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061 – GENERAL NAVIGATION
061-01

BASICS OF NAVIGATION
061-01-01 The Solar System

8260
...
Assuming mid-latitudes (40o to 50o N/S)
...
What is the approximate date of perihelion, when the Earth is nearest to the
Sun?
A – Beginning of January
B – End of December
C – Beginning of July
D – End of March
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

8334
...
Seasons are due to the:
A – Earth’s elliptical orbit around the Sun
B – inclination of the polar axis with the ecliptic plane
C – Earth’s rotation on its polar axis
D – variable distance between Earth and Sun
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

061-01-02 The Earth
8261
...

What is the difference between the great circle track at A and B?
A – it increases by 6o
B – it decreases by 6o
C – it increases by 3o
D – it decreases by 3o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8264
...
5 deg
B – 23
...
3 deg
D – 65
...
Given:
Value for the ellipticity of the Earth is 1/297
...
4 km
...
9
B – 6 378
...
0
D – 6 399
...
At what approximate latitude is the length of one minute of arc along a
meridian equal to one NM (1852 m) correct?
A – 45o
B – 0o
C – 90o
D – 30o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8312
...
What is the UTC time of sunrise in Vancouver, British Columbia, Canada (49N
123 30W) on the 6th December?
A – 2324 UTC
B – 0724 UTC
C – 1552 UTC
D – 0738 UTC
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

8316
...
In order to fly from position A (10o00N, 030o00W) to position B (30o00N),
050o00W), maintaining a constant true course, it is necessary to fly:
A – the great-circle route
B – the constant average drift route
C – a rhumb line track
D – a straight line plotted on a Lambert chart
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8332
...
At what approximate date is the earth closest to the sun (perihelion)?
A – End of June
B – End of March
C – Beginning of July
D – Beginning of January
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

9754
...
What is a line of equal magnetic variation?
A – An isocline
B – An isogonal
C – An isogriv
D – An isovar
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
9778
...
Parallels of latitude, except the equator are:
A – both Rhumb lines and Great circles
B – Great circles
C – Rhumb lines
D – are neither Rhumb lines nor Great circles
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

9818
...
5o
B – 25
...
5o
D – 66
...
Given:
The coordinates of the heliport at Issy les Moulineaux are:
N48o50 E002o16
...
5
B – S48o50 E177o43
...
5
D – S41o10 E177o43
...
An aircraft at latitude 02o20N tracks 180o(T) for 685 km
...
An aircraft departing A(N40o 00’E080o00’) flies a constant true track of 270o
at a ground speed of 120 kt
...
If an aeroplane was to circle around the Earth following parallel 60oN at a
ground speed of 480 kt
...
The angle between the true great-circle track and the true rhumb-line track
joining the following points: A (60oS 165oW) B (60oS 177oE), at the place of
departure A, is:
A – 7
...
6o
D – 5
...
An aircraft flies the following rhumb line tracks and distances from position
04o00N 030o00W: 600 NM South, then 600 NM East, then 600 NM North,
then 600 NM West
...
Which of the following statements concerning the earth’s magnetic field is
completely correct?
A – Dip is the angle between total magnetic field and vertical field component
B – The blue pole of the earth’s magnetic field is situated in North Canada
C – At the earth’s magnetic equator, the inclination varies depending on
whether the geographic equator is north or south of the magnetic equator
D – The earth’s magnetic field can be classified as transient semi-permanent
or permanent
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
16272
...
5o (T)
B – 80
...
You are flying from A (50n 10W) to B (58N 02E)
...
5o
B – 9
...
2o
D – 6
...
Radio bearings:
A – are Rhumb lines
B – cut all meridians at the same angle
C – are Great circles
D – are lines of fixed direction
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

16290
...
The earth may be referred to as:
A – round
B – an oblate spheroid
C – a globe
D – elliptical
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
16317
...
A line which cuts all meridians at the same angle is called a:
A – Line of variation
B – Great circle
C – Rhumb line
D – Agonic line
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

16319
...
The shortest distance between 2 point of the surface of the earth is:
A – a great circle
B – the arc of a great circle
C – half the rhumb line distance
D – Rhumb line
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
16321
...
5 convergency
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
16322
...
The Earth is:
A – A sphere which has a larger polar circumference than equatorial
circumference
B – A sphere whose centre is equidistant (the same distance) from the Poles
and the Equator
C – Considered to be a perfect sphere as far as navigation is concerned
D – None of the above statements is correct
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
25187
...
(Refer to figure 061-14)
When it is 1000 Standard Time in Kuwait, the Standard time in Algeria :
A – 0700
B – 1200
C – 1300
D – 0800
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

8272
...
(Refer to figures 061-13 and 061-15)
An aircraft takes off from Guam at 2300 Standard Time on 30 April local date
...
What is
the Standard Time and local date of arrival (assume summer time rules apply)?
A – 1715 on 30 April
B – 1215 on 1 May
C – 1315 on 1 May
D – 1615 on 30 April
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8693
...
In which months is the difference between apparent noon and mean noon the
greatest?
A – November and February
B – January and July
C – March and September
D – June and December
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
9753
...
Which is the highest latitude listed below at which the sun will rise above the
horizon and set every day?
A – 62o
B – 68o
C – 72o
D – 66o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
9774
...
On
the same day, at 52oS and 035oW, the sunrise is at:
A – 2143 UTC
B – 0243 UTC
C – 0743 UTC
D – 0523 UTC
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

9785
...
What is the local mean time, position 65o25N 123o45W at 2200 UTC?
A – 1345
B – 2200
C – 0615
D – 0815
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
10946
...
The Local Mean Time at longitude 095o20W at 0000 UTC, is:
A – 1738:40 same day
B – 0621:20 same day
C – 1738:40 previous day
D – 0621:20 previous day
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

15423
...
Civil twilight is defined by:
A – sun altitude is 12o below the celestial horizon
B – sun altitude is 18o below the celestial horizon
C – sun upper edge tangential to horizon
D – sun altitude is 6o below the celestial horizon
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
21450
...
(Refer to figure 061-04)
Given:
TAS is 120 kt
ATA ‘X’ 1232 UTC
ETA ‘Y’ 1247 UTC
ATA ‘Y’ is 1250 UTC
What is ETA ‘Z’?
A – 1257 UTC
B – 1302 UTC
C – 1300 UTC
D – 1303 UTC
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
24028
...
Morning Civil twilight begins when:
A – the sun’s upper edge is tangential to the celestial horizon
B – the centre of the sun is 12o below the celestial horizon
C – the centre of the sun is 18o below the celestial horizon
D – the centre of the sun is 6o below the celestial horizon
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

24058
...
When the time is 2000 UTC, it is:
A – 1400 LMT at 90o West
B – 2400 LMT at 120o West
C – 1200 LMT at 60o East
D – 0800 LMT at the Prime meridian
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
24061
...
On 27 Feb, at S5210
...
0, the sunrise is at 0230 UTC
...
0 W03500
...
The UTC of the end of Evening Civil Twilight in position N51000’ W008000’
on 15 August is:
A – 1928 UTC
B – 1944 UTC
C – 2000 UTC
D – 2032 UTC
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
25192
...
The months in which the difference between apparent noon and mean noon is
greatest are:
A – February and November
B – January and July
C – March and September
D – June and December
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

25269
...
If it is 0700 hours Standard Time in Kuwait, what is the Standard Time in
Algeria?
A – 0500
B – 0900
C – 1200
D – 0300
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

061-01-04 Distances
8289
...
The north and south magnetic poles are the only positions on the earth’s
surface where:
A – a freely suspended compass needle will stand horizontal
B – isogonals converge
C – a freely suspended compass needle will stand vertical
D – the value of magnetic variation equals 90o
Ans: C

15426
...
A great circle on the Earth running from the North Pole to the South Pole is
called:
A – a longitude
B – a parallel of latitude
C – a difference of longitude
D – a meridian
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
24013
...
The track followed is a:
A – constant-heading track
B – rhumb line
C – great circle
D – constant-drift track
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
24021
...
How many small circles can be drawn between any two points on a sphere?
A – One
B – None
C – An unlimited number
D – Two
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
24027
...
In which occasions does the rhumb line track and the great circle track
coincide on the surface of the Earth?
A – On East-West tracks in polar areas
B – On high latitude tracks directly East-West
C – On East-West tracks in the northern hemisphere north of the magnetic
equator
D – On tracks directly North-South and on East-West tracks along the Equator
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
24057
...
How many feet are there in 1 sm?
A – 3
...
280 ft
C – 6
...
000 ft
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
16288
...
280 ft
B – 5
...
080 ft
D – 1
...
How many feet are there in a km?
A – 3
...
280 ft
C – 6
...
000 ft
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
16291
...
25 inches?
A – 92
...
014 m
C – 14
...
05 cm
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

16292
...
5 km?
A – 31
...
160 ft
C – 57
...
500 ft
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
16293
...

A – 1
...
652m
C – 1
...
962m
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
24005
...
The distance along a meridian between 63o55’N and 13o47’S is:
A – 3008 NM
B – 7702 NM
C – 5008 NM
D – 4662 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

24055
...
What is the rhumb line distance, in nautical miles, between two positions on
latitude 60oN, that are separated by 10o of longitude?
A – 300 NM
B – 520 NM
C – 600 NM
D – 866 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

061-02

MAGNETISM AND COMPASSES
061-02-01 General Principles

8325
...
What is the value of magnetic dip at the South Magnetic Pole?
A – 360o
B – 180o
C – 090o
D – 0o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8348
...
Isogonic lines connect positions that have:
A – the same variation
B – 0o variation
C – the same elevation
D – the same angle of magnetic dip
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

8354
...
What is the definition of magnetic variation?
A – The angle between the direction indicated by a compass and Magnetic
North
B – The angle between True North and Compass North
C – The angle between Magnetic North and True North
D – The angle between Magnetic Heading and Magnetic North
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8358
...
Isogonals converge at the:
A – Magnetic equator
B – North and South geographic and magnetic poles
C – North magnetic pole only
D – North and South magnetic poles only
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

8375
...
Complete the following statement regarding magnetic variation
...
Which of these is a correct statement about the Earth’s magnetic field?
A – It acts as though there is a large blue magnetic pole in Northern Canada
B – The angle of dip is the angle between the vertical and the total magnetic
force
C – It may be temporary, transient, or permanent
D – It has no effect on aircraft deviation
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8389
...
Isogonals are lines of equal:
A – compass deviation
B – magnetic variation
C – pressure
D – wind velocity
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8391
...
An aircraft is over position HO (55o30N 060o15W), where YYR VOR (53o30N
060o15W) can be received
...
What is the radial from YYR?
A – 031o
B – 208o
C – 028o
D – 332o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8397
...
The angle between True North and Magnetic North is called:
A – compass error
B – deviation
C – variation
D – drift
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8408
...
This is due to:
A – movement of the magnetic poles, causing an increase
B – increase in the magnetic field, causing an increase
C – reduction in the magnetic field, causing a decrease
D – movement of the magnetic poles, which can cause either an increase or a
decrease
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
8414
...
The agonic line:
A – is midway between the magnetic North and South poles
B – follows the geographic equator
C – is the shorter distance between the respective True and Magnetic North
and South poles
D – Follows separate paths out of the North polar regions, one currently
running through Western Europe and the other through theUSA
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
8427
...
Which of the following statements concerning earth magnetism is completely
correct?
A – An isogonal is a line which connects places with the same magnetic
variation; the agonic line is the line of zero magnetic dip
B – An isogonal is a line which connects places with the same magnetic
variation; the aclinic is the line of zero magnetic dip
C – An isogonal is a line which connects places of equal dip; the aclinic is the
line of zero magnetic dip
D – An isogonal is a line which connects places with the same magnetic
variation; the aclinic connects places with the same magnetic field
strength
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

9740
...
The Earth can be considered as being a magnet with the:
A – blue pole near the north pole of the earth and the direction of the magnetic
force pointing straight up from the earth’s surface
B – red pole near the north pole of the earth and the direction of the magnetic
force pointing straight down to the earth’s surface
C – blue pole near the north pole of the earth and the direction of the magnetic
force pointing straight down to the earth’s surface
D – red pole near the north pole of the earth and the direction of the magnetic
force pointing straight up from the earth’s surface
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
9771
...
At the magnetic equator:
A – dip is zero
B – variation is zero
C – deviation is zero
D – the isogonal is an agonic line
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

9783
...
Where is a compass most effective?
A – About midway between the earth’s magnetic poles
B – In the region of the magnetic South pole
C – In the region of the magnetic North pole
D – On the geographic equator
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
9819
...
When accelerating on a westerly heading in the northern hemisphere, the
compass card of a direct reading magnetic compass will turn:
A – clockwise giving an apparent turn towards the north
B – clockwise giving an apparent turn towards the south
C – anti-clockwise giving an apparent turn towards the north
D – anti-clockwise giving an apparent turn towards the south
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

16296
...
When a magnetized compass needle is freely suspended in the Earth’s
magnetic field, and affected by extraneous magnetic influence, it will align
itself with:
A – true North
B – magnetic North
C – compass North
D – relative North
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
16299
...
When the Magnetic Pole is West of the True North pole variation is:
A – + and easterly
B – - and easterly
C – - and westerly
D – + and westerly
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

