Email
info@vayudootaviation.com
Phone No
+91 7276526726
become a pilot
Home
About us
courses
CPL
ATPL
RTR(A)
Airline Cadet Program
IGRUA Entrance Prep
B.Sc. Aviation
Pilot Training
INDIA
USA
CANADA
SOUTH AFRICA
NEW ZEALAND
EUROPE
Aircraft Type-Rating
Airbus A320
Boeing 737
Blogs
contact
CPL Test Series
Question Bank
Questions for R* B*LI HOLDING, RADIAL INTERCEPT BEARING TO PLOT
Answer the following questions
Test Mode
Training Mode
1.
A/C is homing to stn on a hdg 150° with 10°S drift It is asked to hold on holding leg oof 330° in std hold. Entry?
- [SEC 1 = 150° - 70° = 080° TO 330° SEC 2 = 330° TO 150° + 110° = 260° SEC 3 = 260° TO 080° MBT = H + D = 150° + 10° = 160° RADIAL = 160° + 180° = 344° SEC 1 PARALLEL ENTRY ]
- [SEC 1 = 150° - 70° = 080° TO 330° SEC 2 = 330° TO 150° + 110° = 260° SEC 3 = 260° TO 080° MBT = H + D = 150° + 10° = 160° RADIAL = 160° + 180° = 340° SEC 1 PARALLEL ENTRY ]
2. An aero planes RMI shows an NDB indication of 070°, w/v calm. The aero plane is to Join a right hand holding pattern at the NDE, the inbound leg of which is 330°. The aero plane should
fly to the NDB and join the pattern
fly to the NDB then fly outbound on 150° for 1 minute, then turn left to point directly at the NDB joining the pattern overhead.
fly to the NDB then fly a teardrop with an outbound heading of 120° for 1 minute and a rale one turn to join in bound.
3. Holding radial 090° std hold, winds 130°/25kt TAS = 210 kts. Find a) Hdg on inhound and outbount b) Time on outbound
- [A) 265° AND 105°, B) 80 SEC.]
- [A) 265° AND 105°, B) 86 SEC.]
4. An aircraft heading 135° (M) with 13° Right drift intercepts the 082° (M) track outbound from an NDB. The relative bearing of the NDB that confirms track interception is
127° Relative.
132° Relative.
137° Relative.
5.
QTE = 320° in SH, CA = 8°, find BTP on Mercator.
- [BTP = 328°.]
- [BTP = 338°.]
6.
A/C is approaching to stn on radial 300° with 10°S drift it is asked to approach on radial 270°. After reaching stn a/c is to hold on radial 110° non std hold. Entry?
- [MBT = 120° FINAL MBT = 090° HEADING = 120°-10°=110°, (I) DIFF. 30° INTERCEPT = 60° NEW HDG = 090°+60°=150° (II) INTERCEPT = 60° (III) TURN RIGHT BY 40° (150°-110°=40°) (IV) NEW HDG = 150° (V) ADF READS = 300°.]
- [MBT = 120° FINAL MBT = 090° HEADING = 120°-10°=110°, (I) DIFF. 30° INTERCEPT = 60° NEW HDG = 090°+60°=150° (II) INTERCEPT = 60° (III) TURN RIGHT BY 40° (150°-110°=40°) (IV) NEW HDG = 150° (V) ADF READS = 310°.]
7.
A/C is approaching to stn on Hdg 185° with 10°S drift. It is told to intercept radial 340° inbound.
- [MBT = 185°+10°=195° FINAL MBT = 160° DIFFERENCE = 35°, INTERCEPT =90° NEW HEADING = 160°+90° = 250°, TURN = STARBOARD, 160°+90°=250°-185°=65°, ADF = 160°-250°=-90°+360°=270°.]
- [MBT = 185°+10°=195° FINAL MBT = 160° DIFFERENCE = 35°, INTERCEPT =90° NEW HEADING = 160°+90° = 250°, TURN = STARBOARD, 160°+90°=250°-185°=65°, ADF = 160°-250°=-90°+360°=250°.]
8.
ADF RMI = 240° in NH, Var = 10°W, CA = 5°, Find BTP on Mercator.
- [QDM = 240°, QDR = 240°-180° = 060° QTE = 060° - 10° = 050° BTP = 045°.]
- [QDM = 240°, QDR = 240°-180° = 060° QTE = 060° - 10° = 050° BTP = 055°.]
9.
A/C is approaching to stn on radial 230° it is told to app on 270°.
