Kilometre is defined as: cross wind component on landing is:

A) 0,621 Statute mile. A) 26 kts.

B ) 0,454 Nautical mile. B) 15 kts.
C) a 1/10000 part of the meridian length from Equator to the pole. C) 20 kts.
D) the mean length of a 1/40000 part of the Equator. D) 23 kts.
When dealing with heights and altitudes in international aviation, we use the following units: An aircraft travels 100 statute miles in 20 MIN, how long does it take to travel 215 NM?
A) Metre and Foot. A) 100 MIN.
B ) Foot, Kilometre and decimals of Nautical mile. B) 90 MIN.
C) All 3 answers are correct. C) 80 MIN.
D) Foot and Yard. D) 50 MIN.
How many feet are there in a nm? Fuel flow per HR is 22 US-GAL, total fuel on board is 83 IMP GAL. What is the endurance?
A) 6080 ft A) 2 HR 15 MIN
B) 5280 ft B) 3 HR 53 MIN
C) 1000 ft C) 3 HR 12 MIN
D) 3280 ft D) 4 HR 32 MIN
A Nautical mile is defined as: Given: True track 180° Drift 8° R Compass heading 195° Deviation -2° Calculate the variation?
A) The length of a 1 minute arc, measured anywhere on the surface of the Earth. A) 5° W.
B) The average length of a 1' arc of longitude and a 1' arc of latitude. B) 21° W.
C) 1855 metres. C) 25° W.
D) The average length of a 1 minute arc of a meridian. D) 9° W.
How long is 25 Kilometres at 6000N? 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.
A) 13,5 Nautical mile What groundspeed is required to arrive on time at B?
B ) 46,3 Nautical mile A) 360 knots.
C) 40,2 Statute miles B) 330 knots.
D) 27,0 Nautical mile C) 317 knots.
Assuming the Earth being a perfect sphere: D) 342 knots.
A) All 3 answers are correct. Given: GS = 135 kt. Distance from A to B = 433 NM. What is the time from A to B?
B) a 1 minute arc measured on the surface of the Earth will be equally long wherever it is measured. A) 3 HR 25 MIN.
C) distances will vary, dependant on their directions. B) 3 HR 20 MIN.
D) distances will vary, dependant on the latitude. C) 3 HR 12 MIN.
How many feet are equivalent to 9,5 km? D) 3 HR 19 MIN.
A) 9.500 ft
B ) 50.160 ft 730 FT/MIN equals:
C) 57.760 ft A) 5.2 m/sec
D) 31.160 ft B) 2.2 m/sec
How many feet are there in 1 sm? C) 1.6 m/sec
A) 1000 ft D) 3.7 m/sec
B) 6080 ft Given: GS = 345 kt. Distance from A to B = 3560 NM. What is the time from A to B?
C) 5280 ft A) 10 HR 05 MIN.
D) 3280 ft B ) 11 HR 02 MIN.
The International Nautical Mile defined by ICAO is equivalent to … m. C) 11 HR 00 MIN.
A) 1.852 m D) 10 HR 19 MIN.
B ) 1.962 m Given: GS = 236 kt. Distance from A to B = 354 NM What is the time from A to B?
C) 1.652 m A) 1 HR 40 MIN.
D) 1.582 m B) 1 HR 10 MIN.
How many feet are there in a km? C) 1 HR 30 MIN.
A) 6080 ft D) 1 HR 09 MIN.
B) 5280 ft 41. Given: True course 300° drift 8° R variation 10° W deviation -4° Calculate the compass heading?
C) 3280 ft A) 278°
D) 1000 ft B) 306°
An aircraft is landing on runway 23 (QDM 227° ), surface wind 180° /30 kts from ATIS; variation is 13° E. The C) 322°