16301
...
The agonic line is:
A – a line of zero magnetic deviation
B – a line of equal magnetic deviation
C – a line of zero magnetic variation
D – a line of equal magnetic variation
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
16304
...
Deviation is:
A – an error to be added to magnetic headings
B – a correction to be added to magnetic heading to obtain compass heading
C – a correction to be added to compass heading to obtain magnetic heading
D – an error to be added to compass heading to obtain magnetic heading
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

24043
...
The horizontal component of the earth’s magnetic field:
A – weakens with increasing distance from the nearer magnetic pole
B – weakens with increasing distance from the magnetic poles
C – is stronger closer to the magnetic equator
D – is approximately the same at all magnetic latitudes less than 60o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
25196
...
An aircraft is accelerating on a westerly heading in the Northern Hemisphere;
the effect on a Direct Reading Compass will result in:
A – An apparent turn to the West
B – An indication of a turn to the North
C – A decrease in the indicated reading
D – An indication of a turn to the South
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

25198
...
An aircraft, in the Northern Hemisphere, turns right from 330(C) in a Rate 1
Turn for 30 secs
...
An aircraft is accelerating on a westerly heading in the Northern Hemisphere
...
What is the maximum possible value of Dip Angle?
A – 66o
B – 180o
C – 90o
D – 45o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

061-02-02 Aircraft Magnetism
8339
...
5
Drift = 10R
What is Heading (C)?
A – 078 C
B – 346 C
C – 358 C
D – 025 C
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8341
...
When an aircraft on a westerly heading on the northern hemisphere accelerates,
the effect of the acceleration error causes the magnetic compass to:
A – lag behind the turning rate of the aircraft
B – indicate a turn towards the north
C – indicate a turn towards the south
D – to turn faster than the actual turning rate of the aircraft
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

8373
...
Concerning direct reading magnetic compasses, in the northern hemisphere, it
can be said that:
A – on an Easterly heading, a longitudinal acceleration causes an apparent turn
to the South
B – on an Easterly heading, a longitudinal acceleration causes an apparent turn
to the North
C – on a Westerly heading, a longitudinal acceleration causes an apparent turn
to the South
D – on a Westerly heading, a longitudinal deceleration causes an apparent turn
to the North
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8381
...
You are in the Northern hemisphere, heading 135C on a Direct Reading
Magnetic Compass
...
Do you roll
out on an indicated heading of:
A – greater than 225
B – less than 225
C – equal to 225
D – not possible to determine
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8389
...
Compass deviation is defined as the angle between:
A – True North and Magnetic North
B – Magnetic North and Compass North
C – True North and Compass North
D – The horizontal and the total intensity of the earth’s magnetic field
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8401
...
Deviation applied to magnetic heading gives:
A – magnetic course
B – true heading
C – compass heading
D – magnetic track
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8411
...
An aircraft in the northern hemisphere makes an accurate rate one turn to the
right/starboard
...
When accelerating on an easterly heading in the Northern hemisphere, the
compass card of a direct reading magnetic compass will turn:
A – anti-clockwise giving an apparent turn toward the south
B – clockwise giving an apparent turn toward the south
C – anti-clockwise giving an apparent turn toward the north
D – clockwise giving an apparent turn toward the north
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

8423
...

You stop the turn at the correct time
...
Which of the following statements is correct concerning the effect of turning
errors on a direct reading compass?
A – Turning errors are greatest on north/south headings, and are least at high
latitudes
B – Turning errors are greatest on east/west headings, and are least at high
latitudes
C – Turning errors are greatest on north/south headings, and are greatest at
high latitudes
D – Turning errors are greatest on east/west headings, and are greatest at high
latitudes
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
9767
...
One purpose of a compass calibration is to reduce the difference, if any,
between:
A – compass north and magnetic north
B – compass north and true north
C – true north and magnetic north
D – compass north and the lubber line
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

061-02-03 Principles; Direct & Remote Reading Compasses
8343
...
The main advantage of a remote indicating compass over a direct reading
compass is that it:
A – is able to magnify the earth’s magnetic field in order to attain greater
accuracy
B – has less moving parts
C – requires less maintenance
D – senses, rather than seeks, the magnetic meridian
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
8352
...
Which of the following is an occasion for carrying out a compass swing on a
Direct Reading Compass?
A – After an aircraft has passed through a severe electrical storm, or has been
struck by lightning
B – Before an aircraft goes on any flight that involves a large change of
magnetic latitude
C – After any of the aircraft radio equipment has been changed due to
unserviceability
D – Whenever an aircraft carries a large freight load regardless of its content
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8366
...
A direct reading compass should be swung when:
A – there is a large, and permanent, change in magnetic latitude
B – there is a large change in magnetic longitude
C – the aircraft is stored for a long period and is frequently moved
D – the aircraft has made more than a stated number of landings
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

8372
...
The main reason for usually mounting the detector unit of a remote indicating
compass in the wingtip of an aeroplane is to:
A – facilitate easy maintenance of the unit and increase its exposure to the
Earth’s magnetic field
B – reduce the amount of deviation caused by aircraft magnetism and
electrical circuits
C – place it is a position where there is no electrical wiring to cause deviation
errors
D – place it where it will not be subjected to electrical or magnetic
interference from the aircraft
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8405
...
Which one of the following is an advantage of a remote reading compass as
compared with a standby compass?
A – It senses the magnetic meridian instead of seeking it, increasing compass
sensitivity
B – It is lighter than a direct reading compass because it employs, apart from
the detector unit, existing aircraft equipment
C – it eliminates the effect of turning and acceleration errors by pendulously
suspending the detector unit
D – It is more reliable because it is operated electrically and power is always
available from sources within the aircraft
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8460
...
The sensitivity of a direct reading magnetic compass is:
A – inversely proportional to the horizontal component of the earth’s magnetic
field
B – proportional to the horizontal component of the earth’s magnetic field
C – inversely proportional to the vertical component of the earth’s magnetic
field
D – inversely proportional to the vertical and horizontal components of the
earth’s magnetic field
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

9805
...
The main reason for mounting the detector unit of a remote reading compass
in the wingtip of an aeroplane is:
A – to ensure that the unit is in the most accessible position on the aircraft for
ease of maintenance
B – by having detector units on both wingtips, to cancel out the deviation
effects caused by the aircraft structure
C – to minimise the amount of deviation caused by aircraft magnetism and
electrical circuits
D – to maximise the units exposure to the earth’s magnetic field
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
15452
...
If compass HDG is 340o and deviation +3, what is magnetic heading?
A – Deviation is plus therefore East, so compass is least, so magnetic is 343o
B – Deviation is plus therefore West, so compass is least, so magnetic is 343o
C – Deviation is plus therefore East, so compass is best, so magnetic is 337o
D – Deviation is plus therefore East, so compass is best, so magnetic is 343o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

16308
...
In still air, you wish to fly a true of 315o
...
Deviation is 2oE
...
Magnetic compass calibration is carried out to reduce:
A – deviation
B – variation
C – parallax error
D – acceleration errors
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
25132
...
Will a direct reading magnetic compass over-read or under-read
and is the compass indicating a turn to the north or to the south:
A – over-reads north
B – over- reads south
C – under-reads north
D – under-reads south
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

25199
...
The standard parallels of a Lamberts conical orthomorphic projection are
07o40N and 38o20N
...
60
B – 0
...
92
D – 0
...
On a transverse Mercator chart, the scale is exactly correct along the:
A – prime meridian and the equator
B – equator and parallel of origin
C – meridian of tangency and the parallel of latitude perpendicular to it
D – meridians of tangency
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
8439
...
An Oblique Mercator projection is used specifically to produce:
A – plotting charts in equatorial regions
B – radio navigational charts in equatorial regions
C – topographical maps of large east/west extent
D – charts of the great circle route between two points
Ref: AIR: atpl, cpl; HELI: atpl, cpl

Ans: D
8461
...
Scale on a Lamberts conformal chart is:
A – constant along a parallel of latitude
B – constant along a meridian of longitude
C – constant over the whole chart
D – varies with latitude and longitude
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8469
...
The two standard parallels of a conical Lambert projection are at N10o40 and
N41o20
...
18
B – 0
...
66
D – 0
...
The constant of the cone, on a Lambert chart where the convergence angle
between longitudes 010oE and 030oW is 30o, is:
A – 0
...
75
C – 0
...
64
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8502
...
A Mercator chart has a scale at the equator = 1:3 704 000
...
A Lambert conformal conic projection, with two standard parallels:
A – shows lines of longitude as parallel straight lines
B – shows all great circles as straight lines
C – the scale is only correct at parallel of origin
D – the scale is only correct along the standard parallels
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

14651
...
78535
...
The nominal scale of a Lambert conformal conic chart is the:
A – scale at the equator
B – scale at the standard parallels
C – mean scale between pole and equator
D – mean scale between the parallels of the secant cone
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
14669
...
3955
...
On a direct Mercator projection, the distance measured between two meridians
spaced 5o apart at latitude 60oN is 8 cm
...
At 60o N the scale of a direct Mercator chart is 1:
A – 1 : 3 000 000
B – 1 : 3 500 000
C – 1 : 1 500 000
D – 1 : 6 000 000
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
15440
...
A direct Mercator graticule is based on a projection that is:
A – spherical
B – concentric
C – cylindrical
D – conical
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
15459
...
866
B – 0
...
0
D – 1
...
The Earth has been charted using:
A – WGP84
B – WGS84
C – GD84
D – GPS84
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
24007
...
The angular difference between the initial
true track and the final true track of the line is equal to:
A – earth convergency
B – chart convergency
C – conversion angle
D – difference in longitude
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
24022
...
How does the scale vary in a Direct Mercator chart?
A – The scale increases with increasing distance from the Equator
B – The scale decreases with increasing distance from the Equator
C – The scale is constant
D – The scale increases south of the Equator and decreases north of the
Equator
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

24037
...
63
cm
...
What is the constant of the cone for a Lambert conic projection whose
standard parallels are at 50oN and 70oN?
A – 0
...
941
C – 0
...
766
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
25216
...
On a Lambert conformal conic chart the convergence of the meridians:
A – is the same as earth convergency at the parallel of origin
B – is zero throughout the chart
C – varies as the secant of the latitude
D – equals earth convergency at the standard parallels
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

8455
...
On a Direct Mercator chart, meridians are:
A – inclined, equally spaced, straight lines that meet at the nearer pole
B – parallel, equally spaced, vertical straight lines
C – parallel, unequally spaced, vertical straight lines
D – inclined, unequally spaced, curved lines that meet at the nearer pole
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8476
...
On a Direct Mercator chart at latitude 15oS, a certain length represents a
distance of 120 NM on the earth
...
3 NM
B – 117
...
2 NM
D – 118
...
On a Direct Mercator chart at latitude of 45oN, a certain length represents a
distance of 90 NM on the earth
...
5 NM
C – 78 NM
D – 110 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
8511
...
On a Lambert Conformal Conic chart great circles that are not meridians are:
A – curves concave to the parallel of origin
B – straight lines
C – curves concave to the pole of projection
D – straight lines within the standard parallels
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8521
...
On a Direct Mercator chart, great circles are shown as:
A – curves convex to the nearer pole
B – straight lines
C – rhumb lines
D – curves concave to the nearer pole
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
9810
...
Which one of the following, concerning great circles on a Direct Mercator
chart, is correct?
A – They are all curves convex to the equator
B – They are all curves concave to the equator
C – They approximate to straight lines between the standard parallels
D – With the exception of meridians and the equator, they are curves concave
to the equator
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
10970
...
Which one of the following describes the appearance of rhumb lines, except
meridians, on a Polar Stereographic chart?
A – Straight lines
B – Ellipses around the Pole
C – Curves convex to the Pole
D – Curves concave to the Pole
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
10999
...
On a Lambert chart (standard parallels 37oN and 65oN), with respct to the
straight line drawn on the map the between A (N49o W030o) and B (N48o
W040o), the:
A – great circle is to the north, the rhumb line is to the south
B – great circle and rhumb line are to the north
C – great circle and rhumb line are to the south
D – rhumb line is to the north, the great circle is to the south
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
11013
...
On a Direct Mercator, rhumb lines are:
A – straight lines
B – curves concave to the equator
C – ellipses
D – curves convex to the equator
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
11020
...
On a Lambert conformal conic chart, with two standard parallels, the quoted
scale is correct:
A – along the prime meridian
B – along the two standard parallels
C – in the area between the standard parallels
D – along the parallel of origin
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
15419
...
The scale on a Lambert conformal conic chart:
A – is constant along a meridian of longitude
B – is constant across the whole map
C – varies slightly as a function of latitude and longitude
D – is constant along a parallel of latitude
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
15458
...
What is the Rhumb line (RL) direction from 45oN 14o12W to 45oN 12o48E?
A – 270o (T)
B – 090o (T)
C – 090o (M)
D – 270o (M)
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
24006
...
Where on a Direct Mercator projection is the chart convergency correct
compared to the earth convergency?
A – All over the chart
B – At the two parallels of tangency
C – At the poles
D – At the equator
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
25153
...
0 E00213
...
0 W 00713
...
An aircraft starts at position 0411
...
2W and heads True North for
2950nm, then turns 90o left maintaining a rhumb line track for 314 km
...
0N 17412
...
0N 17412
...
0N 17713
...
0N 17713
...
The appearance of a rhumb line on a Mercator chart is:
A – A small circle concave to the nearer pole
B – A straight line
C – A spiral curve
D – A curved line
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