- [MBT = 050° FINAL MBT = 090° DIFF. = 40° INTERCEPT = 90°, NEW HDG = 090°-090°=000°, (I) INTERCEPT = 90° (II) LEFT BY 50° 000° - 050° (III) NEW HDG = 000°, (IV) ADF READS 090° - 000° = 090°.]
- [MBT = 050° FINAL MBT = 090° DIFF. = 40° INTERCEPT = 90°, NEW HDG = 090°-090°=000°, (I) INTERCEPT = 90° (II) LEFT BY 50° 000° - 050° (III) NEW HDG = 000°, (IV) ADF READS 090° - 000° = 050°.]
10.
QTE = 240° in NH, CA = 10°, find BTP on Mercator.
- [BTP = 240°]
- [BTP = 230°]
11.
LSR - 198°, QC = +2°, Hdg (T) = 210°, CA = 5°, in Southern Hemisphere. Find BTP on Mercator.
- [WHILE ADDING HDG AND RB IF YOU EXCEED 360° SUBTRACT 360° FROM IT. RB = 198°+2°=200°, MBT = RB + HDG = 200°+210°=410°-360° = 050°, QTE = 230° BTP = 230°+5° = 235°]
- [WHILE ADDING HDG AND RB IF YOU EXCEED 360° SUBTRACT 360° FROM IT. RB = 198°+2°=200°, MBT = RB + HDG = 200°+210°=410°-360° = 050°, QTE = 230° BTP = 230°+5° = 245°]
12.
A/C is approaching to stn on radial 150°/330° in nil winds, a/c is asked to approach on radial 300°.
- [MBT = 150° FINAL MBT = 120°, INTERCEPT = 30°×2=60° NEW HEADING = 120 + 60° = 180°, DRIFT = 30°, (I) INTERCEPT = 60° (II) TURN 120°+60°=180°-150°=30° RIGHT TURN (III) 180° (IV) FINAL MBT - H = 120°-180°=-60°+360=310°.]
- [MBT = 150° FINAL MBT = 120°, INTERCEPT = 30°×2=60° NEW HEADING = 120 + 60° = 180°, DRIFT = 30°, (I) INTERCEPT = 60° (II) TURN 120°+60°=180°-150°=30° RIGHT TURN (III) 180° (IV) FINAL MBT - H = 120°-180°=-60°+360=300°.]
13.
A/C position 41°N81°W, QDM = 173° from a G/S (VOR) at 59° N 84°W, Var at G/S = 3°E, Var at a/c position = 6°E, mean var = 5°E. Find BTP on Lamberts.
- [QDM = 173°, QUJ = QDM = VAR = 173°+3°=176° QTE = 176° + 180° = 356°.]
- [QDM = 173°, QUJ = QDM = VAR = 173°+3°=176° QTE = 176° + 180° = 346°.]
14.
A/C approaching to station on R-140° it is told to hold on radial 325° for standard hold, what will be the entry?
- [SEC 1 = 255° TO 145° SEC 2 = 145° TO 435° - 360° = 073°]
- [SEC 1 = 255° TO 145° SEC 2 = 145° TO 435° - 360° = 075°]
15.
A/C hdg 170° M. ADF reading 345° Relative. Make 45° Intercept of 355° Trk outbound. Hdg to intercept will be ___________ and ADF at the time of intercept will be__________ .
- [MBT = 345°+170°=515° - 360° = 155° OUT BOUND RADIAL TO INTERCEPT = 335° FINAL MBT = 175° (I) INTERCEPT = 45° (II) NEW HDG = 175°-45°=130° (III) TURN LEFT BY = 170°....130°=040° (IV) ADF READS = 175°-130°=045°. (A) 301°/175° (B) 040°/135° (C) 130°/045° 130°/070°.]
- [MBT = 345°+170°=515° - 360° = 155° OUT BOUND RADIAL TO INTERCEPT = 335° FINAL MBT = 175° (I) INTERCEPT = 45° (II) NEW HDG = 175°-45°=130° (III) TURN LEFT BY = 170°....130°=040° (IV) ADF READS = 175°-130°=045°. (A) 301°/175° (B) 040°/135° (C) 130°/045° 130°/090°.]
16. A/C is homing to stn on Hdg 350° with 15° Port drift. Asked to hold on radial 325° for non std. Entry?
- [SEC 1 = 325° + 70° = 395° - 360° = 035° TO 145°, SEC 2 = 145° TO 325° - 110° = 215° SEC = 215° TO 035°, A/C RADIAL = 330° - 15° - 180° = 135°. SEC A PARALLEL ENTRY]
- [SEC 1 = 325° + 70° = 395° - 360° = 035° TO 145°, SEC 2 = 145° TO 325° - 110° = 215° SEC = 215° TO 035°, A/C RADIAL = 330° - 15° - 180° = 133°. SEC A PARALLEL ENTRY]
17.
