1.
What is the final position after the following rhumb line tracks and distances
have been followed from position 60° 00N 030° 00W?
1.South for 3600 NM
2.East for 3600 NM
3.North for 3600 NM
4.West for 3600 NM
The final position of the aircraft is:
A) 59° 00N 090° 00W
B) 60° 00N 030° 00E
C) 59° 00N 060° 00W
D) 60° 00N 090° 00W
Ans: D
2. A useful method of a pilot resolving, during a visual flight, any uncertainty in
the aircrafts position is to maintain visual contact with the ground and:
A) fly reverse headings and associated timings until the point of departure is
regained.
B) fly the reverse of the heading being flown prior to becoming uncertain until a
pinpoint is obtained.
C) set heading towards a line feature such as a coastline, motorway, river or
railway.
D) fly expanding circles until a pinpoint is obtained.
Ans: C
3. The departure between positions 60° N 160° E and 60sN x is 900 NM. What is the
longitude of x?
A) 170° W
B) 145° E
C) 175° E
D) 140° W
Ans: A
4. An aircraft at latitude 10° South flies north at a GS of 890 km/HR. What will
its latitude be after 1.5 HR?
A) 22° OON
B) 12° 15N
C) 02° 00N
D) 03° 50N
Ans: C
5. The Great Circle bearing of B (70° S 060° E), from A (70° S 030° W), is
approximately?
A) 135° (T)
B) 090° (T)
C) 318° (T)
D) 150° (T)
Ans: A
6. Given: Position A 45° N, ?° E Position B 45° N, 45° 15E Distance A-B=280 NM B is
to the East of A Required: longitude of position A?
A) 38° 39E
B) 49° 57E
C) 40° 33E
D) 51° 51E
Ans: A
7. Refer to chart E(LO)1. What is the aircraft position in lat and long given the
following:
CRN (5318N 00857W) 18 DME
SHA (5243N 00853W) 20 DME Both ranges DME decreasing.
A) 5203N 00843W
B) 5302N 00843W
C) 5301N 00908W
D) 5201N 00908W
Ans: B
8. What is dlat from 30° 39S 20° 20E to 45° 23N 40° 40E:
A) 76° 2 S
B) 14° 44 N
C) 76° 2 S
D) 76° 2 N
Ans: D
9. You are flying from A(30S 20E) to B (30S 20E). At what longitude will the GC
track equal the RL track?
A) 10° E
B) 0° E/W
C) 10° W
D) 20° W
Ans: B
10. Position A is located on the equator at longitude 130° 00E. Position B is
located 100NM from A on a bearing of 225° (T). The coordinates of position B
are:
A) 01° 11N 131° 11E
B) 01° 11S 131° 11E
C) 01° 11S 128° 49E
D) 01° 11N 128° 49E
Ans: C
11. An aircraft at position 60GN 005° W tracks 090° (T) for 315 km. On completion
of the flight the longitude will be:
A) 002° 10W
B) 000° 15E
C) 005° 15E
D) 000° 40E
Ans: D
12. What is the Chlong (in degrees and minutes) from A (45N 16530E) to B (45N
15540W)?
A) 38° 50W
B) 38° 50E
C) 38° 05W
D) 38° 05E
Ans: B
13. A useful method of a pilot resolving, during a visual flight, any uncertainty
in the aircraft’s position is to maintain visual contact with the ground
and:
A) set heading towards a line feature such as a coastline, motorway, river or
railway
B) fly the reverse of the heading being flown prior to becoming undertainuntil a
pinpoint is obtained
C) fly expanding circles until a pinpoint is obtained
D) fly reverse headings and associated timings until the point of departure is
regained.
