT TURN TOTHE NORTH·Consider the following factors that determine the accuracy of a DR position:; 1. The flight time sincethe last position update.; 2. The accuracy of the forecasted wind.; 3. The accuracy of the TAS.; 4. Theaccuracy of the steered heading.; Using the list which of the above statements are correct ?; 1, 2, 3AND 4·Consider the following factors that determine the accuracy of a DR position:; 1. The flight time sincethe last position update.; 2. The accuracy of the forecasted wind.; 3. The accuracy of the TAS.; 4. Theaccuracy of the steered heading.; Using the list above which of the following contains the mostcomplete answer? 1, 2, 3, AND 4·Consider the positions (00ºN/S, 000ºE/W) and (00ºN/S, 180ºE/W) on the ellipsoid. Which statementabout the distances between these positions is correct? THE ROUTE VIA THE NORTH POLE ISSHORTER THAN THE ROUTE ALONG THE EQUATOR.·Contour lines on aeronautical maps and charts connect points: HAVING THE SAME ELEVATIONABOVE SEA LEVEL·Deviation on the standby compass is: DEPENDENT OF THE HEADING OF THE AIRCRAFT·Deviation applied to magnetic heading gives: COMPASS HEADING·Double integration of the output from the east/west accelerometer of an inertial navigation system(INS) in the NAV MODE give: DISTANCE EAST/WEST·During a low level flight 2 parallel roads that are crossed at right angles by an aircraft. The timebetween these roads can be used to check the aircraft: GROUNDSPEED·During approach the following data are obtained:DME 12.0 NM, altitude 3000 ftDME 9.8 NM, altitude2400 ftTAS = 160 kt, GS = 125 ktThe rate of descent is: 570 FT/MIN·During the initial alignment of an inertial navigation system (INS) the equipment: WILL NOT ACCEPTA 10° ERROR IN INITIAL LATITUDE BUT WILL ACCEPT A 10° ERROR IN INITIAL LONGITUDE·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 safereturn (PSR) is: 1685 NM -(1865 NM????)·For a landing on runway 23 (227° magnetic) surface, W/V reported by the ATIS is 180/30kt. ; VAR is13°E. Calculate the cross wind component? 22KT·From Rakovnik (50° 05.9' N, 013° 41.5' E) to Frankfurt FFM (50° 05.9' N, 008° 38.3' E) the True Trackof departure along the straight line is 272.0°. The constant of the cone of this Lambert conformalprojection is: 0.79·From the departure point, the distance to the point of equal time is: INVERSLY PROPORTIONAL TOTHE SUM OF GROUND SPEED OUT AND GROUND SPEED BACK·Fuel flow per HR is 22 US-GAL, total fuel on board is 83 IMP GAL. What is the endurance? 4 HR 32

MIN·Geodetic latitude and geocentric latitude coincide: AT THE POLES AND ON THE EQUATOR·Given :; A is N55° 000°, B is N54° E010°, The average true course of the great circle is 100°. The truecourse of the rhumbline at point A is: 100°·Given the following: Magnetic heading: 060°, Magnetic variation: 8°W, Drift angle: 4° right, What isthe true track? 056°·Given the following: True track: 192°, Magnetic variation: 7°E, Drift angle: 5° left. What is themagnetic heading required to maintain the given track? 190°The standard parallels of a Lambert'sconical orthomorphic projection are 07°40'N and 38°20' N. The constant of the cone for this chart is:0.39·Given: ; FL250, OAT -15 ºC, TAS 250 kt.Calculate the Mach No.? 0.40·Given: ; Maximum allowable crosswind component is 20 kt. Runway 06, RWY QDM 063°(M). ; Winddirection 100°(M). Calculate the maximum allowable windspeed? 33 KT·Given: ; SHA VOR (N5243.3 W00853.1) radial 120°, CRK VOR (N5150.4 W00829.7) radial 033°. Whatis the aircraft position? N5230 W00800·Given: ; SHA VOR (N5243.3 W00853.1) radial 129°, CRK VOR (N5150.4 W00829.7) radial 047°. Whatis the aircraft position? N5220 W00750·Given: ; SHA VOR (N5243.3 W00853.1) radial 143°, CRK VOR (N5150.4 W00829.7) radial 050°. Whatis the aircraft position? N5210 W00800·Given: ; SHA VOR (N5243.3 W00853.1) radial 205°, CRK VOR (N5150.4 W00829.7) radial 317°. Whatis the aircraft position? N5210 W00910·Given: ; SHA VOR (N5243.3 W00853.1) radial 223°,; CRK VOR (N5150.4 W00829.7) radial 322°.; Whatis the aircraft position? N5220 W00920·Given: ; SHA VOR/DME (N5243.3 W00853.1) radial 025°/49 NM. What is the aircraft position? N5330W00830·Given: ; SHA VOR/DME (N5243.3 W00853.1) radial 048°/22 NM. What is the aircraft position? N5300W00830·Given: ; SHA VOR/DME (N5243.3 W00853.1) radial 120°/35 NM. What is the aircraft position? N5230W00800·Given: ; SHA VOR/DME (N5243.3 W00853.1) radial 165°/36 NM. What is the aircraft position? N5210W00830·Given: ; SHA VOR/DME (N5243.3 W00853.1) radial 232°/32 NM. What is the aircraft position? N5220W00930·Given: ; TAS = 135 kt, HDG (°T) = 278, W/V = 140/20kt. Calculate the Track (°T) and GS? 283-150KT·Given: ; TAS = 155 kt, HDG (T) = 216°, W/V = 090/60kt. Calculate the Track (°T) and GS? 231-196KT·Given: ; TAS = 440 kt, HDG (T) = 349°, W/V = 040/40kt. Calculate the drift and GS? 4L- 415 KT·Given: ; TAS 487kt, FL 330, Temperature ISA + 15.Calculate the MACH Number? 0.81·Given: ; True course A to B = 250°, Distance A to B = 315 NM, TAS = 450 kt. W/V = 200°/60kt.; ETDA = 0650 UTC. What is the ETA at B? 0736 UTC·Given: ; True track 180°, Drift 8°R, Compass heading 195°, Deviation -2°, Calculate the variation?21°W·Given: A polar stereographic chart of the northern hemisphere whose grid is aligned with the zeromeridian. Grid track 344°, Longitude 115°00'W, Calculate the true course? 229°·Given: AD = Air distance, GD = Ground distance, TAS = True Airspeed; GS = Groundspeed. Which ofthe following is the correct formula to calculate ground distance (GD) gone? GD= (AD X GS)/TAS·Given: aircraft height 2500 FT, ILS GP angle 3°. At what approximate distance from THR can youexpect to capture the GP? 8.3 NM·Given: Airport elevation is 1000 ft. QNH is 988 hPa. What is the approximate airport pressurealtitude? (Assume 1 hPa = 27 FT): 1760 FT

·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 theaircraft and the station is: THE SAME AS IT WA AT 2254 UTC·Given: An aircraft is flying at FL100, OAT = ISA - 15ºC. The QNH, given by a meteorological stationwith an elevation of 100 ft below MSL is 1032 hPa. 1 hPa = 27 ftCalculate the approximate TrueAltitude of this aircraft.: 9900 FT·Given: An aircraft is on final approach to runway 32R (322°); The wind velocity reported by the toweris 350°/20 kt.; TAS on approach is 95 kt. In order to maintain the centre line, the aircraft's heading(°M) should be: 328°·Given: CAS 120 kt, FL 80, OAT +20°C. What is the TAS? 141 KT·Given: Chart scale is 1: 1 850 000. The chart distance between two points is 4 centimetres. Earthdistance is approximately: 40NM·Given: Compass Heading 090°, Deviation 2°W, Variation 12°E, TAS 160 kt. Whilst maintaining a radial070° from a VOR station, the aircraft flies a ground distance of 14 NM in 6 MIN. What is the W/V °(T)?160°/50 KT·Given: Course 040°(T), TAS is 120 kt, Wind speed 30 kt. Maximum drift angle will be obtained for awind direction of: 130°·Given: course required = 085° (T), Forecast W/V 030/100kt, TAS = 470 kt, Distance = 265 NM.Calculate the true HDG and flight time? 075°, 39 MIN·Given: Direct Mercator chart with a scale of 1: 200 000 at equator; Chart length from 'A' to 'B', in thevicinity of the equator, 11 cm. What is the approximate distance from 'A' to 'B'? 12 NM·Given: Distance 'A' to 'B' 1973 NM Groundspeed 'out' 430 kt Groundspeed 'back' 385 kt The timefrom 'A' to the Point of Equal Time (PET) between 'A' and 'B' is: 130 MIN·Given: Distance 'A' to 'B' 1973 NM Groundspeed 'out' 430 kt Groundspeed 'back' 385 kt Safeendurance 7 HR 20 MIN The distance from 'A' to the Point of Safe Return (PSR) 'A' is: 1490 NM·Given: Distance 'A' to 'B' 2346 NM Groundspeed 'out' 365 kt Groundspeed 'back' 480 kt The timefrom 'A' to the Point of Equal Time (PET) between 'A' and 'B' is: 219 MIN·Given: Distance 'A' to 'B' 2346 NM Groundspeed 'out' 365 kt Groundspeed 'back' 480 kt Safeendurance 8 HR 30 MIN The time from 'A' to the Point of Safe Return is: 290 MIN·Given: Distance 'A' to 'B' 2484 NM Groundspeed 'out' 420 kt Groundspeed 'back' 500 kt The timefrom 'A' to the Point of Equal Time (PET) between 'A' and 'B' is: 193 MIN·Given: Distance 'A' to 'B' 2484 NM Mean groundspeed 'out' 420 kt Mean groundspeed 'back' 500 ktSafe endurance 08 HR 30 MIN The distance from 'A' to the Point of Safe Return (PSR) 'A' is: 1940NM·Given: Distance 'A' to 'B' 3623 NM Groundspeed 'out' 370 kt Groundspeed 'back' 300 kt The timefrom 'A' to the Point of Equal Time (PET) between 'A' and 'B' is: 263 MIN·Given: Distance 'A' to 'B' is 100 NM, Fix obtained 40 NM along and 6 NM to the left of course. Whatheading alteration must be made to reach 'B'? 15° RIGHT·Given: Distance 'A' to 'B' is 325 NM, Planned GS 315 kt, ATD 1130 UTC, 1205 UTC - fix obtained 165NM along track. What GS must be maintained from the fix in order to achieve planned ETA at 'B'? 355KT·Given: Distance 'A' to 'B' is 475 NM, Planned GS 315 kt, ATD 1000 UTC, 1040 UTC - fix obtained 190NM along track. What GS must be maintained from the fix in order to achieve planned ETA at 'B'? 340KT·Given: Distance 'A' to 'B' is 90 NM, Fix obtained 60 NM along and 4 NM to the right of course. Whatheading alteration must be made to reach 'B'? 12° LEFT·Given: Distance 'Q' to 'R' 1760 NM Groundspeed 'out' 435 kt Groundspeed 'back' 385 kt Safeendurance 9 HR The distance from 'Q' to the Point of Safe Return (PSR) between 'Q' and 'R' is: 1838NM·Given: Distance 'Q' to 'R' 1760 NM Groundspeed 'out' 435 kt Groundspeed 'back' 385 kt The timefrom 'Q' to the Point of Equal Time (PET) between 'Q' and 'R' is: 114 MIN·Given: Distance A to B = 120 NM, After 30 NM aircraft is 3 NM to the left of course. What heading

