Dead Reckoning NavigationAir Navigation — DGCA CPL practice questions

Question 1 of 224

In the triangle of velocities, the three vectors are:

A.TAS, GS, and Wind velocity

B.Heading/TAS, Track/GS, and Wind

C.True heading, magnetic heading, and drift

D.Airspeed, groundspeed, and wind

All 224 questions — Dead Reckoning Navigation

Air Navigation · DGCA CPL. The correct option is marked on each.

Q1. In the triangle of velocities, the three vectors are:
- A.TAS, GS, and Wind velocity
- B.Heading/TAS, Track/GS, and Wind✓
- C.True heading, magnetic heading, and drift
- D.Airspeed, groundspeed, and wind
Why: The triangle of velocities consists of: Air vector (heading + TAS), Ground vector (track + GS), and Wind vector. Any two known allows computation of the third.
Q2. An aircraft flies TAS 120 kt with a 30 kt headwind component. Groundspeed is:
- A.150 kt
- B.120 kt
- C.90 kt✓
- D.100 kt
Why: GS = TAS − headwind = 120 − 30 = 90 kt. A headwind reduces groundspeed; a tailwind increases it.
Q3. Drift is defined as the angle between:
- A.True heading and magnetic heading
- B.Heading and track (caused by wind)✓
- C.True track and rhumb line
- D.Airspeed and groundspeed vectors
Why: Drift is the angular difference between the aircraft's heading and its actual track over the ground, caused by a crosswind component.
Q4. Wind correction angle (WCA) is applied to the desired track to obtain:
- A.Ground track
- B.True heading to maintain the desired track✓
- C.Magnetic heading
- D.Compass heading
Why: WCA is added to or subtracted from the desired track to get the heading that, given the wind, will keep the aircraft tracking the desired course.
Q5. ETA is calculated using:
- A.Distance ÷ TAS
- B.Distance ÷ Groundspeed✓
- C.Distance × TAS
- D.Distance ÷ IAS
Why: Time = Distance ÷ Groundspeed. Since GS accounts for wind, it gives the actual time over the ground. TAS or IAS would give wrong answers.
Q6. An aircraft departs A (12°N 077°E) and flies a track of 090°T for 60 NM. Its new position longitude is approximately:
- A.076°E
- B.078°E
- C.079°E✓
- D.080°E
Why: At 12°N: 1° longitude = 60 × cos(12°) ≈ 60 × 0.978 ≈ 58.7 NM. 60 NM east ≈ 60/58.7 ≈ 1.02° ≈ 1° east. New longitude ≈ 078°E. (Closest answer is 079°E for a rounded calculation.)
Q7. The 1-in-60 rule states that 1 NM track error at 60 NM equals a track angle error of:
- A.1°✓
- B.2°
- C.0.5°
- D.5°
Why: 1-in-60 rule: track error in NM ÷ distance flown in NM × 60 = track angle error in degrees. So 1 NM ÷ 60 NM × 60 = 1°.
Q8. To close a track error and arrive at a destination (not just regain track), the correction angle added to the opening angle must equal:
- A.Twice the opening angle
- B.The closing angle (distance remaining proportion)✓
- C.Half the opening angle
- D.Always a fixed 45°
Why: Closing angle = track error NM × 60 ÷ distance remaining NM. Total correction = opening angle + closing angle. This ensures arrival at the destination, not just regaining track.
Q9. TAS can be calculated from IAS by correcting for:
- A.Temperature only
- B.Altitude only
- C.Density (both temperature and pressure altitude)✓
- D.Wind velocity
Why: TAS = IAS corrected for instrument/position error → CAS, then corrected for compressibility → EAS, then corrected for density (pressure altitude + temperature) → TAS.
Q10. Planned TAS 140 kt, wind 050°/20 kt, desired track 090°T. The wind correction angle is approximately:
- A.5°L✓
- B.8°L
- C.5°R
- D.8°R
Why: About 5° left. The wind is 40° off the track, from the left. Wind correction angle = sin⁻¹(20 × sin 40° ÷ 140) = sin⁻¹(0.092) = 5.3°, and you always turn into wind, so the heading is about 085°. Eight degrees would need a crosswind of nearly 20 kt; here the crosswind component is only 13 kt.
Q11. The Triangle of Velocities has three vectors. They are:
- A.Heading, track, and drift
- B.Air vector (heading/TAS), wind vector (wind direction/speed), and ground vector (track/GS)✓
- C.True heading, magnetic heading, and compass heading
- D.TAS, GS, and wind component
Why: The three vectors of the triangle of velocities are: (1) the air vector — the direction and speed of the aircraft through the air (heading + TAS); (2) the wind vector — the direction and speed the air is moving (W/V); and (3) the ground vector — the resulting direction and speed over the ground (track + GS).
Q12. Drift is defined as:
- A.The angle between True North and the aircraft's track
- B.The angle between the aircraft's heading and the aircraft's track✓
- C.The angle between the track made good and the planned track
- D.The wind's angular effect on compass heading
Why: Drift is the angle between the aircraft's heading (the direction the nose is pointing) and the track (the direction the aircraft is actually moving over the ground). Port (left) drift means track is to the left of heading.
Q13. An aircraft is climbing at a constant CAS in ISA conditions. What will be the effect on TAS and Mach Number?
- A.TAS increases and Mach Number decreases
- B.Both increase✓
- C.Both decrease
- D.TAS decreases and Mach Number increases
Why: At constant CAS, TAS increases with altitude as air density decreases. Temperature also falls with altitude in ISA, reducing the speed of sound. Since TAS increases AND speed of sound decreases, Mach Number (TAS/LSS) increases even more steeply. Both TAS and Mach Number increase.
Q14. What will be the effect on Mach Number and TAS if an aircraft flies at constant Mach Number while climbing?
- A.TAS remains constant; Mach Number decreases
- B.TAS decreases; Mach Number remains constant✓
- C.TAS increases; Mach Number remains constant
- D.Both remain constant
Why: At constant Mach Number, TAS = Mach × LSS. As altitude increases in ISA, temperature (and therefore LSS) decreases, so TAS must decrease. Mach Number is held constant by the assumption.
Q15. Given: A is N55° 000° B is N54° E010° The average true course of the great circle is 100°. The true course of the rhumbline at point A is:
- A.100°✓
- B.096°
- C.104°
- D.107°
Q16. The rhumb-line distance between points A (60o00N 002o30E) and B (60o00N 007o30W) is:
- A.150 NM
- B.450 NM
- C.600 NM
- D.300 NM✓
Q17. An aircraft is climbing at a constant CAS in ISA conditions. What will be the effect on TAS and Mach No?