Ans: A
14. Position A is located on the equator at longitude 130o00E. Position B is
located 100 NM from A on a bearing of 225o(T). The co-ordinates of position
B are:
A) 01o11N 128o 49E
B) 01o11S 128o 49E
C) 01o11N 131o 11E
D) 01o11S 131o 11E
Ans: B
15. 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
Ans: A
16. The Great Circle bearing of B (70oS 060oE), from A (70oS 030oW), is
approximately?
A) 150o (T)
B) 090o (T)
C) 318o (T)
D) 135o (T)
Ans: D
17. What is the final position after the following rhumb line tracks and distances
have been followed from position 60o00N 030o00W?
1. South for 3600 NM
2. East for 3600 NM
3. North for 3600 NM
4. West for 3600 NM
The final position of the aircraft is:
A) 59o00N 090o00W
B) 60o00N 090o00W
C) 60o00N 030o00E
D) 59o00N 060o00W
Ans: B
18. An aircraft at positon 60oN 005oW tracks 090o(T) for 315km. On completion of
the flight the longitude will be:
A) 002o 10W
B) 000o 15E
C) 000o 40E
D) 005o 15E
Ans: C
19. The departure between positions 60oN 160oE and 60sN x is 900 NM. What is the
longitude of x?
A) 170oW
B) 140oW
C) 145oE
D) 175oE
Ans: A
20. An aircraft at latitude 10o South flies north at a GS of 890 km/HR. What will
its latitude be after 1.5 HR?
A) 22o00N
B) 03o50N
C) 02o00N
D) 12o15N
Ans: C
21. You are flying from A (30S 20E) to B (30S 20W). At what longitude will the GC
track equal the RL track?
A) 10oE
B) 10oW
C) 0oE/W
D) 20oW
Ans: C
22. What is diat from 30o39S 20o20E to 45o23N 40o40E:
A) 14o44 N
B) 76o2 S
C) 76o2 N
D) 76o4 S
Ans: C
23. What is the Chlong (in degrees and minutes) from A (45N 1630E) to B (45N
15540W)?
A) 38o05E
B) 38o50W
C) 38o05W
D) 38o50E
Ans: D
24. Given: True Track 245o Drift 5o right Variation 3oE Compass Hdg 242o Calculate
the Magnetic Heading:
A) 247o
B) 243o
C) 237o
D) 253o
Ans: C
25. Grid heading is 299o, grid convergency is 55o West and magnetic variation is
90o West. What is the corresponding magnetic heading?
A) 084o
B) 334o
C) 154o
D) 264o
Ans: A
1. What is the final position after the following rhumb line tracks and distances
have been followed from position 60° 00N 030° 00W?
1.South for 3600 NM
2.East for 3600 NM
3.North for 3600 NM
4.West for 3600 NM
The final position of the aircraft is:
A) 59° 00N 090° 00W
B) 60° 00N 030° 00E
C) 59° 00N 060° 00W
D) 60° 00N 090° 00W
Ans: D
2. A useful method of a pilot resolving, during a visual flight, any uncertainty in
the aircrafts position is to maintain visual contact with the ground and:
A) fly reverse headings and associated timings until the point of departure is
regained.
B) fly the reverse of the heading being flown prior to becoming uncertain until a
pinpoint is obtained.
C) set heading towards a line feature such as a coastline, motorway, river or
railway.
D) fly expanding circles until a pinpoint is obtained.
Ans: C
3. The departure between positions 60° N 160° E and 60sN x is 900 NM. What is the
longitude of x?
A) 170° W
B) 145° E
C) 175° E
D) 140° W
Ans: A
4. An aircraft at latitude 10° South flies north at a GS of 890 km/HR. What will
its latitude be after 1.5 HR?
A) 22° OON
B) 12° 15N
C) 02° 00N
D) 03° 50N
Ans: C
5. The Great Circle bearing of B (70° S 060° E), from A (70° S 030° W), is
approximately?
A) 135° (T)
B) 090° (T)
C) 318° (T)
D) 150° (T)
Ans: A
6. Given: Position A 45° N, ?° E Position B 45° N, 45° 15E Distance A-B=280 NM B is
to the East of A Required: longitude of position A?