alteration should be made in order to arrive at point 'B'? 8° RIGHT·Given: Distance A to B is 360 NM. Wind component A - B is -15 kt, Wind component B - A is +15 kt,TAS is 180 kt. What is the distance from the equal-time-point to B? 165 NM·Given: ETA to cross a meridian is 2100 UTC GS is 441 kt TAS is 491 kt At 2010 UTC, ATC requests aspeed reduction to cross the meridian at 2105 UTC. The reduction to TAS will be approximately: 40KT·Given: FL 350, Mach 0.80, OAT -55°C. Calculate the values for TAS and local speed of sound (LSS)?461 KT, LSS 576 KT·Given: FL120, OAT is ISA standard, CAS is 200 kt, Track is 222°(M), Heading is 215°(M), Variation is15°W. Time to fly 105 NM is 21 MIN. What is the W/V? 050° (T)/ 70 KT·Given: GS = 105 kt. Distance from A to B = 103 NM. What is the time from A to B? 00HR 59 MIN·Given: GS = 120 kt. Distance from A to B = 84 NM. What is the time from A to B? 00 HR 42 MIN·Given: GS = 122 kt. Distance from A to B = 985 NM. What is the time from A to B? 8 HR 04 MIN·Given: GS = 135 kt. Distance from A to B = 433 NM. What is the time from A to B? 3 HR 12 MIN·Given: GS = 236 kt. Distance from A to B = 354 NM. What is the time from A to B? 1HR 30 MIN·Given: GS = 345 kt. Distance from A to B = 3560 NM. What is the time from A to B? 10 HR 19 MIN·Given: GS = 435 kt. Distance from A to B = 1920 NM. What is the time from A to B? 4 HR 25 MIN·Given: GS = 480 kt. Distance from A to B = 5360 NM. What is the time from A to B? 11HR 10 MIN·Given: GS = 510 kt. Distance A to B = 43 NM. What is the time (MIN) from A to B? 5·Given: GS = 95 kt. Distance from A to B = 480 NM. What is the time from A to B? 5 HR 03 MIN·Given: Half way between two reporting points the navigation log gives the following information: TAS360 kt, W/V 330°/80kt, Compass heading 237°, Deviation on this heading -5°, Variation 19°W. What isthe average ground speed for this leg? 403 KT·Given: ILS GP angle = 3.5 DEG, GS = 150 kt. What is the approximate rate of descent? 900FT/MIN·Given: Lambert conformal conical projection, scale 1: 1 234 000. Standard parallels 36°N and 60°N..A (53°N, 010°W), B (53°N, 020°W).·Given: M 0.80, CAT -50°C, FL 330, GS 490 kt, VAR 20°W, Magnetic heading 140°, Drift is 11° Right.Calculate the true W/V? 020°/95 KT·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? 020°/95 KT·Given: Magnetic heading = 255° VAR = 40°W GS = 375 kt W/V = 235°(T) / 120 kt Calculate the driftangle? 6° LEFT·Given: Magnetic heading = 255°, VAR = 40°W, GS = 375 kt, W/V = 235°(T) / 120 kt, Calculate thedrift angle? 6° LEFT·Given: Magnetic heading 311°, Drift angle 10° left, Relative bearing of NDB 270°. What is themagnetic bearing of the NDB measured from the aircraft? 221°·Given: Magnetic track = 210°, Magnetic HDG = 215°, VAR = 15°E, TAS = 360 kt, Aircraft flies 64 NMin 12 MIN. Calculate the true W/V? 265°/50 KT·Given: Magnetic track = 315 º, HDG = 301 º(M), VAR = 5ºW, TAS = 225 kt, The aircraft flies 50 NM in12 MIN. Calculate the W/V(°T)? 190°/63 KT·Given: Position 'A' is N00° E100°, Position 'B' is 240°(T), 200 NM from 'A'. What is the position of 'B'?S01°40 E097°07·Given: Position 'A' N60 W020, Position 'B' N60 W021, Position 'C' N59 W020. What are, respectively,the distances from A to B and from A to C? 30 NM AND 60 NM·Given: Position A 45°N, ?°E, Position B 45°N, 45°15'E, Distance A-B = 280 NM, B is to the East of A.Required: longitude of position A? 38°39E·Given: Position NDB (55°10´N, 012°55´E) DR Position (54°53´N, 009°58´E) NDB on the RMI reads090°. Magnetic variation = 10°W. The position line has to be plotted on a Lamberts conformal chartwith standard parallels at 40°N and 48°N. Calculate the direction (T) of the bearing to be plotted fromthe NDB.; 262°·Given: Pressure Altitude 29000 FT, OAT -55°C. Calculate the Density Altitude? 27500 FT·Given: Required course 045°(M); Variation is 15°E; W/V is 190°(T)/30 kt; CAS is 120 kt at FL 55 in

standard atmosphere. What are the heading (°M) and GS? 055° AND 147 KT·Given: Runway direction 083°(M), Surface W/V 035/35kt. Calculate the effective headwindcomponent? 24 KT·Given: Runway direction 230°(T), Surface W/V 280°(T)/40 kt. Calculate the effective cross-windcomponent? 31 KT·Given: TAS = 125 kt, True HDG = 355°, W/V = 320°(T)/30kt. Calculate the true track and GS? 005-102 KT·Given: TAS = 130 kt, Track (T) = 003°, W/V = 190/40kt. Calculate the HDG (°T) and GS? 001-170 KT·Given: TAS = 132 kt, HDG (T) = 053°, W/V = 205/15kt. Calculate the Track (°T) and GS? 050-145 KT·Given: TAS = 140 kt, HDG (T) = 005°, W/V = 265/25kt. Calculate the drift and GS? 10R-146 KT·Given: TAS = 155 kt, Track (T) = 305°, W/V = 160/18kt. Calculate the HDG (°T) and GS? 301- 169KT·Given: TAS = 170 kt, HDG(T) = 100°, W/V = 350/30kt. Calculate the Track (°T) and GS? 109-182 KT·Given: TAS = 190 kt, HDG (T) = 355°, W/V = 165/25kt. Calculate the drift and GS? 1L-215KT·Given: TAS = 190 kt, True HDG = 085°, W/V = 110°(T)/50kt. Calculate the drift angle and GS? 8°L-146 KT·Given: TAS = 197 kt, True course = 240°, W/V = 180/30kt. Descent is initiated at FL 220 andcompleted at FL 40. Distance to be covered during descent is 39 NM. What is the approximate rate ofdescent? 1400 FT/MIN·Given: TAS = 198 kt, HDG (°T) = 180, W/V = 359/25. Calculate the Track(°T) and GS? 180-223 KT·Given: TAS = 200 kt, Track (T) = 073°, W/V = 210/20kt. Calculate the HDG (°T) and GS? 077-214 KT·Given: TAS = 200 kt, Track (T) = 110°, W/V = 015/40kt. Calculate the HDG (°T) and GS? 099-199KT·Given: TAS = 205 kt, HDG (T) = 180°, W/V = 240/25kt. Calculate the drift and GS? 6L-194KT·Given: TAS = 220 kt; Magnetic course = 212 º, W/V 160 º(M)/ 50kt, Calculate the GS? 186 KT·Given: TAS = 225 kt, HDG (°T) = 123°, W/V = 090/60kt. Calculate the Track (°T) and GS? 134-178KT·Given: TAS = 227 kt, Track (T) = 316°, W/V = 205/15kt. Calculate the HDG (°T) and GS? 312-232KT·Given: TAS = 230 kt, HDG (T) = 250°, W/V = 205/10kt. Calculate the drift and GS? 2 R-223 KT·Given: TAS = 250 kt, HDG (T) = 029°, W/V = 035/45kt. Calculate the drift and GS? 1L-205KT·Given: TAS = 270 kt, Track (T) = 260°, W/V = 275/30kt. Calculate the HDG (°T) and GS? 262- 241KT·Given: TAS = 270 kt, True HDG = 145°, Actual wind = 205°(T)/30kt. Calculate the drift angle and GS?6° L-256 KT·Given: TAS = 370 kt, True HDG = 181°, W/V = 095°(T)/35kt. Calculate the true track and GS? 186-370 KT·Given: TAS = 375 kt, True HDG = 124°, W/V = 130°(T)/55kt. Calculate the true track and GS? 123-320KT·Given: TAS = 465 kt, HDG (T) = 124°, W/V = 170/80kt. Calculate the drift and GS? 8L-415 KT·Given: TAS = 465 kt, Track (T) = 007°, W/V = 300/80kt. Calculate the HDG (°T) and GS? 358-428KT·Given: TAS = 472 kt, True HDG = 005°, W/V = 110°(T)/50kt. Calculate the drift angle and GS: 6°L/490 KT·Given: TAS = 480 kt, HDG (°T) = 040°, W/V = 090/60kt. Calculate the Track (°T) and GS? 034-445KT·Given: TAS = 485 kt, HDG (T) = 168°, W/V = 130/75kt. Calculate the Track (°T) and GS? 174-428 KT·Given: TAS = 485 kt, OAT = ISA +10°C, FL 410. Calculate the Mach Number? 0.825·Given: TAS = 485 kt, True HDG = 226°, W/V = 110°(T)/95kt. Calculate the drift angle and GS? 9° R-533 KT·Given: TAS = 90 kt, HDG (T) = 355°, W/V = 120/20kt. Calculate the Track (°T) and GS? 346-102 KT·Given: TAS = 95 kt, HDG (T) = 075°, W/V = 310/20kt. Calculate the drift and GS? 9R-108KT·Given: The coordinates of the heliport at Issy les Moulineaux are N48°50' E002°16.5'. What are thecoordinates of the position directly on the opposite side of the earth? S48°50W177°43.5·Given: True altitude 9000 FT, OAT -32°C, CAS 200 kt. What is the TAS? 220 KT