25204
...
The scale quoted on a Lamberts chart is:
A – The scale at the Standard Parallels
B – The scale at the Equator
C – The mean scale between the Pole and the Equator
D – The mean scale at the Parallel of the Secant of the Cone
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
25212
...
On a Transverse Mercator chart scale is correct at:
A – The 180o meridian
B – The False Meridian
C – The Great Circle of Tangency
D – The Meridian of Tangency
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

25215
...
0N on a Polar Stereographic chart
...
0W, B is at 6000
...
Which of the following differences in latitude will give the biggest difference
in the initial Great Circle track and the mean Great Circle track between two
points separated by 10o change of longitude?
A – 60N and 60S
B – 60N and 55N
C – 30S and 30N
D – 30S and 25S
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
25299
...
On a Polar Stereographic map, a straight line is drawn from position A (70N
102W) to position B (80N 006E)
...
What is the initial straight-line track angle from A
to B, measured at A?
A – 049
B – 077
C – 229
D – 023
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

25575
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
Given:
SHA VOR N5243
...
1
CON VOR N5354
...
1
Aircraft position N5330 W00800
...
(Refer to Jeppesen Student Manual – chart E(lO)1 or figure 061-11)
Given:
SHA VOR N5243
...
1
CRK VOR N5150
...
7
Aircraft position N5220 W00910
Which of the following lists two radials that are applicable to the aircraft
position:
A – SHA 025o CRK 141o
B – SHA 212o CRK 328o
C – SHA 205o CRK 321o
D – SHA 033o CRK 149o
Ref: all
Ans: B

8430
...
At 0035 UTC the radial is 040o and DME distance is 40 NM
...
The true track and ground speed are:
A – 080o – 226 kt
B – 090o – 232 kt
C – 085o – 226 kt
D – 088o – 232 kt
Ref: all
Ans: C
8431
...
3 W00853
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the radial and DME distance from CON VOR/DME (N5354
...
1) to position N5400 W00800?
A – 320o – 8 NM
B – 088o – 29 NM
C – 094o – 64 NM
D – 260o – 30 NM
Ref: all
Ans: B

8434
...
The scale of the chart at that parallel approximates:
A – 1 : 3 750 000
B – 1 : 5 000 000
C – 1 : 2 000 000
D – 1 : 6 000 000
Ref: all
Ans: B
8436
...
What is the initial track
direction (going eastwards) of the line at A?
A – 090 T
B – 030 T
C – 120 T
D – 330 T
Ref: all
Ans: B
8438
...
7
W00613
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
Which of the following lists all the aeronautical chart symbols shown at
position N5318
...
9?
A – VOR: DME: danger area
B – Civil airport: VOR: DME
C – Military airport: VOR: NDB
D – Military airport: VOR: DME
Ref: all
Ans: D
8441
...
4 W00829
...
9 W00931
...
On a Mercator chart, at latitude 60oN, the distance measured between W002o
and E008ois 20 cm
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
Given:
SHA VOR/DME (N5243
...
1)
Birr aerodrome (N5304 W00754)
What is the SHA radial and DME distance when overhead Birr aerodrome?
A – 068o – 41 NM
B – 248o – 42 NM
C – 060o – 42 NM
D – 240o – 41 NM
Ref: all
Ans: A
8444
...
An aircraft starts at position 0410S 17822W and heads true north for 2950 nm,
then turns 90 degrees left, and maintains a rhumb line track for 314 kilometers
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the average track (oT) and distance between CRN NDB (N5318
...
5) and EKN NDB (N5423
...
7)?
A – 044o – 82 NM
B – 042o – 83 NM
C – 036o – 81 NM
D – 035o – 80 NM
Ref: all
Ans: D
8448
...
Given that:
A is N55 E/W 000
B is N54 E 010
If the true great circle track from A to B is 100T, what is the true Rhumb Line
track at A?
A – 096
B – 107
C – 104
D – 100
Ref: all
Ans: C

8451
...
On a Polar Stereographic chart, the initial great circle course from A 70oN
060oW to B 70oN 060oE is approximately:
A – 030o (T)
B – 330o (T)
C – 150o (T)
D – 210o (T)
Ref: all
Ans: A
8454
...
3 W00853
...
4 W00829
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the average track (oT) and distance between BAL VOR (N5318
...
9) and CFN NDB (N5502
...
4)?
A – 335o – 128 NM
B – 327o – 124 NM
C – 325o – 126 NM
D – 320o – 127 NM
Ref: all
Ans: B
8458
...
1 W00856
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
Given:
SHA VOR N5243
...
1
CON VOR N5354
...
1
Aircraft position N5320 W00950
Which of the following lists two radials that are applicable to the aircraft
position?
A – SHA 325o CON 235o
B – SHA 137o CON 046o
C – SHA 317o CON 226o
D – SHA 145o CON 055o
Ref: all
Ans: A

8462
...
3 W00853
...
4 W00829
...
Given:
Chart scale is 1: 850 000
The chart distance between two points is 4 centimetres
Earth distance is approximately:
A – 4 NM
B – 74 NM
C – 100 NM
D – 40 NM
Ref: all
Ans: D
8467
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the average track (oM) and distance between CRN NB (N5318
...
5) and BEL VOR (N5439
...
8)?
A – 229o – 125 NM
B – 089o – 95 NM
C – 057o – 126 NM
D – 237o – 130 NM
Ref: all
Ans: C
8470
...
9
W00931
...
1 W00856
...
On a direct Mercator projection, at latitude 45o North, a certain length
represents 70 NM
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
Given:
SHA VOR/DME (N5243
...
1)
Radial 165o/36 NM
What is the aircraft position?
A – N5210 W00830
B – N5208 W00840
C – N5315 W00915
D – N5317 W00908
Ref: all
Ans: A
8473
...
A course of 120o(T) is drawn between X(61o30N) and Y(58o30N) on a
Lambert Conformal conic chart with a scale of 1: 1 000 000 at 60oN
...
4 cm
B – 66
...
5 cm
D – 36
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
Given:
SHA VOR N5243
...
1
CRK VOR N5150
...
7
Aircraft position N5230 W00820
Which of the following lists two radials that are applicable to the aircraft
position?
A – SHA 131o CRK 017o
B – SHA 304o CRK 189o
C – SHA 312o CRK 197o
D – SHA 124o CRK 009o
Ref: all
Ans: A
8482
...
1 W00856
...
3 W00853
...
On a chart, the distance along a meridian between latitudes 45oN and 46oN is 6
cm
...
The following waypoints are entered into an inertial navigation system (INS)
WPT 1: 60N 30W
WPT 2: 60N 20W
WPT 3: 60N 10W
The intertial navigation is connected to the automatic pilot on the route WP1WP2-WP3
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
Given:
SHA VOR (N5243
...
1) radial 205o
CRK VOR (5150
...
7) radial 317o
What is the aircraft position?
A – N5210 W00910
B – N5118 W00913
C – N5205 W00915
D – N5215 W00917
Ref: all
Ans: A

8486
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the radial and DME distance from CRK VOR/DME (N5150
...
7) to position N5140 W00730?
A – 106o – 38 NM
B – 104o – 76 NM
C – 293o – 39 NM
D – 113o – 38 NM
Ref: all
Ans: D
8488
...
75
inches
...
(Refer to Jeppesen Student Manual – chart (E(LO)1 or figure 061-11)
Which of the following lists all the aeronautical chart symbols shown at
position N5150
...
7?
A – Civil airport: VOR: non-compulsory reporting point
B – Civil airport: VOR: DME: compulsory reporting point
C – VOR: DME: NDB: compulsory reporting point
D – VOR: DME: NDB: ILS
Ref: all
Ans: B
8492
...
H
...
An aircraft at position 6000N 00522WS flies 165 km due East
...
Two positions plotted on a polar stereographic chart, A (80oN 000o) and B
(70oN 102oW) are joined by a straight line whose highest latitude is reached at
035oW
...
Given:
An aircraft is flying a track of 255o(M)
...
At 2300 UTC, it crosses radial 330o from the same
station
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the average track (oT) and distance between WTD NDB (N5211
...
0) and FOY NDB (N5234
...
7)?
A – 075o – 81 NM
B – 294o – 80 NM
C – 286o – 81 NM
D – 277o – 83 NM
Ref: all
Ans: C

8500
...
8
W00849
...
Given:
Waypoint 1
...
60oS 020oW
What will be the approximate latitude shown on the display unit of an inertial
navigation system at longitude 025oW?
A – 060o 11’S
B – 059o 49’S
C – 060o 00’S
D – 060o 06’S
Ref: all
Ans: D
8503
...
0 centimetres
...
(refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the radial and DME distance from SHA VOA/DME (N5243
...
1) to position N5210 W00920?
A – 346o – 34 NM
B – 354o – 34 NM
C – 198o – 37 NM
D – 214o – 37 NM
Ref: all
Ans: D
8507
...
3 W00853
...
4 W00829
...
The distance measured between two points on a navigation map is 42 mm
(millimetres)
...
The actual distance
between these two points is approximately:
A – 3
...
00 NM
C – 67
...
30 NM
Ref: all
Ans: D

8513
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the radial and DME distance from CRK VOR/DME (N5150
...
7) to position N5230 W00750?
A – 039o – 48 NM
B – 024o – 43 NM
C – 023o – 48 NM
D – 017o – 43 NM
Ref: all
Ans: A
8515
...
It is tuned to a
VOR located at 5330N 03613W where the variation is 12W
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
Given:
SHA VOR (N5243
...
1) radial 120o
CRK VOR (N5150
...
7) radial 033o
What is the aircraft position?
A – N5230 W00800
B – N5225 W00805
C – N5220 W00750
D – N5240 W00750
Ref: all
Ans: A
8519
...
From A to B on the chart measures 1
...
54 centimetres), the distance from A to B in NM is:
A – 44
...
1
C – 20
...
2
Ref: all
Ans: C
8520
...
4
W00829
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the average track (oM) and distance between WTD NDB (N5211
...
0) and KER NDB (N5210
...
5)?
A – 270o – 89 NM
B – 090o – 91 NM
C – 278o – 90 NM
D – 098o – 90 NM
Ref: all
Ans: C
8525
...
63 cm and represents 150 NM
...
Route A (44oN 026oE) to B (46oN 024oE) forms an angle of 35o with
longitude 026oE
...
What
is the average magnetic course from A to B?
A – 322o
B – 328o
C – 032o
D – 038o
Ref: all
Ans: A

9751
...
3 W00853
...
4 W00829
...
5 hours 20 minutes and 20 seconds hours time difference is equivalent to which
change of longitude:
A – 81o 30
B – 78o 15
C – 79o 10
D – 80o 05
Ref: all
Ans: D
10205
...
The
total distance travelled is:
A – 2040 NM
B – 1788 NM
C – 5420 NM
D – 3720 NM
Ref: all
Ans: D

10955
...
What is its final
position?
A – 0400N 17000W
B – 0600S 17000W
C – 0400N 16958
...
8W
Ref: all
Ans: C
10957
...
80
...
What is the longitude of B?
A – 011oE
B – 009o36E
C – 008oE
D – 019oE
Ref: all
Ans: A
10959
...
The true
course on departure from position A is approximately:
A – 250o
B – 225o
C – 135o
D – 315o
Ref: all
Ans: B

10961
...
3 W000853
...
Assume a Mercator chart
...
The scale at the parallel is
1: 9 260 000
...
Given:
Magnetic heading 311o
Drift angle 10o left
Relative bearing of NDB 270o
What is the magnetic bearing of the NDB measured from the aircraft?
A – 211o
B – 208o
C – 221o
D – 180o
Ref: all
Ans: C

10964
...
3
W00853
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
At position 5211N 00931W, which of the following denotes all the symbols?
A – Military airport, ILS, NDB
B – Civil airport, VOR, ILS
C – Military airport, VOR, ILS
D – Civil airport, ILS, NDB
Ref: all
Ans: D
10967
...
89 cm long represents 185 NM
...
(Refer to figure 061-10)
What are the average magnetic course and distance between INGO VOR
(N6350 W01640) and Sumburg VOR (N5955 W 00115)?
A – 131o – 494 NM
B – 118o – 440 NM
C – 117o – 494 NM
D – 130o – 440 NM
Ref: all
Ans: A
10972
...
8
W00849
...
On a particular Direct Mercator wall chart, the 180W to 180E parallel of
latitude at 53N is 133 cm long
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the average track (oT) and distance between SLG NDB (N5416
...
0) and CFN NDB (N5502
...
4)?
A – 191o – 45 NM
B – 020o – 46 NM
C – 348o – 46 NM
D – 011o – 47 NM
Ref: all
Ans: D
10975
...
What is the approximate scale of the chart at latitude 30oS?
A – 1 : 25 000 000
B – 1 : 30 000 000
C – 1 : 18 000 000
D – 1 : 21 000 000
Ref: all
Ans: A
10976
...
The scale of the
chart is approximately:
A – 1 : 130 000
B – 1 : 700 000
C – 1 : 1 300 000
D – 1 : 7 000 000
Ref: all
Ans: C
10977
...
3
W00705
...
7 W00836
...
A Lambert conformal conic chart has a constant of the cone of 0
...
The
initial course of a straight line track drawn on this chart from A (40oN
050oW) to B is 043o(T) at A; course at B is 055o(T)
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the average track (oT) and distance between CON VOR (N5354
...
1) and BEL VOR (N5439
...
8)?
A – 293o – 98 NM
B – 071o – 100 NM
C – 113o – 97 NM
D – 063o – 101 NM
Ref: all
Ans: D
10981
...
The length
on the chart between C N60o W008o and D N60o W008o is:
A – 19
...
2 cm
C – 35
...
8 cm
Ref: all
Ans: D