A) 38° 39E
B) 49° 57E
C) 40° 33E
D) 51° 51E
Ans: A
7. Refer to chart E(LO)1. What is the aircraft position in lat and long given the
following:
CRN (5318N 00857W) 18 DME
SHA (5243N 00853W) 20 DME Both ranges DME decreasing.
A) 5203N 00843W
B) 5302N 00843W
C) 5301N 00908W
D) 5201N 00908W
Ans: B
8. What is dlat from 30° 39S 20° 20E to 45° 23N 40° 40E:
A) 76° 2 S
B) 14° 44 N
C) 76° 2 S
D) 76° 2 N
Ans: D
9. You are flying from A(30S 20E) to B (30S 20E). At what longitude will the GC
track equal the RL track?
A) 10° E
B) 0° E/W
C) 10° W
D) 20° W
Ans: B
10. Position A is located on the equator at longitude 130° 00E. Position B is
located 100NM from A on a bearing of 225° (T). The coordinates of position B
are:
A) 01° 11N 131° 11E
B) 01° 11S 131° 11E
C) 01° 11S 128° 49E
D) 01° 11N 128° 49E
Ans: C
11. An aircraft at position 60GN 005° W tracks 090° (T) for 315 km. On completion
of the flight the longitude will be:
A) 002° 10W
B) 000° 15E
C) 005° 15E
D) 000° 40E
Ans: D
12. What is the Chlong (in degrees and minutes) from A (45N 16530E) to B (45N
15540W)?
A) 38° 50W
B) 38° 50E
C) 38° 05W
D) 38° 05E
Ans: B
13. A useful method of a pilot resolving, during a visual flight, any uncertainty
in the aircraft’s position is to maintain visual contact with the ground
and:
A) set heading towards a line feature such as a coastline, motorway, river or
railway
B) fly the reverse of the heading being flown prior to becoming undertainuntil a
pinpoint is obtained
C) fly expanding circles until a pinpoint is obtained
D) fly reverse headings and associated timings until the point of departure is
regained.
Ans: A
14. Position A is located on the equator at longitude 130o00E. Position B is
located 100 NM from A on a bearing of 225o(T). The co-ordinates of position
B are:
A) 01o11N 128o 49E
B) 01o11S 128o 49E
C) 01o11N 131o 11E
D) 01o11S 131o 11E
Ans: B
15. 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
Ans: A
16. The Great Circle bearing of B (70oS 060oE), from A (70oS 030oW), is
approximately?
A) 150o (T)
B) 090o (T)
C) 318o (T)
D) 135o (T)
Ans: D
17. What is the final position after the following rhumb line tracks and distances
have been followed from position 60o00N 030o00W?
1. South for 3600 NM
2. East for 3600 NM
3. North for 3600 NM
4. West for 3600 NM
The final position of the aircraft is:
A) 59o00N 090o00W
B) 60o00N 090o00W
C) 60o00N 030o00E
D) 59o00N 060o00W
Ans: B
18. An aircraft at positon 60oN 005oW tracks 090o(T) for 315km. On completion of
the flight the longitude will be:
A) 002o 10W
B) 000o 15E
C) 000o 40E
D) 005o 15E
Ans: C
19. The departure between positions 60oN 160oE and 60sN x is 900 NM. What is the
longitude of x?
A) 170oW
B) 140oW
C) 145oE
D) 175oE
Ans: A
20. An aircraft at latitude 10o South flies north at a GS of 890 km/HR. What will
its latitude be after 1.5 HR?
A) 22o00N
B) 03o50N
C) 02o00N
D) 12o15N
Ans: C
21. You are flying from A (30S 20E) to B (30S 20W). At what longitude will the GC
track equal the RL track?