A/C is approaching to stn on hdg 225° what 10'S drift. It is told to approach on radial 075°, and after reaching stn a/c is to hold on radial 235° (non standard hold). Find: (i) Hdg to intercept a) 215° b) 195° c) 275° d) 265°
- [MBT = 225° + 10 = 235° FINAL MBT = 255° INTERCEPT = 60°, NEW HDG = 225°-60° = 195° B) (II) HOW MANY DEGREE TO TURN. ANS → TURN LEFT BY 30°. (III) WHAT WILL ADF READ AT THE TIME OF INTERCEPT. ANS → 060°, (I) TYPE OF ENTRY A/C SHOULD FOLLOW. ANS → OFF SET]
- [MBT = 225° + 10 = 235° FINAL MBT = 255° INTERCEPT = 60°, NEW HDG = 225°-60° = 195° B) (II) HOW MANY DEGREE TO TURN. ANS → TURN LEFT BY 30°. (III) WHAT WILL ADF READ AT THE TIME OF INTERCEPT. ANS → 040°, (I) TYPE OF ENTRY A/C SHOULD FOLLOW. ANS → OFF SET]
18.
LSR = 301° in SH, QC = -1°. Hdg (T) = 120° CA = 10°. Find BPT on Mercator.
- [RB = 301°-1° = 300°, MBT = 300°+120°=420°-360°=060°, QTE = 240° BTP = 240°+10°=230°]
- [RB = 301°-1° = 300°, MBT = 300°+120°=420°-360°=060°, QTE = 240° BTP = 240°+10°=250°]
19.
A/C is homing on to stn with 10° P drift on Hdg 060°. It is asked to hold on radial 065°. Entry Std hold?
- [SEC 1 = 065° - 70° = -5° + 360° = 355° TO 245°, SEC 2 = 245° TO 065° + 110° = 175°, SEC 3 = 355° TO 175°, MBT = H-D = 060° - 10° = 050° RADIAL = 231°, SEC 2 OFFSET ENTRY]
- [SEC 1 = 065° - 70° = -5° + 360° = 355° TO 245°, SEC 2 = 245° TO 065° + 110° = 175°, SEC 3 = 355° TO 175°, MBT = H-D = 060° - 10° = 050° RADIAL = 230°, SEC 2 OFFSET ENTRY]
20.
A/C magnetic brg to the sstn (VOR) = 310°, A/C position 30°S 45°45'E, ground station 19°20'S, 30°30'E, Variation at stn =+7°, at a/c position = +14°, mean var +10°. Find BTP on Mercator.
- [QDM = 310° QUJ = 310 +7° = 317, QTE = 317° - 180° = 137°, CA = 1/2 × 15° 15' × SIN(24°40') (MEAN LAT) CA = 4° BTP = 133°.]
- [QDM = 310° QUJ = 310 +7° = 317, QTE = 317° - 180° = 137°, CA = 1/2 × 15° 15' × SIN(24°40') (MEAN LAT) CA = 4° BTP = 134°.]
21.
A/C true brg. from stn is 050° in NH. CA = 5° Find BTP on Mercator.
- [QTE = 050°, BTP = 050° + 5° = 055°]
- [QTE = 050°, BTP = 050° + 5° = 054°]
22. An aircraft heading 200° (M) has an ADF reading of 160° Relative, The heading to steer to intercept the 150° track outbound from the NDB at 30° is
130°(M)
140°(M)
120°(M)
23. An aircraft heading 040°(M) has an ADF reading of 060° Relative, is to intercept the 120° (M) track inbound to an NDB at 50°. The relative bearing of the NDB that confirms track interception is
060 Relative.
050 Relative
080 Relative.
24.
VOR RMI = 120° in SH, Var at a/c position = 5°E, variation at ground station = 10°E, CA = 7°, BTP on Mercator?
- [QDM = 120° QUJ = 120°+10° = 130°, QTE = 130° + 180° BTP = 310° + 7° = 317°.]
- [QDM = 120° QUJ = 120°+10° = 130°, QTE = 130° + 180° BTP = 310° + 7° = 327°.]
25.
VOR RMI = 120° in SH, Var at a/c position = 5°E, variation at ground station = 10°E, CA = 7°, BTP on Mercator?
