General Navigation • Chapter 32

Revision Questions

187 Questions • Tap any question to reveal the answer

Appendix Figures

Questions that reference these appendices are marked with ↑ Refer to Figure N above.

Figure 1
Figure 1 — Appendix A — Chart Symbols Reference
Figure 2
Figure 2 — Appendix B — Navigation Chart
Figure 3
Figure 3 — Appendix C — Flight Navigation Log

Questions

1

Pressure Altitude is 28 000 feet, OAT = -45°C, Mach No = 0.46, W/V = 270/85, Track = 200°T. What is the drift and groundspeed?

A18L / 252 knots
B15R / 310 knots
C17L / 228 knots
D17R / 287 knots
Show Answer
Correct Answer: C
A common mistake here is to set the W/V, then put 200 up at the 12 o’clock index and just read the drift and ground speed off. But that is the answer for a HEADING of 200. The question asks for a TRACK of 200. You now have to lay off for drift, which will give you a heading of 217 at the top, balanced by 17 degrees of port drift, giving a track of 200.
2

G/S = 240 knots, Distance go = 530 NM. What is time to go?

A27 minutes
B29 minutes
C2 h 05 m
D2 h 12 m
Show Answer
Correct Answer: D
3

‘ISA Dev = +30°C, Pressure alt = 5000 feet. What is true alt?

A4550 feet
B5600 feet
C4290 feet
D5320 feet
Show Answer
Correct Answer: B
We recommend that you use the CRP-5 for true altitude problems, especially if the question gives you the actual SAT instead of ISA deviation. But if you want to use the formula, the ISA deviation is +30 degrees: True Alt = 5000 + (+30 × 4/1000 × 5000) = 5600 feet which is pretty close to answer (b).
4

Course 040°(T), TAS 120 kt, Wind speed = 30 knots. From which direction will the wind give the greatest drift?

A220°(T)
B230°(T)
C235°(T)
D245°(T)
Show Answer
Correct Answer: D
You would expect the greatest drift when the wind direction makes the greatest angle with from the track or reciprocal direction (the JAA use the word ‘course’ to mean ‘desired track’). The reciprocal track of 040° is 220° and answer (d) is the furthest from 220°.
5

Required course 045°(T), W/V = 190 /30, FL = 55 @ ISA, Variation = 15°W. CAS = 120 knots. What is mag heading and G/S?

A052°(M) 154
B067°(M) 154
C037°(M) 154
D037°(M) 113
Show Answer
Correct Answer: B
6

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

A3720 NM
B5420 NM
C1788 NM
D2040 NM
Show Answer
Correct Answer: A
You should note that these longitudes are meridian and anti-meridian. This is the over-the-pole case.
7

You are flying 090°(C) heading. Deviation is 2W and Variation is 12E. Your TAS is 160 knots. You are flying the 070 radial outbound from a VOR and you have gone 14 NM in 6 minutes. What is the W/V?

A158°(T) / 51
B060°(T) / 50
C340°(T) / 25
D055°(T) / 25
Show Answer
Correct Answer: A
If your compass heading is 090°(C), you can fill in the following table: T V M D. C 12E 2W So your true heading is 100. VOR radials are always magnetic, so you are flying a true track of 082. Your TAS is 160. If you have gone 14 NM in 6 min, your G/S is 140. Now use your CRP-5 as in Chapter 6.
8

Please refer to Appendix B (attached at back). Assume a North Polar Stereographic chart whose grid is aligned with the Greenwich meridian. An aircraft flies from the geographic North Pole for a distance of 480 NM along the 110° E meridian, then follows a grid track of 154° for a distance of 300 NM. Its position is now approximately:

↑ Refer to Figure 2 (Appendix B) above

A78°45’N 087°E
B70°15’N 080°E
C79°15’N 074°E
D80°00’N 080°E
Show Answer
Correct Answer: D
This question is solved by scale drawing. Use the scale to draw a line 480 NM along the 110°E meridian from the North Pole, then plot a line 300 NM in a direction of 154°Grid.
9

The sensitivity of a direct reading magnetic compass is:

AInversely proportional to the horizontal component of the Earth’s magnetic field.
BInversely proportional to the vertical and horizontal components of the Earth’s magnetic field.
CInversely proportional to the vertical component of the Earth’s magnetic field.
DProportional to the horizontal component of the Earth’s magnetic field.
Show Answer
Correct Answer: D
10

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

A010°40’W
B000°15’E
C000°40’E
D002°10’W
Show Answer
Correct Answer: C
This is a departure question. A very similar example is given in Chapter 15
11

What is the definition of magnetic variation?

AThe angle between the direction indicated by a compass and Magnetic North.
BThe angle between True North and Compass North.
CThe angle between Magnetic North and True North.
DThe angle between Magnetic Heading and Magnetic North.
Show Answer
Correct Answer: C
12

At the magnetic equator:

ADip is zero
BVariation is always maximum
CDeviation is zero
DThe isogonal is an agonic line
Show Answer
Correct Answer: A
13

Which of these is a correct statement about the Earth’s magnetic field?

AIt has no effect on aircraft deviation.
BThe angle of dip is the angle between the vertical and the total magnetic force.
CIt may be temporary, transient, or permanent.
DIt acts as though there is a large blue magnetic pole in Northern Canada
Show Answer
Correct Answer: D
14

Where is a compass most effective?

AAbout midway between the equator and the Magnetic North Pole
BIn the region of the magnetic South Pole
CIn the region of the magnetic North Pole
DOn the geographic equator
Show Answer
Correct Answer: A
15

The value of variation:

Ais zero at the magnetic equator
Bcannot exceed 180°
Chas a maximum value of 45° E or 45° W
Dcannot exceed 180°
Show Answer
Correct Answer: B
16

Grivation is the combination of

AVariation and Deviation
BDeviation and the Agonic value
CVariation and Grid Convergence
DGrid Convergence and Deviation
Show Answer
Correct Answer: C
17

An aircraft leaves at 0900UTC on a 250 NM journey with a planned ground speed of 115 knots. After 74 NM the aircraft is 1.5 minutes behind the planned schedule. What is the revised ETA at the destination?

A1100
B1110
C1115
D1054
Show Answer
Correct Answer: C
74 NM @ the planned G/S of 115 knots = 38.6 min 250 NM @ the planned G/S of 115 knots = 130.4 min After 74 NM, the actual elapsed time = 40.1 min Therefore the flight time for 250 NM at the revised G/S = 130.4 × 40.1 / 38/6 This comes to 135 min. 0900UTC + 135 min = 1115UTC
18

A B C 30 NM 20 NM ATA A is 1010. ETA B is 1030. ETA C is 1043. ATA B is 1027. What is revised ETA C?

A1040
B1043
C1038
D1036
Show Answer
Correct Answer: C
ETA B and ETA C are not relevant. You fly 30 NM in 17 min. How long will it take you to fly the remaining 20 NM at the same ground speed?
19

Isogrivs are lines that connect positions that have:

Athe same grivation
Bthe same variation
C0°(M)agnetic dip
Dthe same horizontal magnetic field strength
Show Answer
Correct Answer: A
20

What is the average magnetic track and distance between WTD NDB (5211.3N 00705.0W) and FOY NDB (5234.0N 00911.7W)? - use your Jeppesen E(LO)1 Track Dist

A294 76
B286 76
C294 81
D286 81
Show Answer
Correct Answer: C
Plotting question. Solve by measurement on the chart.
21

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

Aset heading towards a line feature such as a coastline, river or motorway
Bfly the reverse of the heading being flown prior to becoming uncertain until a pinpoint is obtained
Cfly expanding circles until a pinpoint is obtained
Dfly reverse headings and associated timings until the point of departure is reached
Show Answer
Correct Answer: A
22

An aircraft is at FL140 with an IAS of 210 and a true OAT of -5°(C). The wind component is -35 knots. When the aircraft is at 150 NM from a reporting point, ATC request the crew to lose 5 minutes by the time they get to the beacon. How much do they need to reduce IAS?

A15 knots
B25 knots
C30 knots
D20 knots
Show Answer
Correct Answer: D
Find the present TAS, apply the wind and get the present G/S (230 knots) 150 NM @ 230 kt G/S is 39 min. You need to arrive 5 minutes later, so your new time to go is 44 min. The new required G/S will be 150 NM in 44 min, which is 205 kt G/S The wind should not change, so the new required TAS will be 240 Then either use FL140/-5°(C) in the airspeed window to convert the TAS to 190 kt CAS, or put the old TAS (264) against the old IAS (210), then against 240 TAS, you will see 190. 210 to 190 is a 20 knot reduction in IAS (strictly CAS, but the PEC should not change much in 20 knots). Answer (d).
23

X Y Z 30 NM 20 NM ATA X is 1420. ETA Y is 1447. ATA Y is 1450. What is new ETA Z?

