555.6km = 300nm and this is the departure distance between the two
meridians 10º apart.
Departure = Ch.long(') x cos lat so cos lat = departure/ch.long(') =
300/600 = 0.5.
0.5 is cos 60º so this is at 60º N or S.
Question 281:
On a Lambert chart (standard parallels 37°N and 65°N), with respect to
the straight line drawn on the map between A ( N49° W030°) and B (N48°
W040°), the:
A great circle and rhumb line are to the south
B great circle and rhumb line are to the north
C rhumb line is to the north, the great circle is to the south
D great circle is to the north, the rhumb line is to the south
Explanation: Work out where the line is and it turns out to be south of
the parallel of origin (1/2 way between std parallels), then apply the
fact that the GC/RL curve concave to the parallel of origin on a Lamberts
and therefore they are to the south of the straight line drawn on the
map.
Note the parallel of origin is at about 51ºN.
Question 282:
Given: ETA to cross a meridian is 2100 UTC GS is 441 kt TAS is 491 kt At
2010 UTC, ATC requests a speed reduction to cross the meridian at 2105
UTC. The reduction to TAS will be approximately:
A 40 kt
B 75 kt
C 60 kt
D 90 kt

Explanation: At 2110; 50 minutes from meridian at 441 kts = 367½ nm to
run.
Now need to cover 367½ nm in 55 minutes = 401 kts; reduction 40 kts.
Question 283:
The flight log gives the following data: "True track, Drift, True
heading, Magnetic variation, Magnetic heading, Compass deviation, Compass
heading" The right solution, in the same order, is:
A 119°, 3°L, 122°, 2°E, 120°, +4°, 116°
B 117°, 4°L, 121°, 1°E, 122°, -3°, 119°
C 125°, 2°R, 123°, 2°W, 121°, -4°, 117°
D 115°, 5°R, 120°, 3°W, 123°, +2°, 121°
Explanation: Tr ºT
Drift
Hdg ºT
Var
Hdg ºM
Dev.
Hdg ºC
119º
3º L
122º
2ºE
120º
+4º(E)
116º
Question 284:
At 0020 UTC an aircraft is crossing the 310° radial at 40 NM of a VOR/DME
station. At 0035 UTC the radial is 040° and DME distance is 40 NM.
Magnetic variation is zero. The true track and ground speed are:
A 085° - 226 kt
B 080° - 226 kt
C 088° - 232 kt
D 090° - 232 kt
Explanation: Draw a diagram.
Consider the triangle made up by the VOR, the aircraft's 0020 position
and the aircraft's 0035 position.
The angle at the VOR = 50º + 40º = 90º.
Since the two sides VOR-0020 and VOR-0035 are both 40nm it is an
isosceles triangle and the other two angles are 45º each. At 0020 the
bearing from the aircraft to the VOR is 310º - 180º = 130º so the
aircraft's track is 130º - 45º = 085º. Since the four answers are 085º,
090º, 080º and 088º it must be 085º - 226 kt.
There is absolutely no need to work out the groundspeed.
You can work out the ground speed by considering the VOR-0020-0035
triangle again.
It is a right angle triangle and we can work out the 0020-0035 distance
by trigonometry = 40/sin45º = 56.6nm.
This could also be done using Pythagoras.
This distance has been covered in 15 minutes so the groundspeed is 226
kts.
The problem is that we still have two possible answers and still have to
work out the track angle.
Top tip on a question like this is to calculate the unique value (in this
case track) and not the duplicated value (in this case groundspeed)
because it will save time.

Question 285:
A straight line on a chart 4.89 cm long represents 185 NM. The scale of
this chart is approximately:
A 1: 7 000 000
B 1: 3 500 000
C 1: 5 000 000
D 1: 6 000 000