10983
...
The
scale of the chart at 47o North approximates:
A – 1: 2 500 000
B – 1 : 8 000 000
C – 1 : 3 000 000
D – 1 : 6 000 000
Ref: all
Ans: D
10984
...
3 W00853
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the average track (oM) and distance between WTD NDB (N5211
...
0) and BAL VOR (N5318
...
9)?
A – 206o – 71 NM
B – 018o – 153 NM
C – 026o – 71 NM
D – 198o – 72 NM
Ref: all
Ans: C

10988
...
3
W00853
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the average track and distance between BAL VOR (N5318
...
9) and CRN NDB (N5318
...
5)?
A – 278o – 89 NM
B – 270o – 90 NM
C – 268o – 91 NM
D – 272o – 89 NM
Ref: all
Ans: B
10990
...
7 W00836
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the radial and DME distance from BEL VOR/DME (N5439
...
8) to position N5440 W00730?
A – 090o – 46 NM
B – 278o – 44 NM
C – 278o – 10 NM
D – 098o – 45 NM
Ref: all
Ans: B
10992
...
Waypoint 2 is 60N 20W
...
What is the latitude on passing 25W?
A – 6005N
B – 6011N
C – 6032N
D – 5949M
Ref: all
Ans: A
10997
...
What is the Great Circle
track of B from A, measured at A?
A – 132T
B – 048T
C – 090T
D – 228T
Ref: all
Ans: A
11000
...
8
W00849
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
Which of the following lists all the aeronautical chart symbols shown at
position N5211 W00705?
A – NDB: ILS
B – VOR: NDB
C – civil airport: ILS
D – civil airport: NDB
Ref: all
Ans: D
11002
...
7
W00613
...
(Refer to figure 061-10)
What are the initial true course and distance between positions N5800
W01300 and N6600 E00200?
A – 032o – 470 NM
B – 029o – 570 NM
C – 042o – 635 NM
D – 036o – 638 NM
Ref: all
Ans: D

11005
...
3 W00853
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the radial and DME distance from SHA VOR/DME (N5243
...
1) to position N5300 W00940?
A – 057o – 27 NM
B – 309o – 33 NM
C – 293o – 33 NM
D – 324o – 17 NM
Ref: all
Ans: B
11009
...
3 W00853
...
4 W00829
...
An aircraft at latitude 0220N tracks 180T for 685 kilometres
...
Contour lines on aeronautical maps and charts connect points:
A – of equal latitude
B – with the same variation
C – having the same longitude
D – having the same elevation above sea level
Ref: all
Ans: D
11016
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What feature is shown on the chart at position N5211 W00931?
A – KERRY/Farranfore aerodrome
B – Waterford NDB
C – Connemara aerodrome
D – Punchestown aerodrome
Ref: all
Ans: A

11019
...
8 W00849
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What feature is shown on the chart at position N5417 W01005?
A – Cammore aerodrome
B – Belmullet aerodrome
C – EAGLE ISLAND LT
...
NDB
D – Clonbullogue aerodrome
Ref: all
Ans: C
11023
...
1 W00856
...
3 W00853
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
Given:
SHA VOR/DME (N5243
...
1) radial 048o/22 NM
What is the aircraft position?
A – N5228 00920
B – N5300 W0830
C – N5258 W00825
D – N5225 W00917
Ref: all
Ans: B
21452
...
3 W00853
...
4 W00829
...
(Refer to figure 061-09)
At 1215 UTC LAJES VORTAC (38o 46’N 027o 05’W) RMI reads 178o,
range 135 NM
...
(Refer to figure 061-07)
Assume a North polar stereographic chart whose grid is aligned with the
Greenwich meridian
...
Its position is now approximately:
A – 70o 15’N 080o E
B – 80o 00’N 080oE
C – 78o 45’N 087oE
D – 79o 15’N 074oE
Ref: all
Ans: B
21669
...
8 W00849
...
(Refer to figure 061-10)
An aircraft on radial 110o at a range of 120 NM from SAXAVORD VOR
(N6050 W00050) is at position:
A – N6127 W00443
B – N6010 E00255
C – N6109 E00255
D – N6027 E00307
Ref: all
Ans: D

21672
...
3
W000853
...
8 W00849
...
(Refer to figure 061-06)
Complete line 5 of the ‘FLIGHT NAVIGATION LOG’ positions ‘J’ to ‘K’
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the average track (oM) and distance between BAL VOR (N5318
...
9) and SLG NDB (N5416
...
0)?
A – 262o – 86 NM
B – 128o – 99 NM
C – 308o – 98 NM
D – 316o – 96 NM
Ref: all
Ans: D

21676
...

What is the HDGo (M) and ETA?
A – HDG 344o – ETA 1303 UTC
B – HDG 344o – ETA 1336 UTC
C – HDG 354o – ETA 1326 UTC
D – HDG 034o – ETA 1336 UTC
Ref: all
Ans: B
21677
...
4
WS00829
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the average track (oM) and distance between RK VOR (N5150
...
7) and CRN NDB (N5318
...
5)?
A – 177o – 92 NM
B – 357o – 89 NM
C – 169o – 91 NM
D – 349o – 90 NM
Ref: all
Ans: B

21679
...
(Refer to figure 061-03)
Which of the aeronautical chart symbols indicates a DME?
A–2
B–1
C–6
D–3
Ref: all
Ans: A
21684
...
1
00856
...
/3 W00705
...
(Refer to figure 061-03)
Which of the aeronautical chart symbols indicates a VOR?
A–1
B–2
C–3
D–7
Ref: all
Ans: C
21686
...
(Refer to figure 061-03)
Which of the aeronautical chart symbols indicates a basic, non-specified,
navigation aid?
A–5
B–4
C–2
D–6
Ref: all
Ans: A

21688
...
(Refer to figure 061-01)
Which aeronautical chart symbol indicates a flight Information Region (FIR)
boundary?
A–1
B–3
C–4
D–5
Ref: all
Ans: A
21690
...
(Refer to figure 061-03)
Which of the aeronautical chart symbols indicates a TACAN?
A–6
B–7
C–3
D–1
Ref: all
Ans: A
21693
...
(Refer to figure 061-01)
Which aeronautical chart symbol indicates a group of lighted obstacles?
A–9
B – 10
C – 11
D – 12
Ref: all
Ans: D

21695
...
(Refer to figure 061-01)
Which aeronautical chart symbol indicates an exceptionally high lighted
obstacle?
A – 14
B – 13
C – 12
D – 16
Ref: all
Ans: A
21697
...
(Refer to figure 061-01)
Which aeronautical chart symbol indicates a lightship?
A – 12
B – 10
C – 15
D – 16
Ref: all
Ans: D
21701
...
(Refer to figure 061-01)
Which aeronautical chart symbol indicates an unlighted obstacle?
A–9
B – 10
C–8
D – 15
Ref: all
Ans: A
21704
...
(Refer to figure 061-01)
Which aeronautical chart symbol indicates a compulsory reporting point?
A–8
B – 15
C–6
D–7
Ref: all
Ans: D
21706
...
(Refer to figure 061-01)
Which aeronautical chart symbol indicates the boundary of advisory airspace?
A–2
B–3
C–4
D–5
Ref: all
Ans: D
21708
...
(Refer to figure 061-01)
What is the meaning of aeronautical chart symbol number 15?
A – Aeronautical ground light
B – Visual reference point
C – Hazard to aerial navigation
D – Lighthouse
Ref: all
Ans: A
24001
...
Determine the distance between points A (N4500
...
0) and B
(N4500
...
0) is:
A – 300 nm
B – 636
...
1 nm
Ref: all
Ans: B

25225
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
SHA VOR (5243N 00853W) DME 41 nm
CRK VOR (5150N 00829W) DME 30 nm
What is the position of the aircraft?
A – 5215N 00805W
B – 5205N 00915W
C – 5215N 00915W
D – 5225N 00810W
Ref: all
Ans: A
25255
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
Your radial from the SHA VOR (5243N 00853W) is 120o M
...
What is your position?
A – 5320N 00800W
B – 5240N 00821W
C – 5220N 00821W
D – 5230N 00800W
Ref: all
Ans: D
25301
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What is the radial and DME distance from SHA VOR (5243N 00853W) to
BIRR airport (5311N 00754W)?
A – 068 M 42 nm
B – 060 M 40 nm
C – 068 M 40 nm
D – 060 M 42 nm
Ref: all
Ans: A

25303
...
(Refer to figure 061-10)
Which of the following beacons is 185 NM from AKRABERG (N6124
W00640)?
A – KIRKWALL (N5858 W00254)
B – STORNOWAY (N5815 W00617)
C – SUMBURGH (N5955 W00115)
D – SAXAVORD (N6050 W00050)
Ref: all
Ans: C

061-04

DEAD RECKONING NAVIGATION (DR)
061-04-01 Basics of dead reckoning

8297
...

The true course of the rhumbline at point A is:
A – 100o
B – 096o
C – 104o
D – 107o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8299
...
An aircraft is climbing at a constant CAS in ISA conditions
...
Heading is 156oT, TAS is 320 knots, W/V is 130o/45
...
The ICAO definition of ETA is the:
A – actual time of arrival at a point or fix
B – estimated time of arrival at destination
C – estimated time of arrival at an en-route point or fix
D – estimated time en route
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8601
...
Given:
True course A to B = 250o
Distance A to B = 315 NM
TAS = 450 kt
W/V = 200o/60 kt
ETD A – 0650 UTC
What is the ETA at B?
A – 0730 UTC
B – 0736 UTC
C – 0810 UTC
D – 0716 UTC
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
9734
...
What is the great circle track on departure from A?
A – 261o
B – 288o
C – 279o
D – 270o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
9736
...
What is the longitude of a position 6 NM to the east of 58o42N 094o00W?
A – 093o53
...
0W
C – 093o48
...
0W
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
9815
...
The rhumb line track between position A (45o00N, 010o00W) and position B
(48o30N, 015o00W) is approximately:
A – 345
B – 300
C – 330
D – 315
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
10945
...
Given the following:
Magnetic heading: 060o
Magnetic variation: 8oW
Drift angle: 4o right
What is the true track?
A – 048o
B – 064o
C – 056o
D – 072o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
11067
...
If the wind component is 60 knots head, what is the
distance from the first airfield to the critical point?
A – 250 nm
B – 200 nm
C – 300 nm
D – 280 nm
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
14649
...
It then flies westward along the parallel
of latitude for 382 NM to position B
...
An aircraft in the northern hemisphere is making an accurate rate one turn to
the right
...
5 HR 20 MIN 20 SEC corresponds to a longitude difference of:
A – 75o00
B – 78o45
C – 80o05
D – 81o10
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
15422
...
55 litres
B – 1 litre equals 4
...
78 litres
D – 1 litre equals 3
...
What is the ISA temperature value at FL 330?
A - -56oC
B - -66oC
C - -81oC
D - -51oC
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

16274
...
What is its position?
A – South pole
B – North pole
C – 30oS
D – 45oS
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
16275
...
What is its position as a true bearing from the south pole?
A – 30oT
B – 000oT
C – 45oT
D – 60oT
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
16278
...
If initial Great circle track
is 047oT what is Final Great circle track?
A – 57o
B – 52o
C – 43o
D – 29o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
16280
...
What is the RL track from
A to B?
A – 250o (T)
B – 270o (T)
C – 290o (T)
D – 300o (T)
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

16281
...
What is the initial GC
track?
A – 260o (T)
B – 270o (T)
C – 290o (T)
D – 300o (T)
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
24012
...
If the
CAS is 150 kt, what is the TAS?
A – 115 kt
B – 195 kt
C – 180 kt
D – 145 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
24015
...
If the Compass Heading is 265o variation is 33oW and deviation is 3oE, what is
the True Heading?
A – 229o
B – 235o
C – 301o
D – 295o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

24026
...
5 km
C – 35 000 m
D – 0
...
In the Northern Hemisphere the rhumb line track from position A to B is 230o,
the convergency is 6o and the difference in longitude is 10o
...
On a Direct Mercator projection a particular chart length is measured at 30oN
...
The Great Circle bearing from A (70oS 030oW) to B (70oS 060oE) is
approximately:
A – 090o (T)
B – 048o (T)
C – 132o (T)
D – 312o (T)
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

24046
...
If the Conversion Angle is 4o, what is the great circle
bearing of A from B?
A – 228o
B – 212o
C – 220o
D – 224o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
24047
...
The initial great circle track from A to B is 080o and the rhumb line track is
083o
...
A flight is planned from A (N37000’ E/W000000’) to B (N46000’
E/W000000’)
...
Given:
Variation 7oW
Deviation 4oE
If the aircraft is flying a Compass heading of 270, the True and Magnetic
Headings are:
A – 274o (T) 267o (M)
B – 267o (T) 274o (M)
C – 277o (T) 281o (M)
D – 263o (T) 259o (M)
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
25188
...
On a chart, 49 nm is represented by 7
...
The distance Q to R is 3016 nm; TAS is 480 kts
...
Leaving Q at 1320 UTC, what is the ETA at the point of Equal
Time:
A – 1631 UTC
B – 1802 UTC
C – 1702 UTC
D – 1752 UTC
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

061-04-02 Use of the navigational computer
8255
...
The QNH is 988
...