A) 10oE
B) 10oW
C) 0oE/W
D) 20oW
Ans: C
22. What is diat from 30o39S 20o20E to 45o23N 40o40E:
A) 14o44 N
B) 76o2 S
C) 76o2 N
D) 76o4 S
Ans: C
23. What is the Chlong (in degrees and minutes) from A (45N 1630E) to B (45N
15540W)?
A) 38o05E
B) 38o50W
C) 38o05W
D) 38o50E
Ans: D
24. Given: True Track 245o Drift 5o right Variation 3oE Compass Hdg 242o Calculate
the Magnetic Heading:
A) 247o
B) 243o
C) 237o
D) 253o
Ans: C
25. Grid heading is 299o, grid convergency is 55o West and magnetic variation is
90o West. What is the corresponding magnetic heading?
A) 084o
B) 334o
C) 154o
D) 264o
Ans: A
1. What is the final position after the following rhumb line tracks and distances
have been followed from position 60° 00N 030° 00W?
1.South for 3600 NM
2.East for 3600 NM
3.North for 3600 NM
4.West for 3600 NM
The final position of the aircraft is:
A) 59° 00N 090° 00W
B) 60° 00N 030° 00E
C) 59° 00N 060° 00W
D) 60° 00N 090° 00W
Ans: D
2. A useful method of a pilot resolving, during a visual flight, any uncertainty in
the aircrafts position is to maintain visual contact with the ground and:
A) fly reverse headings and associated timings until the point of departure is
regained.
B) fly the reverse of the heading being flown prior to becoming uncertain until a
pinpoint is obtained.
C) set heading towards a line feature such as a coastline, motorway, river or
railway.
D) fly expanding circles until a pinpoint is obtained.
Ans: C
3. The departure between positions 60° N 160° E and 60sN x is 900 NM. What is the
longitude of x?
A) 170° W
B) 145° E
C) 175° E
D) 140° W
Ans: A
4. An aircraft at latitude 10° South flies north at a GS of 890 km/HR. What will
its latitude be after 1.5 HR?
A) 22° OON
B) 12° 15N
C) 02° 00N
D) 03° 50N
Ans: C
5. The Great Circle bearing of B (70° S 060° E), from A (70° S 030° W), is
approximately?
A) 135° (T)
B) 090° (T)
C) 318° (T)
D) 150° (T)
Ans: A
6. Given: Position A 45° N, ?° E Position B 45° N, 45° 15E Distance A-B=280 NM B is
to the East of A Required: longitude of position A?
A) 38° 39E
B) 49° 57E
C) 40° 33E
D) 51° 51E
Ans: A
7. Refer to chart E(LO)1. What is the aircraft position in lat and long given the
following:
CRN (5318N 00857W) 18 DME
SHA (5243N 00853W) 20 DME Both ranges DME decreasing.
A) 5203N 00843W
B) 5302N 00843W
C) 5301N 00908W
D) 5201N 00908W
Ans: B
8. What is dlat from 30° 39S 20° 20E to 45° 23N 40° 40E:
A) 76° 2 S
B) 14° 44 N
C) 76° 2 S
D) 76° 2 N
Ans: D
9. You are flying from A(30S 20E) to B (30S 20E). At what longitude will the GC
track equal the RL track?
A) 10° E
B) 0° E/W
C) 10° W
D) 20° W
Ans: B
10. Position A is located on the equator at longitude 130° 00E. Position B is
located 100NM from A on a bearing of 225° (T). The coordinates of position B
are:
A) 01° 11N 131° 11E
B) 01° 11S 131° 11E
C) 01° 11S 128° 49E
D) 01° 11N 128° 49E
Ans: C
11. An aircraft at position 60GN 005° W tracks 090° (T) for 315 km. On completion
of the flight the longitude will be:
A) 002° 10W
B) 000° 15E
C) 005° 15E
D) 000° 40E
Ans: D
12. What is the Chlong (in degrees and minutes) from A (45N 16530E) to B (45N
15540W)?