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D) 294° A) 26 kts.
An aircraft has a groundspeed of 510 kts and a True Air Speed of 440 kts. If the distance from A to B is 43 NM, B) 15 kts.
the time in minutes from A to B will be: C) 20 kts.
A) 6 D) 23 kts.
B) 7 An aircraft travels 100 statute miles in 20 MIN, how long does it take to travel 215 NM?
C) 4 A) 100 MIN.
D) 5 B) 90 MIN.
Given: GS = 435 kt. Distance from A to B = 1920 NM. What is the time from A to B? C) 80 MIN.
A) 4 HR 10 MIN D) 50 MIN.
B) 4 HR 25 MIN The relative bearing to a beacon is 270° R. Three minutes later, at a groundspeed of 180 knots, it has changed to
C) 3 HR 25 MIN 225° R. What was the distance of the closest point of approach of the aircraft to the beacon?
D) 3 HR 26 MIN A) 9 nm.
Given: GS = 122 kt. Distance from A to B = 985 NM. What is the time from A to B? B) 45 nm.
A) 8 HR 10 MIN. C) 18 nm.
B) 8 HR 04 MIN. D) 3 nm.
C) 7 HR 48 MIN. Given:
D) 7 HR 49 MIN. Fuel flow per HR is 22 US-GAL, total fuel on board is 83 IMP GAL. What is the endurance?
Given: GS = 480 kt. Distance from A to B = 5360 NM. What is the time from A to B? A) 2 HR 15 MIN
A) 11 HR 06 MIN. B) 3 HR 53 MIN
B) 11 HR 15 MIN. C) 3 HR 12 MIN
C) 11 HR 07 MIN. D) 4 HR 32 MIN
D) 11 HR 10 MIN. Given: True track 180° Drift 8° R Compass heading 195° Deviation -2° Calculate the variation?
An airfield has two runways, 05/23 and 30/12. The surface wind is given as 250° /30. The headwind component A) 5° W.
on 23 and the crosswind component on runway 30 will be: B) 21° W.
A) 28 kts headwind and 23 kts crosswind. C) 25° W.
B) 28 kts headwind and 19 kts crosswind. D) 9° W.
C) 10 kts headwind and 23 kts crosswind. 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.
D) 10 kts headwind and 19 kts crosswind. What groundspeed is required to arrive on time at B?
A) 360 knots.
How many NM would an aircraft travel in 1 MIN 45 SEC if GS is 135 kt? B) 330 knots.
A) 2.36 C) 317 knots.
B) 3.94 D) 342 knots.
C) 39.0 Given: GS = 135 kt. Distance from A to B = 433 NM. What is the time from A to B?
D) 3.25 A) 3 HR 25 MIN.
Given: GS = 510 kt. Distance A to B = 43 NM What is the time (MIN) from A to B? B) 3 HR 20 MIN.
A) 5 C) 3 HR 12 MIN.
B) 7 D) 3 HR 19 MIN.
C) 6 730 FT/MIN equals:
D) 4 A) 5.2 m/sec
Given: true track 070° variation 30° W deviation +1° drift 10° R Calculate the compass heading? B) 2.2 m/sec
A) 089° C) 1.6 m/sec
B) 091° D) 3.7 m/sec
C) 100° Given: GS = 345 kt. Distance from A to B = 3560 NM. What is the time from A to B?
D) 101° A) 10 HR 05 MIN.
An aircraft travels 2.4 statute miles in 47 seconds. What is its groundspeed? B ) 11 HR 02 MIN.
A) 209 kt. C) 11 HR 00 MIN.
B) 160 kt. D) 10 HR 19 MIN.
C) 131 kt. Given:
D) 183 kt. FL250
An aircraft is landing on runway 23 (QDM 227° ), surface wind 180° /30 kts from ATIS; variation is 13° E. The OAT -15 ° C
cross wind component on landing is: TAS 250 kt