·Given: True course from A to B = 090°, TAS = 460 kt, W/V = 360/100kt, Average variation = 10°E,Deviation = -2°. Calculate the compass heading and GS? 070°-450 KT·Given: True HDG = 002°, TAS = 130 kt, Track (T) = 353°, GS = 132 kt. Calculate the W/V? 095/20·Given: True HDG = 035°, TAS = 245 kt, Track (T) = 046°, GS = 220 kt. Calculate the W/V? 340/50KT·Given: True HDG = 054°, TAS = 450 kt, Track (T) = 059°, GS = 416 kt. Calculate the W/V? 010/50KT·Given: True HDG = 074°, TAS = 230 kt, Track (T) = 066°, GS = 242 kt. Calculate the W/V? 180/35KT·Given: True HDG = 133°, TAS = 225 kt, Track (T) = 144°, GS = 206 kt. Calculate the W/V? 070/45KT·Given: True HDG = 145°, TAS = 240 kt, Track (T) = 150°, GS = 210 kt. Calculate the W/V? 115/35KT·Given: True HDG = 206°, TAS = 140 kt, Track (T) = 207°, GS = 135 kt. Calculate the W/V? 180/05KT·Given: True HDG = 233°, TAS = 480 kt, Track (T) = 240°, GS = 523 kt. Calculate the W/V? 110/75KT·Given: True HDG = 307°, TAS = 230 kt, Track (T) = 313°, GS = 210 kt. Calculate the W/V? 260/30KT·Given: True Heading = 090° TAS = 180 kt GS = 180 kt Drift 5° right Calculate the W/V? 005°/15KT·Given: True Heading = 090°, TAS = 200 kt, W/V = 220° / 30 kt. Calculate the GS? 220 KT·Given: True Heading = 180°, TAS = 500 kt, W/V 225° / 100 kt, Calculate the GS? 435 KT·Given: True heading = 310°, TAS = 200 kt, GS = 176 kt, Drift angle 7° right. Calculate the W/V?270°/33KT·Given: True Track = 095°, TAS = 160 kt, True Heading = 087°, GS = 130 kts; Calculate W/V:057°/36KT·Given: True Track 239° True Heading 229° TAS 555 kt G/S 577 kt Calculate the wind velocity:130°/100KT·Given: True Track 245° Drift 5° right Variation 3° E Compass Hdg 242° Calculate the MagneticHeading. 237°·Given: True Track 245° Drift 5° right Variation 3° E Compass Hdg 242° Calculate the deviation: 5° W·Given: true track is 348°, drift 17° left, variation 32° W, deviation 4°E. What is the compass heading?033°·Given: value for the flattening of the Earth is 1/298. Earth's semi-major axis, as measured at theequator, equals 6378.4 km. What is the semi-minor axis (km) of the earth at the axis of the poles?6357.0·Given: Variation is 6°W. Isogonic lines Jan 2002. Average annual increase 10'. Calculate: Variation in2005.; 6.5°W·Given: Waypoint 1. 60°S 030°W, Waypoint 2. 60°S 020°W. What will be the approximate latitudeshown on the display unit of an inertial navigation system at longitude 025°W? 060° 06S·Given:; Aircraft at FL 150 overhead an airport. Elevation of airport 720 FT. QNH is 1003 hPa.; OAT atFL150 -5°C. What is the true altitude of the aircraft? (Assume 1 hPa = 27 FT) 15300 FT·Given:; CAS 120 kt, FL 80, OAT +20°C. What is the TAS? 141 KT·Given:; CON VOR (N5354.8 W00849.1) DME 30 NM, CRN VOR (N5318.1 W00856.5) DME 25 NM,;Aircraft heading 270°(M), Both DME distances decreasing. What is the aircraft position? N5330W00820·Given:; CON VOR/DME (N5354.8 W00849.1), Abbey Shrule aerodrome (N5335 W00739), What is theCON radial and DME distance when overhead Abbey Shrule aerodrome? 123°-46 NM·Given:; CON VOR/DME (N5354.8 W00849.1), Castlebar aerodrome (N5351 W00917), What is the CONradial and DME distance when overhead Castlebar aerodrome? 265°-17NM·Given:; CRK VOR/DME (N5150.4 W00829.7) Kerry aerodrome (N5210.9 W00931.4). What is the CRKradial and DME distance when overhead Kerry aerodrome? 307°-43NM·Given:; CRN VOR (N5318.1 W00856.5) DME 18 NM, SHA VOR (N5243.3 W00853.1) DME 30 NM,;Aircraft heading 270°(M), Both DME distances decreasing. What is the aircraft position? N5310

W0080·Given:; CRN VOR (N5318.1 W00856.5) DME 34 NM, SHA VOR (N5243.3 W00853.1) DME 26 NM,;Aircraft heading 090°(M), Both DME distances increasing. What is the aircraft position? N5255W00915·Given:; For take-off an aircraft requires a headwind component of at least 10 kt and has a cross-windlimitation of 35 kt. The angle between the wind direction and the runway is 60°. Calculate theminimum and maximum allowable wind speeds? 20 KT AND 40 KT·Given:; Magnetic track = 075°, HDG = 066°(M), VAR = 11°E, TAS = 275 kt. Aircraft flies 48 NM in 10MIN. Calculate the true W/V °? 340° 45 KT·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 themaximum allowable windspeed that can be accepted without exceeding the tailwind limit? 10 KT·Given:; Position NDB (55°10´N, 012°55´E); DR Position (54°53´N, 009°58´E); NDB on the RMI reads090°; Magnetic variation = 10°W; The position line has to be plotted on a Lamberts conformal chartwith standard parallels at 40°N and 48°N. Calculate the direction (T) of the bearing to be plotted fromthe NDB. 262°·Given:; Runway direction 210°(M), Surface W/V 230°(M)/30kt.Calculate the cross-wind component?10 KT·Given:; Runway direction 305°(M),Surface W/V 260°(M)/30 kt. Calculate the cross-wind component?21 KT·Given:; SHA VOR (N5243.3 W00853.1) DME 41 NM, CRK VOR (N5150.4 W00829.7) DME 30 NM,Aircraft heading 270°(M), Both DME distances decreasing. What is the aircraft position? N5215 W00805·Given:; SHA VOR (N5243.3 W00853.1) DME 50 NM, CRK VOR (N5150.4 W00829.7) DME 41 NM,Aircraft heading 270°(M), Both DME distances increasing. What is the aircraft position? N5200 W00935·Given:; SHA VOR N5243.3 W00853.1, CON VOR N5354.8 W00849.1, Aircraft position N5330 W00800;Which of the following lists two radials that are applicable to the aircraft position? SHA 042° CON138°·Given:; SHA VOR N5243.3 W00853.1, CON VOR N5354.8 W00849.1. Aircraft position N5320 W00950;Which of the following lists two radials that are applicable to the aircraft position? SHA 325° CON235°·Given:; SHA VOR N5243.3 W00853.1, CRK VOR N5150.4 W00829.7. Aircraft position N5230 W00820;Which of the following lists two radials that are applicable to the aircraft position? SHA 131° CRK017°·Given:; SHA VOR N5243.3 W00853.1, CRK VOR N5150.4 W00829.7. Aircraft position N5220 W00910;Which of the following lists two radials that are applicable to the aircraft position? SHA 214°; CRK330°·Given:; SHA VOR N5243.3 W00853.1, CRK VOR N5150.4 W00829.7. Aircraft position N5230 W00930;Which of the following lists two radials that are applicable to the aircraft position? SHA248° CRK325°·Given:; SHA VOR/DME (N5243.3 W00853.1), Birr aerodrome (N5304 W00754). What is the SHA radialand DME distance when overhead Birr aerodrome? 068°-41NM·Given:; SHA VOR/DME (N5243.3 W00853.1), Connemara aerodrome (N5314 W00928), What is theSHA radial and DME distance when overhead Connemara aerodrome? 333°-37 NM·Given:; TAS = 132 kt, True HDG = 257°, W/V = 095°(T)/35kt. Calculate the drift angle and GS? 4°R-165 KT·Given:; TAS = 140 kt, True HDG = 302°, W/V = 045°(T)/45kt. Calculate the drift angle and GS? 16°L-156 KT·Given:; TAS = 235 kt, HDG (T) = 076°, W/V = 040/40kt. Calculate the drift angle and GS? 7L-204 KT·Given:; TAS = 270 kt, True HDG = 270°, Actual wind 205°(T)/30kt. Calculate the drift angle and GS?