A) 38° 50W
B) 38° 50E
C) 38° 05W
D) 38° 05E
Ans: B
13. A useful method of a pilot resolving, during a visual flight, any uncertainty
in the aircraft’s position is to maintain visual contact with the ground
and:
A) set heading towards a line feature such as a coastline, motorway, river or
railway
B) fly the reverse of the heading being flown prior to becoming undertainuntil a
pinpoint is obtained
C) fly expanding circles until a pinpoint is obtained
D) fly reverse headings and associated timings until the point of departure is
regained.
Ans: A
14. Position A is located on the equator at longitude 130o00E. Position B is
located 100 NM from A on a bearing of 225o(T). The co-ordinates of position
B are:
A) 01o11N 128o 49E
B) 01o11S 128o 49E
C) 01o11N 131o 11E
D) 01o11S 131o 11E
Ans: B
15. 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
Ans: A
16. The Great Circle bearing of B (70oS 060oE), from A (70oS 030oW), is
approximately?
A) 150o (T)
B) 090o (T)
C) 318o (T)
D) 135o (T)
Ans: D
17. What is the final position after the following rhumb line tracks and distances
have been followed from position 60o00N 030o00W?
1. South for 3600 NM
2. East for 3600 NM
3. North for 3600 NM
4. West for 3600 NM
The final position of the aircraft is:
A) 59o00N 090o00W
B) 60o00N 090o00W
C) 60o00N 030o00E
D) 59o00N 060o00W
Ans: B
18. An aircraft at positon 60oN 005oW tracks 090o(T) for 315km. On completion of
the flight the longitude will be:
A) 002o 10W
B) 000o 15E
C) 000o 40E
D) 005o 15E
Ans: C
19. The departure between positions 60oN 160oE and 60sN x is 900 NM. What is the
longitude of x?
A) 170oW
B) 140oW
C) 145oE
D) 175oE
Ans: A
20. An aircraft at latitude 10o South flies north at a GS of 890 km/HR. What will
its latitude be after 1.5 HR?
A) 22o00N
B) 03o50N
C) 02o00N
D) 12o15N
Ans: C
21. You are flying from A (30S 20E) to B (30S 20W). At what longitude will the GC
track equal the RL track?
A) 10oE
B) 10oW
C) 0oE/W
D) 20oW
Ans: C
22. What is diat from 30o39S 20o20E to 45o23N 40o40E:
A) 14o44 N
B) 76o2 S
C) 76o2 N
D) 76o4 S
Ans: C
23. What is the Chlong (in degrees and minutes) from A (45N 1630E) to B (45N
15540W)?
A) 38o05E
B) 38o50W
C) 38o05W
D) 38o50E
Ans: D
24. Given: True Track 245o Drift 5o right Variation 3oE Compass Hdg 242o Calculate
the Magnetic Heading:
A) 247o
B) 243o
C) 237o
D) 253o
Ans: C
25. Grid heading is 299o, grid convergency is 55o West and magnetic variation is
90o West. What is the corresponding magnetic heading?
A) 084o
B) 334o
C) 154o
D) 264o
Ans: A
Name range specifics of maximum range and radius of action:
1. An aircraft was over A at 1435 hours flying direct to B. Given:
1.Distance A to B 2900 NM
2.True airspeed 470 kt
3.Mean wind component OUT +55 kt
4.Mean wind component BACK -75 kt
5.Safe endurance 9 HR 30 MIN
The distance from A to the Point of Safe Return (PSR) A is:
A) 1611 NM.
B) 1759 NM.
C) 2844 NM.
D) 2141 NM.