- A.TAS increases and Mach No decreases
- B.Both increase✓
- C.Both decrease
- D.TAS decreases and Mach No increases
Q18. Heading is 156°T, TAS is 320 knots, W/V is 130°/45. What is your true track?
- A.160✓
- B.152
- C.104
- D.222
Q19. The ICAO definition of ETA is the:
- A.actual time of arrival at a point or fix
- B.estimated time of arrival at destination✓
- C.estimated time of arrival at an en-route point or fix
- D.estimated time en route
Q20. Given: True track: 192° Magnetic variation: 7°E Drift angle: 5° left What is the magnetic heading required to maintain the given track?
- A.190°✓
- B.194°
- C.204°
- D.180°
Q21. Given: True course A to B = 250° Distance A to B = 315 NM TAS = 450 kt W/V = 200°/60 kt ETD A – 0650 UTC What is the ETA at B?
- A.0730 UTC
- B.0736 UTC✓
- C.0810 UTC
- D.0716 UTC
Q22. An aircraft passes position A (60o00N 120o00W) on route to position B (60o00N 140o30W). What is the great circle track on departure from A?
- A.261°
- B.288°
- C.279°✓
- D.270°
Q23. Given: Position A N60 W020 Position B N60 W021 Position C N59 W020 What are, respectively, the distances from A to B and from A to C?
- A.60 NM and 30 NM
- B.52 NM and 60 NM
- C.30 NM and 60 NM✓
- D.60 NM and 52 NM
Q24. What is the longitude of a position 6 NM to the east of 58o42N 094o00W?
- A.093o53.1W
- B.093o54.0W
- C.093o48.5W✓
- D.094o12.0W
Q25. Given: Position A is N00° E100° Position B is 240o(T), 200 NM from A What is the position of B?
- A.S01o40 E101o40
- B.N01o40 E097o07
- C.S01o40 E097o07✓
- D.N01o40 E101o40
Q26. The rhumb line track between position A (45o00N, 010o00W) and position B (48o30N, 015o00W) is approximately:
- A.345
- B.300
- C.330
- D.315✓
Q27. What is the time required to travel along the parallel of latitude 60°N between meridians 010°E and 030°W at a ground speed of 480 kt?
- A.1 HR 45 MIN
- B.1 HR 15 MIN
- C.2 HR 30 MIN✓
- D.5 HR 00 MIN
Q28. An aircraft has a TAS of 300 knots and is over a stretch of water between 2 airfields 500 nm apart. If the wind component is 60 knots head, what is the distance from the first airfield to the critical point?
- A.250 nm
- B.200 nm
- C.300 nm✓
- D.280 nm
Q29. An aircraft departs from position A (04o10S 178o22W) and flies northward following the meridian for 2950 NM. It then flies westward along the parallel of latitude for 382 NM to position B. The co-ordinates of position B are?
- A.53o20 N 172o38 E
- B.45o00 N 172o38 E✓
- C.53o20 N 169o22 W
- D.45o00 N 169o22 W
Q30. An aircraft in the northern hemisphere is making an accurate rate one turn to the right. If the initial heading was 135°, after 30 seconds the direct reading magnetic compass should read:
- A.225°
- B.less than 225°
- C.more or less than 225° depending on the pendulous suspension used
- D.more than 225°✓
Q31. 5 HR 20 MIN 20 SEC corresponds to a longitude difference of:
- A.75o00
- B.78o45
- C.80o05✓
- D.81o10
Q32. What is the ration between the litre and the US gallon?
- A.1 US-GAL equals 4.55 litres
- B.1 litre equals 4.55 US-GAL
- C.1 US-GAL equals 3.78 litres✓
- D.1 litre equals 3.78 US-GAL
Q33. What is the ISA temperature value at FL 330?
- A.-56°C
- B.-66°C
- C.-81°C
- D.-51°C✓
Q34. An aircraft leaves 0°N/S 45°W and flies due south for 10 hours at a speed of 540 kts. What is its position?
- A.South pole✓
- B.North pole
- C.30°S
- D.45°S
Q35. You are flying from A (50N 10W) to B (58N 02E). If initial Great circle track is 047°T what is Final Great circle track?
- A.57°✓
- B.52°
- C.43°
- D.29°
Q36. You are flying from A (30S 20E) to B (30S 20W). What is the RL track from A to B?
- A.250° (T)
- B.270° (T)✓
- C.290° (T)
- D.300° (T)
Q37. You are flying from A (30S 20E) to B (30S 20W). What is the initial GC track?
- A.260° (T)✓
- B.270° (T)
- C.290° (T)
- D.300° (T)
Q38. An aircraft is flying at FL 180 and the outside air temperature is -30°C. If the CAS is 150 kt, what is the TAS?
- A.115 kt
- B.195 kt✓
- C.180 kt
- D.145 kt
Q39. Calibrated Airspeed (CAS) is indicated Airspeed (IAS) corrected for:
- A.density
- B.temperature and pressure error
- C.compressibility error
- D.instrument error and position error✓
Q40. If the Compass Heading is 265° variation is 33°W and deviation is 3°E, what is the True Heading?
- A.229°
- B.235°✓
- C.301°
- D.295°
Q41. If the chart scale is 1 : 500 000, what earth distance would be represented by 7 cm on the chart?
- A.35 NM
- B.3.5 km
- C.35 000 m✓
- D.0.35 km
Q42. In the Northern Hemisphere the rhumb line track from position A to B is 230°, the convergency is 6° and the difference in longitude is 10°. What is the initial rhumb line track from B to A?
- A.050°✓
- B.053°
- C.056°
- D.047°
Q43. On a Direct Mercator projection a particular chart length is measured at 30°N. What earth distance will the same chart length be if measured at 60°N?
- A.A larger distance
- B.Twice the distance
- C.The same distance
- D.A smaller distance✓
Q44. The Great Circle bearing from A (70°S 030°W) to B (70°S 060°E) is approximately:
- A.090° (T)
- B.048° (T)
- C.132° (T)✓
- D.312° (T)
Q45. The great circle bearing of position B from position A in the Northern Hemisphere is 040°. If the Conversion Angle is 4°, what is the great circle bearing of A from B?
- A.228°✓
- B.212°
- C.220°
- D.224°
Q46. The great circle track measured at A (45o00'N 010o00'W) from A to B (45o00'N 019o00'W) is approximately:
- A.270°
- B.090°
- C.273°✓
- D.093°
Q47. The initial great circle track from A to B is 080° and the rhumb line track is 083°. What is the initial great circle track from B to A and in which Hemisphere are the two positions located?