6R-259KT·Given:; TAS = 290 kt, True HDG = 171°, W/V = 310°(T)/30kt. Calculate the drift angle and GS? 4° L-314 KT·Given:; TAS = 470 kt, True HDG = 317°, W/V = 045°(T)/45kt. Calculate the drift angle and GS? 5° L -470 KT·Given:; TAS = 472 kt, True HDG = 005°, W/V = 110°(T)/50kt. Calculate the drift angle and GS?6°L/490KT·Given:; TAS is120 kt. ATA 'X' 1232 UTC, ETA 'Y' 1247 UTC, ATA 'Y' is 1250 UTC. What is ETA 'Z'? 1302UTC·Given:; true track 070°, variation 30°W, deviation +1°, drift 10°R, Calculate the compass heading?089°·Given:; true track 352°, variation 11° W, deviation is -5°, drift 10°R. Calculate the compass heading?358°·Given:A (56°N , 145°E)B (57°N , 165°W)What is the difference in longitude between A and B? 050°·Given:Compass Heading = 233°True Track = 256°Drift Angle = 10°R Deviation = -3°What is thevariation? 16°E·Given:Hdg 265°, TAS 290 kt, W/V 210°/35 kt. Calculate Track and Groundspeed.; 271° AND 272 KT·Grid heading is 299°, grid convergency is 55° West and magnetic variation is 90° West. What is thecorresponding magnetic heading? 084°·Gyrocompassing of an inertial reference system (IRS) is accomplished with the mode selectorswitched to: ALIGN·How does the chart convergency change with latitude in a Lambert Conformal projection? IT ISCONSTANT AND DOES NOT CHANGE WITH LATITUDE·How does the convergency of any two meridians on the Earth change with varying latitude? ITCHANGES AS SINE OF LATITUDE·How does the scale vary in a Direct Mercator chart? THE SCALE INCREASES WITH INCREASINGDISTANCE FROM THE EQUATOR·How is the radio position determined by the FMC in the B737-400 Electronic Flight InstrumentSystem? DME/DME OR VOR/DME·How long will it take to fly 5 NM at a groundspeed of 269 Kt ? 1 MIN 07 SEC·How many degrees has the mean sun moved along the celestial equator in 8 hours and 8 minutes?122°·How many NM would an aircraft travel in 1 MIN 45 SEC if GS is 135 kt? 3.94·How many small circles can be drawn between any two points on a sphere? AN UNLIMITEDNUMBER·If an aeroplane was to circle around the Earth following parallel 60°N at a ground speed of 480 kt. Inorder to circle around the Earth along the equator in the same amount of time, it should fly at aground speed of: 960 KT·If the acceleration of an aircraft is zero, its velocity:; IS CONSTANT·If the chart scale is 1: 500 000, what earth distance would be represented by 7 cm on the chart?35.000M·If the Compass Heading is 265°, variation is 33° W and deviation is 3°E, what is the True Heading?235°·If you are flying along a parallel of latitude, you are flying: A RHUMB LINE TRACK·In a Flight Management System (FMS), control Display Units (CDUs) are used pre-flight to:MANUALLY INITIALIZE THE IRS AND FMC WITH DISPATCH INFORMATION·In a Flight Management System (FMS), control Display Units (CDUs) are used preflight to:MANUALLY INITIALIZE THE IRSS AND FMC WITH DISPATCH INFORMATION·In a remote indicating compass system the amount of deviation caused by aircraft magnetism andelectrical circuits may be minimised by: MOUNTING THE DETECTOR UNIT (FLUX VALVE) IN THEWINGTIP

·In a sunrise/sunset table given for the 28th of June at a certain latitude, sunrise is gven as 0239 andsunset is given as 2127.What is the latitude? 60° N·In an Inertial Navigation System (INS), Ground Speed (GS) is calculated: BY INTEGRATINGMEASURED ACCELERATION·In an Inertial Reference System, accelerations are measured in relation to: AIRCRAFT AXIS·In northern hemisphere, during an acceleration in an easterly direction, the magnetic compass willindicate: A DECREASE IN HEADING·In order to fly from position A (10°00'N, 030°00'W) to position B (30°00'N, 050°00'W), maintaining aconstant true course, it is necessary to fly: A RHUMB LINE TRACK·In order to maintain an accurate vertical using a pendulous system, an aircraft inertial platformincorporates a device: WITH DAMPING AND A PERIOD OF 84.4 MIN·In the Northern Hemisphere the rhumb line track from position A to B is 230°, the covergency is 6°and the difference in longitude is 10°. What is the initial rhumb line track from B to A? 050°·In which occasions does the rhumb line track and the great circle track coincide on the surface of theEarth? ON TRACKS DIRECTLY NORTH-SOUTH AND ON EAST- WEST TRACKS ALONG THEEQUATOR·In which statement is the "Mean Sun" best described? THE MEAN SUN IS A FICTITIOUS SUNCOINCIDING EACH YEAR WITH THE APPARENT SUN AT THE SPRING EQUINOX ANDTRAVELLING ALONG THE CELESTIAL EQUATOR AT UNIFORM SPEED·In which two months of the year is the difference between the transit of the Apparent Sun and MeanSun across the Greenwich Meridian the greatest? FEBRUARY AND NOVEMBER·Inertial Reference System sensors include: ACCELEROMETERS, AND LASER GYROS, MOUNTEDIN THE DIRECTION OF THE AIRCRAFT AXIS·Isogonals are lines of equal: MAGNETIC VARIATION·Isogonic lines connect positions that have: THE SAME VARIATION·Isogrives are lines that connect positions that have: THE SAME GRIVATION·Keplar's second law states: EACH PLANET REVOLVES SO THAT RADIUS VECTOR, SUN TOEARTH, SWEEPS OUT EQUAL ARE IN EQUAL INTERVALS OF TIME·Kepler's second law states that: THE RADIUS VECTOR SUN-EARTH SWEEPS OUT EQUAL AREASIN EQUAL TIME·Location A is at 50ºN 030ºW and location B is at 50ºS 030ºW. On 27th November it is noted that:SUNRISE WILL BE LATER AT A THAN IT IS AT B AND SUNSET WILL BE EARLIER AT A THAN ITIS AT B·Magnetic compass calibration is carried out to reduce: DEVIATION·Magnetic heading of an aircraft is 040 degrees. On the airborne weather radar (AWR) display therelative bearing of the distance to the must southerly part of Lands End, (approximate position: 50 03N 005 40 W) are 030 degrees R and 80 NM. What is the position of the aircraft based on theseobservations? The slant range correction and the map convergency between aircraft position andLands End may be neglected: (49 25N 007 30W)·Morning Civil twilight begins when: THE CENTRE OF THE SUN IS 6° BELOW THE CELESTIALHORIZON·Mu'a, Tonga Islands, is situated at (21°11'S, 175°07'W); In the Air Almanac the standard time ofTonga Islands is listed as UTC +13.; For August 21st the sunrise table in the Air Almanac shows:;20°S: 06:18; 30°S: 06:28; What is the Standard Time of sunrise at Mu'a? 06:59 ON AUGUST 22ND·On a chart a straight line is drawn between two points and has a length of 4.63 cm. What is the chartscale if the line represents 150 NM? 1:6.000.000·On a chart, the distance along a meridian between latitudes 45°N and 46°N is 6 cm. The scale of thechart is approximately: 1: 1.850.000Given: ; True course 300°, drift 8°R, variation 10°W, deviation -4°. Calculate the compass heading? 306°·On a Direct Mercator chart a great circle will be represented by a: CURVE CONCAVE TO THEEQUATOR

·On a Direct Mercator chart at latitude 15°S, a certain length represents a distance of 120 NM on theearth. The same length on the chart will represent on the earth, at latitude 10°N, a distance of: 122.3NM·On a Direct Mercator chart at latitude 45°N, a certain chart length along 45°N represents a distanceof 90nm on the surface of the earth. The same length on a chart along latitude 30°N will represent adistance on the earth of: 110 NM·On a Direct Mercator chart, a rhumb line appears as a: STRAIGHT LINE·On a Direct Mercator chart, meridians are: PARALLEL, EQUALLY SPACED, VERTICAL STRAIGHTLINES·On a Direct Mercator projection a particular chart length is measured at 30°N. What earth distancewill the same chart length be if measured at 60°N? A SMALLER DISTANCE·On a direct Mercator projection, at latitude 45° North, a certain length represents 70 NM. At latitude30° North, the same length represents approximately: 86NM·On a direct Mercator projection, the distance measured between two meridians spaced 5° apart atlatitude 60°N is 8 cm. The scale of this chart at latitude 60°N is approximately: 1: 3500000·On a Direct Mercator, rhumb lines are: STRAIGHT LINES·On a Lambert chart (standard parallels 37°N and 65°N), with respect to the straight line drawn on themap between A ( N49° W030°) and B (N48° W040°), the: GREAT CIRCLE AND RHUMB LINE ARETO THE SOUTH·On a Lambert Conformal chart the distance between meridians 5° apart along latitude 37° North is 9cm. The scale of the chart at that parallel approximates: 1:5.000.000·On a Lambert Conformal Conic chart earth convergency is most accurately represented at the:PARALLEL OF ORIGIN·On a Lambert Conformal Conic chart great circles that are not meridians are: CURVES CONCAVE TOTHE PARALLEL OF ORIGIN·On a Lambert conformal conic chart the convergence of the meridians: IS THE SAME AS EARTHCONVERGENCY AT THE PARALLEL OF ORIGIN·On a Lambert conformal conic chart, the distance between parallels of latitude spaced the samenumber of degrees apart: IS SMALLER BETWEEN THE STANDARD PARALLELS THAN OUTSIDETHEM·On a Lambert conformal conic chart, with two standard parallels, the quoted scale is correct: ALONGTHE TWO STANDARD PARALLELS·On a Mercator chart, at latitude 60°N, the distance measured between W002° and E008° is 20 cm.The scale of this chart at latitude 60°N is approximately: 1: 2.780.000·On a Mercator's projection the distance between (17°N, 035°E) and (17°N, 040°E) is 5 cm. The scaleat 57°N is approximately: 1: 6.052.030·On a Polar Stereographic chart, the initial great circle course from A 70°N 060°W to B 70°N 060°E isapproximately: 030° (T)·On a polar stereographic projection chart showing the South Pole, a straight line joins position A(70°S 065°E) to position B (70°S 025°W). The true course on departure from position A isapproximately: 225°·On a Transverse Mercator chart, scale is exactly correct along the: MERIDIAN OF TANGENCY·On a transverse Mercator chart, with the exception of the Equator, parallels of latitude appear as:ELLIPSES·On a True Heading of 090° the aircraft experiences drift of 5°S. On a True Heading of 180° theaircraft experiences no drift. On both headings the TAS is 200 kt and it is assumed that the wind is thesame.What is the experienced wind speed and direction? 360°/17 KT·On an oblate spheroid representing the earth's shape: 1 MINUTE OF ARC ALONG THE EQUATORMEASURES A GREATER DISTANCE THAN 1 MINUTE OF ARC ALONG THE MERIDIAN AT ALATITUDE OF 45° N/S·On the 27th of February, at 52°S and 040°E, the sunrise is at 0243 UTC. On the same day, at 52°S