What is pressure altitude?
A – 675
B – 325
C – 1675
D – 825
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8526
...
265 US-GAL equals? (Specific gravity 0
...
Ground speed is 540 knots
...
What is time to go?
A – 8 mins
B – 9 mins
C – 18 mins
D – 12 mins
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8543
...
Three minutes later, at a ground
speed of 180 knots, it has changed to 225oR
...
An aeroplane flying at 180 kts TAS on a track of 090o
...
The distance the aeroplane can fly out and return in one hour is:
A – 88 NM
B – 85 NM
C – 56 NM
D – 176 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

8549
...
Given:
IAS 120 kt
FL 80
OAT +20oC
What is the TAS?
A – 132 kt
B – 102 kt
C – 120 kt
D – 141 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
8557
...
An aircraft is heading 180 at a TAS of 198 knots
...
What is its track and ground speed?
A – 180
...
220
C – 180
...
223
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

8558
...
You are flying 090oC heading
...
Your
TAS is 160 knots
...
What is the W/V?
A – 158oT/51
B – 060oT/50
C – 340oT/25
D – 055oT/25
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8564
...
Given:
GS = 510 kt
Distance A to B = 43 NM
What is the time (MIN) from A to B?
A–6
B–4
C–5
D–7
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8568
...
On a particular take-off, you can accept up to 10 knots tailwind
...

What is the maximum wind strength you can accept?
A – 18 knots
B – 11 knots
C – 8 knots
D – 4 knots
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

8584
...
G/S = 240 knots, Distance to go = 500 nm
...
Given:
True track 070o
Variation 30oW
Deviation +1o
Drift 10oR
Calculate the compass heading?
A – 100o
B – 091o
C – 089o
D – 101o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
ANS: C

8594
...
45
W/V = 270/85, Track = 200T
What is drift and ground speed?
A – 18L/252 knots
B – 15R/310 knots
C – 17L/228 knots
D – 17R/287 knots
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8596
...

A – 069o and 448 kts
B – 068o and 460 kts
C – 078o and 450 kts
D – 070o and 453 kts
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8597
...
Given:
GS = 345 kt
Distance from A to B = 3560 NM
What is the time from A to B?
A – 10 HR 19 MIN
B – 10 HR 05 MIN
C – 11 HR 00 MIN
D – 11 HR 02 MIN
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8614
...
4 statute miles in 47 seconds
...
At 1000 hours an aircraft is on the 310 radial from a VOR/DME, at 10 nautical
miles range
...
What is the aircraft’s
track and ground speed?
A – 080 / 85 knots
B – 085 / 85 knots
C – 080 / 80 knots
D – 085 / 90 knots
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8618
...
Given:
GS = 135 kt
Distance from A to B = 433 NM
What is the time from A to B?
A – 3 HR 20 MIN
B – 3 HR 25 MIN
C – 3 HR 19 MIN
D – 3 HR 12 MIN
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
10930
...
The cross wind component on landing is:
A – 26 kts
B – 23 kts
C – 20 kts
D – 15 kts
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
11037
...
Given:
GS – 95 kt
Distance from A to B =- 480 NM
What is the time from A to B?
A – 4 HR 59 MIN
B – 5 HR 03 MIN
C – 05 HR 00 MIN
D – 5 HR 08 MIN
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
11043
...
Your ETA at B is 1130
...
What ground speed is required to arrive on
time at B?
A – 317 knots
B – 330 knots
C – 342 knots
D – 360 knots
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
11046
...
The equivalent of 70 m/sec is approximately:
A – 145 kt
B – 136 kt
C – 210 kt
D – 35 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

11060
...
Given:
GS = 435 kt
Distance from A to B = 1920 NM
What is the time from A to B?
A – 4 HR 10 MIN
B – 3 HR 25 MIN
C – 3 HR 26 MIN
D – 4 HR 25 MIN
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
11069
...
Given:
True course from A to B = 090o
TAS = 460 kt
W/V = 360/100 kt
Average variation = 10oE
Deviation = -2o
Calculate the compass heading and GS?
A – 078o – 450 kt
B – 068o – 460 kt
C – 069o – 448 kt
D – 070o – 453 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
11080
...
7 m/sec
B – 5
...
6 m/sec
D – 2
...
Fuel flow per HR is 22 US-GAL, total fuel on board is 83 IMP GAL
...
How many NM would an aircraft travel in 1 MIN 45 SEC if GS is 135 kt?
A – 39
...
36
C – 3
...
94
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

15427
...
44
B – 0
...
39
D – 0
...
Given:
TAS = 225 kt
HDG (oT) – 123o
W/V – 090/60 kt
Calculate the Track (oT) and GS?
A – 134 – 178 kt
B – 134 – 188 kt
C – 120 – 190 kt
D – 123 – 180 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
15430
...
Given:
TAS – 230 kt
HDG (T) – 250o
W/V m 205/10 kt
Calculate the drift and GS?
A – 1L – 225 kt
B – 1R – 221 kt
C – 2R – 223 kt
D – 2L – 224 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
15433
...
Given:
True altitude 9000 FT
OAT -32oC
CAS 200 kt
What is the TAS?
A – 215 kt
B – 200 kt
C – 210 kt
D – 220 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

24016
...
Given:
True Track
True Heading
TAS
G/S

239o
229o
555 kt
577 kt

Calculate the wind velocity
...
Given:
True Track
Drift
Variation
Compass Hdg

245o
5o right
3o E
242o

Calculate the deviation
...
True Heading of an aircraft is 265o and TAS is 290 kt
...
Course 040oT, TAS 120 kt, Wind speed 30 knots
...
Required course 045oT, W/V = 190/30, FL 55, ISA, Variation 15oE, CAS 120
knots
...
Given:
Pressure Altitude = 5000 ft
OAT = +35C
What is true altitude:
A – 4550 ft
B – 5550 ft
C – 4290 ft
D – 5320 ft
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

25228
...
If the headwid component is 50 kt, the FL is 330, temperature JSA -7oC and
the ground speed is 496 kt, the Mach No
...
98
B – 0
...
95
D – 0
...
Given:
True Heading = 090o
TAS = 180 kt
GS = 180 kt
Drift 5o right
Calculate the W/V?
A – 360o / 15 kt
B – 190o / 15 kt
C – 010o / 15 kt
D – 180o / 15 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

8529
...
What is the W/V?
A – 050o(T) / 70 kt
B – 040o(T) / 105 kt
C – 055o(T) / 105 kt
D – 065o(T) / 70 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8530
...
The angle between the wind direction
and the runway is 60o
...
Given:
Maximum allowable tailwind component for landing 10 kt
Planned runway 05 (047o magnetic)
The direction of the surface wind reported by ATIS 210o
Variation is 17oE
Calculate the maximum allowable windspeed that can be accepted without
exceeding the tailwind limit?
A – 15 kt
B – 18 kt
C – 8 kt
D – 11 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

8534
...
An aircraft is following a true track of 048o at a constant TAS of 210 kt
...
The GS and drift angle are:
A – 192 kt, 7o left
B – 200 kt – 3
...
Given:
Runway direction 230o(T)
Surface W/V 280o(T)/40 kt
Calculate the effective cross-wind component?
A – 21 kt
B – 36 kt
C – 31 kt
D – 26 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

8538
...
Given:
TAS = 198 kt
HDG (oT) = 180
W/V = 359/25
Calculate the Track (oT) and GS?
A – 180 – 223 kt
B – 179 – 220 kt
C – 181 – 180 kt
D – 180 – 183 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8546
...
Given:
Magnetic track = 210o
Magnetic HDG = 215o
VAR = 15oE
TAS = 360 kt
Aircraft flies 64 NM in 12 MIN
Calculate the true W/V?
A – 265o/50 kt
B – 195o/50 kt
C – 235o/50 kt
D – 300o/30 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8551
...
Given:
TAS = 95 kt
HDG (T) = 075o
W/V = 310/20 kt
Calculate the drift and GS?
A – 9R – 108 kt
B – 10L – 104 kt
C – 9L – 105 kt
D – 8R – 104 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

8559
...
Given:
TAS = 250 kt
HDG (T) = 029o
W/V = 035/45kt
Calculate the drift and GS?
A – 1L – 205 kt
B – 1R – 205 kt
C – 1L – 265 kt
D – 1R – 295 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8569
...
Given:
TAS = 235 kt
HDG (T) = 076o
W/V = 040/40kt
Calculate the drift angle and GS?
A – 5R – 207 kt
B – 7L – 269 kt
C – 5L – 255 kt
D – 7R – 204 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
8573
...
Given:
True HDG = 133o
TAS = 225 kt
Track (T) = 144o
GS = 206 kt
Calculate the W/V?
A – 070/40 kt
B – 075/45 kt
C – 070/45 kt
D – 075/50 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

8581
...
Given:
True Heading = 090o
TAS = 200 kt
W/V = 220o/30 kt
Calculate the GS?
A – 180 kt
B – 230 kt
C – 220 kt
D – 200 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8585
...
What is the W/V (oT)?
A – 165o/25 kt
B – 340o/25 kt
C – 340o/98 kt
D – 160o/50 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

8586
...
Given:
TAS = 140 kt
True HDG = 302o
W/V = 045o(T)/45 kt
Calculate the drift angle and GS?
A – 9oR – 143 kt
B – 16oL – 156 kt
C – 9oL – 146 kt
D – 18oR – 146 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8595
...
Given:
Maximum allowabl crosswind component is 20 kt
Runway 06
RWY QDM 063o(M)
Wind direction 100o(M)
Calculate the maximum allowable windspeed?
A – 26 kt
B – 31 kt
C – 33 kt
D – 25 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8602
...
Given:
Course required = 085o (T)
Forecast W/V 030/100 kt
TAS = 470 kt
Distance = 265 NM
Calculate the true HDG and flight time?
A – 096o, 29 MIN
B – 076o, 34 MIN
C – 075o, 39 MIN
D – 095o, 31 MIN
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

8607
...
Given:
TAS = 132 kt
HDG (T) = 053o
W/V = 205/15 kt
Calculate the track (oT) and GS?
A – 057 – 144 kt
B – 050 – 145 kt
C – 052 – 143 kt
D – 051 – 144 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8609
...
Given:
TAS = 227 kt
Track (T) = 316o
W/V = 205/15 kt
Calculate the HDG (oT) and GS?
A – 313 – 235 kt
B – 311 – 230 kt
C – 312 – 232 kt
D – 310 – 233 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8616
...
Given:
Runway direction 305o(M)
Surface W/V 260o(M)/30 kt
Calculate the cross-wind component?
A – 18 kt
B – 24 kt
C – 27 kt
D – 21 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

11024
...
Given:
True HDG = 074o
TAS = 230 kt
Track (T) = 066o
GS = 242 kt
Calculate the W/V
A – 180/30 kt
B – 180/35 kt
C – 185/35 kt
D – 180/40 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
11026
...
An aeroplane is flying at TAS 180 kt on a track of 090o
...
How far can the aeroplane fly out from its base and return in one hour?
A – 56 NM
B – 88 NM
C – 85 NM
D – 176 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
11030
...
Given:
Magnetic heading = 255o
VAR = 40oW
GS = 375 kt
W/V = 235o(T)/120 kt
Calculate the drift angle?
A – 7o left
B – 7o right
C – 9o left
D – 16o right
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

11033
...
Given:
TAS = 485 kt
HDG (T) = 168o
W/V = 130/75 kt
Calculate the Track (oT) and GS?
A – 175 – 432 kt
B – 173 – 424 kt
C – 175 – 420 kt
D – 174 – 428 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
11036
...
Given:
Magnetic track = 315o
HDG = 301o (M)
VAR = 5oW
TAS = 225 kt
The aircraft flies 50 NM in 12 MIN
...
Given:
TAS = 155 kt
Track (T) = 305o
W/V = 160/18 kt
Calculate the HDG (oT) and GS?
A – 301 – 169 kt
B – 305 – 169 kt
C – 309 – 170 kt
D – 309 – 141 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
11042
...
For a given track the:
Wind component = 45 kt
Drift angle = 15o left
TAS = 240 kt
What is the wind component on the reverse track?
A - -55 kt
B - -65 kt
C - -45 kt
D - -35 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
11048
...
The wind velocity
reported by the tower is 350o/20 kt
...
In order to
maintain the centre line, the aircrafts heading (oM) should be:
A – 322o
B – 328o
C – 316o
D – 326o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
11049
...
Given:
True HDG = 233o
TAS = 480 kt
Track (T) = 240o
GS = 523 kt
Calculate the W/V?
A – 115/70 kt
B – 110/75 kt
C – 110/80 kt
D – 105/75 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
11053
...
Given:
True HDG = 035o
TAS = 245 kt
Track (T) = 046o
GS = 220 kt
Calculate the W/V?
A – 335/55 kt
B – 335/45 kt
C – 340/50 kt
D – 340/45 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