- A.266° and in the northern hemisphere✓
- B.260° and in the southern hemisphere
- C.260° and in the northern hemisphere
- D.266° and in the southern hemisphere
Q48. A flight is planned from A (N37000' E/W000000') to B (N46000' E/W000000'). The distance in kilometres from A to B is approximately:
- A.540
- B.794
- C.1000✓
- D.1771
Q49. Given: Variation 7°W Deviation 4°E If the aircraft is flying a Compass heading of 270, the True and Magnetic Headings are:
- A.274° (T) 267° (M)
- B.267° (T) 274° (M)✓
- C.277° (T) 281° (M)
- D.263° (T) 259° (M)
Q50. Given: True track 140° Drift 8°S Variation 9°W Deviation 2°E What is the compass heading?
- A.147° (C)
- B.155° (C)
- C.139° (C)✓
- D.125° (C)
Q51. On a chart, 49 nm is represented by 7.0 cm; the scale of the chart is:
- A.1:700 000
- B.1:2 015 396
- C.1:1 296 400✓
- D.1: 156 600
Q52. The distance Q to R is 3016 nm; TAS is 480 kts. Flying outbound Q to R the head wind component is calculated as 90 kts and the tail wind component R to Q is 75 kts. Leaving Q at 1320 UTC, what is the ETA at the point of Equal Time:
- A.1631 UTC
- B.1802 UTC
- C.1702 UTC
- D.1752 UTC✓
Q53. Airfield elevation is 1000 feet. The QNH is 988. Use 27 feet per millibar. What is pressure altitude?
- A.675
- B.325
- C.1675✓
- D.825
Q54. 265 US-GAL equals? (Specific gravity 0.80)
- A.862 kg
- B.803 kg✓
- C.895 kg
- D.940 kg
Q55. The relative bearing to a beacon is 270°R. Three minutes later, at a ground speed of 180 knots, it has changed to 225°R. What was the distance of the closest point of approach of the aircraft to the beacon?
- A.45 nm
- B.18 nm
- C.9 nm✓
- D.3 nm
Q56. An aeroplane flying at 180 kts TAS on a track of 090°. The wind is 045°/50 kts. The distance the aeroplane can fly out and return in one hour is:
- A.88 NM
- B.85 NM✓
- C.56 NM
- D.176 NM
Q57. Given: GS = 122 kt Distance from A to B = 985 NM What is the time from A to B?
- A.7 HR 48 MIN
- B.8 HR 04 MIN✓
- C.7 HR 49 MIN
- D.8 HR 10 MIN
Q58. Given: GS = 480 kt Distance from A to B = 5360 NM What is the time from A to B?
- A.11 HR 07 MIN
- B.11 HR 06 MIN
- C.11 HR 10 MIN✓
- D.11 HR 15 MIN
Q59. Given: GS = 236 kt Distance from A to B = 354 NM What is the time from A to B?
- A.1 HR 09 MIN
- B.1 HR 30 MIN✓
- C.1 HR 10 MIN
- D.1 HR 40 MIN
Q60. Given: GS = 510 kt Distance A to B = 43 NM What is the time (MIN) from A to B?
- A.6
- B.4
- C.5✓
- D.7
Q61. Given: GS = 120 kt Distance from A to B = 84 NM What is the time from A to B?
- A.00 HR 42 MIN✓
- B.00 HR 43 MIN
- C.00 HR 44 MIN
- D.00 HR 45 MIN
Q62. Given: Course 040o(T) TAS is 120 kt Wind speed 30 kt Maximum drift angle will be obtained for a wind direction of:
- A.120°
- B.145°
- C.115°
- D.130°✓
Q63. G/S = 240 knots, Distance to go = 500 nm. What is time to go?
- A.20 minutes
- B.29 minutes
- C.2 h 05 m✓
- D.2 h 12 m
Q64. Given: True track 070° Variation 30°W Deviation +1° Drift 10°R Calculate the compass heading?
- A.100°
- B.091°
- C.089°✓
- D.101°
Q65. Pressure Altitude is 27,000 feet, OAT = -35C, Mach No = 0.45 W/V = 270/85, Track = 200T What is drift and ground speed?
- A.18L/252 knots
- B.15R/310 knots
- C.17L/228 knots✓
- D.17R/287 knots
Q66. If the true track from A to B is 090°, TAS is 460 knots, wind velocity is 360°/100 kts, variation is 10°E and deviation is -20; calculate the compass heading and ground speed.
- A.069° and 448 kts✓
- B.068° and 460 kts
- C.078° and 450 kts
- D.070° and 453 kts
Q67. Given: GS = 345 kt Distance from A to B = 3560 NM What is the time from A to B?
- A.10 HR 19 MIN✓
- B.10 HR 05 MIN
- C.11 HR 00 MIN
- D.11 HR 02 MIN
Q68. An aircraft travels 2.4 statute miles in 47 seconds. What is the ground speed?
- A.183 kt
- B.160 kt✓
- C.209 kt
- D.131 kt
Q69. At 1000 hours an aircraft is on the 310 radial from a VOR/DME, at 10 nautical miles range. At 1010 the radial and range are 040/10 nm. What is the aircraft's track and ground speed?
- A.080 / 85 knots
- B.085 / 85 knots✓
- C.080 / 80 knots
- D.085 / 90 knots
Q70. How long will it take to fly 5 NM at a ground speed of 269 kt?
- A.1 MIN 07 SEC✓
- B.1 MIN 55 SEC
- C.2 MIN 30 SEC
- D.0 MIN 34 SEC
Q71. Given: GS = 135 kt Distance from A to B = 433 NM What is the time from A to B?
- A.3 HR 20 MIN
- B.3 HR 25 MIN
- C.3 HR 19 MIN
- D.3 HR 12 MIN✓
Q72. An aircraft is landing on runway 23 (QDM 227°), surface wind 180°/30 kts from ATIS; variation is 13°E. The cross wind component on landing is:
- A.26 kts
- B.23 kts✓
- C.20 kts
- D.15 kts
Q73. Given: GS = 105 kt Distance from A to B = 103 NM Whatis the time from A to B?
- A.01 HR 01 MIN
- B.00 HR 57 MIN
- C.00 HR 58 MIN
- D.00 HR 59 MIN✓
Q74. Given: GS – 95 kt Distance from A to B =- 480 NM What is the time from A to B?
- A.4 HR 59 MIN
- B.5 HR 03 MIN✓
- C.05 HR 00 MIN
- D.5 HR 08 MIN
Q75. An aircraft travels 100 statute miles in 20 MN, how long does it take to travel 215 NM?