and 035°W, the sunrise is at: 0743 UTC·On the earth's ellipsoid one degree of latitude near the equator is: LESS THAN 60 NM·On which of the following chart projections is it NOT possible to represent the north or south poles?DIRECT MERCATOR·One of the errors inherent in a ring laser gyroscope occurs at low input rotation rates tending towardszero when a phenomenon known as 'lock-in' is experienced. What is the name of the technique,effected by means of a piezo-electric motor, that is used to correct this error? DITHER·One purpose of a compass calibration is to reduce the difference, if any, between: COMPASSNORTH AND MAGNETIC NORTH·One purpose of compass calibration is to determine the deviation: ON ANY HEADING·Parallels of latitude on a Direct Mercator chart are: PARALLEL STRAIGHT LINES UNEQUALLYSPACED·Parallels of latitude, except the equator, are: RHUMB LINES·Permanent magnetism in aircraft arises chiefly from: HAMMERING, AND THE EFFECT OF THEEARTHS MAGNETIC FIELD, WHILST UNDER CONTRUCTION·Position "Elephant Point" is situated at (58°00'N, 135°30'W).Standard time for this location is listed inthe Air Almanac as UTC -8.If sunset occurs at 00:57 UTC on 21st January, what is the time of Sunset inLMT? 15:55 ON JANUARY 20TH·Position A is (31º00'S, 176º17'W)Rhumb line track (T) from A to B is 270º.Initial great circle track (T)from A to B is 266.2º.The Approximate position of B is: 31°00S, 168°58E·Position A = (30°00.0''N, 175°23.2''W) Position B = (30°00.0''N, 173°48.1''E) For the route from A toB the: RHUMB LINE DISTANCE IS 578NM·Position A = (56°00.0''S, 163°57.2''E) Position B = (56°00.0''S, 171°47.4''W) For the route from A to Bthe; GREAT CIRCLE DIRECTION AT B IS 080.7°·Position A is located on the equator at longitude 130°00E. Position B is located 100 NM from A on abearing of 225°(T). The coordinates of position B are: 01° 11S 128°49E·Route 'A' (44°N 026°E) to 'B' (46°N 024°E) forms an angle of 35° with longitude 026°E. Variation at Ais 3°E. What is the initial magnetic track from A to B? 322°·Route A - B is drawn on a Polar Stereographic chart with the grid aligned with the Greenwichmeridian. The true track of the straight line at A (75°S, 010°W) is 080°. What is the Grid Track whenpassing the meridian of 050°E? 070° (G)·Route A - B is drawn on a Polar Stereographic chart with the grid aligned with the Greenwichmeridian. The true track of the straight line at A is 060°. When passing the meridian 100°E, the truetrack is 090°. The grid track of this route on the chart is: 350° (G)·Route A - B is drawn on a Polar Stereographic chart with the grid aligned with the Greenwichmeridian. The True Track of the straight line at A (75°N, 010°W) is 080°. What is the Grid Track whenpassing the meridian 050°E? 090° (G)·Route A - B is drawn on a Southern Polar Stereographic chart whose grid is aligned with theGreenwich meridian. The true track of the straight line at A is 120°. When passing the meridian of100°E the true track is 090°. The grid track of this route on the chart is: 190° (G)·Seasons are due to the: INCLINATION OF THE POLAR AXIS WITH THE ECLIPTIC PLANE·Some inertial reference and navigation systems are known as "strapdown". This means that: THEGYROSCOPES AND ACCELEROMETERS BECOME PART OF THE UNITS FIXTURE TO THEAIRCRAFT STRUCTURE·Standard time for some areas is listed in the Air Almanac as UTC +13 instead of UTC -11. The reasonfor this is KEEPING THE SAME DATE AS THE POLITICAL AND OR ECONOMICAL ENTILY TOWHICH THEY BELONG·The 'departure' between positions 60°N 160°E and 60°N 'x' is 900 NM. What is the longitude of 'x'?170°W·The accuracy of the manually calculated DR-position of an aircraft is, among other things, affectedby: THE FLIGHT TIME SINCE THE LAST POSITION UPDATE

·The accuracy of the, manually calculated, DR-position of an aircraft is, among other things, affectedby: THE ACCURACY OF THE FORECASTED WIND·The accuracy of the, manually calculated, DR-position of an aircraft is, among other things, affectedby: THE FLIGHT TIME SINCE THE LAST POSITION UPDATE·The alignment time, at mid-latitudes, for an Inertial Reference System using laser ring gyros isapproximately: 10 MIN·The angle between Magnetic North and Compass North is called: COMPASS DEVIATION·The angle between the plane of the ecliptic and the plane of equator is approximately: 23.5°·The angle between the true great-circle track and the true rhumb-line track joining the followingpoints: A (60° S 165° W) B (60° S 177° E), at the place of departure A, is: 7.8°·The angle between True North and Magnetic North is called: VARIATION·The annunciator of a remote indicating compass system is used when: SYNCHRONISING THEMAGNETIC AND GYRO COMPASS ELEMENTS·The automatic flight control system (AFCS) in an aircraft is coupled to the guidance outputs from aninertial navigation system (INS) and the aircraft is flying from waypoint No. 2 (60°00'S 070°00'W) toNo. 3 (60°00'S 080°00'W). Comparing the initial track (°T) at 070°00'W and the final track (°T) at080°00'W, the difference between them is that the initial track is approximately: 9° LESS THAN THEFINAL ONE·The automatic flight control system (AFCS) in an aircraft is coupled to the guidance outputs of aninertial navigation system (INS).The aircraft is flying between waypoints No. 4 (45 00''N 040 00''W)and No.5 (45 00''N 030 00''W). On arrival over waypoint No. 5, compared to the TT at waypoint No.4,the TT has:; INCREASED BY LESS THAN 10°·The automatic flight control system (AFCS) in an aircraft is coupled to the guidance outputs from aninertial navigation system (INS). The aircraft is flying between inserted waypoints No. 3 (55°00'N020°00'W) and No.4 (55°00'N 030°00'W). With DSRTK/STS selected on the CDU, to the nearest wholedegree, the initial track read-out from waypoint No. 3 will be: 274°·The automatic flight control system is coupled to the guidance outputs from an inertial navigationsystem. Which pair of latitudes will give the greatest difference between initial track read-out and theaverage true course given, in each case, a difference of longitude of 10°? 60°N TO 60°N·The chart distance between meridians 10° apart at latitude 65° North is 9.5 cm. The chart scale atthis latitude approximates: 1: 5.000.000·The chart that is generally used for navigation in polar areas is based on a: STEREOGRAPHICALPROJECTION·The circumference of the earth is approximately: 21600NM·The circumference of the parallel of latitude at 60°N is approximately: 10800NM·The constant of cone of a Lambert conformal conic chart is quoted as 0.3955. At what latitude on thechart is earth convergency correctly represented? 23° 18·The constant of the cone in a Lambert chart is 0.8666500. The angle between the north directions ofthe meridian in position A (65°00'N, 018°00'W) and the meridian of position B (75°00'N, 023°00'W) onthe chart is: 4.3°·The constant of the cone, on a Lambert chart where the convergence angle between longitudes010°E and 030°W is 30°, is: 0.75·The convergence factor of a Lambert conformal conic chart is quoted as 0.78535. At what latitude onthe chart is earth convergency correctly represented? 51° 45·The data that needs to be inserted into an Inertial Reference Sysytem in order to enable the systemto make a successful alignment of navigation is: AIRCRAFT POSITION IN LATITUDE ANDLONGITUDE·The declination of the sun is defined as: THE ANGULAR DISTANCE OF THE SUN NORTH ORSOUTH OF THE CELESTIAL EQUATOR·The definition of True North for any observer is: THE DIRECTION OF THE OBSERVERS MERIDIANTO THE NORTH POLE

·The diameter of the Earth is approximately: 12700 KM·The direct reading magnetic compass is made aperiodic (dead beat) by: KEEPING THE MAGNETICASSEMBLY MASS CLOSE TO THE COMPASS POINT AND BY USING DAMPING WIRES·The direction "magnetic north" at a position on the earth is: THE DIRECTION OF THE HORIZONTALCOMPONENT OF THE EARTHS MAGNETIC FIELD AT THAT POSITION·The direction of Magnetic North at a certain position coincides with the direction of: THEHORIZONTAL DISTANCE COMPONENT OF THE EARTHS MAGNETIC FIELD·The Directive Force: IS THE COMPONENT OF THE EARTHS MAGNETIC FIELD WHICH ALIGNSTHE COMPASS NEEDLE·The distance along a meridian between 63°55'N and 13°47'S is: 4662 NM·The distance between A and B is 90 NM. At a distance of 15 NM from A the aircraft is 4 NM right ofcourse. To reach destination B, the correction angle on the heading should be: 19°·The distance between A and B is 90 NM. At a distance of 75 NM from A the aircraft is 4 NM right ofcourse. The track angle error (TKE) is: 3° RIGHT·The distance between point of departure and destination is 340 NM and wind velocity in the wholearea is 100°/25kt. TAS is 140kt, True Track is 135° and safe endurance 3 hr and 10 min. How long willit take to reach the Point of Safe Return? 1 HR AND 49 MIN·The distance between positions A and B is 180 NM. An aircraft departs position A and after havingtravelled 60 NM, its position is pinpointed 4 NM left of the intended track. Assuming no change in windvelocity, what alteration of heading must be made in order to arrive at position B? 6° RIGHT·The distance between two waypoints is 200 NM, To calculate compass heading, the pilot used 2°Emagnetic variation instead of 2°W. Assuming that the forecast W/V applied, what will the off trackdistance be at the second waypoint? 14NM·The distance measured between two points on a navigation map is 42 mm (millimetres). The scale ofthe chart is 1:1 600 000. The actual distance between these two point is approximately: 36.30 NM·The distance on the map between position A and position B measured along the rhumb line: IS LESSTHAN 54.19 CM·The drift of the azimuth gyro on an inertial unit induces an error in the position given by this unit. "t"being the elapsed time. The total error is: PROPORTIONAL TO T·The duration of civil twilight is the time: BETWEEN SUNSET AND WHEN THE CENTRE OF THESUN IS 6° BELOW THE CELESTIAL HORIZON·The equivalent of 70 m/sec is approximately: 136 KT·The first law of Kepler states: ; PLANETS MOVE IN ELLIPTICS ORBITS WITH THE SUN IN ONE OFTHE FOCI·The fix of the aircraft position is determined by radials from three VOR-stations. The measurementscontain small random errors, known systematic errors and unknown systematic errors. The measuredradials are corrected for known systematic errors and are plotted on a navigation chart. The result isshown at the reference. What is the most probable position of the aircraft? 1·The flight log gives the following data: "True track, Drift, True heading, Magnetic variation, Magneticheading, Compass deviation, Compass heading" The right solution, in the same order, is: 119°, 3°L,122°, 2°E, 120°, +4°, 116°·The following information is displayed on an Inertial Navigation System: GS 520 kt, True HDG 090°,Drift angle 5° right, TAS 480 kt. SAT (static air temperature) -51°C. The W/V being experienced is:320°/60KT·The following points are entered into an inertial navigation system (INS).; WPT 1: 60°N 30°W; WPT 2:60°N 20°W; WPT 3: 60°N 10°W; The inertial navigation system is connected to the automatic pilot onroute (1-2-3). The track change when passing WPT 2 will be approximately: A 9° DECREASE·The force acting on the needle of a direct reading compass varies: DIRECTLY WITH THEHORIZONTAL COMPONENT OF THE EARTHS MAGNETIC FIELD·The full alignment of the stable platform of an Inertial Navigation System: IS ONLY POSSIBLE ONTHE GROUND WHEN THE AIRCRAFT IS AT A COMPLETE STOP