Ans: D
2. An aircraft takes-off from an airport 2 hours before sunset. The pilot flies a
track of 090° (T), W/V 130° / 20 kt, TAS 100 kt. In order to return to the point of
departure before sunset, the furthest distance which may be travelled is:
A) 115 NM
B) 97 NM
C) 105 NM
D) 84 NM
Ans: B
3. Given: AD = Air distance GD = Ground distance TAS = True Airspeed GS =
Groundspeed Which of the following is the correct formula to calculate ground
distance (GD) gone?
A) GD = TAS/(GS X AD)
B) GD = (AD X GS)/TAS
C) GD = (AD – TAS)/TAS
D) GD = AD X (GS -TAS)/GS
Ans: B
4. Given: Distance A to B is 360 NM Wind component A – B is -15 kt Wind component B
– A is +15 kt TAS is 180 kt. What is the distance from the
equal-time-point to B?
A) 195 NM.
B) 165 NM.
C) 170 NM.
D) 180 NM.
Ans: B
5. 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.
Ans: C
6. Given: Distance Q to R 1760 NM, Groundspeed out 435 kt, Groundspeed back 385 kt
The time from Q to the Point of Equal Time (PET) between Q and R is:
A) 110 MIN.
B) 106 MIN.
C) 102 MIN.
D) 114 MIN.
Ans: D
7. An aircraft has a TAS of 300 knots and a safe endurance of 10 hours. If the wind
component on the outbound leg is 50 knots head, what is the distance to
the point of safe endurance?
A) 1458 nm.
B) 1622 nm.
C) 1544 nm.
D) 1500 nm
Ans: A
8. 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) 1838 NM.
B) 1467 NM.
C) 1313 NM.
D) 1642 NM.
Ans: A
9. You have calculated Point of No Return (PNR) on a flight, having all negative
WCs in the flight plan. During the flight you experience that the W/V is
stronger but coming from the same direction as in the flight plan. Consider the
following statements:
A) The PNR will, if recalculated, move toward the no-wind PNR.
B) The PNR will not change because the neither TAS nor Fuel Flow has changed.
C) A recalculated PNR will move toward the place of departure.
D) You will arrive at the PNR at a later time than flight planned.
Ans: C
10. An aircraft was over A at 1435 hours flying direct to B. Given: Distance A to B
2.900 NM True airspeed 470 kt Mean wind component OUT +55 kt Mean wind
component BACK -75 kt The ETA for reaching the Point of Equal Time (PET) between A
and B is:
A) 1744
B) 1846
C) 1657
D) 1721
Ans: C
Question 747 of 995
11.An aircraft was over Q at 1320 hours flying direct to R. Given:
1.Distance Q to R 3016 NM
2.True airspeed 480 kt
3.Mean wind component out 90 kt
4.Mean wind component back 75 kt
5.Safe endurance 10:00 HR
The distance from Q to the Point of Safe Return (PSR) Q is:
A) 1510 NM
B) 2370 NM
C) 2290 NM
D) 1310 NM
Ans: C
12. Given: Distance A to B 1973 NM Groundspeed OUT 430 kt Groundspeed BACK 385 kt
The time from A to the Point of Equal Time (PET) between A and B is:
A) 181 MIN.
B) 130 MIN.
C) 145 MIN.
D) 162 MIN.
Ans: B
13. 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) 290 MIN.
B) 209 MIN.
C) 219 MIN.
D) 197 MIN.
Ans: A
14. From the departure point, the distance to the point of equal time is:
A) inversely proportional to the total distance to go.
B) inversely proportional to the sum of ground speed out and ground speed back.
C) proportional to the sum of ground speed out and ground speed back.
D) inversely proportional to ground speed back.
Ans: B
15. Two points A and B are 1000 NM apart. TAS = 490 kt. On the flight between A and
B the equivalent headwind is -20 kt. On the return leg between B and A,
the equivalent headwind is +40 kt. What distance from A, along the route A to B, is
the Point of Equal Time (PET)?
A) 455 NM.
B) 530 NM.
C) 470 NM.
D) 500 NM.