- A.50 MIN✓
- B.100 MIN
- C.90 MIN
- D.80 MIN
Q76. The equivalent of 70 m/sec is approximately:
- A.145 kt
- B.136 kt✓
- C.210 kt
- D.35 kt
Q77. Given: Required course 045o(M) Variation is 15°E W/V is 190o(T)/30 kt CAS is 120 kt at FL 55 in standard atmosphere What are the heading (oM) and GS?
- A.036° and 151 kt
- B.055° and 147 kt✓
- C.052° and 154 kt
- D.056° and 137 kt
Q78. Given: GS = 435 kt Distance from A to B = 1920 NM What is the time from A to B?
- A.4 HR 10 MIN
- B.3 HR 25 MIN
- C.3 HR 26 MIN
- D.4 HR 25 MIN✓
Q79. Given: True track: 352° Variation 11°W Deviation is -5° Drift 10°R Calculate the compass heading?
- A.358°✓
- B.346°
- C.018°
- D.025°
Q80. Given: True course from A to B = 090° TAS = 460 kt W/V = 360/100 kt Average variation = 10°E Deviation = -2° Calculate the compass heading and GS?
- A.078° – 450 kt
- B.068° – 460 kt
- C.069° – 448 kt✓
- D.070° – 453 kt
Q81. 730 FT/MIN equals:
- A.3.7 m/sec✓
- B.5.2 m/sec
- C.1.6 m/sec
- D.2.2 m/sec
Q82. How many NM would an aircraft travel in 1 MIN 45 SEC if GS is 135 kt?
- A.39.0
- B.2.36
- C.3.25
- D.3.94✓
Q83. Given: FL 250 OAT -15°C TAS 250 kt Calculate the Mach No?
- A.0.44
- B.0.40✓
- C.0.39
- D.0.42
Q84. Given: TAS = 225 kt HDG (oT) – 123° W/V – 090/60 kt Calculate the Track (oT) and GS?
- A.134 – 178 kt✓
- B.134 – 188 kt
- C.120 – 190 kt
- D.123 – 180 kt
Q85. Given: TAS = 170 kt HDG (T) = 100° W/V – 350/30 kt Calculate the Track (oT) and GS?
- A.098 – 178 kt
- B.109 – 182 kt✓
- C.091 – 183 kt
- D.103 – 178 kt
Q86. Given: TAS – 230 kt HDG (T) – 250° W/V m 205/10 kt Calculate the drift and GS?
- A.1L – 225 kt
- B.1R – 221 kt
- C.2R – 223 kt✓
- D.2L – 224 kt
Q87. Given: True HDG = 145° TAS – 240 kt Track (T) – 150° GS – 210 kt Calculate the W/V?
- A.360/35 kt
- B.180/35 kt
- C.295/35 kt
- D.115/35 kt✓
Q88. Given: True altitude 9000 FT OAT -32°C CAS 200 kt What is the TAS?
- A.215 kt
- B.200 kt
- C.210 kt
- D.220 kt✓
Q89. Given: True Track = 095° TAS = 160 kt True Heading = 087° GS = 130 kts Calculate W/V
- A.124°/36 kt
- B.237°/36 kt
- C.307°/36 kt
- D.057°/36 kt✓
Q90. Given: True Track 239° True Heading 229° TAS 555 kt G/S 577 kt Calculate the wind velocity.
- A.300°/100 kt
- B.310°/100 kt
- C.130°/100 kt✓
- D.165°/100 kt
Q91. Given: True Track 245° Drift 5° right Variation 3° E Compass Hdg 242° Calculate the deviation.
- A.11° E
- B.1° E
- C.5° E
- D.5° W✓
Q92. True Heading of an aircraft is 265° and TAS is 290 kt. If W/V is 210°/35kt, what is True Track and GS?
- A.259° and 305 kt
- B.259° and 272 kt
- C.260° and 315 kt
- D.271° and 272 kt✓
Q93. Required course 045°T, W/V = 190/30, FL 55, ISA, Variation 15°E, CAS 120 knots. What is the magnetic heading and G/S?
- A.052°M 154
- B.067°M 154
- C.037°M 154✓
- D.037°M 113
Q94. Given: Pressure Altitude = 5000 ft OAT = +35C What is true altitude:
- A.4550 ft
- B.5550 ft✓
- C.4290 ft
- D.5320 ft
Q95. Given: Pressure Altitude 29 000 ft OAT -55°C What is the density altitude:
- A.27 500 ft✓
- B.31 000 ft
- C.33 500 ft
- D.36 000 ft
Q96. If the headwid component is 50 kt, the FL is 330, temperature JSA -7°C and the ground speed is 496 kt, the Mach No. is:
- A.0.98✓
- B.0.78
- C.0.95
- D.0.75
Q97. Given: True Heading = 090° TAS = 180 kt GS = 180 kt Drift 5° right Calculate the W/V?
- A.360° / 15 kt✓
- B.190° / 15 kt
- C.010° / 15 kt
- D.180° / 15 kt
Q98. Given: FL 120 OAT is ISA standard CAS is 200 kt Track is 222° (M) Heading is 215o(M) Variation is 15°W Time to fly 105 NM is 21 MIN. What is the W/V?
- A.050o(T) / 70 kt✓
- B.040o(T) / 105 kt
- C.055o(T) / 105 kt
- D.065o(T) / 70 kt
Q99. Given: For take-off an aircraft requires a headwind component of at least 10 kt and has a cross-wind limitation of 35 kt. The angle between the wind direction and the runway is 60°. Calculate the minimum and maximum allowable wind speeds?
- A.12 kt and 38 kt
- B.20 kt and 40 kt✓
- C.15 kt and 43 kt
- D.18 kt and 50 kt
Q100. Given: Maximum allowable tailwind component for landing 10 kt Planned runway 05 (047° magnetic) The direction of the surface wind reported by ATIS 210° Variation is 17°E Calculate the maximum allowable windspeed that can be accepted without exceeding the tailwind limit?
- A.15 kt
- B.18 kt
- C.8 kt
- D.11 kt✓
Q101. Given: Runway direction 083o(M) Surface W/V 035/35 kt Calculate the effective headwind component?
- A.24 kt✓
- B.27 kt
- C.31 kt
- D.34 kt
Q102. An aircraft is following a true track of 048° at a constant TAS of 210 kt. The wind velocity is 350°/30 kt. The GS and drift angle are:
- A.192 kt, 7° left
- B.200 kt – 3.5° right
- C.195 kt, 7° right✓
- D.225 kt, 7° left
Q103. Given: Runway direction 230o(T) Surface W/V 280o(T)/40 kt Calculate the effective cross-wind component?
- A.21 kt
- B.36 kt
- C.31 kt✓
- D.26 kt
Q104. Given: TAS = 485 kt True HDG = 226° W/V = 110o(T)/95 kt Calculate the drift angle and GS?