- [QDM = 120° QUJ = 120°+10° = 130°, QTE = 130° + 180° BTP = 310° + 7° = 327°.]
- [QDM = 120° QUJ = 120°+10° = 130°, QTE = 130° + 180° BTP = 310° + 7° = 317°.]
26.
Present Hdg = 090° to the station in nil winds. Final inbound trk = 070°.
- [MBT = 090° FINAL MBT = 070° DRIFF =090° - 070° = 20°, INTERCEPT ANGLE = 2 × 20° = 40° ROUNDED TO 60°, HDG TO INTERCEPT NEW TRK = 070°+060°=130° TURN RIGHT THROUGH = 40°, ADF READING AT THE TIME OF INTERCEPT (RB) = FINAL MBT - H = 070° - 130°, = -60°+360° = 330°]
- [MBT = 090° FINAL MBT = 070° DRIFF =090° - 070° = 20°, INTERCEPT ANGLE = 2 × 20° = 40° ROUNDED TO 60°, HDG TO INTERCEPT NEW TRK = 070°+060°=130° TURN RIGHT THROUGH = 40°, ADF READING AT THE TIME OF INTERCEPT (RB) = FINAL MBT - H = 070° - 130°, = -60°+360° = 300°]
27.
A/C is approaching stn on radial 090° and told to approach on radial 082°.
- [MBT = 270° FINAL MBT = 262° INTERCEPT = 30°, NEW HDG = 262°+30°=292°, TURN BY 292°-270°=22°, ADF = 330° (I) 30° (II) RIGHT AND 22° (III) 292° (IV) 330°.]
- [MBT = 270° FINAL MBT = 262° INTERCEPT = 30°, NEW HDG = 262°+30°=292°, TURN BY 292°-270°=22°, ADF = 330° (I) 30° (II) RIGHT AND 22° (III) 292° (IV) 333°.]
28.
A/C is homing on to stn on inbound heading 300° with 10°S drift He is asked to hold on radial 305° (STD Hold). Entry?
- [SEC 1 = 305° - 70° = 235° TO 125°, SEC 2 = 125° TO 305° + 110° = 415° - 360° = 055°, AIRCRAFT RADIAL = 300° + 010° - 180° = 130°. ENTRY IS PARALLEL, SEC 1 ]
- [SEC 1 = 305° - 70° = 235° TO 125°, SEC 2 = 125° TO 305° + 110° = 415° - 360° = 055°, AIRCRAFT RADIAL = 300° + 010° - 180° = 130°. ENTRY IS PARALLEL, SEC 2 ]
29.
A/C is approaching to stn on Hdg 330° and a/c is told to intercept radial 170° inbound.
- [MBT = 330° FINAL MBT = 355°, DIFF = 25°, (I) INTERCEPT = 60°, (II) TURN LEFT BY 35° (335°-60°=295°) (III) NEW HDG = 295°, ADF READS = 335°-295° = 040°.]
- [MBT = 330° FINAL MBT = 355°, DIFF = 25°, (I) INTERCEPT = 60°, (II) TURN LEFT BY 35° (335°-60°=295°) (III) NEW HDG = 295°, ADF READS = 335°-295° = 060°.]
30.
Holding leg 210° left hand hold, winds 270°/30kts. TAS = 180 kts, Find (a) Hdg on outbound and inbound (b) Time on outbound
- [DRIFT = 8°S ON OB. IN BOUND HEADING = 210 DEG + 8 DEG = 218 DOG. OUT BOUND HEADING = 030 - 8 × 3 = 006 DEG, G/S = 193 KTS, TIME = 60 SEC - TAIL WIND = 60 - 13 = 47 SEC (A) 006° AND 218° (B) 43 SEC]
- [DRIFT = 8°S ON OB. IN BOUND HEADING = 210 DEG + 8 DEG = 218 DOG. OUT BOUND HEADING = 030 - 8 × 3 = 006 DEG, G/S = 193 KTS, TIME = 60 SEC - TAIL WIND = 60 - 13 = 47 SEC (A) 006° AND 218° (B) 47 SEC]
31.
RMI = 055° in NH, Var 5°E, CA = 10°, find BTP on Mercator.
- [QDM = 055° QUJ = 055° + 5°E = 060°, QTE = 060°+180° = 240°, BTP = 231°.]
- [QDM = 055° QUJ = 055° + 5°E = 060°, QTE = 060°+180° = 240°, BTP = 230°.]
Submit
Make New Set