Ans: B
16. You fly from C to D, a distance of 450 NM. The WC C – D is +30, and the WC D –
C is -40. TAS is 160 Kt and reduced TAS is 130 Kt. The Fuel Flow is
165 kg/hr, and the Safe endurance when overhead C is 4 hours. Calculate PET between
C and D, based on reduced TAS for the flight from PET to C/D. What
is the flying time from C to PET?
A) 0:51
B) 1:48
C) 1:04
D) 1:21
Ans: A
17. 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) 1630 NM.
D) 1736 NM.
Ans: B
18. An aircraft w as 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) 1752
B) 1756
C) 1820
D) 1742
Ans: A
19. Given: Distance A to B 2484 NM Groundspeed OUT 420 kt Groundspeed BACK 500 kt
The time from A to the Point of Equal Time (PET) between A and B is:
A) 163 MIN.
B) 173 MIN.
C) 193 MIN.
D) 183 MIN.
Ans: C
20. For a distance of 1860 NM between Q and R, a ground speed OUT of 385 kt, a
ground speed BACK of 465 kt and an endurance of 8 HR (excluding reserves) the
distance
from Q to the point of safe return (PSR) is:
A) 930 NM
B) 1685 NM
C) 1532 NM
D) 1865 NM
Ans: B
21. You fly from C to D, a distance of 450 NM. The WC C – D is +30, and the WC D –
C is -40. TAS is 160 Kt and reduced TAS is 130 Kt. The Fuel Flow is
165 kg/hr, and the Safe endurance when overhead C is 4 hours. Calculate PNR for
return to C. What is the distance from PNR to D?
A) 180,0 NM.
B) 155,5 NM.
C) 243,0 NM.
D) 95,5 NM.
Ans: B
22. The distance from A to B is 2368 nautical miles. 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) 190 minutes.
C) 209 minutes.
D) 219 minutes.
Ans: A
23. Why do we normally overlook the descend phase when calculating Point of Equal
Time (PET)?
A) Because we never know what kind of descend clearance we will get from ATC.
B) Because there are so many uncertain factors in the descend phase.
C) Because the descend will have an equal effect, whatever destination we decide to
proceed to.
D) Because the W/V during the descend is not known in academic situations.
Ans: C
32. 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
Ans: B
33. Given: Distance A to B 2484 NM Groundspeed OUT 420 kt Groundspeed BACK 500 kt
The time from A to the Point of Equal Time (PET) between A and B is:
A) 173 min
B) 163 min
C) 193 min
D) 183 min
Ans: C
34. Given:
1. AD = Air distance
2. GD = Ground distance
3. TAS = True airspeed
4. 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)
Ans: A
35. Given: Distance A to B is 360 NMWind component A – B is -15 kt Wind component B
– A is +15 kt TAS is 180 kt What is the distance from the equal-time-point to B?
A) 170 NM
B) 195 NM
C) 180 NM
D) 165 NM
Ans: D
36. 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
Ans: C
37. An aircraft has a TAS of 300 knots and a safe endurance of 0 hours. 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
Ans: B
38. The distance from A to B is 2368 nautical miles. 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
Ans: A
39. For a distance of 1860 NM between Q and R, a ground speed OUT of 385 kt, a
ground speed BACK of 465 kt and an endurance of 8 hr (excluding reserves) the
distance from Q to the point of safe return (PSR) is:
A) 930 NM
B) 1532 NM
C) 1685 NM
D) 1865 NM
Ans: C
40. 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
Ans: B
41. 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
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
Ans: B
42. An aircraft takes off from an airport 2 hours before sunset. The pilot flies a
track of 090o(T). W/V 130o/20 kt, TAS 100 kt. 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
Ans: A
43. The distance between point of departure and destination is 340 NM and wind
velocity in the whole area is 100o/25 kt. TAS is 140 kt. True Track is 135o and
safe endurance 3 hr and 10
min. 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
Ans: A