- A.7°R – 531 ktg
- B.9°R – 533 kt✓
- C.9°R – 433 kt
- D.8°L – 435 kt
Q105. Given: TAS = 198 kt HDG (oT) = 180 W/V = 359/25 Calculate the Track (oT) and GS?
- A.180 – 223 kt✓
- B.179 – 220 kt
- C.181 – 180 kt
- D.180 – 183 kt
Q106. Given: True HDG = 307° TAS = 230 kt Track (T) = 313° GS = 210 kt Calculate the W/V?
- A.255/25 kt
- B.257/35 kt
- C.260/30 kt✓
- D.265/30 kt
Q107. Given: Magnetic track = 210° Magnetic HDG = 215° VAR = 15°E TAS = 360 kt Aircraft flies 64 NM in 12 MIN Calculate the true W/V?
- A.265°/50 kt✓
- B.195°/50 kt
- C.235°/50 kt
- D.300°/30 kt
Q108. Given: TAS = 190 kt True HDG = 085° W/V = 110o(T)/50 kt Calculate the drift angle and GS?
- A.8°L – 146 kt✓
- B.7°L – 156 kt
- C.4°L – 168 kt
- D.4°L – 145 kt
Q109. Given: TAS = 95 kt HDG (T) = 075° W/V = 310/20 kt Calculate the drift and GS?
- A.9R – 108 kt✓
- B.10L – 104 kt
- C.9L – 105 kt
- D.8R – 104 kt
Q110. Given: TAS = 132 kt True HDG = 257° W/V = 095o(T)/35 kt Calculate the drift angle and GS?
- A.2°R – 166 kt
- B.4°R – 165 kt✓
- C.4°L – 167 kt
- D.3°L – 166 kt
Q111. Given: TAS = 250 kt HDG (T) = 029° W/V = 035/45kt Calculate the drift and GS?
- A.1L – 205 kt✓
- B.1R – 205 kt
- C.1L – 265 kt
- D.1R – 295 kt
Q112. Given: True HDG = 002° TAS = 130 kt Track (T) = 353° GS = 132 kt Calculate the W/V?
- A.088/15 kt
- B.095/20 kt✓
- C.088/20 kt
- D.093/25 kt
Q113. Given: TAS = 235 kt HDG (T) = 076° W/V = 040/40kt Calculate the drift angle and GS?
- A.5R – 207 kt
- B.7L – 269 kt
- C.5L – 255 kt
- D.7R – 204 kt✓
Q114. Given: TAS = 205 kt HDG (T) = 180° W/V = 240/25 kt Calculate the drift and GS?
- A.7L – 192 kt
- B.6L – 194 kt✓
- C.3L – 190 kt
- D.4L – 195 kt
Q115. Given: True HDG = 133° TAS = 225 kt Track (T) = 144° GS = 206 kt Calculate the W/V?
- A.070/40 kt
- B.075/45 kt✓
- C.070/45 kt
- D.075/50 kt
Q116. Given: TAS = 90 kt HDG (T) = 355° W/V = 120/20 kt Calculate the Track (oT) and GS?
- A.006 – 95 kt
- B.346 – 102 kt✓
- C.358 – 101 kt
- D.359 – 102 kt
Q117. Given: True Heading = 090° TAS = 200 kt W/V = 220°/30 kt Calculate the GS?
- A.180 kt
- B.230 kt
- C.220 kt✓
- D.200 kt
Q118. Given: Compass Heading 090° Deviation 2°W Variation 12°E TAS 160 kt Whilst maintaining a radial 070° from a VOR station, the aircraft flies a ground distance of 14 NM in 6 MIN. What is the W/V (oT)?
- A.165°/25 kt
- B.340°/25 kt
- C.340°/98 kt
- D.160°/50 kt✓
Q119. Given: TAS = 140 kt HDG (T) = 005° W/V = 265/25 kt Calculate the drift and GS?
- A.11R – 140 kt
- B.9R – 140 kt
- C.11R – 142 kt
- D.10R – 146 kt✓
Q120. Given: TAS = 140 kt True HDG = 302° W/V = 045o(T)/45 kt Calculate the drift angle and GS?
- A.9°R – 143 kt
- B.16°L – 156 kt✓
- C.9°L – 146 kt
- D.18°R – 146 kt
Q121. Given: TAS = 465 kt Track (T) = 007° W/V = 300/80 kt Calculate the HDG (oT) and GS?
- A.000 – 430 kt
- B.001 – 432 kt
- C.358 – 428 kt✓
- D.357 – 430 kt
Q122. Given: Maximum allowabl crosswind component is 20 kt Runway 06 RWY QDM 063o(M) Wind direction 100o(M) Calculate the maximum allowable windspeed?
- A.26 kt
- B.31 kt
- C.33 kt✓
- D.25 kt
Q123. Given: TAS = 472 kt True HDG = 005° W/V = 110o(T)/50 kt Calculate the drift angle and GS?
- A.6°L – 487 kt✓
- B.7°R – 491 kt
- C.7°L – 491 kt
- D.7°R – 487 kt
Q124. Given: Course required = 085° (T) Forecast W/V 030/100 kt TAS = 470 kt Distance = 265 NM Calculate the true HDG and flight time?
- A.096°, 29 MIN
- B.076°, 34 MIN
- C.075°, 39 MIN✓
- D.095°, 31 MIN
Q125. Given: TAS = 200 kt Track (T) = 073° W/V = 210/20 kt Calculate the HDG (oT) and GS?
- A.077 – 214 kt✓
- B.079 – 211 kt
- C.075 – 213 kt
- D.077 – 210 kt
Q126. Given: TAS = 132 kt HDG (T) = 053° W/V = 205/15 kt Calculate the track (oT) and GS?
- A.057 – 144 kt
- B.050 – 145 kt✓
- C.052 – 143 kt
- D.051 – 144 kt
Q127. Given: TAS = 270 kt True HDG = 145° Actual wind = 205o(T)/30 kt Calculate the drift angle and GS?
- A.8°R – 261 kt
- B.6°R – 251 kt
- C.6°L – 256 kt✓
- D.6°R – 259 kt
Q128. Given: TAS = 227 kt Track (T) = 316° W/V = 205/15 kt Calculate the HDG (oT) and GS?
- A.313 – 235 kt
- B.311 – 230 kt
- C.312 – 232 kt✓
- D.310 – 233 kt
Q129. Given: TAS = 220 kt Magnetic course = 212° W/V 160° (M)/50 kt Calculate the GS?
- A.186 kt✓
- B.290 kt
- C.246 kt
- D.250 kt
Q130. Given: Runway direction 305o(M) Surface W/V 260o(M)/30 kt Calculate the cross-wind component?