·The GMT of Morning Civil Twilight at (66º48'N, 095º26'W) on 27th of January is? 1436 GMT·The GMT of Sunrise at (66º48'N, 095º26'W) on 27th of January is? 1549 GMT·The Great Circle bearing from A (70°S 030°W) to B (70°S 060°E) is approximately: 132° (T)·The great circle bearing of position B from position A in the Northern Hemisphere is 040°. If theConversion Angle is 4°, what is the great circle bearing of A from B? 228°·The great circle distance between position A (59°34.1'N 008°08.4'E) and B (30°25.9'N 171°51.6'W)is: 5400 NM·The great circle track measured at A (45 00'N 010 00'W) from A to B (45 00'N 019 00'W) isapproximately: 273°·The great circle track measured at A (45°00'N 010°00'W) from A to B (45°00'N 019°00'W) isapproximately: 273°Which aeronautical chart symbol indicates a group of lighted obstacles? 11·The horizontal component of the earth's magnetic field: IS STRONGER CLOSER TO THEMAGNETIC EQUATOR·The horizontal component of the earth's magnetic field: IS VERY SMALL CLOSE TO THEMAGNETIC POLES·The ICAO definition of ETA is the: ESTIMATED TIME OF ARRIVAL AT DESTINATION·The initial great circle track from A to B is 080° and the rhumb line track is 083°. What is the initialgreat circle track from B to A and in which Hemisphere are the two positions located? 266° AND INTHE NORTHERN HEMISPHERE·The length of the apparent solar day varies continuously throughout a year. This is caused by: THETILT OF THE EARTHS AXIS AND THE ELLIPTICAL ORBIT OF THE EARTH AROUND THE SUN·The lines on the earth's surface that join points of equal magnetic variation are called: ISOGONALS·The local hour angle of the mean sun at 1200 LMT is: 000·The local hour angle of the mean sun at 1200 LMT is: THE DATE WILL INCREASE CROSSING ON AWESTERLY HEADING·The Local Mean Time at longitude 095°20'W, at 0000 UTC, is: 1738:40 PREVIOUS DAY·The long term periodic change in the Earth's Magnetic Field: IS REFLECTED IN THE SLOWMOVEMENT OF THE MAGNETIC POLES·The main reason for mounting the detector unit of a remote reading compass in the wingtip of anaeroplane is: TO MINIMISE THE AMOUNT OF DEVIATION CAUSED BY AIRCRAFT MAGNETISMAND ELECTRICAL CIRCUITS·The main reason for the occurrence of seasons on earth is: THE INCLINATION OF THE EARTHAXIS WITH REGARD TO THE PLANE OF THE ECLIPTIC·The main reason for usually mounting the detector unit of a remote indicating compass in the wingtipof an aeroplane is to: REDUCE THE AMOUNT OF DEVIATION CAUSED BY AIRCRAFTMAGNETISM AND ELECTRICAL CIRCUITS·The main reason that day and night, throughout the year, have different duration, is due to the:INCLINATION OF THE ECLIPTIC TO THE EQUATOR·The maximum difference between geocentric and geodetic latitude occurs at about: 45° NORTHAND SOUTH·The maximum difference in distance when proceeding along the great circle between two positions,in stead of the rhumb line, will occur: ON EAST-WEST TRACKS AT HIGH LATITUDES·The nominal scale of a Lambert conformal conic chart is the: SCALE AT THE STANDAR PARALLELS·The north and south magnetic poles are the only positions on the earth's surface where: A FREELYSUSPENDED COMPASS NEEDLE WOULD STAND VERTICAL·The outer marker of an ILS with a 3° glide slope is located 4.6 NM from the threshold. Assuming aglide slope height of 50 FT above the threshold, the approximate height of an aircraft passing theouter marker is: 1450 FT·The parallels on a Lambert Conformal Conic chart are represented by: ARCS OF CONCENTRICCIRCLES·The platform of an inertial navigation system (INS) is maintained at right angles to the local vertical

by applying corrections for the effects of: AIRCRAFT MANOEUVRES, EARTH ROTATION,TRANSPORT WANDER AND CORIOLIS·The positions A (30°00'N, 017°30'E) and B at longitude (30°00'N, 023°30'E) are plotted on a Lambertchart with a constant of the cone of 0.5. A and B are connected by a straight line. The True Trackmeasured at A is 088.5°. ; What is the True Track measured at B? 091.5°·The principle of 'Schuler Tuning' as applied to the operation of Inertial Navigation Systems/ InertialReference Systems is applicable to: BOTH GYRO-STABLISED PLATFORM AND STRAPDOWNSYSTEMS·The purpose of the TAS input, from the air data computer, to the Inertial Navigation System is for:THE CALCULATION OF WIND VELOCITY·The QNH, given by a station at 2500 ft, is 980hPa.The elevation of the highest obstacle along a routeis 8 000 ft and the OAT = ISA -10°C. ; When an aircraft, on route has to descend the minimumindicated altitude (QNH on the subscale of the altimeter) to maintain a clearance of 2000 ft, will be:10400 FT·The reason that the solar day lasts longer than the sidereal day is that; BOTH THE DIRECTION OFROTATION OF THE EARTH AROUND ITS AXIS AND ITS ORBITAL ROTATION AROUND THE SUNARE THE SAME·The reported surface wind from the Control Tower is 240°/35 kt. Runway 30 (300°). What is thecross-wind component? 30 KT·The resultant of the first integration from the north/south accelerometer of an inertial navigationsystem (INS) in the NAV MODE is: VELOCITY ALONG THE LOCAL MERIDIAN·The resultant of the first integration of the output from the east/west accelerometer of an inertialnavigation system (INS) in NAV MODE is: VELOCITY ALONG THE LOCAL PARALLEL OF LATITUDE·The rhumb line track between position A (45°00'N, 010°00'W) and position B (48°30'N, 015°00'W) isapproximately: 315·The rhumb-line distance between points A (60°00'N 002°30'E) and B (60°00'N 007°30'W) is: 300 NM·The scale on a Lambert conformal conic chart: IS CONSTANT ALONG THE PARALLEL OFLATITUDE·The sensors of an INS measure: ACCELERATION·The SR/SS table for the 23rd of February at latitude 40ºN gives: SR = 06:44 SS = 17:44 At 12:00Central European Time (UTC+1) at 40ºN: THE SUN RISES AT 64°W·The standard parallels of a Lambert chart are 26°N and 48°N and the stated scale is 1:2 500 000.Which statement is correct? THE SCALE AT 28°N IS SMALLER THAN THE SCALE AT 24° N·The term drift refers to the wander of the axis of a gyro in: THE HORIZONTAL PLANE·The time difference in Local Mean Time between sunset at positions A (50°N, 120°E) and B (50°S,120°E) on the 21st of November is: SOME HOURS AND THE SUN RISES EARLIER IN B THAN IN A·The time interval between sunrise and sunset is dependent on: THE DECLINATION OF THE SUNAND THE LATITUDE OF THE OBSERVER·The total length of the 53°N parallel of latitude on a direct Mercator chart is 133 cm. What is theapproximate scale of the chart at latitude 30°S? 1: 25.000.000·The True course in the flight log is 270º, the forecast wind is 045º(T)/15kt and the TAS is 120kt. After15 minutes of flying with the planned TAS and TH the aircraft is 3 NM South of the intended track and2.5 NM ahead of the dead reckoning position: 5°L·The two standard parallels of a conical Lambert projection are at N10°40'N and N41°20'. The coneconstant of this chart is approximatively: 0.44·The value of magnetic variation: HAS A MAXIMUM OF 180°·Thule VOR is located at (76°32'N, 68°15'W). A Polar Stereographic chart with the grid aligned withthe Greenwich meridian is to be used. The local variation is 75°W. Which grid track must bemaintained to track radial 210(M) inbound? 023° (G)·Transverse Mercator projections are used for: MAPS OF LARGE NORTH/SOUTH EXTENT·True Heading of an aircraft is 265° and TAS is 290 kt. If W/V is 210°/35kt, what is True Track and GS?