A1503
B1508
C1510
D1512
Show Answer
Correct Answer: C
ETA Y is irrelevant. If it takes you 30 minutes to cover 30 NM, it will take you a further 20 minutes to cover the remaining 20 NM at the same ground speed.
24

Given: Airport elevation is 1000 feet. QNH is 988 hPa What is the approximate airport pressure altitude?

A320
B1680
C-320
D680
Show Answer
Correct Answer: B
The QNH is 988 hPa. The datum for Pressure Altitude is 1013 hPa, which is 25 hPa greater. Using 27 feet to one hPa, this equates to a distance of 675 feet. A greater static pressure occurs at a lower pressure level. Therefore the 1013 hPa pressure level is below sea level by 675 feet. The airport is 1000 feet above sea level, which means that it is 1675 feet above the 1013 hPa pressure level.
25

An aircraft starts at position 0410S 17822W and heads true north for 2950 NM, then turns 90 degrees right, and maintains a rhumb line track for 314 kilometres. What is its final position?

A5500N 17422W
B4500N 17422W
C5500N 17738E
D4500N 17738E
Show Answer
Correct Answer: B
The aircraft flies 2950 NM N, which takes it to 4500N 17822W. Now convert 314 km to NM and then use the departure formula.
26

You are heading 080°(T) when you get a range and bearing fix from your AWR on

Aheadland at 185 NM 30° left of the nose. What true bearing do you plot on the chart? / 050 from the headland, using the headland’s meridian
B050 from the headland, using the aircraft’s meridian
C230 from the headland, using the headland’s meridian
D230 from the headland, using the aircraft’s meridian
Show Answer
Correct Answer: D
The true bearing from you TO the headland is 050°(T). However, you are going to have to plot FROM the headland to the aircraft (the headland is on your map – the aircraft’s position is not!) The aircraft’s heading has been measured relative to the direction of True North at the aircraft’s position. If there is significant chart convergence between the aircraft’s position and the headland’s position, the direction of True North will be different at the headland. We must therefore plot the reciprocal bearing using the aircraft’s meridian paralleled through the headland’s position.
27

By what amount must you change your rate of descent given a 10 knot decrease in headwind on a 3° glide slope?

A50 feet per minute increase
B30 feet per minute increase
C50 feet per minute decrease
D30 feet per minute decrease
Show Answer
Correct Answer: A
28

You are on a heading of 105°(C), deviation 3°E. WTD NDB (5211.3N 00705.0W) bears 013°R, CRK VOR (5150.4N 00829.7W) QDM is 211°. What is your position? (Use Jeppesen E(LO)1)

A5245N 00757W
B5217N 00745W
C5412N 00639W
D5228N 00802W
Show Answer
Correct Answer: D
Plotting question. Solve by measurement on the chart.
29

The airport at 5211N 00706W is: (use Jeppesen E(LO)1)

AKerry
BCork
CShannon
DWaterford
Show Answer
Correct Answer: D
Plotting question. Solve by measurement on the chart.
30

In which months is the difference between apparent noon and mean noon the greatest?

ANovember and February
BJanuary and July
CMarch and September
DJune and December
Show Answer
Correct Answer: A
31

5 hours 20 minutes and 20 seconds time difference is equivalent to which change of longitude?

A81°30’
B78°15’
C79°10’
D80°05’
Show Answer
Correct Answer: D
Multiply by 15° per hour. The easiest way of dealing with minutes and seconds is by using the DMS function of a scientific calculator.
32

The main reason that day and night, throughout the year, have different durations is due to the:

AEarth’s rotation
Brelative speed of the Sun along the ecliptic
Cinclination of the ecliptic to the Equator
Dgravitational effect of the Sun and the Moon on the speed of rotation of the Earth
Show Answer
Correct Answer: C
The combination of the inclination of the ecliptic to the Equator and the orbit of the Earth round the Sun produces a constantly changing declination. The length of the hours of daylight at a given latitude varies with the declination of the Sun.
33

A Lambert’s Conical conformal chart has standard parallels at 63N and 41N. What is the convergence factor?

A.891
B.788
C.656
D.707
Show Answer
Correct Answer: B
On a Lambert chart, the Parallel of Origin is midway between the 2 Standard Parallels. The ‘constant of the cone’ is simply another term for the sine of the parallel of origin.
34

On a chart, 90.74 km is represented by 7.0 centimetres. What is the scale?

A1 / 700 000
B1 / 2 015 396
C1 / 1 296 400
D1 / 1 156 600
Show Answer
Correct Answer: C
See Example 1 in Chapter 16, which is the same type of question.
35

On a direct Mercator chart, great circles are shown as:

ACurves convex to the nearer pole
BStraight lines
CRhumb lines
DCurves concave to the nearer pole
Show Answer
Correct Answer: A
36

The scale on a Lambert’s conformal conic chart

Ais constant along a meridian of longitude
Bis constant along a parallel of latitude
Cvaries slightly as a function of latitude and longitude
Dis constant across the whole map
Show Answer
Correct Answer: B
The scale on a Lambert chart alters as latitude changes but, for a fixed latitude (i.e. along a parallel of latitude), it remains constant.
37

Please refer to Appendix A. What is the symbol for a DME?

↑ Refer to Figure 1 (Appendix A) above

A9
B10
C12
D4
Show Answer
Correct Answer: D
38

Reference Jeppesen E(LO)1, position 5211N 00706W, which of the following denotes all the symbols?

Amilitary airport, ILS, NDB
Bcivil airport, VOR, ILS
Cmilitary airport, VOR, ILS
Dcivil airport, ILS, NDB
Show Answer
Correct Answer: D
The Jeppesen conventions differ slightly from the ICAO ones. The key is given in the introduction to the Jeppesen Student Pilots’ Manual.
39

Heading is 156°(T), TAS is 320 knots, W/V is 130/45 and the Variation is 10°W. What is your magnetic track?

A170
B150
C160
D222
Show Answer
Correct Answer: A
40

You are heading 345°(M), the variation is 20°E, and you take a radar bearing of 30° left of the nose from an island. What bearing do you plot from the island?

A160°(T)
B155°(T)
C140°(T)
D180°(T)
Show Answer
Correct Answer: B
Apply 20°E variation to 345°(M) to get 005°(T). 30° left of this is a true bearing of 335°(T) from you TO the headland. However, you are going to have to plot FROM the headland to the aircraft (the headland is on your map – the aircraft’s position is not!) You therefore plot the reciprocal of 155°(T).
41

Your pressure altitude is FL55, the QNH is 998, and the SAT is +30°(C). What is density altitude?

A6980 feet
B7750 feet
C8620 feet
D10 020 feet
Show Answer
Correct Answer: C
As per the reply to Q3.
42

On a particular take-off, you can accept up to 10 knots tailwind. The runway QDM is 047, the variation is 17°E and the ATIS gives the wind direction as 210. What is the maximum wind strength you can accept?

A18 knots
B4 knots
C8 knots
D11 knots
Show Answer
Correct Answer: D
43

The agonic line:

Ais midway between the magnetic North and South poles
Bfollows the geographic equator
Cis the shorter distance between the respective True and Magnetic North and South poles
Dindicates zero variation
Show Answer
Correct Answer: D
44

On a 5% glide slope your groundspeed is 150 kt. What should be your rate of descent to maintain the glide slope?

A750 feet/min
B450 feet/min
C900 feet/min
D650 feet/min
Show Answer
Correct Answer: A
A 5% glide slope means that for every 100 knots you go forward horizontally, you go down 5 knots vertically. 5% of your forward speed of 150 is 7.5 knots vertically. Convert this to feet per min and you will see that option (a) is the nearest
45

At 65 NM from a VOR you commence a descent from FL330 in order to arrive over the VOR at FL80. Your mean ground speed in the descent is 240 knots. What rate of descent is required?

A1540 feet/min
B1630 feet/min
C1270 feet/min
D1830 feet/min
Show Answer
Correct Answer: A
65 NM @ 240 knots will take 16.25 minutes. I need to lose 25 000 feet, also in 16.25 minutes. The required ROD is 25 000/16.25 = 1338 feet per minute.
46

On the Jeppesen E(LO)1 chart, what are the symbols at Sligo (5354.8N 00849.1W)?

AVOR, NDB, DME, compulsory reporting point
Bcivil airport, NDB, DME, compulsory reporting point
Ccivil airport, VOR, DME, non-compulsory reporting point
DVOR, NDB, DME, non-compulsory reporting point
Show Answer
Correct Answer: B
The Jeppesen conventions differ slightly from the ICAO ones. The key is given in the introduction to the Jeppesen Student Pilots’ Manual.
47

In which month does aphelion occur?