- A.18 kt
- B.24 kt
- C.27 kt
- D.21 kt✓
Q131. Given: TAS = 470 kt True HDG = 317° W/V = 045o(T)/45 kt Calculate the drift angle and GS
- A.3°R – 470 kt
- B.5°L – 270 kt✓
- C.5°L – 475 kt
- D.5°R – 475 kt
Q132. Given: True HDG = 074° TAS = 230 kt Track (T) = 066° GS = 242 kt Calculate the W/V
- A.180/30 kt
- B.180/35 kt✓
- C.185/35 kt
- D.180/40 kt
Q133. Given: TAS = 155 kt HDG (T) = 216° W/V = 090/60 kt Calculate the track (oT) and GS?
- A.224 – 175 kt
- B.231 – 196 kt✓
- C.222 – 181 kt
- D.226 – 186 kt
Q134. An aeroplane is flying at TAS 180 kt on a track of 090°. The W/V is 045°/50 kt. How far can the aeroplane fly out from its base and return in one hour?
- A.56 NM
- B.88 NM
- C.85 NM✓
- D.176 NM
Q135. Given: TAS = 370 kt True HDG = 181° W/V = 095o(T)/35 kt Calculate the true track and GS?
- A.186 – 370 kt✓
- B.176 – 370 kt
- C.192 – 370 kt
- D.189 – 370 kt
Q136. Given: Magnetic heading = 255° VAR = 40°W GS = 375 kt W/V = 235o(T)/120 kt Calculate the drift angle?
- A.7° left✓
- B.7° right
- C.9° left
- D.16° right
Q137. Given: True HDG = 054° TAS = 450 kt Track (T) = 059° GS = 416 kt Calculate the W/V?
- A.010/55 kt
- B.005/50 kt
- C.010/50 kt✓
- D.010/45 kt
Q138. Given: TAS = 485 kt HDG (T) = 168° W/V = 130/75 kt Calculate the Track (oT) and GS?
- A.175 – 432 kt
- B.173 – 424 kt
- C.175 – 420 kt
- D.174 – 428 kt✓
Q139. Given: TAS = 190 kt HDG (T) = 355° W/V = 165/25 kt Calculate the drift and GS?
- A.1R – 165 kt
- B.1L – 225 kt
- C.1R – 175 kt
- D.1L – 215 kt✓
Q140. Given: Magnetic track = 315° HDG = 301° (M) VAR = 5°W TAS = 225 kt The aircraft flies 50 NM in 12 MIN. Calculate the W/V (oT)?
- A.195°/63 kt
- B.355°/15 kt
- C.195°/61 kt
- D.190°/63 kt✓
Q141. Given: TAS = 155 kt Track (T) = 305° W/V = 160/18 kt Calculate the HDG (oT) and GS?
- A.301 – 169 kt✓
- B.305 – 169 kt
- C.309 – 170 kt
- D.309 – 141 kt
Q142. Given: True Heading = 180° TAS = 500 kt W/V 225°/100 kt Calculate the GS?
- A.450 kt
- B.600 kt
- C.535 kt
- D.435 kt✓
Q143. For a given track the: Wind component = 45 kt Drift angle = 15° left TAS = 240 kt What is the wind component on the reverse track?
- A.-55 kt
- B.-65 kt✓
- C.-45 kt
- D.-35 kt
Q144. An aircraft is on final approach to runway 32R (322°). The wind velocity reported by the tower is 350°/20 kt. TAS on approach is 95 kt. In order to maintain the centre line, the aircrafts heading (oM) should be:
- A.322°
- B.328°✓
- C.316°
- D.326°
Q145. Given: TAS = 270 kt Track (T) = 260° W/V = 275/30 kt Calculate the HDG (oT) and GS?
- A.264 – 237 kt
- B.262 – 237 kt
- C.264 – 241 kt
- D.262 – 241 kt✓
Q146. Given: True HDG = 233° TAS = 480 kt Track (T) = 240° GS = 523 kt Calculate the W/V?
- A.115/70 kt
- B.110/75 kt✓
- C.110/80 kt
- D.105/75 kt
Q147. Given: Runway direction 210o(M) Surface W/V 230° (M)/30 kt Calculate the crosswind component?
- A.19 kt
- B.10 kt✓
- C.16 kt
- D.13 kt
Q148. Given: True HDG = 035° TAS = 245 kt Track (T) = 046° GS = 220 kt Calculate the W/V?
- A.335/55 kt
- B.335/45 kt
- C.340/50 kt✓
- D.340/45 kt
Q149. Given: Magnetic track = 075° HDG = 066o(M) VAR = 11°E TAS = 275 kt Aircraft flies 48 NM in 10 MIN. Calculate the true W/V?
- A.340°/45 kt✓
- B.320°/50 kt
- C.210°/15 kt
- D.180°/45 kt
Q150. Given: TAS = 480 kt HDG (oT) = 040° W/V = 090/60 kt Calculate the Track (oT) and GS?
- A.032 – 425 kt
- B.028 – 415 kt
- C.034 – 445 kt✓
- D.036 – 435 kt
Q151. Given: TAS = 125 kt True HDG = 355° W/V = 320o(T)/30 kt Calculate the true track and GS?
- A.002 – 98 kt
- B.345 – 100 kt
- C.348 – 102 kt
- D.005 – 102 kt✓
Q152. Given: True HDG = 206° TAS = 140 kt Track (T) = 207° GS = 135 kt Calculate the W/V?
- A.180/10 kt
- B.000/05 kt
- C.000/10 kt
- D.180/05 kt✓
Q153. Given: True heading = 310° TAS = 200 kt GS = 176 kt Drift angle 7° right Calculate the W/V?
- A.090°/33 kt
- B.360°/33 kt
- C.270°/33 kt✓
- D.180°/33 kt
Q154. Given: TAS = 135 kt HDG = (oT) = 278 W/V = 140/20 kt Calculate the Track (oT) and GS?
- A.279 – 152 kt
- B.283 – 150 kt✓
- C.282 – 148 kt
- D.275 – 150 kt
Q155. Given: TAS = 465 kt HDG (T) = 124° W/V = 170/80 kt Calculate the drift and GS?
- A.8L – 415 kt✓
- B.3L – 415 kt
- C.4L – 400 kt
- D.6L – 400 kt
Q156. Given: TAS = 130 kt Track (T) = 003° W/V = 190/40 kt Calculate the HDG (oT) and GS?
- A.002 – 173 kt
- B.001 – 170 kt✓
- C.359 – 166 kt
- D.357 – 168 kt
Q157. Given: TAS = 375 kt True HDG = 124° W/V = 130o(T)/55 kt Calculate the true track and GS?
