Indigo Call letters for Freshers

[AditR]

I had emailed S.Pinto about it and this is the reply i got..

Please be informed that the IR will not be required to be current during the time of the exam. Only the ME-IR needs to be endorsed on the CPL.

Thanks & Regards
Sandra Pinto
Recruiter-Pilot Provisioning

[flyingfreak]

same here, neither she replies to my emails, nor receives the calls. Please let me know what step you take.

[maverick11]

question about confirmation mail...........

hey guys i got the first mail long time back and wanted to know is there any chance to get the second mail now.....or i have to wait for the next attempt.

and is there anyone who got the call for 17th april...........

[gAMbl3]

The great circle track X - Y measured at x is 319° , and Y 325° Consider the following statements



319 = westerly track

The direction increases from X (319) to Y (325).

In Southern Hemisphere direction of a westerly great circle increases.

The difference = convergency = 325- 319 = 6
Conversion angle = 6/2 = 3

Rhumb line track = 319 + 3 = 322


Answer : D) Southern hemisphere, Rhumb line track is 322° .




*In Northern Hemisphere direction of a westerly great circle decreases.

In Northern Hemisphere direction of a easterly great circle increases.

In Southern Hemisphere direction of a easterly great circle decreases.

In Southern Hemisphere direction of a westerly great circle increases.

[AditR]

@gAMbl3

Thanks

[ITAviator]

@KKAARR

P.3.A hyperbola cuts the base line 60 Km from the Master end and 150 Km from the Slave end. When on the same hyperbola at a range of 90 Km from the Master, the range from the Slave will be:
A)300 km B)150 km C)240 km D)180 km

Soln:
Let one arm of hyperbola be Master(M) and other arm as Slave(S).
Center(O)
Given: MO=60 and SO= 150.
Hence the distance between the two arms of hyperbola i.e. MS=150-60=90 (which is a constant)
Same parabola, new location Mi and Si. Although with same center and MS=90
MiO= 90
Thus, MiO + 90 = SiO = Range of Slave = 180

[airplane.pilot]

Aircraft A departs VOR X (55n 176w) to VOR Y (55N 174 E) maintaining radials throughout .If local variation at X is 4 E and Y is 6 E and drift experienced is 6 Deg S , the Hdg maintained on dep from X : (Sin 55 : 0.8 ; Cos 55 : 0.6)
a. 280 M
b. 270 M
c. 274 M
d. 264 M

Also the heading on arrival at Y :
a. 254 M
b. 260 M
c. 274 M
d. 270 M

Dont have the answers for it.

[3greens 1inthemirror]

Aircraft A departs VOR X (55n 176w) to VOR Y (55N 174 E) maintaining radials throughout .If local variation at X is 4 E and Y is 6 E and drift experienced is 6 Deg S , the Hdg maintained on dep from X : (Sin 55 : 0.8 ; Cos 55 : 0.6)
a. 280 M
b. 270 M
c. 274 M
d. 264 M

Also the heading on arrival at Y :
a. 254 M
b. 260 M
c. 274 M
d. 270 M

Dont have the answers for it.

Hdg on departure from X - 264 M

Hdg on arrival at Y - 254 M

Is anybody else getting these answers?

[ITAviator]

Yup, same answers..

[KKAARR]

i appreciate your eply ITaviator..

Here's another problem i am unable to match answer with... if anyone can..

Q. A is N55*000*E/W and B is N54*E010*. The Average true GC is 100*. The true course of Rhumb line at point A is ?
a) 096
b) 104
c) 107
d) 100

[KKAARR]

AND one mORE...