AJanuary
BMarch
CJuly
DNovember
Show Answer
Correct Answer: C
48

What is the highest latitude listed below at which the Sun will rise above the horizon and set every day?

A68°N
B66°N
C62°N
D72°N
Show Answer
Correct Answer: B
The lowest latitude at which there is at least one day a year without a Sunset (Mid-summer Day) and one day a year without a Sunrise (Mid-winter Day) is the Arctic Circle (which is 66½°N). Therefore the Sun will rise and set every day at 62N and 66N. The higher of these two is 66N.
49

The pressure alt is 29 000 feet and the SAT is -55°(C). What is density altitude?

A27 500 feet
B26 000 feet
C30 000 feet
D31 000 feet
Show Answer
Correct Answer: A
(Pressure alt + (ISA dev × 120) gives 27 560; CRP5 gives 27 000)
50

What is the UTC time of Sunrise in Vancouver, British Columbia, Canada (49N 123 30W) on the 6th December? (In the exam, tables were supplied. The answers given below are based on the tables in your Gen Nav notes).

A2324 UTC
B0724 UTC
C1552 UTC
D0738 UTC
Show Answer
Correct Answer: C
Look up the LMT of Sunrise at 49N on the 6th December. There is 3 minutes change between the 4th Dec and the 7th Dec. So, interpolating, the times at 50N and 45N are 0742 and 0723 respectively. That is a difference of 19 minutes. One-fifth of that is about 4 minutes, so the LMT of Sunrise at 49N on the 6th December is 0738. Now set it out in a table: LMT sunrise at Vancouver 6 Dec 07 LMT Arc/time 123° 30’W (long. west, UTC best) - +08 - UTC 6 Dec 15 UTC Day Hour Minute Time To calculate arc/time, enter the degrees and minutes in DMS format into your calculator and divide by 15.
51

How does scale change on a normal Mercator chart?

AExpands as the cosine of the latitude
BExpands directly with the secant of the latitude
CCorrect on the standard parallels, expands outside them, contracts within them
DExpands as the secant of the E/W great circle distance
Show Answer
Correct Answer: B
52

You are on an ILS 3-degree glide slope which passes over the runway threshold at 50 feet. Your DME range is 25 NM from the threshold. What is your height above the runway threshold elevation? (Use the 1 in 60 rule and 6000 feet = 1 nautical mile)

A7450 feet
B[Option not in source]
C6450 feet
D7550 feet / 8010 feet
Instructor's Note: Source formatting error — one or more options appear duplicated or missing in the original text. Options shown as extracted; verify against printed copy.
Show Answer
Correct Answer: A
This is just the 1 in 60 rule. A 3-degree glide path gives you 300 feet per nautical mile. 25 NM @ 300 feet per NM is 7500 feet. That would be for a touch-down at zero feet at the runway threshold. However, height over the threshold is 50 feet, which has to be added.
53

When it is 0600 Standard Time in Queensland, Australia, what is the Standard Time in Hawaii, USA? (Disregard Summer Time)

A0200 ST
B0600 ST
C1000 ST
D1200 ST
Show Answer
Correct Answer: C
As always, use a table: Standard Time at Queensland Today 00 ST STD (long. east, UTC least) - -10 00 - UTC Yesterday 00 UTC STD (long. west, UTC best) - -10 00 - Standard Time at Hawaii Yesterday 1000 00 ST Hour Minute Time Day
54

Please refer to Appendix A. Which of the following is the symbol for an exceptionally high (over 1000 feet AGL) unlit obstruction?

↑ Refer to Figure 1 (Appendix A) above

A6
B9
C10
D15
Show Answer
Correct Answer: A
55

You are flying at a True Mach No. of .82 in a SAT of -45°(C). At 1000 hours you are 100 NM from the POL DME and your ETA at POL is 1012. ATC ask you to slow down to be at POL at 1016. What should your new TMN be if you reduce speed at 100 NM distance to go?

AM .76
BM .72
CM .68
DM .61
Show Answer
Correct Answer: D
56

The relative bearing to a beacon is 090°R. Three minutes later, at a ground speed of 180 knots, it has changed to 135°R. What was the distance of the closest point of approach of the aircraft to the beacon?

A45 NM
B18 NM
C9 NM
D3 NM
Show Answer
Correct Answer: C
57

Ground speed is 540 knots. 72 NM to go. What is time to go?

A8 min
B9 min
C18 min
D12 min
Show Answer
Correct Answer: A
58

An aircraft at position 2700N 17000W travels 3000 km on a track of 180°(T), then 3000 km on a track of 090°(T), then 3000 km on a track of 000°(T), then 3000 km on

Atrack of 270°(T). What is its final position? / 2700N 17318W
B0000N/S 17000W
C2700N 17000W
D2700N 14300W
Show Answer
Correct Answer: A
The first leg takes it down to the Equator at 170W, the second leg takes it to the Equator at 143W, the third leg takes it to 27N 143W and you then work out the change of longitude of 1620 NM due west departure at 27N, which is 30°18’ W, i.e. 17318W. However, you do not need to do the full calculation. You should realize that a departure of 3000 km will give a greater change of longitude at 27N than it will at the Equator. Therefore, the answer must be at 27N, but WEST of the 170W meridian. There is only one option which fits.
59

On the Jeppesen E(LO)1, Baldonnel (5318.0N 00626.9W) is 91 NM DME, Galway (5318.1N 00856.5W) is 50 NM DME. What is your position?

A5242N 00827W
B5230N 00834W
C5255N 00819W
D5219N 00809W
Show Answer
Correct Answer: B
Plotting question. Solve by measurement on the chart.
60

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

A920 ft/min
B890 ft/min
C860 ft/min
D960 ft/min
Show Answer
Correct Answer: D
You have 120 NM horizontal distance to travel at 288 knots. This will take 25 minutes. You need to lose 24 000 feet in 25 minutes, which is 960 feet per minute.
61

You are homing to overhead a VORTAC and will descend from 7500 QNH to be 1000 AMSL by 6 NM DME. Your ground speed is 156 knots and the ROD will be 1000 feet/min. At what range from the VORTAC do you commence the descent?

A22.9 NM
B15.8 NM
C16.9 NM
D30.2 NM
Show Answer
Correct Answer: A
You need to lose 6500 feet at a rate of descent of 1000 feet per minute. This will take 6.5 minutes. Flying for this time at a ground speed of 156 knots gives a distance of 16.9 NM. However, this is 6 NM before the DME, so the total distance to start the descent is 22.9 NM.
62

A rhumb line is:

Athe vertex of a conformal polyformic projection / straight line on a Lambert’s conformal chart / line on the Earth which cuts all meridians at the same angle
B[Option not in source]
C[Option not in source]
Dthe shortest distance between two points on the Earth’s surface
Instructor's Note: Source formatting error — one or more options appear duplicated or missing in the original text. Options shown as extracted; verify against printed copy.
Show Answer
Correct Answer: C
63

You fly from 49N to 58N along the 180 E/W meridian. What is the distance in kilometres?

A540 km
B804 km
C1222 km
D1000 km
Show Answer
Correct Answer: D
This is a 9-degree change of latitude. You could work this out as 540 NM and then convert it to km, but there is a quicker way. 90 degrees of change of latitude is 10 000 km (see pages 28/29, so a 9 degree change of latitude will be 1000 km.
64

On a particular direct Mercator wall chart, ‘the full length of the parallel of latitude at 53N is 133 cm long. What is the scale of the chart at 30S?

A1: 30 000 000
B1: 18 000 000
C1: 21 000 000
D1: 26 000 000
Show Answer
Correct Answer: D
The thing to realize is that on a Mercator chart, meridians are drawn as parallel lines. Therefore the total length of the 53N parallel of latitude will be 133 cm as will be the length of the 30S parallel of latitude. So now find the departure from 180E to 180W at 30S. Dep = ch.long × cos 30 = 360 × 60 × cos 30 = 18 706 NM You now have a simple scale problem. The Chart length is 133 cm, the Earth Distance is 18 706. This gives you a scale of approximately 1:26 million.
65

What is the highest latitude on the Earth at which the Sun can be vertically overhead?

A23½°
B66½°
C45°
D90°
Show Answer
Correct Answer: A
The Sun would be at its Zenith.
66

Track = 090°(T), TAS = 460 knots, W/V = 360°(T) / 100, Variation = 12°E, Deviation = -2. What is compass heading and ground speed?