11057
...
Calculate the true W/V?
A – 340o/45 kt
B – 320o/50 kt
C – 210o/15 kt
D – 180o/45 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
11062
...
Given:
TAS = 125 kt
True HDG = 355o
W/V = 320o(T)/30 kt
Calculate the true track and GS?
A – 002 – 98 kt
B – 345 – 100 kt
C – 348 – 102 kt
D – 005 – 102 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

11064
...
Given:
True heading = 310o
TAS = 200 kt
GS = 176 kt
Drift angle 7o right
Calculate the W/V?
A – 090o/33 kt
B – 360o/33 kt
C – 270o/33 kt
D – 180o/33 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
11068
...
Given:
TAS = 465 kt
HDG (T) = 124o
W/V = 170/80 kt
Calculate the drift and GS?
A – 8L – 415 kt
B – 3L – 415 kt
C – 4L – 400 kt
D – 6L – 400 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
11076
...
Given:
TAS = 375 kt
True HDG = 124o
W/V = 130o(T)/55 kt
Calculate the true track and GS?
A – 125 – 322 kt
B – 123 – 320 kt
C – 126 – 320 kt
D – 125 – 318 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

11084
...
Given:
TAS = 270 kt
True HDG = 270o
Actual wind 205o(T)/30 kt
Calculate the drift angle and GS?
A – 6R – 259 kt
B – 6L – 256 kt
C – 6R – 251 kt
D – 8R – 259 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
11088
...
The following information is displayed on an Inertial Navigation System: GS
520 kt
...
The W/V being experienced is:
A – 225o/60 kt
B – 320o/60 kt
C – 220o/60 kt
D – 325o/60 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
11090
...
Given:
M 0
...
The reported surface wind from the control tower is 240o/35 kt
...
What is cross-wind component?
A – 30 kt
B – 24 kt
C – 27 kt
D – 21 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
15434
...
VAR is 13oE
...
How long will it take to travel 284 nm at a speed of 526 KPH?
A – 1
...
9 h
C – 45 min
D–1h
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
16295
...
4 mins to travel 840 nm, what is your speed in kmh?
A – 705 kmh
B – 290 kmh
C – 120 kmh
D – 966 kmh
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

061-04-04 List elements required for establishing DR position
8582
...

Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
14654
...
Position B is
located 100 NM from A on a bearing of 225o(T)
...
Given:
Position A 45oN, ?oE
Position B 45oN, 45o15E
Distance A-B = 280 NM
B is to the East of A
Required: longitude of position A?
A – 38o39E
B – 49o57E
C – 51o51E
D – 40o33E
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

14670
...
What is the final position after the following rhumb line tracks and distances
have been followed from position 60o00N 030o00W?
South for 3600 NM
East for 3600 NM
North for 3600 NM
West for 3600 NM
The final position of the aircraft is:
A – 59o00N 090o00W
B – 60o00N 090o00W
C – 60o00N 030o00E
D – 59o00N 060o00W
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
15435
...
On completion
of the flight the longitude will be:
A – 002o 10W
B – 000o 15E
C – 000o 40E
D – 005o 15E
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

15436
...
What is
the longitude of x?
A – 170oW
B – 140oW
C – 145oE
D – 175oE
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
15437
...
What will
its latitude be after 1
...
You are flying from A (30S 20E) to B (30S 20W)
...
What is diat from 30o39S 20o20E to 45o23N 40o40E:
A – 14o44 N
B – 76o2 S
C – 76o2 N
D – 76o4 S
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

16285
...
Given:
True Track
Drift
Variation
Compass Hdg

245o
5o right
3oE
242o

Calculate the Magnetic Heading:
A – 247o
B – 243o
C – 237o
D – 253o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
24020
...
What is the corresponding magnetic heading?
A – 084o
B – 334o
C – 154o
D – 264o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

061-04-05 Calculate DR elements
8540
...
Given:
Airport elevation is 1000 ft
QNH is 988 hPa
What is the approximate airport pressure altitude?
(Assume 1 hPa = 27 FT)
A – 680 FT
B – 320 FT
C – 1680 FT
D - -320 FT
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8561
...
What
is Density Altitude?
A – 6980 feet
B – 7750 feet
C – 8620 feet
D – 10020 feet
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

8565
...
Your DME range is 25 nm from the threshold
...
Given:
FL 350
Mach 0
...
The pressure alt is 29000 feet and the SAT is -55C
...
You are flying at a True Mach No of 0
...
At 1000 hours
you are 100 nm from the POL DME and your ETA at POL is 1012
...
What should your new TMN be if you
reduce speed at 100 nm distance to:
A – M
...
72
C – M
...
61
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
8604
...
85
B – 0
...
825
D – 0
...
Given:
TAS 487 kt
FL 330
Temperature ISA + 15
Calculate the Mach Number?
A – 0
...
84
C – 0
...
78
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

9752
...
The
distance is kilometres from A to B is approximately:
A – 1222
B – 1000
C – 540
D – 804
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
9790
...
After 3 hours the
latitude is:
A – 10S
B – 02N
C – 02S
D – 0N/S
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
11027
...
An aircraft takes off from the aerodrome of BRIOUDE (altitude 1 483 ft, QFE
= 963 hPa, temperature = 32oC)
...
An aircraft maintaining a 5
...
Given:
Pressure Altitude 29,000 ft, OAT -55C
...
An aircraft leaves point A (75N 50W) and flies due North
...

What is the total distance covered?
A – 1,650 nm
B – 2,000 nm
C – 2,175 nm
D – 1,800 nm
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

25144
...
What
is Density Altitude:
A – 7080 feet
B – 8120 feet
C – 9280 feet
D – 9930 feet
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
25218
...
Given:
M0
...
A Lamberts Conical conformal chart has standard parallels at 63N and 41N
...
891
B – 0
...
656
D – 0
...
Given:
A polar stereographic chart whose grid is aligned with the zero meridian
...
The great circle distance between position A (59o34
...
4E) and B
(30o25
...
6W) is:
A – 5,400 NM
B – 10,800 NM
C – 2,700 NM
D – 10,800 NM
Ref: AIR: atpl, cpl;
Ans: A

11073
...
Calculate
the ETA at PORTO SANTO NDB:
A – 1341
B – 1344
C – 1348
D - 1354
Ref: AIR: atpl, cpl;
Ans: C
16277
...
At what longitude will the
Great Circle track equal the Rhumb Line (RL) track between A and B:
A – 06oW
B – 0oW
C – 04oW
D – 04oE
Ref: AIR: atpl, cpl;
Ans: C
25202
...
0 on a Mercator chart the scale is 1:5 000 000; the length of a line
on the chart between C N6010
...
0 and D N6010
...
0 is:
A – 19
...
2 cm
C – 17
...
6 cm
Ref: AIR: atpl, cpl;
Ans: C

25206
...
0 E14000
...
An aircraft was over Q at 1320 hours flying direct to R
Given:
Distance Q to R 3016 NM
True airspeed 480 kt
Mean wind component OUT -90 kt
Mean wind component BACK +75 kt
The ETA for reaching the Point of Equal Time (PET) between Q and R is:
A – 1820
B – 1756
C – 1752
D – 1742
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: C
8537
...
Given: Distance A to B
2,900 NM True airspeed 470 kt Mean wind component OUT +55 kt Mean wind
component BACK -75 kt
...
Given:
Distance A to B 2346 NM
Groundspeed OUT 365 kt
Groundspeed BACK 480 kt
Safe endurance 8 HR 30 MIN
The time from A to the Point of Safe Return (PSR) A is:
A – 197 min
B – 219 min
C – 290 min
D – 209 min
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: C
8544
...
TAS = 490 kt
...
On the return leg between B and A,
the equivalent headwind is +40 kt
...
An aircraft was over A at 1435 hours flying direct to B
Given:
Distance A to B 2900 NM
True airspeed 470 kt
Mean wind component OUT +55 kt
Mean wind component BACK -75 kt
Safe endurance 9 HR 30 MIN
The distance from A to the Point of Safe Return (PSR) A is:
A – 2844 NM
B – 1611 NM
C – 1759 NM
D – 2141 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: D

8579
...
Given:
Distance A to B 2484 NM
Mean groundspeed out 420 kt
Mean groundspeed back 500 kt
Safe endurance 08 Hr 30 min
The distance from A to the Point of Safe Return (PSR) A is:
A – 1908 NM
B – 1940 NM
C – 1736 NM
D – 1630 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: B
8593
...
From the departure point, the distance to the point of equal time is:
A – proportional to the sum of ground speed out and ground speed back
B – inversely proportional to the sum of ground speed out and ground speed
back
C – inversely proportional to the total distance to go
D – inversely proportional to ground speed back
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: B
8611
...
Given:
AD = Air distance
GD = Ground distance
TAS = True airspeed
GS = Ground speed
Which of the following is the correct formula to calculate ground distance
(GD) gone?
A – GD = (AD X GS)/TAS
B – GD = (AD – TAS)/TAS
C – GD = AD X (GS – TAS)/GS
D – GD = TAS/(GS X AD)
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: A

8622
...
Given:
Distance A to B 3623 NM
Groundspeed out 370 kt
Groundspeed back 300 kt
The time from a to the Point of Equal Time (PET) between A and B is:
A – 323 min
B – 288 min
C – 263 min
D – 238 min
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: C
11040
...
If the
wind component on the outbound leg is 50 knots head, what is the distance to
the point of safe endurance?
A – 1500 nm
B – 1458 nm
C – 1544 nm
D – 1622 nm
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: B

11045
...
If outbound groundspeed in
365 knots and homebound groundspeed is 480 knots and safe endurance is 8
hours 30 minutes, what is the time to the PNR?
A – 290 minutes
B – 209 minutes
C – 219 minutes
D – 190 minutes
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: A
11058
...
Given:
Distance Q to R 1760 NM
Groundspeed out 435 kt
Groundspeed back 385 kt
Safe endurance 9 hr
The distance from Q to the Point of Safe Return (PSR) between Q and R is:
A – 1313 NM
B – 1838 NM
C – 1467 NM
D – 1642 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: B

11087
...

Given:
Distance Q to R 3016 NM
True airspeed 480 kt
Mean wind component out – 90 kt
Mean wind component back +75 kt
Safe endurance 10:00 hr
The distance from Q to the Point of Safe Return (PSR) Q is:
A – 2370 NM
B – 2290 NM
C – 1310 NM
D – 1510 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: B
11091
...
The pilot flies a
track of 090o(T)
...
In order to return to the point
of departure before sunset, the furthest distance which may be travelled is:
A – 97 NM
B – 115 NM
C – 105 NM
D – 84 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: A
24042
...
TAS is 140 kt
...
How long will it take to reach the Point
of Safe Return?
A – 1 hr and 44 min
B – 1 hr and 37 min
C – 1 hr and 21 min
D – 5 hr and 30 min
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: A

061-04-08 Miscellaneous DR uncertainties and
practical means of correction
16314
...
Calculate the dlong from N001 15 E090 00 to N001 15 E015 15:
A – 74o45E
B – 74o15E
C – 74o45W
D – 105o15N
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: C

061-05

IN-FLIGHT NAVIGATION

061-05-01 Use of visual observations and application to in-flight
navigation
8627
...
An island appears 45o to the right of the centre line on an airborne weather
radar display
...
An island is observed to be 15o to the left
...
The bearing (oT) from the aircraft to the island is:
A – 122
B – 088
C – 268
D – 302
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: B

8664
...
The W/V is calm; aircraft GS 180 kt
...
An island is observed by weather radar to be 15o to the left
...
What is the true bearing
of the aircraft from the island?
A – 122o
B – 302o
C – 088o
D – 268o
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: D
8669
...
An island appears 30o to the right of the centre line on an airborne weather
radar display
...
(Refer to figure 061-01)
Which of the following is the symbol for an exceptionally high (over 1000 feet
AGL) lighted obstruction?
A – 13
B – 10
C – 14
D – 12
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: C
8677
...
The time between these roads can be used to check the aircraft:
A – groundspeed
B – position
C – track
D – drift
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: A

8700
...
What is the true bearing of the aircraft from the island if at the time of
observation the aircraft was on a magnetic heading (MH) of 276o with the
magnetic variation (VAR) 10oE?
A – 046o
B – 086o
C – 226o
D – 026o
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: A
8701
...
What is the true bearing of the aircraft from the island if at the time of
observation the aircraft was on a magnetic heading (MH) of 020o with the
magnetic variation (VAR) 25o W?
A – 145o
B – 195o
C – 205o
D – 325o
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: A
8713
...

Which is the best course of action?
A – Set heading towards a line feature – coastline, river, or motorway
B – Turn round and fly your flight plan tracks in reverse until you see
something you recognised before
C – Fly a series of ever-expanding circles from your present position till you
find your next check point
D – Turn round and fly your flight plan in reverse back to base
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: A

11106
...
If the heading of the aircraft is 355o (M) and the
magnetic variation is 15o East, the true bearing of the aircraft from the feature
is:
A – 160o
B – 220o
C – 310o
D – 130o
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: A

061-05-02 Navigation in climb and descent
8629
...
5 NM
B – 7
...
1 NM
D – 8
...
An aircraft is descending down a 12% slope whilst maintaining a GS of 540 kt
...
On a 12% glide slope, your ground speed is 540 knots
...
An aircraft at FL 350 is required to commence descent when 85 NM from a
VOR and to cross the VOR at FL 80
...