Q. Two positions plotted on a Polar Stereographic A 80N000E/W B 70N102W are joined by a Straight line whose highest lattitude is reached at 035W. At point B, the true course is ?
a) 230
b) 203
c) 023
d) 030

[KKAARR]

An aircraft has a climb gradient of 7 percent after take-off. There is an obstacle having a height of 320ft at a distance of 5400ft from the screen. The aircraft will clear the obstacle by:
A)108 ft B)58 ft C)93 ft D)378 ft

[gAMbl3]

Q. A is N55*000*E/W and B is N54*E010*. The Average true GC is 100*. The true course of Rhumb line at point A is ?


Convergency = d long X sin mean lat = 10 x Sin {(55+54)/2} = 10 x Sin 54.5 = 8

C angle = 8/2 = 4

Average true GC = 100

At point A : RL = GC + CA = 100 + 4 = 104

[ITAviator]

Q. A is N55*000*E/W and B is N54*E010*. The Average true GC is 100*. The true course of Rhumb line at point A is ?
a) 096
b) 104
c) 107
d) 100

Soln:
I feel,
Average True GC=100. This also means this is the GC track at the centre of the course which is same as Rhumb line track.
Hence at a Rhumb line track will be 100*

@KKAARR: whats the answer marked?

Q. Two positions plotted on a Polar Stereographic A 80N000E/W B 70N102W are joined by a Straight line whose highest lattitude is reached at 035W. At point B, the true course is ?
a) 230
b) 203
c) 023
d) 030

Soln:
Diagram has to be drawn, which i cannot here..
A = 80N000E/W lies on smaller circle of 80N intersecting at 000 E/W
B = 70N102W lies on bigger circle of 70N intersecting at 102W
Centre of circle is North Pole (N)

From given data: we get a triangle ANB, Angle ANB=102 (Long diff between A and B),
Required is angle NBA, which is True Course from B to A.
Highest lat=35W, a line drawn from N perpendicular
to AB at point P.
Hence Angle ANP = 35 deg.

In triangle ANP, Ang ANP=35, ang NAP=90, hence ang PAN = 180-(90+35) = 55
Now with this data, in triangle ANB, ang ANB=102, ang PAN/BAN = 55, hence ang NBA = 180-(102+55) = 023

[amritsar]

QDM QUESTIONS

Hi I am new here,
T was wondering wether you guys can help me with the following questions wid explanation please:
1.8863
act QDM 330° act HDG 060° req QDM 350° Intercept HDG?
a 260°
b 315°
c 305° ans
d 350°
2 .act QDM 140° act HDG 020° req QDM 110° First turn?
a Left HDG 155° ans
b Left HDG 190°
c Right HDG 065°
d Right HDG 110°

3.Which interception do you choose with a difference of angle of 15° and an EET of 4
minutes?
a Homing
b 90°/45° Interception
c Correction
d 45° Interception ans

4act QDM 210° act HDG 060° req QDM 260° First turn?
a left HDG 170°
b right HDG 215°
c right HDG 170° ans
d right HDG 260°

5act QDR 285° act HDG 290° req QDR 290° Time passed the station 2Min.
Correction HDG?
a 305°
b 290°
c 285°
d 300°


Thanks

[gAMbl3]

My bad! Average GC track is parallel to the RL track so answer is 100

[KKAARR]

@ ITaviator

your answer is correct bro.. its 100...

Can you give a look on that Climb Gradient Question i posted there..
just not able to get the answer !

[KKAARR]

& here, the another one

Q. Still air gradient of climb: 5%
TAS: 250 kts
G/S: 200 kts
The rate of climb is: A)2000 fpm B)1250 fpm C)500 fpm D)1000 fpm

[gAMbl3]

An aircraft has a climb gradient of 7 percent after take-off. There is an obstacle having a height of 320ft at a distance of 5400ft from the screen. The aircraft will clear the obstacle by:


For obstacle clearance, screen height = 50 ft

Climb gradient = 7% after take off i.e. after reaching screen height the a/c will climb 7 ft for every 100 ft horizontal.

At 5400 ft from screen height, height of a/c =50 + [(7 x 5400)/100] = 50 + 378 = 428 ft

obstacle height = 320 ft

clearance = 428 - 320 = 108 ft

[KKAARR]

gAMbl3

Thank you very much bro ! You Rock

But the problem above... the one with Answer 500ft... Isn't the answer 1250 ft/min.. ??