A079° 470 knots
B067° 450 knots
C068° 460 knots
D070° 455 knots
Show Answer
Correct Answer: B
Find the true heading in the normal way, then apply variation and deviation. First, check that you have balanced the drift. When deviation is given as East or West then you use the DEVIATION EAST COMPASS LEAST RULE. However, deviation (not variation, only deviation) is sometimes quoted as plus or minus-. In this case the rule is that deviation is what you apply to the compass to get magnetic (not the other way round). So EAST deviation is plus and WEST deviation is minus.
67

Please refer to Appendix A. What symbol is used to show multi unlit obstacles on a map/chart?

↑ Refer to Figure 1 (Appendix A) above

A5
B7
C13
D11
Show Answer
Correct Answer: D
68

The angle between True North and Magnetic North is known as:

Adeviation
Bdip
Calignment error
Dvariation
Show Answer
Correct Answer: D
69

An aircraft is at 10°N and is flying North at 444 km/hour. After 3 hours the latitude is:

A10°S
B02°S
C22°N
D00°N/S
Show Answer
Correct Answer: C
3 × 444 is 1332 km. Divide by 1.852 to get 719.2 NM. This has obviously been approximated to 720 to give 12° change of latitude, which takes us to latitude 02°S.
70

Given that: A is N55° E/W 000° B is N54° E 010°, If the average true course of the great circle is 100°(T), the true course of the rhumb line at point A is:

A096°
B107°
C104°
D100°
Show Answer
Correct Answer: D
The average great circle track is the rhumb line track.
71

The circumference of the Earth is approximately:

A40 000 NM
B10 800 NM
C5400 NM
D21 600 NM
Show Answer
Correct Answer: D
You should remember that the distance from the Equator to a Pole is 5400 NM (90° change of latitude × 60 NM per degree). So the Earth’s circumference is 4 × 5400, which is 21 600 NM. Alternatively, multiply 360° (a meridian plus the associated anti-meridian) by 60.
72

The angle between the plane of the Equator and the plane of the Ecliptic is:

A66.5°
B23.5°
C25.3°
D65.6°
Show Answer
Correct Answer: B
73

Position A is at 70S 030W, position B is 70S 060E. What is the great circle track of B from A, measured at A?

A132°(T)
B048°(T)
C090°(T)
D228°(T)
Show Answer
Correct Answer: A
The rhumb line track from 70S to 70S is along a parallel of latitude, so it will be 090(T). However, the great circle track will ‘cut the corner’ and take the shortest route. This will be to the south of 090(T). Draw it out on a Lambert’s style projection, or try it on a globe. The conversion angle will be 42 degrees, so this makes the initial great circle track 132(T).
74

The value of magnetic variation on a chart changes with time. This is due to:

AMovement of the magnetic poles, causing an increase
BIncrease in the magnetic field, causing an increase
CReduction in the compass deviation, causing a decrease
DMovement of the magnetic poles, which can cause either an increase or a decrease
Show Answer
Correct Answer: D
75

Isogonal lines converge as follows:

AAt the North Magnetic Pole
BAt the North and South Magnetic and both Geographical Poles
CAt the North and South Magnetic Poles
DAt the Magnetic equator.
Show Answer
Correct Answer: B
76

Position A is 55N 30W. Position B is 54N 20W. The great circle track from A to B, measured at A, is 100°(T). What is the rhumb line bearing from A to B?

A104°(T)
B090°(T)
C100°(T)
D284°(T)
Show Answer
Correct Answer: A
77

An aircraft departs a point 0400N 17000W and flies 240 NM South, followed by 240 NM East, then 240 NM North, then 240 NM West. What is its final position?

A0400N 17000W
B0600S 17000W
C0400N 170°35.9’W
D0400N 170°01.8’W
Show Answer
Correct Answer: C
6°S is a greater value of latitude than 4°N. Therefore, for a given departure, you get a greater change of longitude eastwards. When you come back to 4°N again, 600 NM does not take you so far westwards. So you will finish up east of where you started.
78

At 1000 hours an aircraft is on the 310 radial from a VOR/DME, at 40 nautical miles range. At 1015 the radial and range are 040/40 NM. What is the aircraft’s track and ground speed?

A080° / 226 knots
B085° / 226 knots
C080° / 56 knots
D085° / 90 knots
Show Answer
Correct Answer: B
79

A straight line is drawn on a North Polar Stereographic chart joining Point A (7000N 06000W) to Point B (7000N 06000E). What is the initial track direction (going eastwards) of the line at A ?

A090°(T)
B030°(T)
C120°(T)
D330°(T)
Show Answer
Correct Answer: B
80

What is the maximum possible value of Dip Angle at either Pole?

A66°
B180°
C90°
D45°
Show Answer
Correct Answer: C
81

Given: Magnetic heading 311° Drift is 10° left Relative bearing of NDB 270 What is the magnetic bearing of the NDB measured from the aircraft?

A221°
B208°
C211°
D180°
Show Answer
Correct Answer: A
82

The initial straight track from A(75N 60E) to B(75N 60W) on a Polar Stereographic chart is:

A030°
B360°
C[Option not in source]
D060° / 330°
Instructor's Note: Source formatting error — one or more options appear duplicated or missing in the original text. Options shown as extracted; verify against printed copy.
Show Answer
Correct Answer: D
83

Given: Runway direction 083°(M), Surface W/V 035/35 kt. Calculate the effective crosswind component.

A24 kt
B26 kt
C31 kt
D34 kt
Show Answer
Correct Answer: B
CRP-5
84

Given: For take-off an aircraft requires a headwind component of at least 10 kt and has a cross-wind limit of 35 kt. The angle between the wind direction and the runway is 45°. Calculate the maximum and minimum allowable wind speeds.

A15 kt and 50 kt
B15 kt and 43 kt
C12 kt and 38 kt
D18 kt and 50 kt
Show Answer
Correct Answer: A
85

What is the weight in kilograms of 380 US Gallons at a Specific Gravity of 0.78?

A1123
B2470
C5434
D543
Show Answer
Correct Answer: A
Either use the CRP-5 or remember that there are 5 imperial gallons to 6 US.gal and the imp.gal to litres conversion is 4.55. Then multiply by .78 to get the weight in kg, i.e. 380 × 5/6 × 4.55 × .78 = 1123.85 It is probably easier with the CRP-5.
86

You leave A to fly to B, 475 NM away, at 1000 hours. Your ETA at B is 1130. At 1040, you are 190 NM from A. What ground speed is required to arrive on time at B?

A342 knots
B330 knots
C317 knots
D360 knots
Show Answer
Correct Answer: A
Distance still remaining = 475 – 190 = 285 NM. Time to go = 1130 - 1040 = 50 min
87

What is the aircraft position in lat and long given the following (use Jeppesen chart E(LO)1): CRN (5318N 00857W) 18 DME SHA (5243N 00853W) 20 DME Heading 270(M) Both ranges DME increasing

A5201N 00908W
B5301N 00908W
C5302N 00843W
D5203N 00843W
Show Answer
Correct Answer: B
Plotting question. Solve by measurement on the chart. An explanation of the point about ‘both DME distances decreasing’ is given in the Plotting chapter.
88

What is the NDB frequency and ident at 5211N 00932W? - Use E(LO)1)

AKerry NDB - 113MHz and KER
BCork NDB - 343KHz and OC
CWaterford NDB - 368MHz and WTD
DKerry NDB - 334KHz and KER
Show Answer
Correct Answer: D
The Jeppesen conventions differ slightly from the ICAO ones. The key is given in the introduction to the Jeppesen Student Pilots’ Manual.
89

What is the mean magnetic track and distance from the BAL VOR (5318N 00627W) to CFN NDB (5502N 00820W)? - (use E(LO)1)

A335° 125
B148° 125
C328° 134
D148° 134
Show Answer
Correct Answer: A
Plotting question. Solve by measurement on the chart.
90

You are at position 5340N 00800W. What is the QDR from the SHA VOR (5243N 00853W)? - (Use E(LO)1)

A217
B037
C209
D029
Show Answer
Correct Answer: B
Plotting question. Solve by measurement on the chart.
91

Your radial from the SHA VOR (5243N 00853W) is 120°(M). From the CRK VOR (5151N 00830W), the radial is 033°(M). What is your position? - (use E(LO)1)

A5230N 00820W
B5240N 00821W
C5220N 00821W
D5230N 00800W
Show Answer
Correct Answer: D
Plotting question. Solve by measurement on the chart.
92

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 10° change of longitude?

A60N and 60S
B60N and 60N
C30S and 30N
D30S and 25S
Show Answer
Correct Answer: B
The mean great circle is the same as the rhumb line track. The question is asking which pairs of latitudes will give the greatest difference between great circle and rhumb line track, i.e. which will give the greatest conversion angle? The ca formula is 1/2 ch.long sin mean lat. If you keep the 1/2 ch.long constant for all cases, then you are asking which pairs of latitudes will give you the greatest sine mean lat.
93

An aircraft is at 5530N 03613W, where the variation is 15W. It is tuned to a VOR located at 5330N 03613W, where the variation is 12W. What VOR radial is the aircraft on?