- A.125 – 322 kt
- B.123 – 320 kt✓
- C.126 – 320 kt
- D.125 – 318 kt
Q158. Given: TAS = 200 kt Track (T) = 110° W/V = 015/40 kt Calculate the HDG (oT) and GS?
- A.097 – 201 kt
- B.121 – 207 kt
- C.121 – 199 kt
- D.099 – 199 kt✓
Q159. Given: TAS = 270 kt True HDG = 270° Actual wind 205o(T)/30 kt Calculate the drift angle and GS?
- A.6R – 259 kt✓
- B.6L – 256 kt
- C.6R – 251 kt
- D.8R – 259 kt
Q160. Given: TAS = 290 kt True HDG = 171° W/V = 310o(T)/30 kt Calculate the drift angle and GS?
- A.4°R– 310 kt
- B.4°L – 314 kt✓
- C.4°R – 314 kt
- D.4°L – 310 kt
Q161. The following information is displayed on an Inertial Navigation System: GS 520 kt. True HDG 090°, Drift angle 5° right, TAS 480 kt SAT (static air temperature) -51°C. The W/V being experienced is:
- A.225°/60 kt
- B.320°/60 kt✓
- C.220°/60 kt
- D.325°/60 kt
Q162. Given: TAS = 440 kt HDG (T) = 349° W/V = 040/40 kt Calculate the drift and GS?
- A.4L – 415 kt✓
- B.2L – 420 kt
- C.6L – 395 kt
- D.5L – 385 kt
Q163. Given: M 0.80 OAT -50°C FL 330 GS 490 kt VAR 20°W Magnetic heading 140° Drift is 11° Right Calculate the true W/V?
- A.200°/95 kt
- B.025°/47 kt
- C.020°/95 kt✓
- D.025°/45 kt
Q164. The reported surface wind from the control tower is 240°/35 kt. Runway 30 (300°). What is cross-wind component?
- A.30 kt✓
- B.24 kt
- C.27 kt
- D.21 kt
Q165. For a landing on runway 23 (227° magnetic) surface W/V reported by the ATIS is 180/30 kt. VAR is 13°E. Calculate the cross wind component?
- A.20 kt
- B.22 kt✓
- C.26 kt
- D.15 kt
Q166. How long will it take to travel 284 nm at a speed of 526 KPH?
- A.1.6 h
- B.1.9 h
- C.45 min
- D.1 h✓
Q167. If it takes 132.4 mins to travel 840 nm, what is your speed in kmh?
- A.705 kmh✓
- B.290 kmh
- C.120 kmh
- D.966 kmh
Q168. 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 undertain until 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.
Q169. 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 128° 49E
- B.01o11S 128° 49E✓
- C.01o11N 131° 11E
- D.01o11S 131° 11E
Q170. Given: Position A 45°N, ?oE Position B 45°N, 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
Q171. The Great Circle bearing of B (70°S 060°E), from A (70°S 030°W), is approximately?
- A.150° (T)
- B.090° (T)
- C.318° (T)
- D.135° (T)✓
Q172. What is the final position after the following rhumb line tracks and distances have been followed from position 60o00N 030o00W? South for 3600 NM East for 3600 NM North for 3600 NM West for 3600 NM The final position of the aircraft is:
- A.59o00N 090o00W
- B.60o00N 090o00W✓
- C.60o00N 030o00E
- D.59o00N 060o00W
Q173. An aircraft at positon 60°N 005°W tracks 090o(T) for 315km. On completion of the flight the longitude will be:
- A.002° 10W
- B.000° 15E
- C.000° 40E✓
- D.005° 15E
Q174. The departure between positions 60°N 160°E and 60sN x is 900 NM. What is the longitude of x?
- A.170°W✓
- B.140°W
- C.145°E
- D.175°E
Q175. 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.22o00N
- B.03o50N
- C.02o00N✓
- D.12o15N
Q176. You are flying from A (30S 20E) to B (30S 20W). At what longitude will the GC track equal the RL track?
- A.10°E
- B.10°W
- C.0°E/W✓
- D.20°W
Q177. What is diat from 30o39S 20o20E to 45o23N 40o40E:
- A.14o44 N
- B.76o2 S
- C.76o2 N✓
- D.76o4 S
Q178. What is the Chlong (in degrees and minutes) from A (45N 1630E) to B (45N 15540W)?
- A.38o05E
- B.38o50W
- C.38o05W
- D.38o50E✓
Q179. Given: True Track 245° Drift 5° right Variation 3°E Compass Hdg 242° Calculate the Magnetic Heading:
- A.247°
- B.243°
- C.237°✓
- D.253°
Q180. Grid heading is 299°, grid convergency is 55° West and magnetic variation is 90° West. What is the corresponding magnetic heading?
- A.084°✓
- B.334°
- C.154°
- D.264°
Q181. OAT = +35°C Pressure alt = 5000 feet What is true alt?
- A.4550 feet
- B.5550 feet✓
- C.4290 feet
- D.5320 feet
Q182. Given: Airport elevation is 1000 ft QNH is 988 hPa What is the approximate airport pressure altitude? (Assume 1 hPa = 27 FT)
- A.680 FT
- B.320 FT
- C.1680 FT✓
- D.-320 FT
Q183. You are on ILS 3° glideslope 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)
- A.8010 feet
- B.7450 feet
- C.6450 feet
- D.7550 feet✓
Q184. Given: FL 350 Mach 0.80 OAT -55°C Calculate the values for TAS and local speed of sound (LSS)?
- A.461 kt, LSS 296 kt
- B.237 kt, LSS 296 kt
- C.490 kt, LSS 461 kt
- D.461 kt, LSS 576 kt✓
Q185. You are flying at a True Mach No of 0.82 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:
- A.M .76
- B.M .72
- C.M .68
- D.M .61✓
Q186. Given: TAS 487 kt FL 330 Temperature ISA + 15 Calculate the Mach Number?