271° AND 272 KT·Two places are situated on the same parallel in the Southern Hemisphere. The great circle, rhumbline and the straight line between these places are drawn on a Polar Stereographic Projection. Whichstatement is correct? THE GREAT CIRCLE IS SITUAATED BETWEEN THE PARALLEL AND THESTRAIGHT LINE, BACAUSE THE CONCAVE SIDE OF TH GREAT CIRCLE IS IS ALWAYS POINTEDTOWARDS THE POLE·Two places on the parallel of 47ºS lie 757.8 km apart. Calculate the difference in longitude: 10°00·Two points A and B are 1000 NM apart. TAS = 490 kt.; On the flight between A and B the equivalentheadwind is -20 kt.; On the return leg between B and A, the equivalent tailwind is +40 kt.; Whatdistance from A, along the route A to B, is the the Point of Equal Time (PET)? 530 NM·Two positions plotted on a polar stereographic chart, A (80°N 000°) and B (70°N 102°W) are joinedby a straight line whose highest latitude is reached at 035°W. At point B, the true course is: 203°·Use Europe Low Altitude Enroute Chart E(LO) 1A); Two consecutive waypoints of a flight plan areStornoway VORDME (N58°12.4', W006°11.0') and Glasgow VORDME (N55°52.2', W 004°26.7'). ;During the flight the Actual Time Over Stornoway is 11:15 UTC and the Estimated Time Over Glasgowis 11:38 UTC. At 11:21 UTC the fix of the aircraft is exactly over reporting point RONAR.; What is theRevised UTC over Glasgow, based on this last fix? 11:36·Use the Air Almanac Tables.; The UTC of sunrise on 6 December at WINNIPEG (Canada) (49°50'N097°30'W) is: 1413·Use the Air Almanac Tables.; When it is 0600 Standard Time in Queensland (Australia) the StandardTime in Hawaii (USA) is: 1000·Use the Air Almanac Tables.; When it is 1000 Standard Time in Kuwait, the Standard Time in Algeriais: 0800·Variation at an NDB is 9W. Variation at the aircraft is 11W. The true track of the great circle to theNDB from the aircraft, at the aircraft, is 101.5. The magnetic bearing of the NDB from the aircraft is:112.5·Waypoints can be entered in an INS memory in different formats. In which of the following formatscan waypoints be entered into all INSs? GEOGRAPHIC COORDINATES·What additional information is required to be input to an Inertial Navigation System (INS) in order toobtain an W/V readout? TAS·What are, in order of highest priority followed by lowest, the two levels of message produced by theCDU of the B737-400 Electronic Flight Instrument System? ALERTING AND ADVISORY·What feature is shown on the chart at position N5211 W00931? KERRY/FARRANFOREAERODROME·What feature is shown on the chart at position N5212 W00612? TUSKAR ROCK LT.H. NDB·What feature is shown on the chart at position N5311 W00637? PUNCHESTOWN AERODROME·What feature is shown on the chart at position N5351 W00917? CASTLEBAR AERODROME·What feature is shown on the chart at position N5417 W01005? EAGLE ISLAND LT.H NDB·What is meant by "Aphelion"? ; THE POINT OF THE EARTHS ORBIT FURTHERST AWAY FROMTHE SUN·What is the average track (°M) and distance between BAL VOR (N5318.0 W00626.9) and SLG NDB(N5416.7 W00836.0)? 316°-96NM·What is the average track (°M) and distance between CRK VOR (N5150.4 W00829.7) and CRN NDB(N5318.1 W00856.5)? 357°-89 NM·What is the average track (°M) and distance between CRN NDB (N5318.1 W00856.5) and WTD NDB(N5211.3 W00705.0)? 142°-95NM·What is the average track (°M) and distance between CRN NDB (N5318.1 W00856.5) and BEL VOR(N5439.7 W00613.8)? 058°- 128 NM·What is the average track (°M) and distance between KER NDB (N5210.9 W00931.5) and CRN NDB(N5318.1 W00856.5)? 025°- 70 NM·What is the average track (°M) and distance between WTD NDB (N5211.3 W00705.0) and KER NDB

(N5210.9 W00931.5)? 278°-90 NM·What is the average track (°M) and distance between WTD NDB (N5211.3 W00705.0) and BAL VOR(N5318.0 W00626.9)? 026°-71NM·What is the average track (°T) and distance between BAL VOR (N5318.0 W00626.9) and CRN NDB(N5318.1 W00856.5)? 270°-90 NM·What is the average track (°T) and distance between BAL VOR (N5318.0 W00626.9) and CFN NDB(N5502.6 W00820.4)? 327°-124 NM·What is the average track (°T) and distance between CON VOR (N5354.8 W00849.1) and BEL VOR(N5439.7 W00613.8)? 063°-101 NM·What is the average track (°T) and distance between CRN NDB (N5318.1 W00856.5) and EKN NDB(N5423.6 W00738.7)? 035°-80 NM·What is the average track (°T) and distance between SHA VOR (N5243.3 W00853.1) and CON VOR(N5354.8 W00849.1)? 002°-72NM·What is the average track (°T) and distance between SLG NDB (N5416.7 W00836.0) and CFN NDB(N5502.6 W00820.4)? 011°-47 NM·What is the average track (°T) and distance between WTD NDB (N5211.3 W00705.0) and FOY NDB(N5234.0 W00911.7)? 286°-81NM·What is the average track (°T) and distance between WTD NDB (N5211.3 W00705.0) and SLG NDB(N5416.7 W00836.0)? 336°-137 NM·What is the chart distance between longitudes 179°E and 175°W on a direct Mercator chart with ascale of 1: 5 000 000 at the equator? 133 NM·What is the constant of the cone for a Lambert conic projection whose standard parallels are at 50°Nand 70°N? 0.866·What is the correct definition of latitude of a position on the earth? LATITUDE IS THE ANGLEBETWEEN THE PLANE OF THE EQUATOR AND THE LINE FROM THE CENTRE OF THE EARTHTO THE POSITION·What is the duration of morning Civil Twilight at (66º48'N, 095º26'W) on 27th of January? 01H 13MIN·What is the effect on the Mach number and TAS in an aircraft that is climbing with constant CAS?MACH NUMBER INCREASES, TAS INCREASES·What is the final position after the following rhumb line tracks and distances have been followed fromposition 60°00'N 030°00'W? South for 3600 NM, East for 3600 NM, North for 3600 NM, West for 3600NM. The final position of the aircraft is: 60° 00 N 090° 00 W·What is the highest latitude listed below at which the sun will reach an altitude of 90° above thehorizon at some time during the year? 23.5°·What is the ISA temperature value at FL 330? -51°C·What is the length of one degree of longitude at latitude 60° South? 30 NM·What is the local mean time, position 65°25'N 123°45'W at 2200 UTC? 1345·What is the longitude of a position 6 NM to the east of 58°42'N 094°00'W? 093° 48.5W·What is the meaning of aeronautical chart symbol No. 15? AREONAUTICAL GROUND LIGHT·What is the meaning of aeronautical chart symbol No. 16? LIGHTSHIP·What is the meaning of the term "standard time" ? IT IS THE TIME SET BY THE LEGALAUTHORITIES FOR A COUNTRY OR PART OF A COUNTRY·What is the name given to an Inertial Reference System (IRS) which has the gyros andaccelerometers as part of the unit's fixture to the aircraft structure? STRAPDOWN·What is the radial and DME distance from BEL VOR/DME (N5439.7 W00613.8) to position N5410W00710? 236°- 44NM·What is the radial and DME distance from BEL VOR/DME (N5439.7 W00613.8) to position N5440W00730? 278°- 44 NM·What is the radial and DME distance from BEL VOR/DME (N5439.7 W00613.8) to position N5500W00700? 315°-34 NM

·What is the radial and DME distance from CON VOR/DME (N5354.8 W00849.1) to position N5340W00820? 140°-23 NM·What is the radial and DME distance from CON VOR/DME (N5354.8 W00849.1) to position N5400W00800? 088°- 29 NM·What is the radial and DME distance from CON VOR/DME (N5354.8 W00849.1) to position N5430W00900? 358° - 36 NM·What is the radial and DME distance from CON VOR/DME (N5354.8 W00849.1) to position N5330W00930? 233°-35NM·What is the radial and DME distance from CRK VOR/DME (N5150.4 W00829.7) to position N5140W00730? 113°-38 NM·What is the radial and DME distance from CRK VOR/DME (N5150.4 W00829.7) to position N5230W00750? 039°-48 NM·What is the radial and DME distance from CRK VOR/DME (N5150.4 W00829.7) to position N5220W00810? 030°-33NM·What is the radial and DME distance from CRK VOR/DME (N5150.4 W00829.7) to position N5210W00920? 311°-38 NM·What is the radial and DME distance from SHA VOR/DME (N5243.3 W00853.1); to position N5210W00920? 214°- 37 NM·What is the radial and DME distance from SHA VOR/DME (N5243.3 W00853.1) to position N5220W00810? 139°-35 NM·What is the radial and DME distance from SHA VOR/DME (N5243.3 W00853.1) to position N5300W00940? 309°-33 NM·What is the radial and DME distance from SHA VOR/DME (N5243.3 W00853.1) to position N5310W00830? 035°-30NM·What is the ratio between the litre and the US-GAL ? 1 US-GAL EQUALS 3.78 LITRES·What is the rhumb line distance, in nautical miles, between two positions on latitude 60°N, that areseparated by 10° of longitude? 300 NM·What is the source of magnetic variation information in a Flight Management System (FMS)?MAGNETIC VARIATION INFORMATION IS STORED IS EACH IRS MEMORY, IT IS APPLIED TOTHE TRU HEADING CALCULATED BY THE RESPECTIVE IRS·What is the time required to travel along the parallel of latitude 60° N between meridians 010° E and030° W at a groundspeed of 480 kt? 2 HR 30 MIN·What is the validity period of the 'permanent' data base of aeronautical information stored in the FMCIn the B737-400 Flight Management System? 28 DAYS·What is the value of the convergence factor on a Polar Stereographic chart? 1.0·What is the value of the magnetic dip at the magnetic south pole ? 90°·What may cause a difference between a DR-position and a Fix? THE DIFFERENCE BETWEEEN THEACTUAL WIND AND THE FORESCASTED WIND·When accelerating on a westerly heading in the northern hemisphere, the compass card of a directreading magnetic compass will turn: ANTI-CLOCKWISE GIVING AND APPARENT TURN TOWARDSTHE NORTH·When accelerating on an easterly heading in the Northern hemisphere, the compass card of a directreading magnetic compass will turn: CLOCKWISE GIVING AN APPARENT TURN TOWARD THENORTH·When an aircraft on a westerly heading on the northern hemisphere accelerates, the effect of theacceleration error causes the magnetic compass to: INDICATE A TURN TOWARDS THE NORTH·When decelerating on a westerly heading in the Northern hemisphere, the compass card of a directreading magnetic compass will turn: CLOCKWISE GIVING AN APPARENT TURN TOWARD THESOUTH·When flying on a westerly great circle track in the Southern Hemisphere you will: EXPERIENCE ANINCREASE IN THE VALUE OF TRUE TRACK