A348
B012
C165
D015
Show Answer
Correct Answer: B
The longitudes are the same so, from the given lats and longs, the true bearing of the aircraft from the VOR is 000/360. For VOR, you take the variation at the VOR, because the bearing is measured at the beacon. (For an ADF bearing, it would be the variation at the aircraft, because that is where the variation is measured).
94

The wind velocity is 359/25. An aircraft is heading 180 at a TAS of 198 knots. (All directions are True). What is its track and ground speed?

A179 223
B179 220
C180 220
D180 223
Show Answer
Correct Answer: D
The air vector is 5 times as long as the wind vector, so the direction will be nearer 180 than 179. Try drawing the triangle of velocities out as a diagram.
95

An aircraft’s compass must be swung:

AIf the aircraft has been in the hangar for a long time and has been moved several times.
BIf the aircraft has been subjected to hammering.
CEvery maintenance inspection
DAfter a change of theatre of operations at the same magnetic latitude.
Show Answer
Correct Answer: B
96

Civil Twilight occurs between:

ASunset and 6° below the horizon
B6° and 12° below the horizon
C12° and 18° below the horizon
DSunrise and Sunset
Show Answer
Correct Answer: A
97

What is the dip angle at the North Magnetic Pole?

A
B90°
C180°
D64°
Show Answer
Correct Answer: B
98

What is a line of equal grivation?

AAn isocline
BAn isogonal
CAn isogriv
DAn isovar
Show Answer
Correct Answer: C
99

What is the reason for seasonal changes in climate?

ABecause the Earth’s spin axis is inclined to the plane of its orbit round the Sun
BBecause the distance between the Earth and the Sun varies over a year
CBecause the Earth’s orbital speed round the Sun varies according to the time of the year
DBecause of the difference between the Tropical Year and the Calendar Year
Show Answer
Correct Answer: A
The combination of the inclination of the ecliptic to the Equator and the orbit of the Earth round the Sun produces a constantly changing declination. The predominant cause of the seasons is the inclination (tilt) of the Earth.
100

The aircraft position is at 5330N 00800W. The VORs are tuned to Shannon (SHA, 5243N 00853W) and Connaught (CON, 5355N 00849W). Which radials will be indicated? SHA. CON

A033 130
B221 318
C042 138
D213 310
Show Answer
Correct Answer: C
Radials radiate. That is what the word means. In other words they are, by definition, the magnetic bearing FROM the VOR. On the VOR in the aircraft you have a needle with 2 ends. The sharp end points to the VOR and is the QDM, or mag track to the VOR and the other end points to the QDR, or radial, which is the mag bearing from the VOR.
101

You are on the 205 radial from the Shannon VOR (SHA, 5243N 00853W) and on the 317 radial from Cork VOR (CRK, 5150N 00830W). What is the aircraft position?

A5205N 00915W
B5215N 00917W
C5118N 00913W
D5210N 00909W
Show Answer
Correct Answer: D
Plotting question. Solve by measurement on the chart.
102

What is the radial and DME distance from Connaught VOR/DME (CON, 5355N 00849W) to overhead Abbey Shrule aerodrome (5336N 00739W)?

A304 47 NM
B124 47 NM
C296 46 NM
D116 46 NM
Show Answer
Correct Answer: B
Plotting question. Solve by measurement on the chart.
103

What is the average magnetic track and distance between Kerry NDB (KER, 5211N 00932W) and CarNMore NDB (CRN, 5318N 00856W)?

A025 70 NM
B197 71 NM
C017 70 NM
D205 71 NM
Show Answer
Correct Answer: A
Plotting question. Solve by measurement on the chart.
104

What is the approximate course (T) and distance between Waterford NDB (WTD, 5212N 00705W) and Sligo NDB (SLG, 5417N 00836W)?

A344 139 NM
B164 138 NM
C156 136 NM
D336 137 NM
Show Answer
Correct Answer: D
Plotting question. Solve by measurement on the chart.
105

What is the rhumb line track from A (4500N 01000W) to B (4830N 00500W)?

A045°T
B030°T
C225°T
D150°T
Show Answer
Correct Answer: A
106

What is the effect on the Mach number and TAS in an aircraft that is climbing with constant CAS?

AMach number decreases; TAS decreases
BMach number increases; TAS remains constant
CMach number increases; TAS increases
DMach number remains constant; TAS increases
Show Answer
Correct Answer: C
As the air becomes less dense, the aircraft has to fly faster through it (TAS) to experience the same dynamic pressure (CAS). So the TAS is increasing, which means that the Mach No. would also increase even if there were no temperature change. However, in addition, temperature normally decreases with increasing altitude. This means that the speed of sound will decrease so, for a given TAS, the Mach No. will increase, giving an additional effect.
107

Please refer to Appendix A. What is the chart symbol for a lightship?

↑ Refer to Figure 1 (Appendix A) above

A6
B8
C9
D12
Show Answer
Correct Answer: B
108

Given: Track 198°, Heading 184°, TAS 427 kt and GS 453kt, what are the W/V and Drift Angle?

A280°/110kt and 14°P
B087°/109kt and 14°S
C116°/110kt and 14°S
D294°/110kt and 14°P
Show Answer
Correct Answer: B
109

An aircraft is on the 025 radial from Shannon VOR (SHA, 5243N 00853W) at 49 DME. What is its position?

A5329N 00930W
B5239N 00830W
C5229N 00930W
D5329N 00830W
Show Answer
Correct Answer: D
Plotting question. Solve by measurement on the chart.
110

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

A268
B302
C088
D122
Show Answer
Correct Answer: C
Apply 17°W variation to 120°(M) to get 103°(T) heading. The island is 15°(T)o the left, which makes the true bearing TO the island 088°(T).
111

An aircraft is flying around the Earth eastwards along the 60N parallel of latitude at a ground speed of 360 knots. At what ground speed would another aircraft have to fly eastwards along the Equator to fly once round the Earth in the same journey time?

A600 knots
B240 knots
C720 knots
D120 knots
Show Answer
Correct Answer: C
This is a departure problem. In one hour, the aircraft covers 360 NM. The departure formula is: Departure = change of longitude (minutes) × cosine latitude 360 NM = change of longitude (minutes) × cosine 60 (which is 0.5) Change of longitude = 720 minutes. At the Equator, 720 minutes = 720 NM, which also has to be covered in one hour.
112

If it is 0700 hours Standard Time in Kuwait, what is the Standard Time in Algeria?

A0500 hours
B0900 hours
C1200 hours
D0300 hours
Show Answer
Correct Answer: A
As always, use a table: Standard Time at Kuwait Today 00 ST STD (long. east, UTC least) - -3 00 - UTC Today 00 UTC STD (long. east, UTC least) +1 00 - Standard Time at Algeria Today 0500 00 ST Hour Minute Time Day
113

If variation is East, then:

ATrue North is West of Magnetic North
BCompass North is West of Magnetic North
CTrue North is East of Magnetic North
DMagnetic North is West of Compass North
Show Answer
Correct Answer: A
114

At what latitude does the maximum difference between geodetic and geocentric latitude occur?

A
B45°
C60°
D90°
Show Answer
Correct Answer: B
115

At what times of the year does the length of the hours of daylight change most rapidly?

ASpring Equinox and Autumn Equinox
BSummer Solstice and Winter Solstice
CSpring Equinox and Summer Solstice
DAutumn Equinox and Winter Solstice
Show Answer
Correct Answer: A
116

Given: Aircraft height = 2500 feet, ILS GP angle = 3°, at what approximate distance from the threshold can you expect to intercept the glide-path?

A8.0 NM
B14.5 NM
C13.1 NM
D7.0 NM
Show Answer
Correct Answer: A
This is just the 1 in 60 rule. A 3-degree glide path gives you 300 feet per nautical mile. Divide 2500 feet by 300 feet and the answer is just over 8 nautical miles.
117

Convert 80 metres/sec into knots.