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Calculate the Mach No.? D) 10 kts headwind and 19 kts crosswind.
A) 0.39 How many NM would an aircraft travel in 1 MIN 45 SEC if GS is 135 kt?
B ) 0.44 A) 2.36
C) 0.40 B) 3.94
D) 0.42 C) 39.0
Given: GS = 236 kt. Distance from A to B = 354 NM What is the time from A to B? D) 3.25
A) 1 HR 40 MIN. Given: GS = 510 kt. Distance A to B = 43 NM What is the time (MIN) from A to B?
B) 1 HR 10 MIN. A) 5
C) 1 HR 30 MIN. B) 7
D) 1 HR 09 MIN. C) 6
D) 4
If TAS is 472 kts, heading is 005° T, wind is 110° /50 kts the drift and groundspeed will be: Given: true track 070° variation 30° W deviation +1° drift 10° R Calculate the compass heading?
A) 7° Stbd and 491 kts. A) 089°
B) 6° Stbd and 487 kts. B) 091°
C) 6° Port and 487 kts. C) 100°
D) 7° Port and 491 kts. D) 101°
An aircraft travels 2.4 statute miles in 47 seconds. What is its groundspeed?
41. Given: True course 300° drift 8° R variation 10° W deviation -4° Calculate the compass heading? A) 209 kt.
A) 278° B) 160 kt.
B) 306° C) 131 kt.
C) 322° D) 183 kt.
D) 294° Given: GS = 120 kt. Distance from A to B = 84 NM. What is the time from A to B?
A) 00 HR 44 MIN.
An aircraft has a groundspeed of 510 kts and a True Air Speed of 440 kts. If the distance from A to B is 43 NM, B) 00 HR 43 MIN.
the time in minutes from A to B will be: C) 00 HR 42 MIN.
A) 6 D) 00 HR 45 MIN.
B) 7 The tank capacity of an aircraft is 310 US GAL. Fuel specific gravity is 0,78 kg/litre. The tanks are now 3/4 full.
C) 4 You want to refuel so that total fuel will be 850 kg. How much fuel will you have to refuel? Answer in pounds.
D) 5 A) 410 LB.
B ) 320 LB.
Given: GS = 435 kt. Distance from A to B = 1920 NM. What is the time from A to B? C) 360 LB.
A) 4 HR 10 MIN D) 164 LB.
B) 4 HR 25 MIN Given: GS = 95 kt. Distance from A to B = 480 NM. What is the time from A to B?
C) 3 HR 25 MIN A) 5 HR 03 MIN
D) 3 HR 26 MIN B ) 4 HR 59 MIN
Given: GS = 122 kt. Distance from A to B = 985 NM. What is the time from A to B? C) 5 HR 08 MIN
A) 8 HR 10 MIN. D) 5 HR 00 MIN
B) 8 HR 04 MIN. Given: true track 352° variation 11° W deviation is -5° drift 10° R. Calculate the compass heading?
C) 7 HR 48 MIN. A) 358°
D) 7 HR 49 MIN. B) 346°
C) 018°
Given: GS = 480 kt. Distance from A to B = 5360 NM. What is the time from A to B? D) 025°Given: aircraft height 2500 FT, ILS GP angle 3° . At what approximate distance from THR can you expect to
A) 11 HR 06 MIN. capture
B) 11 HR 15 MIN. the GP?
C) 11 HR 07 MIN. A) 13.1 NM.
D) 11 HR 10 MIN. B ) 8.3 NM.
An airfield has two runways, 05/23 and 30/12. The surface wind is given as 250° /30. The headwind component C) 14.5 NM.
on 23 and the crosswind component on runway 30 will be: D) 7.0 NM.
A) 28 kts headwind and 23 kts crosswind.
B) 28 kts headwind and 19 kts crosswind. At 0422 an aircraft at FL370, GS 320kt, is on the direct track to VOR X 185 NM distant. The aircraft is required to
C) 10 kts headwind and 23 kts crosswind. cross VOR X at FL80. For a mean rate of descent of 1800 FT/MIN at a mean GS of 232 kt, the latest time at which

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to commence descent is: D) 1830 feet/min.
A) 0445 An aircraft at FL370 is required to commence descent at 120 NM from a VOR and to cross the facility at
B) 0454 FL130. If the mean GS for the descent is 288 kt, the minimum rate of descent required is:
C) 0451 A) 960 FT/MIN.
D) 0448 B) 860 FT/MIN.
An aircraft at FL350 is required to cross a VOR/DME facility at FL110 and to commence descent when 100 NM from C) 920 FT/MIN.
the facility. If the mean GS for the descent is 335 kt, the minimum rate of descent required is: D) 890 FT/MIN.
A) 1290 FT/MIN.
B ) 1340 FT/MIN.
C) 1240 FT/MIN.
D) 1390 FT/MIN.You are required to descend from FL 230 to FL 50 over a distance of 32 NM in 7 minutes. What will
the
glideslope be when you expect WC-25 during the descend?
A) 4,68°
B ) 5,29°
C) 4,07°
D) 6,25°
On a 12% glide slope with a groundspeed of 540 kts the required rate of descent is:
A) 6500 feet/min.
B) 4800 feet/min.
C) 8700 feet/min.
D) 3100 feet/min.
An aircraft is descending down a 12% slope whilst maintaining a GS of 540 kt. The rate of descent of the aircraft
is approximately:
A) 6500 FT/MIN.
B) 3900 FT/MIN.
C) 4500 FT/MIN.
D) 650 FT/MIN.
An aircraft at FL290 is required to commence descent when 50 NM from a VOR and to cross that VOR at FL80.
Mean GS during descent is 271kt. What is the minimum rate of descent required?
A) 1800 FT/MIN.
B) 2000 FT/MIN.
C) 1900 FT/MIN.
D) 1700 FT/MIN.
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) 38.4 NM.
B) 16.0 NM.
C) 26.7 NM.
D) 19.2 NM.
You are required to descend from FL 230 to FL 50 over a distance of 32 NM in 7 minutes. What is the required TAS
when you expect WC-25 during the descent?
A) 317 Kt.
B) 300 Kt.
C) 329 Kt.
D) 308 Kt.
21. At 65 nm from a VOR you commence a descent from FL330 in order to arrive over the VOR at FL 100.
Your mean groundspeed in the descent is 240 knots. What rate of descent is required?
A) 1270 feet/min.
B) 1420 feet/min.
C) 1630 feet/min.

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