·When is the magnetic compass most effective? ABOUT MIDWAY BETWEEN THE MAGNETICPOLES·When proceeding, on a given date, along a parallel towards the East, the moment of sunrise willoccur one hour earlier every 15° difference in longitude when it is expressed in: UTC·When the time is 1400 LMT at 90° West, it is: 1200 LMT AT 120° WEST·When the time is 2000 UTC, it is: 1400 LMT AT 90° WEST·When turning right from 330°(C) to 040°(C) in the northern hemisphere, the reading of a directreading magnetic compass will: UNDER-INDICATE THE TURN AND LIQUID SWIRL WILLINCREASE THE EFFECT·Where and when are the IRS positions updated? ONLY IN THE GROUND DURING THE ALIGMENTPROCEDURE·Where on a Direct Mercator projection is the chart convergency correct compared to the earthconvergency? AT THE EQUATOR·Which aeronautical chart symbol indicates a compulsory reporting point? 7·Which aeronautical chart symbol indicates a Control Zone boundary? 2·Which aeronautical chart symbol indicates a Flight Information Region (FIR) boundary? 1·Which aeronautical chart symbol indicates a group of unlighted obstacles? 10·Which aeronautical chart symbol indicates a lighted obstacle? 9·Which aeronautical chart symbol indicates a lightship? 16·Which aeronautical chart symbol indicates a non-compulsory reporting point? 6·Which aeronautical chart symbol indicates a Way-point? 8·Which aeronautical chart symbol indicates an aeronautical ground light? 15·Which aeronautical chart symbol indicates an exceptionally high lighted obstacle? 13·Which aeronautical chart symbol indicates an exceptionally high unlighted obstacle? 12·Which aeronautical chart symbol indicates an uncontrolled route? 4·Which aeronautical chart symbol indicates an unlighted obstacle? 8·Which aeronautical chart symbol indicates the boundary of advisory airspace? 5·Which component of the B737-400 Flight Management System (FMS) is used to enter flight planrouteing and performance parameters? MULTI-FUNCTION CONTROL DISPLAY UNIT·Which definition describes best the notion "Poles"? THE POLES ARE THE POINTS OFINTERSECTION BETWEEN THE EARTHS AXIS AND THE SURFACE OF THE EARTH·Which definition of the equator is correct? The ecuator is a great circle with its planeperpendicular TO THE EARTH ROTATIONAL AXIS·Which figure in the Appendix represents the geocentric latitude of position P, which is situated abovethe surface of the ellipsoid? FIGURE B·Which figure in the Appendix represents the geographic latitude of position P, which is situated abovethe surface of the ellipsoid? FIGURE A·Which FMC/CDU page normally appears on initial power application to the B737-400 Electronic FlightInstrument System? IDENT·Which is the highest latitude listed below at which the sun will rise above the horizon and set everyday? 62°·Which of the following statements about hard and soft iron in relation to magnetism is correct?;HARD IRON MAGNETISM IS OF A PERMANENT NATURE AND SOFT IRON IS OF A NON-PERMANENT NATURE·Which of the aeronautical chart symbols indicates a basic, non-specified, navigation aid? 1·Which of the aeronautical chart symbols indicates a DME? 4·Which of the aeronautical chart symbols indicates a TACAN? 6·Which of the aeronautical chart symbols indicates a VOR? 3·Which of the aeronautical chart symbols indicates a VOR/DME? 5·Which of the aeronautical chart symbols indicates a VORTAC? 7·Which of the aeronautical chart symbols indicates an NDB? 2

·Which of the following alternatives is correct when you cross the international date line? THE DATEWILL INCREASE IF YOU ARE CROSSING ON A WESTERLY HEADING·Which of the following correctly lists the order of available selections of the Mode Selector switches ofan inertial reference system (IRS) mode panel? QFF-ALIGN-NAV-ATT·Which of the following is an occasion for carrying out a compass swing on a Direct Reading Compass?AFTER AN AIRCRAFT HAS PASSED THROUGH A SEVERE ELECTRICAL STORM OR HAS BEENSTRUCK BY LIGHTNING·Which of the following lists all the aeronautical chart symbols shown at position N5318.1 W00856.5?CIVIL AIRPORT: NDB: DME:NON-COMPOLSURY REPORTING POINT·Which of the following lists all the aeronautical chart symbols shown at position N5318.0 W00626.9?MILITARY AIRPORT:VOR:DME·Which of the following lists all the aeronautical chart symbols shown at position N5211 W00705?CIVIL AIRPORT: NDB·Which of the following lists all the aeronautical chart symbols shown at position N5416.7 W00836.0?CIVIL AIRPORT: NDB: DME: COMPOLSURY REPORTING POINT·Which of the following lists all the aeronautical chart symbols shown at position N5150.4 W00829.7?CIVIL AIRPORT: VOR: DME: COMPULSORY REPORTING POINT·Which of the following lists the first three pages of the FMC/CDU normally used on initial start-up ofthe B737-400 Electronic Flight Instrument System? IDENT-POST INIT-RTE·Which of the following lists the order of available selections on the Mode Selector switches of a 737-400 Inertial Reference System? OFF-ALIGN-NAV-ATT·Which of the following lists, which compares an Inertial Reference System that utilises Ring LaserGyroscopes (RLG) instead of conventional gyroscopes, is completely correct? THERE IS LITTLE ORNO SPIN UP TIME AND IT IS INSENSITIVE TO GRAVITIONAL (G) FORCES·Which of the following statement is correct concerning gyro-compassing of an inertial navigationsystem (INS)? GYRO-COMPASSING OF INS IS NOT POSSIBLE IN FLIGHT BECAUSE IT CANNOTDIFERENCIATE BETWEEN MOVEMENT INDUCED AND MISALGMENT INDUCEDACCELERATIONS·Which of the following statements concerning the aircraft positions indicated on a triple fit InertialNavigation System (INS)/ Inertial Reference System (IRS) on the CDU is correct? THE POSITIONSARE IKELY TO DIFFER VECAUSE THEY ARE CALCULATED FROM DIFFERENT SOURCES·Which of the following statements concerning the alignment procedure for Inertial NavigationSystems(INS)/Inertial Reference Systems (IRS) at mid-latitudes is correct? INS/IRS CAN BE ALIGNEDIN EITHER THE ALIGN OR NAV MODE·Which of the following statements concerning the loss of alignment by an Inertial Reference System(IRS) in flight is correct? THE NAVIGATION MODE, INCLUDING PRESENT POSITION ANDGROUND SPEED OUTPUTS, ITS INOPERATIVE FOR THE REMAINDER OF THE FLIGHT·Which of the following statements concerning the operation of an Inertial Navigation System(INS)/Inertial Reference System (IRS) is correct? NAV MODE MUST BE SELECTED PRIOR TOMOVEMENT OF THE AIRCRAFT OF THE GATE·Which of the following statements concerning the position indicated on the Inertial Reference System(IRS) display is correct? IS NOR UPDATED ONCE THE IRS MODE IS SET TO NAV·Which of the following statements is correct concerning the effect of turning errors on a directreading compass? TURNING ERRORS ARE GREATEST ON NORTH/SOUTH HEADINGS, AND AREGREATEST AT HIGH LATITUDES·Which of the following variables affect deviation?1. magnetic latitude2. aircraft heading3. aircraftaltitude4. aircraft electronic equipment: 1, 2 AND 4·Which one of the following describes the appearance of rhumb lines, except meridians, on a PolarStereographic chart? CURVES CONCAVE TO THE POLE·Which one of the following statements is correct concerning the appearance of great circles, with theexception of meridians, on a Polar Stereographic chart whose tangency is at the pole ? THE HIGUER

THE LATITUDE THE CLOSER THEY APPROXIMATE TO A STRAIGHT LINE·Which one of the following, concerning great circles on a Direct Mercator chart, is correct? WITH THEEXEPTION OF MERIDIANS AND THE EQUATOR, THEY ARE CURVES CONCAVE TO THEEQUATOR·Which statement about meridians is correct?; A MERIDIAN AND ITS ANTI-MERIDIAN FORM ACOMPLETE GREAT CIRCLE·Which statement about ST is true? STANDARD TIME IS DETERMINED BY THE GOVERNMENT OFTHE APPROPIATE STATE AND DOES NOT NECESSARALY FOLLOW THE BORDERS OF 15°WIDE LONGITUDE ZONES·Which statement about ST is true?; STANDARD TIME IS DETERMINED BY THE GOVERNMENT OFTHE AEROPLANE STATE AND DOES NOT NECESARILY FOLLOW THE BORDERS OF 15° WIDELONGITUDE ZONES·Which statement about the duration of daylight is true? CLOSE TO THE EQUINOXES THEINFLUENCE OF LATITUDE ON THE DURATION OF DAYLIGHT IS AT ITS SMALLEST·Which statement about the orbit of the earth is correct? THE ORBIT OF THE EARTH AROUND THESUN IS AN ELLIPSE WITH THE SUN AT ONE OF THE FOCI·Which statement is correct about the apparent solar day? THE APPARENT SOLAR DAY IS THEPERIOD BETWEEN TWO SUCCESIVE TRANSITS OF THE TRUE SUN THROUGH THE SAMEMERIDIAN·Which statement is correct about the scale of a Lambert projection? THE SCALE REACHES ITSMINIMUM VALUE AT THE PARALLEL OF ORIGIN·Which statement is correct about the scale of a Polar Stereographic projection of the Northern polararea? THE SCALE REACHES ITS MINIMUM VALUE AT THE NORTH POLE·Which statement is correct? THE EARTH IS ONE OF THE PLANETS WHICH ARE ALL MOVING INELLIPTICAL ORBIT AROUND THE SUN·Which statement is true about the parallel of origin of a conformal chart? THE PARALLEL OFORIGIN IS THE PARALLEL AT WHICH THE SCALE REACHES ITS MINIMIM VALUE·Which statement is true? THE DECLINATION OF THE SUN AND THE LATITUDE OF THEOBSERVER WILL AFFECT THE DURATION OF THE CIVIL TWILIGHT·Which statement is true? THE DECLINATION OF THE SUN AND THE LATITUDE OF THEOBSERVER WILL AFFECT THE DURATION OF CIVIL TWILIGHT·Which statement regarding the apparent sun and the mean sun is correct? THE APARENT SUN ISTHE VISIBLE SUN, THE MEAN SUN IS A FICTITIOUS SUN·With an increase in magnetic latitude there will be a decrease in the: DIRECTIVE FORCE·With reference to an inertial navigation system (INS), the initial great circle track between computerinserted waypoints will be displayed when the control display unit (CDU) is selected to: DSRTK/STS·With reference to inertial navigation systems, a TAS input is: REQUIRED TO PROVIDE A W/V READOUT·You are departing from an airport which has an elevation of 1500 ft. The QNH is 1003 hPa.15 NMaway there is a waypoint you are required to pass at an altitude of 7500 ft.Given a groundspeed of120 kt, what is the minimum rate of climb? 800FT/MIN·You are departing from an airport which has an elevation of 2000 ft. The QNH is 1013 hPa. 10 NMaway there is a waypoint you are required to pass at an altitude of 7500 ft. Given a groundspeed of100 kt, what is the minimum rate of climb?; 920 FT/MIN·You are tracking the 200° radial inbound to a VOR and your true heading is 010°. At the VOR youthen track the 090° radial outbound and are showing a heading of 080°M The variation is +5° and theTAS is 240 kts. What is the wind (°T) has affected the aircraft ?; 310°/65·Your on an airfield elevation 2000ft, QNH 1003. You want to climb to FL50, your rate of climb is1000ft/min, your TAS is 100 and you have a headwind of 20. What is the distance it takes to get toFL50 ? 3. 6 NM