A155 knots
B55 knots
C160 knots
D16 knots
Show Answer
Correct Answer: A
To convert metres into NM, divide by 1852. This gives you NM per second. Now multiply by 60 × 60 to convert to NM per hour, i.e. knots. × 3600/1852 = 155 knots
118

The chart that is generally used for navigation in polar areas is based on a:

ADirect Mercator Projection
BGnomonic projection
CLambert conformal projection
DStereographic projection
Show Answer
Correct Answer: D
Polar navigation can be (and often is) carried out on a Polar Stereographic, a Transverse Mercator or an Oblique Mercator, in fact, but the JAA consider Polar Stereographic to be the preferred answer.
119

Which of the following conversions from True to Compass is the correct one? T V M D. C

A130 2W 132 -1 131
B130 2E 132 -1 133
C130 2W 132 -1 133
D130 2E 132 -1 131
Show Answer
Correct Answer: C
120

Your position is 5833N 17400W. You fly exactly 6 NM westwards. What is your new position?

A5833N 17411.5W
B5833N 17355W
C5833N 17340W
D5833N 17348.5W
Show Answer
Correct Answer: A
This is a departure problem. The departure formula is: Departure = change of longitude (minutes) × cosine latitude 6 NM = change of longitude (minutes) × cosine 58°33’ Change of longitude = 6/.5218 = 11.5 minutes (westwards) from 174°00’W
121

TAS = 240 knots. Track is 180°(T). The relative bearing from an NDB is 315(R) at 1410. At 1420 the bearing has changed to 270(R). What is your distance from the NDB at 1420?

A40 NM
B50 NM
C60 NM
D70 NM
Show Answer
Correct Answer: A
This is covered in detail in the ‘General Navigation Problems’ Chapter. The numbers are different but the principle is the same. If the question does not give you any wind information, you can only assume that TAS = G/S.
122

Given: True Track = 352 Variation = 11W Deviation = - 5 Drift = 8°R What is Heading (C)?

A078°(C)
B346°(C)
C000°(C)
D025°(C)
Show Answer
Correct Answer: C
Do not forget that a deviation of -5° is equivalent to 5°W.
123

What is the radial and DME distance from CRK VOR (5151N 00830W) to position 5220N 00910W?

A322(M) 39 NM
B330(M) 41 NM
C330(M) 39 NM
D322(M) 41 NM
Show Answer
Correct Answer: C
Plotting question. Solve by measurement on the chart.
124

What is the radial and DME distance from SHA VOR (5243N 00853W) to Birr airport (5304N 00755W)?

A068(M) 42 NM
B060(M) 40 NM
C068(M) 40 NM
D060(M) 42 NM
Show Answer
Correct Answer: A
Plotting question. Solve by measurement on the chart.
125

What is the lat and long of the SHA VOR (5243N 00853W) 239(M)/36 NM radial/ range?

A5215N 00930W
B5220N 00937W
C5212N 00930W
D5212N 00915W
Show Answer
Correct Answer: B
Plotting question. Solve by measurement on the chart.
126

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

A019°E
B008°E
C009°36’E
D011°E
Show Answer
Correct Answer: D
‘Constant of the cone’ is just another term for ‘sine of the parallel of origin’. Chart convergence = change of longitude × sin parallel of origin 12° = change of longitude × 0.80 Change of longitude = 15° 15° eastwards from 004°W is 011°E.
127

An aircraft at position 0000N/S 16327W flies a track of 225°(T) for 70 NM. What is its new position?

A0049N 16238W
B0049S 16238W
C0049N 16416W
D0049S 16416W
Show Answer
Correct Answer: D
Resolve 70 NM along a track of 225 into their change of latitude and change of longitude, i.e. 70 sin 45 and 70 cos 45. This gives 49 NM south and 49 NM west. As you are at the Equator, you can take the 49 NM west as 49 min change of longitude - you do not need the departure formula. So the final position is 49 min south and 48 min west of the original position.
128

On a Polar Stereographic map, a straight line is drawn from position A (70N 102W) to position B (80N 006E). The point of highest latitude along this line occurs at longitude 035W. What is the straight-line track angle from B to A, measured at B?

A023°(T)
B077°(T)
C229°(T)
D131°(T)
Show Answer
Correct Answer: D
129

Given that the value of ellipticity of the Earth is 1/297 and that the semi-major axis of the Earth, measured at the axis of the Equator is 6378.4 km, what is the semi- minor axis of the Earth measured at the axis of the Poles?

A6399.9 km
B6367.0 km
C6378.4 km
D6356.9 km
Show Answer
Correct Answer: D
The only problem with this question is the language, not the concept. The question is telling us that the polar diameter and the equatorial diameter of the Earth are in the ratio 296:297. The ‘semi-major axis of the Earth, measured at the axis of the Equator’ is simply half the equatorial diameter and the ‘semi- minor axis of the Earth measured at the axis of the Poles’ is half the Polar diameter. We need to find half the Polar diameter, or the Polar radius. Polar radius / 6378.4 = 296 / 297 So Polar radius = 6356.9 km Alternatively, you could remember from page 3 that the Polar diameter is 43 km shorter than the equatorial diameter. So the radius or semi-diameter will be 21.5 km shorter. 6378.4 – 21.5 = 6356.9 km. Either method gives you the right answer.
130

On a chart, meridians at 45N are shown every 10 degrees apart. This is shown on the chart by a distance of 14 cm. What is the scale?

A1: 2 000 000
B1: 4 000 000
C1: 5 000 000
D1: 5 600 000
Show Answer
Correct Answer: D
This is a combination of a departure and a representative fraction (scale) problem. Departure = change of longitude × cos latitude = 10° (× 60, to get minutes) × 0.7071 = 424.3 NM Scale = Chart Length / Earth Distance = 14 cm / (424.3 × 1852 × 100) = 1 / 5.6 Million (approx), which is closest to (d).
131

Please refer to Appendix A. Which is the symbol for a NDB?

↑ Refer to Figure 1 (Appendix A) above

A4
B5
C2
D14
Show Answer
Correct Answer: C
132

How do rhumb lines (with the exception of meridians) appear on a Polar Stereographic chart?

Aconcave to the nearer pole
Bconvex to the nearer pole
Cellipses round the pole
Dstraight lines
Show Answer
Correct Answer: A
133

Please refer to Appendix A. What does symbol 9 represent?

↑ Refer to Figure 1 (Appendix A) above

Alit obstacle
Blighthouse
CVRP
Daeronautical ground light
Show Answer
Correct Answer: A
134

What is the chart convergence factor on a Polar Stereographic chart?

A0
B1.0
C0.866
D0.5
Show Answer
Correct Answer: B
135

At 0422 you are 185 NM from a VOR at FL370. You need to descend at a mean descent rate of 1800’/min to be at FL80 overhead the VOR. Your ground speed in the level cruise is currently 320 knots. In the descent your mean G/S will be 232 knots. What is the latest time to commence descent?

A0437
B0441
C0445
D0451
Show Answer
Correct Answer: C
There are 2 parts to this profile. You are going to continue to fly at FL370 until the top of descent, and then descend to FL80. The question is asking for the ETA for top of descent. From FL370 to FL80 is 29 000 feet. At 1800 feet per minute, that is 16.1 minutes. Your mean ground speed in the descent is 232 knots so, for 16.1 min, that is 62 NM in the descent. Your total distance to run is 185 NM. If there are 62 NM is the descent, the high level distance to top of descent is 123 NM. 123 NM @ your level ground speed of 320 is 23 min. Add 23 to 0422 to get 0445.
136

Given: Heading 165(M), Variation 25W, Drift 10°R, G/S 360 knots. At ‘A’ your relative bearing to an NDB is 325R. Five minutes later, at ‘B’, the relative bearing is 280(R). What is the true bearing and distance from ‘B’ to the NDB?

A060°(T) 40 NM
B105°(T) 30 NM
C060°(T) 30 NM
D105°(T) 40 NM
Show Answer
Correct Answer: C
True heading is 140, true track is 150. Relative bearing at A is 35° left of the nose, ie a true bearing of 105 to the NDB. At B, the NDB is 80 left of the nose, true bearing to the NDB is 060. So you now have 3 lines. A track line of 150, and a bearing from A of 105 and a bearing from B of 060. This gives a right angles isosceles triangle, with the 2 short sides being 30 NM long (5 minutes at 360 knots G/S). Draw it out and it become obvious.
137

What is the diameter of the Earth?

A40 000 km
B12 732 km
C21 600 km
D6366 km
Show Answer
Correct Answer: B
You are not supposed to remember the diameter of the Earth. But you should know that the circumference of the Earth is 40 000 km. Circumference = 2 π r (or π d). So divide 40 000 km by π.
138

An aircraft flies 100 stat.m in 20 minutes. How long does it take to fly 215 NM?

A50 min
B37 min
C57 min
D42 min
Show Answer
Correct Answer: A
Use the CRP-5.
139

What is the duration of civil twilight?