What is the minimum rate of descent required?
A – 1900 ft/min
B – 1800 ft/min
C – 1600 ft/min
D – 1700 ft/min
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: B
8654
...
The mean GS during the descent is 330
kt
...
An aircraft at FL 350 is required to descend to cross a DME facility at FL80
...

The minimum range from the DME at which descent should start is:
A – 79 NM
B – 69 NM
C – 49 NM
D – 59 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: B

8656
...
Assuming zero wind, what distance will be covered by an aircraft descending
15000 FT with a TAS of 320 kt and maintaining a rate of descent of 3000
ft/min?
A – 26
...
2 NM
C – 38
...
0 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: A
8665
...
Your mean groundspeed in the descent is 240 knots
...
An aircraft at FL 370 is required to commence descent when 100 NM from a
DME facility and to cross the station at FL 120
...
Given:
ILS GP angle = 3
...
At 0422 an aircraft at FL 370, GS 320 kt, is on the direct track to VOR X 185
NM distant
...
For a mean rate
of descent of 1800 ft/min at a mean GS of 232 kt, the latest time at which to
commence descent is:
A – 0448
B – 0445
C – 0451
D – 0454
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: B
8697
...
If the mean GS for the
descent is 335 kt, the minimum rate of descent required is:
A – 1390 ft/min
B – 1340 ft/min
C – 1240 ft/min
D – 1290 ft/min
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: B

8698
...

Maximum rate of descent is 2500 ft/min, mean GS during descent is 248 kt
...
An aircraft at FL 290 is required to commence descent when 50 NM from a
VOR and to cross that VOR at FL 80
...

What is the minimum rate of descent required?
A – 1700 ft/min
B – 2000 ft/min
C – 1900 ft/min
D – 1800 ft/min
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: C
11101
...
If the mean GS for the descent is 288
kt, the minimum rate of descent required is:
A – 960 ft/min
B – 860 ft/min
C – 890 ft/min
D – 920 ft/min
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: A

11102
...
Distance to be covered
during descent is 39 NM
...
An aircraft is descending down a 6% slope whilst maintaining a G/S of 300 kt
...
The outer marker of an ILS with a 3o glide slope is located 4
...
Assuming a glide slope height of 50 ft above the threshold, the
approximate height of an aircraft passing the outer marker is:
A – 1400 ft
B – 1450 ft
C – 1350 ft
D – 1300 ft
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: B

25143
...
Isogrivs on a chart indicate lines of:
A – zero magnetic variation
B – equal magnetic dip
C – equal horizontal directive force
D – equal grivation
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: D
8626
...
The reduction to TAS will be approximately:
A – 60 kt
B – 90 kt
C – 75 kt
D – 40 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: D

8628
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
You are on a heading of 105o, deviation 3 E WTD NDB (5211
...
0W)
bears 013R, CRK VOR (5150
...
7W) QDM is 211
...
An aircraft at FL 140, IAS 210 kt, OAT -5oC and wind component minus 35 kt,
is required to reduce speed in order to cross a reporting point 5 min later than
planned
...
An island appears 30o to the left of the centre line on an airborne weather radar
display
...
An aircraft is planned to fly from position A to position B, distance 480 NM at
an average GS of 240 kt
...
After flying 150 NM
along track from A, the aircraft is 2 min behind planned time
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
Kerry (5210
...
0W) is 41 nm DME, Galway 5318
...
5W) is 50
nm DME
...
Given:
Half way between two reporting points the navigation log gives the following
information:
TAS 360 kt
W/V 330o/80 kt
Compass heading 237o
Deviation on this heading -5o
Variation 19oW
What is the average ground speed for this leg?
A – 360 kt
B – 354 kt
C – 373 kt
D – 403 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: D
8650
...
83, temperature -30oC, is required to reduce speed in
order to cross a reporting point five minutes later than planned
...
76
B – M 0
...
78
D – M 0
...
TAS = 240 knots
The relative bearing from an NDB is 315R at 1410
...
What is your distance from the NDB at 1420?
A – 40 nm
B – 50 nm
C – 60 nm
D – 70 nm
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: A

8657
...
What is its position?
A – 5329N 00930W
B – 5239N 00830W
C – 5229N 00930W
D – 5329N 00830W
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: D
8660
...
What is its final position?
A – 2700N 17000W
B – 0000N 17000W
C – 2700N 17318W
D – 2700N 14300W
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: C
8661
...
A pilot receives the following signals from a VOR DME station: radial 180o+/1o, distance = 200 NM
...
5 NM
B - +/- 1 NM
C - +/- 2 NM
D - +/- 7 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: A
8670
...
Assuming flight conditions do not change, when 100 NM from the
reporting point IAS should be reduced to:
A – 169 kt
B – 165 kt
C – 159 kt
D – 174 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: C
8681
...
It departs A at 1200 UTC
...
Using the actual GS
experienced, what is the revised ETA at B?
A – 1401 UTC
B – 1333 UTC
C – 1347 UTC
D – 1340 UTC
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: B

8685
...
What is the QDR from the SHA VOR
(5243N 00853W)?
A – 217
B – 037
C – 209
D – 029
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: B
8694
...
It departs A at 0900 UTC
...
5 min behind planned time
...
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?
A – 8o left
B – 6o right
C – 4o right
D – 8o right
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: D

8702
...
A ground feature was observed on a relative bearing of 325o and five minutes
later on a relative bearing of 280o
...
When the relative bearing was 280o the
distance and true bearing of the aircraft from the feature was:
A – 30 NM and 240o
B – 40 NM and 110o
C – 40 NM and 290o
D – 30 NM and 060o
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: A
8711
...
An aircraft obtains a relative bearing of 315o from an NDB at 0830
...
Assuming no drift and a
GS of 240 kt, what is the approximate range from the NDB at 0840?
A – 50 NM
B – 40 NM
C – 60 NM
D – 30 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: B
11099
...
Given:
Distance A to B is 90 NM
Fix obtained 60 NM along and 4 NM to the right of course
What heading alteration must be made to reach B?
A – 4o Left
B – 16o Left
C – 12o Left
D – 8o Left
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: C

11103
...
An aircraft departs
position A and after having travelled 60 NM, its position is pinpointed 4 NM
left of the intended track
...
The distance between two waypoints is 200 NM
...
Assuming that
the forecast W/V applied, what will the off track distance be at the second
waypoint?
A – 0 NM
B – 7 NM
C – 14 NM
D – 21 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: C
11109
...
86, OAT -44oC, headwind component 110 kt, is
required to reduce speed in order to cross a reporting point 5 min later than
planned
...
79
B – M 0
...
75
D – M 0
...
Given: Distance A to B is 475 NM, Planned GS 315 kt, ATD 1000 UTC, 1040
UTC – fix obtained 190 NM along track
...
As the INS position of the departure aerodrome, co-ordinates 35o32
...
3W are input instead of 35o32
...
3E
...
4W
B – 099o32
...
6 E
D – 080o27
...
Given:
Distance A to B 1973 NM
Groundspeed out 430 kt
Groundspeed back 385 kt
Safe endurance 7 hr 20 min
The distance from A to the Point of Safe Return (PSR) A is:
A – 1664 nm
B – 1698 nm
C – 1422 nm
D – 1490 nm
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: D

15442
...
You are flying from A (30S 20E) to B (30S 20W)
...
Refer to figure 061-02)
What is the True bearing of point A from point B?
A – 000o
B – 090o
C – 270o
D – 360o
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: C

21692
...

What will be its latitude after 3 hrs?
A – 03o 50’S
B – 02o 00’S
C – 12o 15’S
D – 22o 00’S
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: B
24003
...
8 W00849
...
1 W00856
...
An island is observed to be 30o to the right of the nose of the aircraft
...
You are heading 080oT when you get a range and bearing fix from your AWR
on a headland at 185 nm 30o left of the nose
...
An aircraft starts from (S0400
...
2) and flies north for 2950 nm along
the meridian, then west for 382 nm along the parallel of latitude
...
An aircraft at latitude S0612
...
On completion of
the flight the latitude will be:
A – S2112
...
5
C – N0848
...
0
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: C
25190
...
0 E0450
...
On completion of the flight the latitude will be:
A – S1112
...
0
C – S0357
...
5
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: C

25191
...
0 W01000
...
0 W03000
...
An aircraft at position 6010
...
2W flies 165 km due East
...
0N 00812
...
0N 00212
...
0N 00110
...
0N 00110
...
An aircraft at position 0000N/S 16327W flies a track of 225oT for 70 nm
...
You are heading 345M, the variation is 20E, and you take a radar bearing of 30
left of the nose from an island
...
(Refer to Jeppesen Student Manual – chart E(LO)1 or figure 061-11)
What are the symbols at Galway Carnmore (5318
...
5W)?
A – VOR, NDB, DME, compulsory reporting point
B – Civil airport, NDB, DME, non-compulsory reporting point
C – Civil airport, VOR, DME, non-compulsory reporting point
D – VOR, NDB, DME, non-compulsory reporting point
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: B
8692
...
The flight log gives the following data: True track, Drift, True heading,
Magnetic variation, Magnetic heading, Compass deviation, Compass heading
...
(Refer to figure 061-06)
Complete line 6 of the ‘FLIGHT NAVIGATION LOG’, positions ‘L’ to ‘M’
...
(Refer to figure 061-06)
Complete line 3 of the ‘FLIGHT NAVIGATION LOG’, positions ‘E’ to ‘F’
...
(Refer to figure 061-06)
Complete line 2 of the ‘FLIGHT NAVIGATION LOG’, positions ‘C’ to ‘D’
...
(Refer to figures 061-06 and 061-05)
Complete line 1 of the ‘FLIGHT NAVIGATION LOG’; positions ‘A’ to ‘B’
...
Which of the following lists the first three pages of the FMC/CDU normally
used to enter data on initial start-up of te B737-400 Electronic Flight Intrument
System?
A – IDENT – RTE – DEPARTURE
B – POS INIT – RTE – IDENT
C – IDENT – POS INIT – RTE
D – POS INIT – RTE – DEPARTURE
Ref: AIR: atpl, cpl; HELI: atpl, cpl;
Ans: C
8645
...
In which of the following situations is the FMC present position of a B737-400
Electronic Flight Instrument System likely to be least accurate?
A – At top of descent
B – At top of climb
C – Just after take-off
D – On final approach
Ref: AIR: atpl; HELI: atpl, cpl;
Ans: C
8675
...
What is the validity period of the permanent data base of aeronautical
information stored in the FMC in the B737-400 Flight Management System?
A – 28 days
B – One calendar month
C – 3 calendar months
D – 14 days
Ref: AIR: atpl; HELI: atpl, cpl;
Ans: A
8689
...
Given:
Distance A to B is 325 NM
Planned GS 315 kt
ATD 1130 UTC
1205 UTC – fi obtained 165 NM along track
What GS must be maintained from the fix in order to achieve planned ETA at
B?
A – 335 kt
B – 375 kt
C – 395 kt
D – 355 kt
Ref: AIR: atpl; HELI: atpl, cpl;
Ans: D
8707
...
Which of the following can all be stored as five letter waypoint identifiers
through the CDU of a B737-400 Electronic Flight Instrument System?
A – Waypoint names; navaid frequencies; runway codes; airport ICAO
identifiers
B – Airway names; navaid identifiers; airport names; waypoint code numbers
C – Waypoint names; navaid identifiers; runway numbers; airport ICAO
identifiers
D – Waypoint names; navaid positions; airport ICAO identifiers; airport
names
Ref: AIR: atpl; HELI: atpl, cpl;
Ans: C

8710
...
What are the levels of message on the Boeing 737-400 FMC?
A – Urgent and Routine
B – Priority and Alerting
C – Alert and Advisory
D – Urgent and Advisory
Ref: AIR: atpl; HELI: atpl, cpl;
Ans: C
11096
...
What indication, if any, is given in the B737-400 Flight Management System
if radio updating is not available?
A – A warning message is displayed on the IRS displays
B – A warning message is displayed on the EHSI and MFDU
C – A warning message is displayed on the Flight Director System
D – No indication is given so long as the IRS positions remain within limits
Ref: AIR: atpl; HELI: atpl, cpl;
Ans: B
11112
...
An aeroplane flies from A (59oS 142oW) to B (61oS 148oW) with a TAS of
480 kt
...
On route AB, the true track:
A – varies by 10o
B – decreases by 6o
C – varies by 4o
D – increases by 5o
Ref: AIR: atpl; HELI: atpl, cpl;
Ans: D

24029
...
When can a pilot change the data in the FMS data base?
A – Every 28 days
B – When deemed necessary
C – When there is a fault
D – He can’t; for the pilot the FMS data base is read only
Ref: AIR: atpl; HELI: atpl, cpl;
Ans: D
25134
...
Which of the following can be input to the FMC using a maximum of 5
alphanumerics:
A – Waypoints, latitude and longitude and SIDs/STARs
B – ICAO aerodrome indicators, navigation facilities and SIDs/STARs
C – Waypoints, airway designators and latitude and longitude
D – Navigation facilities, reporting points and airway designators
Ref: AIR: atpl; HELI: atpl, cpl;
Ans: D