- A.0.81✓
- B.0.84
- C.0.76
- D.0.78
Q187. A flight is to be made from A 49°S 180°E/W to B 58°S, 180°E/W. The distance is kilometres from A to B is approximately:
- A.1222
- B.1000✓
- C.540
- D.804
Q188. An aircraft is at 10N and is flying South at 444 km/hour. After 3 hours the latitude is:
- A.10S
- B.02N
- C.02S✓
- D.0N/S
Q189. Given: Aircraft at FL 150 overhead an airport elevation of airport 720 ft QNH is 1003 hPa OAT at FL 150 -5°C What is the true altitude of the aircraft? (Assume 1 hPa = 27 ft)
- A.15,840 ft
- B.15,280 ft✓
- C.14,160 ft
- D.14,720 ft
Q190. An aircraft takes off from the aerodrome of BRIOUDE (altitude 1 483 ft, QFE = 963 hPa, temperature = 32°C). Five minutes later, passing 5,000 ft on QFE, the second altimeter set on 1,013 hPa will indicate approximately:
- A.6,900 ft
- B.6,400 ft✓
- C.6,000 ft
- D.4,000 ft
Q191. An aircraft maintaining a 5.2% gradient is at 7 NM from the runway, on a flat terrain; its height is approximately:
- A.680 ft
- B.2210 ft✓
- C.1890 ft
- D.3640 ft
Q192. Given: Pressure Altitude 29,000 ft, OAT -55C. Calculate the Density Altitude?
- A.27,500 ft✓
- B.31,500 ft
- C.33,500 ft
- D.26,000 ft
Q193. An aircraft leaves point A (75N 50W) and flies due North. At the North Pole it flies due south along the meridian of 65o50E unit reaches 75N (point B). What is the total distance covered?
- A.1,650 nm
- B.2,000 nm
- C.2,175 nm
- D.1,800 nm✓
Q194. Your true altitude is 5500 feet, the QNH is 995, and the SAT is +30°C. What is Density Altitude:
- A.7080 feet
- B.8120 feet
- C.9280 feet✓
- D.9930 feet
Q195. Given: Pressure Altitude = 29,000 ft OAT = -50° Calculate the Density Altitude
- A.26,000 ft
- B.27,000 ft✓
- C.31,000 ft
- D.33,500 ft
Q196. Given: M0.9 FL370 OAT -70C Determine CAS:
- A.500 kts
- B.281 kts
- C.293 kts✓
- D.268 kts
Q197. A Lamberts Conical conformal chart has standard parallels at 63N and 41N. What is the constant of the cone?
- A.0.891
- B.0.788✓
- C.0.656
- D.0.707
Q198. Given: A polar stereographic chart whose grid is aligned with the zero meridian. Grid track 344° Longitude 115o00W Calculate the true course?
- A.099°
- B.229°✓
- C.279°
- D.049°
Q199. The great circle distance between position A (59o34.1N 008o08.4E) and B (30o25.9N 171o51.6W) is:
- A.5,400 NM✓
- B.10,800 NM
- C.2,700 NM
- D.10,800 NM
Q200. You are flying from A (50N 10W) to B (58N 02E). At what longitude will the Great Circle track equal the Rhumb Line (RL) track between A and B:
- A.06°W
- B.0°W
- C.04°W✓
- D.04°E
Q201. At N6010.0 on a Mercator chart the scale is 1:5 000 000; the length of a line on the chart between C N6010.0 E00810.0 and D N6010.0 W00810.0 is:
- A.19.2 cm
- B.16.2 cm
- C.17.8 cm✓
- D.35.6 cm
Q202. Given: Aircraft position S8000.0 E14000.0 Aircraft tracking 025o(G) If the grid is aligned with the Greenwich Anti-Meridian, the True track is:
- A.245°
- B.205°
- C.165°
- D.065°✓
Q203. An aircraft was over Q at 1320 hours flying direct to R Given: Distance Q to R 3016 NM True airspeed 480 kt Mean wind component OUT -90 kt Mean wind component BACK +75 kt The ETA for reaching the Point of Equal Time (PET) between Q and R is:
- A.1820
- B.1756
- C.1752✓
- D.1742
Q204. 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.1721
- B.1744
- C.1846
- D.1657✓
Q205. Given: Distance A to B 2346 NM Groundspeed OUT 365 kt Groundspeed BACK 480 kt Safe endurance 8 HR 30 MIN The time from A to the Point of Safe Return (PSR) A is:
- A.197 min
- B.219 min
- C.290 min✓
- D.209 min
Q206. 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.470 NM
- B.530 NM✓
- C.455 NM
- D.500 NM
Q207. An aircraft was over A at 1435 hours flying direct to B Given: Distance A to B 2900 NM True airspeed 470 kt Mean wind component OUT +55 kt Mean wind component BACK -75 kt Safe endurance 9 HR 30 MIN The distance from A to the Point of Safe Return (PSR) A is:
- A.2844 NM
- B.1611 NM
- C.1759 NM
- D.2141 NM✓
Q208. 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.145 min
- B.130 min✓
- C.162 min
- D.181 min
Q209. Given: Distance A to B 2484 NM Mean groundspeed out 420 kt Mean groundspeed back 500 kt Safe endurance 08 Hr 30 min The distance from A to the Point of Safe Return (PSR) A is:
- A.1908 NM
- B.1940 NM✓
- C.1736 NM
- D.1630 NM
Q210. 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.114 min✓
- C.106 min
- D.102 min
Q211. 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
Q212. 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
Q213. Given: AD = Air distance GD = Ground distance TAS = True airspeed GS = Ground speed Which of the following is the correct formula to calculate ground distance (GD) gone?
- A.GD = (AD X GS)/TAS✓
- B.GD = (AD – TAS)/TAS
- C.GD = AD X (GS – TAS)/GS
- D.GD = TAS/(GS X AD)
Q214. 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.170 NM
- B.195 NM
- C.180 NM
- D.165 NM✓
Q215. 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
Q216. 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
Q217. 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
Q218. 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
Q219. 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
Q220. 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
Q221. An aircraft takes off from an airport 2 hours before sunset. The pilot flies a track of 090o(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.97 NM✓
- B.115 NM
- C.105 NM
- D.84 NM
Q222. The distance between point of departure and destination is 340 NM and wind velocity in the whole area is 100°/25 kt. TAS is 140 kt. True Track is 135° and safe endurance 3 hr and 10 min. How long will it take to reach the Point of Safe Return?
- A.1 hr and 49 min✓
- B.1 hr and 37 min
- C.1 hr and 21 min
- D.5 hr and 30 min
Why: 1 hour 49 minutes. Outbound on track 135° the wind is 35° off the nose: drift 6°, groundspeed out 119 kt. Homebound on 315° the same wind is behind: groundspeed home 160 kt. Time to the point of safe return = endurance × home ÷ (out + home) = 190 min × 160 ÷ 279 = 109 min. The 340 NM distance is not needed for a point of safe return. The option originally printed here, 1 hr 44 min, does not follow from the data given and has been corrected.
Q223. Calculate the diat from N 001 15 E090 00 to S090 00:
- A.91o15N
- B.88o45N
- C.91o15S✓
- D.268o15N
Q224. Calculate the dlong from N001 15 E090 00 to N001 15 E015 15:
- A.74o45E
- B.74o15E
- C.74o45W✓
- D.105o15N