AFrom the moment when the centre of the Sun is on the sensible horizon until the centre reaches a depression angle of 6° from the sensible horizon.
BFrom the moment when the tip of the Sun disappears below the sensible horizon until the centre reaches a depression angle of 6° from the sensible horizon.
CFrom the moment when the centre of the Sun is on the visual horizon until the centre reaches a depression angle of 6° from the sensible horizon.
DFrom the moment when the tip of the Sun disappears below the visual horizon until the centre reaches a depression angle of 6° from the sensible horizon.
Show Answer
Correct Answer: D
140

Please refer to Appendix B (or your Jeppesen E(LO)1). From the Connaught (CON, 5355N 00849W) VOR / DME, you plot a radial of 048°(M) and a range of 22 NM. What is the aircraft position?

↑ Refer to Figure 2 (Appendix B) above

A5410N 00844W
B5350N 00821W
C5407N 00837W
D5411N 00824W
Show Answer
Correct Answer: D
Plotting question. Solve by measurement on the chart.
141

What is the shortest distance between Point ‘A’ (3543N 00841E) and Point ‘B’ (5417N 17119W)?

A5400 NM
B6318 NM
C6557 NM
D6000 NM
Show Answer
Correct Answer: A
Plotting question. Solve by measurement on the chart.
142

On a conformal chart, the standard parallels are 41° 20’N and 11° 40’N. What is the constant of the cone?

A.660
B.202
C.446
D.895
Show Answer
Correct Answer: C
On a Lambert chart, the parallel of origin is midway between the 2 Standard Parallels, which in this case is 26°30’. Sine 26°30’ is 0.446.
143

Given: Runway direction 083°(M), Surface W/V 045/35 kt. Calculate the effective headwind component.

A29 kt
B27 kt
C31 kt
D34 kt
Show Answer
Correct Answer: A
CRP-5 problem.
144

Given: TAS=375 Trk=335°(T) W/V=340°(T)/50 What is heading and Ground speed?

A335°(T) 322
B335°(T) 318
C336°(T) 326
D333°(T) 326
Show Answer
Correct Answer: C
145

Lines of latitude on a chart are always:

AGreat circles
BSmall circles except for the Equator
CVertices
DMeridians
Show Answer
Correct Answer: B
146

On a Lambert chart, the convergence factor is .78585. What is the parallel of tangency?

A51°02’
B51°36’
C51°15’
D51°48’
Show Answer
Correct Answer: D
The ‘constant of the cone’ is the sine of the parallel of origin (or tangency). Simply use your calculator to find the arc sine of 0.78585. Use the DMS button if you need to convert the answer from degrees and decimal degrees to degrees and minutes.
147

On a Lambert’s chart the constant of the cone is 0.80. A is at 53N 04W. You plan to fly to B. The initial Lambert’s chart straight-line track is 070(T) and the rhumb line track from A to B is 080(T). What is the longitude of B?

A021E
B034W
C011E
D015E
Show Answer
Correct Answer: A
The ‘constant of the cone’ is another term for the sine of the parallel of origin. The initial straight-line track is 070 and the mean is 080. Therefore half chart convergence is 10 degrees, so the chart convergence between A and B is 20 degrees. Chart convergence = change of longitude × sine parallel of origin Therefore change of longitude = chart convergence/sin parallel of origin Change of longitude = 20/.8 = 25 degrees. 25 degrees east of 004W is 021E.
148

On which chart projection is it not possible to show the North Pole?

ADirect Mercator
BLambert’s
CTransverse Mercator
DPolar Stereographic
Show Answer
Correct Answer: A
The North Pole is in the hole at the end of the cylinder.
149

You are at FL150 and the SAT is -5°(C). You are over an airport with an elevation of 720 feet. The QNH is 1003. Assume 27 feet = 1 hPa. What is your true height?

A14 300 feet
B15 300 feet
C14 700 feet
D15 600 feet
Show Answer
Correct Answer: C
(The correct answer is 14 610, but 14 700 was the nearest option).
150

What is the formula for conversion angle?

AChange of longitude × Sine latitude
BChange of longitude / 2 × Sine mean longitude
CChange of longitude / 2 × Sine mean latitude
DChange of longitude × Cosine latitude
Show Answer
Correct Answer: C
151

On the Polar Stereographic projection, a great circle appears as:

Astraight line / curve which becomes more near to a straight line as the latitude increases / curve convex to the nearer pole
B[Option not in source]
C[Option not in source]
Da curve which can be concave or convex to the nearer pole, depending on the latitude
Instructor's Note: Source formatting error — one or more options appear duplicated or missing in the original text. Options shown as extracted; verify against printed copy.
Show Answer
Correct Answer: B
152

An aircraft departs Guam (13N 145E) at 2300 Standard Time on 30th April local date. Flight time to Los Angeles, California, USA (34N 118W) is 11 hours 15 minutes. What is the California Standard Time and local date of arrival? Assume Summer Time is being kept.

A1715 ST 30 Apr
B1215 ST 01 May
C1315 ST 01 May
D1615 ST 30 Apr
Show Answer
Correct Answer: A
As always, use a table: Standard Time at Guam 30 Apr 00 ST STD (long east, UTC least) -10 00 UTC 30 Apr 00 UTC Flight Time +11 UTC arrival at Los Angeles 01 May 00 UTC STD (long west, UTC best) -8 00 Standard Time at LA 30 Apr 16* ST Hour Minute Time Day * But California Summer Time rules apply. Add an hour to get 1715.
153

What rate of descent is required to maintain a 3.5° glide slope at a ground speed of 150 knots?

A850 fpm
B800 fpm
C600 fpm
D875 fpm
Show Answer
Correct Answer: D
154

What is the meaning of the term ‘standard time’?

AIt is another term for UTC
BIt is the time zone system applicable only in the USA.
CIt is an expression for local mean time.
DIt is the time set by the legal authorities for a country or part of a country.
Show Answer
Correct Answer: D
155

On 27 Feb at 52°S 040°E Sunrise is at 0243UTC. On the same day at 52°S 035°W the time of Sunrise is:

A0743 UTC
B0243 UTC
C2143 UTC
D0543 UTC.
Show Answer
Correct Answer: A
The change in longitude between 040°E and 035°W is 75°. The Earth rotates at 15° per hour, so this is equivalent to 5 hours of time. The Earth rotates eastwards, so a westerly longitude will have a later UTC Sunrise time.
156

A compass swing is performed in order to correct for:

Aacceleration
Bdeviation
Cvariation
Daperiodicity
Show Answer
Correct Answer: B
157

Isogonals are lines of equal:

Acompass deviation
Bmagnetic variation
Cwind velocity
Dpressure
Show Answer
Correct Answer: B
158

On a direct Mercator chart, a rhumb line appears as a:

Asmall circle concave to the nearer pole
Bstraight line
Ccurve convex to the nearer pole
Dspiral curve
Show Answer
Correct Answer: B
159

Given: IAS 120 kt FL80 OAT +20°(C) What is the TAS?

A141 kt
B102 kt
C120 kt
D132 kt
Show Answer
Correct Answer: A
160

The distance between two waypoints is 200 NM. To calculate compass heading the pilot used 2°E magnetic variation instead of 2°W. Assuming that the forecast W/V applied, what will the off track distance be at the second waypoint?

A14 NM
B7 NM
C0 NM
D21 NM
Show Answer
Correct Answer: A
This is just the one in sixty rule. The variation is 4° in error. For each 60 NM between waypoints, the aircraft will be 4 NM off track. Dist off = × 200 / 60 = 13⅓ NM
161

Given: True course 300° Drift 8°R Variation 10°W Deviation -4° Calculate the compass heading.

A322°
B306°
C278°
D294°
Show Answer
Correct Answer: B
The JAA use the term ‘course’ to mean ‘desired track’. If the drift is 8° starboard, the true heading is 292°(T). With 10°W variation, this is 302°(M). Negative deviation must be applied to the compass to give magnetic heading, so the compass heading is 306°(M). In other words, -4° deviation is the same as 4°W.
162

Given: True track 180° Drift 8°R Compass Heading 195° Deviation -2° Calculate the variation.

A21°W
B25°W
C5°W
D9°W
Show Answer
Correct Answer: A
If the compass heading is 195C, apply -2 degrees deviation to get 193 Mag heading. If you have a True Track of 180 and 8 degrees starboard drift, the true heading must be 172(T). With a heading of 172(T) and 195(M), you have 21 degrees West variation.
163

Given the following: Magnetic heading: 060° Magnetic variation: 8°W Drift angle: 4° right What is the true track?

A064°
B056°
C072°
D048°
Show Answer
Correct Answer: B
Apply 8°W variation to 060°(M) to get 052°(T). With 4° starboard drift, the true track is 056°(T).
164

Given: W/V 262/90 kt, Track 234° and TAS 305 kt, what are the Heading and Groundspeed?