061-06

INERTIAL NAVIGATION SYSTEMS (INS)
061-06-01 Principles and Practical application

8715
...
An INS platform is kept at right angles to local gravity by applying corrections
for the effects of:
i
...

iii
...

v
...
The term drift refers to the wander of the axis of a gyro in:
A – the vertical and horizontal plane
B – the vertical plane
C – the horizontal plane
D – any plane
Ref: AIR: atpl;
Ans: C

8718
...
In an IRS:
A – the accelerometers are strapped down but the platform is gyro stabilised
B – the platform is strapped down but the accelerometers are gyro-stabilised
C – accelerometers and platform are both gyro-stabilised
D – accelerometers and platform are both strapped down
Ref: AIR: atpl;
Ans: D
8725
...
4 min
B – with damping and a period of 84
...
4 sec
D – with damping and a period of 84
...
With reference to inertial navigation systems, a TAS input is:
A – not required
B – required to provide a W/V read out
C – required for Polar navigation
D – required for rhumb line navigation
Ref: AIR: atpl;
Ans: B

8735
...
In a ring laser gyro, the purpose of the dither motor is to:
A – enhance the accuracy of the gyro at all rotational rates
B – overcome laser lock
C – compensate for transport wander
D – stabilise the laser frequencies
Ref: AIR: atpl;
Ans: B
8749
...
What measurement is used to carry out alignment of an Inertial Navigation
System?
A – Acceleration sensed by the east gyro horizontal accelerometer
B – Acceleration sensed by the north gyro horizontal accelerometer
C – Acceleration sensed by the north gyro ertical accelerometer
D – Difference in magnitude of the value of gravity compared with the gravity
at the last known position
Ref: AIR: atpl;
Ans: A

8764
...
What is the name given to an Inertial Reference System (IRS) which has the
gyros and accelerometers as part of the units fixture to the aircraft structure?
A – Solid state
B – Rigid
C – Strapdown
D – Ring laser
Ref: AIR: atpl;
Ans: C
8778
...
What is the name of the technique, effected by means of a piezoelectric motor, that is used to correct this error?
A – Dither
B – Cavity rotation
C – Zero drop
D – Beam lock
Ref: AIR: atpl;
Ans: A
8782
...
The resultant of the first integration from the north/south accelerometer of an
inertial navigation system (INS) in the NAV MODE is:
A – latitude
B – groundspeed
C – change latitude
D – velocity along the local meridian
Ref: AIR: atpl;
Ans: D
8792
...
Double integration of the output from the east/west accelerometer of an inertial
navigation system (INS) in the NAV MODE give:
A – distance north/south
B – vehicle longitude
C – distance east/west
D – velocity east/west
Ref: AIR: atpl;
Ans: C
8803
...
This means:
A – the system is mounted on a stabilised platform
B – the system is mounted and fixed to the aircraft structure
C – the accelerometers are fixed bu the gyros are stabilised
D – the gyros are fixed but the accelerometers are stabilised
Ref: AIR: atpl;
Ans: B

14665
...
This
means that:
A – only the gyros and not the accelerometers, become part of the units fixture
to the aircraft structure
B – gyros, and accelerometers are mounted on a stabilised platform in the
aircraft
C – gyros and accelerometers need satellite information input to obtain a
vertical reference
D – the gyroscopes and accelerometers become part of the units fixture to the
aircraft structure
Ref: AIR: atpl;
Ans: D
15450
...
After alignment of the stable platform of an Inertial Navigation System, the
output data from the platform is:
A – acceleration north/south and east/west and true heading
B – latitude, longitude and attitude
C – acceleration north/south and east/west, attitude and true heading
D – latitude, longitude and true heading
Ref: AIR: atpl;
Ans: C

24030
...
Inertial Reference System sensors include:
A – one east-west and one north-south gyro; one east-west and one northsouth accelerometer
B – accelerometers mounted in the direction of the aircraft axis
C – laser gyros mounted in the direction of the aircraft axis
D – accelerometers, and laser gyros, mounted in the direction of the aircraft
axis
Ref: AIR: atpl;
Ans: D
24051
...
The purpose of the TAS input, from the air data computer, to the Inertial
Navigation System is for:
A – position update in Attitude mode
B – the calculation of wind velocity
C – position update in Navigation mode
D – the calculation of drift
Ref: AIR: atpl;
Ans: B

25139
...
Which of the following statements concerning the aircraft positions indicated
on a triple fit Inertial Navigation System (INS)/Inertial Reference System (IRS)
on the CDU is correct?
A – The positions will only differ if one of the systems has been decoupled
because of a detected malfunction
B – The positions will be the same because they are an average of three
difference positions
C – The positions are likely to differ because they are calculated from
different sources
D – The positions will only differ if an error has been made when inputting the
present position at the departure airport
Ref: AIR: atpl;
Ans: C
8739
...
Alignment of INS and IRS equipments can take place in which of the following
modes?
A – ATT and ALIGN
B – NAV and ALIGN
C – ALIGN and ATT
D – NAV and ATT
Ref: AIR: atpl;
Ans: B
8759
...
During initial alignment an inertial navigation system is north aligned by inputs
from:
A – horizontal accelerometers and the east gyro
B – the aircraft remote reading compass system
C – computer matching of measured gravity magnitude to gravity magnitude
of initial alignment
D – vertical accelerometers and the north gyro
Ref: AIR: atpl;
Ans: A

8762
...
When initial position is put into an FMS, the system:
A – rejects initial latitude error, but it will accept longitude error
B – rejects initial longitude error, but it will accept latitude error
C – rejects initial latitude or longitude error
D – cannot detect input errors, and accepts whatever is put in
Ref: AIR: atpl;
Ans: C
8772
...
The alignment time, at mid-latitudes, for an Inertial Reference System using
laser ring gyros is approximately:
A – 5 min
B – 20 min
C – 2 min
D – 10 min
Ref: AIR: atpl;
Ans: D
8794
...
Following this incident:
A – everything returns to normal and is usable
B – no useful information can be obtained from the INS
C – it can only beused for attitude reference
D – the INS is usable in NAV MODE after a position update
Ref: AIR: atpl;
Ans: C
8795
...
When and where are IRS positions updated?
A – During all phases of flight
B – Only on the ground during the alignment procedure
C – When the FMS is in IRS ONLY NAV operation
D – When the VHF Nav Radios are selected to AUTO
Ref: AIR: atpl;
Ans: B

24009
...
The data that needs to be inserted into an Inertial Reference System in order to
enable the system to make a successful alignment for navigation is:
A – airport ICAO identifier
B – aircraft heading
C – the position of an in-range DME
D – aircraft position in latitude and longitude
Ref: AIR: atpl;
Ans: D
24044
...
The drift of the azimuth gyro on an inertial unit induces an error in the position
given by this unit
...
The total error is:
A – sinusoidal
B – proportional to the square of time, t?
C – proportional to t/2
D – proportional to t
Ref: AIR: atpl;

Ans: D
8799
...
01o/hr
...
The platform of an inertial navigation system (INS) is maintained at right
angles to the local vertical by applying corrections for the effects of:
A – gyroscopic inertia, earth precession and pendulous oscillation
B – vertical velocities, earth precession, centrifugal forces and transport drift
C – movements in the yawing plane, secondary precession and pendulous
oscillation
D – aircraft manoeuvres, earth rotation, transport wander and coriolis
Ref: AIR: atpl;
Ans: D
15449
...
Comparing the Present Position display on the Boeing 737-400 FMC, you
note that there is a 10-mile difference between the left IRS and the right IRS
positions
...
Within the platform levelling loop of an earth-vertical referenced INS:
A – The levelling signals are unbounded, with a period of 84
...
4 minutes
C – The levelling signals are unbounded, with a period of 84
...
4 seconds
Ref: AIR: atpl;
Ans: B

061-06-04 Flight deck equipment and operation
8720
...
Both inertial navigation systems are navigating from
waypoint A to B
...
2 is drifting
C – only inertial navigation system No
...
ATT Mode of the Inertial Reference System (IRS) is a back-up mode
providing:
A – only attitude and heading information
B – only attitude information
C – navigation information
D – altitude, heading and position information
Ref: AIR: atpl;
Ans: A
8734
...
In the Boeing 737-400 FMS, the CDU is used to:
A – manually initialise the IRS and FMC with dispatch information
B – automatically initialise the IRS and FMC with dispatch information
C – manually initialise the Flight Director System and FMC with dispatch
information
D – manually initialise the Flight Director System, FMC and Autothrottle with
dispatch information
Ref: AIR: atpl;
Ans: A

8737
...
What are the positions (in the order left to right) on the Boeing 737-400 IRS
MSU mode selector?
A – OFF STBY ALIGN NAV
B – OFF ON ALIGN NAV
C – OFF STBY ATT NAV
D – OFF ALIGN NAV ATT
Ref: AIR: atpl;
Ans: D
8744
...
On a triple-fit IRS system, present positions on the CDU:
A – will only differ if one IRS has been decoupled due to a detected
malfunction
B – will only differ if an initial input error of aircraft position has been made
C – are likely to differ as the information comes from different sources
D – will not differ as the information is averaged
Ref: AIR: atpl;
Ans: C

8750
...
Waypoints can be entered in an INS memory in different formats
...
When is the last point at which an INS or IRS may be selected to NAV mode?
A – After passengers and freight are aboard
B – Immediately prior to push back or taxi from the gate
C – At the holding point
D – On operation of the TOGA switch when opening the throttles for the takeoff
Ref: AIR: atpl;
Ans: B
8758
...
On the IRS, selection of ATT mode gives?
A – attitude and heading
B – altitude, heading, and groundspeed
C – altitude, attitude, and heading
D – attitude information only
Ref: AIR: atpl;
Ans: A
8777
...
Gyro-compassing of an inertial reference system (IRS) is accomplished with the
mode selector switched to:
A – ATT/REF
B – STBY
C – ALIGN
D – ON
Ref: AIR: atpl;
Ans: C
8791
...
What method of entering waypoints can be used on all INS equipments?
A – Distance and bearing
B – Waypoint name
C – Navaid identifier
D – Latitude and longitude
Ref: AIR: atpl;
Ans: D

061-06-05 INS operation
8719
...
In what formats can created waypoints be entered into the scratch pad of the
B737-400 FMS?
A – Place Bearing/Distance, Place Distance/Place Distance, Along-Track
Displacement, Latitude and Longitude
B – Place Bearing/Distance, Place Bearing/Place Bearing, Across-Track
Displacement, Latitude and Longitude
C – Place Bearing/Distance, Place Bearing/Place Bearing, Along-Track
Displacement, Latitude and Longitude
D – Place, Place Bearing/Distance, Along-Track Displacement, Latitude and
Longitude
Ref: AIR: atpl;
Ans: C

8724
...

WPT 1:60oN 30oW; WPT 2:60oN 20oW; WPT 3:60oN 10oW
The inertial navigation system is connected to the automatic pilot on route (12-3)
...
The automatic flight control system (AFCS) in an aircraft is coupled to the
guidance outputs from an inertial navigation system (INS)
...
3 (55o00N 020o00W) and No
...
With DSRTK/STS selected on the CDU, to the nearest
whole degree, the initial track read-out from waypoint No
...
The sensors of an INS measure:
A – precession
B – velocity
C – the horizontal component of the earth’s rotation
D – acceleration
Ref: AIR: atpl;
Ans: D

8738
...
Where and when are the IRS positions updated?
A – During flight IRS positions are automatically updated by the FMC
B – Only on the ground during the alignment procedure
C – IRS positions are updated by pressing the Take-off/Go-around button at
the start of the take-off roll
D – Updating is normally carried out by the crew when over-flying a known
position (VOR station or NDB)
Ref: AIR: atpl;
Ans: B
8747
...
Assuming that flight conditions do not change, when 150 nm from the
reporting point the speed must be reduced by:
A – 15 knots
B – 25 knots
C – 30 knots
D – 20 knots
Ref: AIR: atpl;
Ans: D

8753
...
Which pair of latitudes will give the greatest
difference between initial track read-out and the average true course given, in
each case, a difference of longitude of 10o?
A – 30oS to 25oS
B – 60oN to 50oN
C – 30oS to 30oN
D – 60oN to 80oN
Ref: AIR: atpl;
Ans: D
8769
...
The co-ordinates of A (45oS 010oW) and B (45oS
030oW) have been entered
...
The automatic flight control system (AFCS) in an aircraft is coupled to the
guidance outputs from an inertial navigation system (INS) and the aircraft is
flying from waypoint No
...
3 (60o00S 080o00W)
...
Which of the following statements concerning the position indicated on the
Inertial Reference System (IRS) display is correct?
A – It is updated when go-around is selected on take-off
B – It is constantly updated from information obtained by the FMC
C – It is not updated once the IRS mode is set to NAV
D – The positions from the two IRSs are compared to obtain a best position
which is displayed on the IRS
Ref: AIR: atpl;
Ans: C
8791
...
An aircraft is flying with the aid of an inertial navigation system (INS)
connected to the autopilot
...
0’N
B – 59o 49
...
0’N
D – 60o 05
...
What is the sequence of pages on start-up of the Boeing 737-400 FMS?
A – POS INIT, IDENT, DEPARTURES
B – IDENT, POS INIT, RTE
C – POS INIT, RTE, IDENT
D – IDENT, POS INIT, DEPARTURES
Ref: AIR: atpl;
Ans: B



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