A226° and 224 kt
B252° and 214 kt
C242° and 224 kt
D250° and 224 kt
Show Answer
Correct Answer: C
165

Given: Half way between two reporting points the navigation log gives the following information: TAS 360 kt W/V 330°/80 kt Compass heading 237° Deviation on this heading -5° Variation 19°W What is the average ground speed for this leg?

A403 kt
B354 kt
C373 kt
D360 kt
Show Answer
Correct Answer: A
166

(For this question use Appendix C) Complete line 5 of the ‘FLIGHT NAVIGATION LOG’, position ‘J’ to ‘K’. What is the HDG°(M) and ETA?

↑ Refer to Figure 3 (Appendix C) above

AHDG 337° - ETA 1422 UTC
BHDG 320° - ETA 1412 UTC
CHDG 337° - ETA 1322 UTC
DHDG 320° - ETA 1432 UTC
Show Answer
Correct Answer: A
167

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

Atrack
Bdrift
Cground speed
Dheading
Show Answer
Correct Answer: C
Ground speed is the only of these options which depends on time difference.
168

The angle between the true great circle track and the true rhumb line track joining the following points: A (60S 165W) and B (60S 177E) at the place of departure A, is:

A
B15.6°
C5.2°
D7.8°
Show Answer
Correct Answer: D
This is a conversion angle problem: Conversion angle = ½ change of longitude × sine latitude = ½ × × 0.866 = 7.8°
169

Which of the following indicates an advisory airspace (ADA) boundary?

A[Chart symbol — see printed exam paper]
B[Chart symbol — see printed exam paper]
C[Chart symbol — see printed exam paper]
D[Chart symbol — see printed exam paper]
Instructor's Note: Options for this question are visual chart symbols that cannot be reproduced as text. Refer to the printed examination paper or syllabus chart legend.
Show Answer
Correct Answer: C
170

Given the following: True track: 192° Magnetic variation: 7°E Drift angle: 5° left What is the magnetic heading required to maintain the given track?

A180°
B190°
C194°
D204°
Show Answer
Correct Answer: B
192°(T) + 7E Var = 185°(M)ag Track. With 5° of left drift, you need to aim off 5°(T)o the right, onto a Mag heading of 190°(M).
171

Given: A Polar Stereographic chart whose grid is aligned with the zero meridian. Grid track 344°, longitude 115°00’W, calculate the true course. (Assume N hemisphere).

A099°
B279°
C049°
D229°
Show Answer
Correct Answer: D
The datum meridian is Greenwich, so the aircraft is 115° West of the datum in the Northern hemisphere. Therefore convergence is 115°E. Convergence East – True Least. So True track will be 115 degrees less than Grid Track. 344°(G) – 115°(C)onvergence = 229°(T).
172

The rhumb line distance between points A (60°00’N 002°30’E) and B (60°00’N 007°30’W) is:

A300 NM
B450 NM
C600 NM
D150 NM
Show Answer
Correct Answer: A
Rhumb line distance between 2 points at the same latitude is the Departure. Departure = change of longitude (in minutes) × cos latitude = 10° × 60 (to get minutes) × cos 60° = 600 × 0.5 = 300 NM
173

Given: TAS = 485 kt, OAT = ISA +10°C, FL410. Calculate the Mach Number.

A0.87
B0.825
C0.90
D0.85
Show Answer
Correct Answer: B
Do not forget that ISA at FL410 is still -56.5°(C), even though you are higher than 36 090 feet pressure alt. So the SAT is -46.5°(C). Use the CRP-5.
174

Fuel flow per hr is 22 US.gal, total fuel on board is 83 imp.gal. What is the endurance?

A2 hr 15 min
B4 hr 32 min
C3 hr 12 min
D3 hr 53 min
Show Answer
Correct Answer: B
Convert 22 US.gal per hour to imperial gallons on the CRP-5. The answer is about 18.3 imp.gal/hour. Then use either the CRP-5 or your calculator to get the total endurance.
175

Given: Position A is 60N 020W, Position B is 60N 021W, and Position C is 59N 020W, what are, respectively, the distances from A to B and from A to C?

A60 NM and 30 NM
B30 NM and 60 NM
C52 NM and 60 NM
D60 NM and 52 NM
Show Answer
Correct Answer: B
A and B are at the same latitude but different longitudes. So it is a departure problem. Departure = change of longitude (in minutes) × cosine latitude Departure = 1 × 60 × cos 60 = 30 NM A and C are at different latitudes but the same longitude. So it is a simple change of latitude problem. One degree is 60 NM.
176

Given: FL350, Mach 0.80, OAT -55°(C), calculate values for TAS and local speed of sound?

A461 kt, LSS 296 kt
B461 kt, LSS 576 kt
C237 kt, LSS 296 kt
D490 kt, LSS 461 kt
Show Answer
Correct Answer: B
Use the CRP-5.
177

How many nautical miles are travelled in 1 minute 45 seconds at a ground speed of 135 knots?

A2.36
B3.25
C39.0
D3.94
Show Answer
Correct Answer: D
135 × 1.75/60 = 3.94
178

The distance A to B is 90 NM in a straight line. You are 60 NM from A when you fix your position 4 NM to the left of track. What correction do you need to make to arrive at B?

A
B
C12°
D10°
Show Answer
Correct Answer: C
You need 4° right to correct for track error angle and parallel track. Then, with 60 NM gone, there are 30 NM to go. Four NM off in 30 gives a closing angle of 8°. Total change = 12° right.
179

A Great Circle crosses the Equator at longitude 030°W. The direction of the GC at Equator is 035°(T). An aircraft following this Great Circle will reach its highest latitude (N or S) at position:

A35S 120W
B55S 060E
C35N 120W
D55N 060E
Show Answer
Correct Answer: D
180

The heading is 299°(G). Magnetic variation is 90°W and chart convergence is 55°W. What is magnetic heading?

A154°(M)
B084°(M)
C264°(M)
D334°(M)
Show Answer
Correct Answer: B
181

The orbit of the Earth round the Sun is elliptical. An ellipse has 2 foci. Which of the following is a correct statement?

AThe Earth is positioned at one of the foci.
BThe Sun is positioned at the mid-point of the 2 foci.
CThe Sun is positioned at one of the foci.
DThe Earth is positioned at the mid-point of the 2 foci.
Show Answer
Correct Answer: C
182

An aircraft homing to a VOR/DME loses 2500 feet in 11.1 NM change of DME range. What is the gradient of the slope?

A4.1%
B3.5%
C3.9%
D3.7%
Show Answer
Correct Answer: D
183

An aircraft at position 8500N 02000E flies a rhumb line track of 075°(T). What will be its path over the Earth?

AThe shortest route to a destination
BA spiral path leading towards the North Pole
CA great circle route continuing over the pole and then southwards over the other side
DAn increasing track angle
Show Answer
Correct Answer: B
184

Given: For take-off an aircraft requires a headwind component of at least 15 kt and has a cross-wind limit of 35 kt. The angle between the wind direction and the runway is 60°. Calculate the maximum and minimum allowable wind speeds.

A30 kt and 40 kt
B15 kt and 43 kt
C12 kt and 38 kt
D18 kt and 50 kt
Show Answer
Correct Answer: A
185

An aircraft is cruising at FL350, Temp -50°C and is told to descend to FL80, Temp -10°C. If the IAS for the descent was 188 kt, what would be the appropriate TAS?

A260 kt
B188 kt
C335 kt
D224 kt
Show Answer
Correct Answer: A
To calculate the TAS for the descent you need to have the mean altitude and mean temperature with the IAS. Mean altitude is 21 500 ft and mean temperature is -30°C.
186

An aircraft has to climb from FL50 -10°C to FL260 -25°C. The IAS for the climb is 180 kt and the WC is +30 kt. If the ROC is 900 ft/min, how many miles will the climb take?

A96 NM
B106 NM
C83 NM
D120 NM
Show Answer
Correct Answer: B
For a climb the TAS is calculated using the 2/3 change in altitude and temperature. 2/3 altitude is 21 000 ft and temperature is -20°C so the TAS is 242 kt. GS is 272 kt and the time is 23.3 min
187

An aircraft is flying at FL200, the OAT is 0°C. When the actual air pressure on an airfield at MSL is placed on the subscale of the altimeter the indicated altitude is 19 300 ft. What is the aircraft’s True Altitude?

A17 300 ft
B19 300 ft
C20 000 ft
D21 300 ft
Show Answer
Correct Answer: D
ISA Dev = +25°C, Temperature Correction = 4 ft × 20 × +25 = 2000 ft. TA = IA + Temp Corr = 19 300 + 2000 = 21 300 ft
Capt. Pankaj Pahil
www.ghostaviator.com