Indigo Call letters for Freshers
Join Date: Jul 2011
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I had a word with Parul in the evening around 7PM IST and she said that we can expect our hall tickets tonight and tomorrow as well. BTW even I didnt get the hall ticket as well. All the guys who didnt get the hall ticket today will definitely get it tomorrow. Dont worry and study hard. Best of luck to all of you........ 
Join Date: Apr 2008
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even i haven't got the hall ticket yet .....please guys let us know if any one gets the hall ticket for 29th ....i stay in mumbai have a train ticket booked for today now confused whether leave or not ....
:confu sed::confu sed::confused
:confu sed::confu sed::confused
Join Date: Jan 2011
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Re:
Hi,
I too cleared my written n flunked in the intrvw back in 2008. So does that mean i can Never apply for Indigo ever again? I have heard that for such ppl who cudn clr intervws they dnt need to give writns again bt will b directly called for intrvw.. I dnt knw how much of it is true.. Got any clue abt this?
Awaiting your reply!!
I too cleared my written n flunked in the intrvw back in 2008. So does that mean i can Never apply for Indigo ever again? I have heard that for such ppl who cudn clr intervws they dnt need to give writns again bt will b directly called for intrvw.. I dnt knw how much of it is true.. Got any clue abt this?
Awaiting your reply!!
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Join Date: Jul 2011
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Questions
Hey can anyone help me with the answers to the following...
1. (this is from indigo question bank 1)
Any acceleration in climb, with a constant power setting:
A) decreases the rate of climb and the angle of climb
B) improves the rate of climb if the airspeed is below VY
C) improves the climb gradient if the airspeed is below VX
D) decreases rate of climb and increase angle of climb
2. An aircraft heading 325° (M) has an ADF reading of 330° Relative. The heading to steer to intercept the 280° track inbound to the NDB at 50° is:
A) 340(M)
B) 330(M)
C) 310(M)
D) 320(M)
3. At 0900 hrs relative brg of NDB is 045 and at 0910 hrs relative brg is 090. if Ground speed is 150 knots, what is the distance of aircraft from beacon????
4. An aircraft pinpoints itself at 300 nm from destination at 0800 hrs. Gnd Speed is 180 kts. it is required to delay its ETA by 15 mins. After how much time and what distance should the speed be reduced to 150 kts.
Help would be appreciated, also if someone can explain how to solve.
thanks
1. (this is from indigo question bank 1)
Any acceleration in climb, with a constant power setting:
A) decreases the rate of climb and the angle of climb
B) improves the rate of climb if the airspeed is below VY
C) improves the climb gradient if the airspeed is below VX
D) decreases rate of climb and increase angle of climb
2. An aircraft heading 325° (M) has an ADF reading of 330° Relative. The heading to steer to intercept the 280° track inbound to the NDB at 50° is:
A) 340(M)
B) 330(M)
C) 310(M)
D) 320(M)
3. At 0900 hrs relative brg of NDB is 045 and at 0910 hrs relative brg is 090. if Ground speed is 150 knots, what is the distance of aircraft from beacon????
4. An aircraft pinpoints itself at 300 nm from destination at 0800 hrs. Gnd Speed is 180 kts. it is required to delay its ETA by 15 mins. After how much time and what distance should the speed be reduced to 150 kts.
Help would be appreciated, also if someone can explain how to solve.
thanks
Join Date: Jul 2011
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dude lets not get down,,lets see till evening. The guys who got it for 28th,, received hall ticket yesterday evening itself. So assuming we are on 29th..we ill get.. should get it by today evening. Even though all this situation is highly fishy and weird and unprofessional. 
anyways good luck
anyways good luck
Join Date: Apr 2009
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@flyhigh27
1) To accelerate in a climb, without increasing the power setting, the only way will be to lower the nose. This will lead to a decrease in ROC and also angle of climb. Hence answer A.
2) A/c is HDG 325. NDB is on relative bearing 330. Hence the QDM to NDB is 295 i.e. radial 115 from NDB. To track inbound on 280 means intercept radial 100 inbound to NDB. For this we have to turn right. To intercept at angle 50 would mean heading 280+50 = 330.
3) This is a sum of 'doubling the angle at the bow'. Basically the angle to the NDB is increased fromo 45 to 90 degrees. This forms a 45-45-90 isoceles triangle, so the distance travelled to double the angle (150 x (10/60) = 25 Nm) is the same as the distance from beacon when it is abeam. So the answer is 25Nm. Difficult to explain here without a diagram, but I hope you get the idea.
4) I find the easiest way to do this is with algebraic equation because it is fool proof and I am comfortable with them, but there are simpler ways. In this case you would assume x to the distance you would reduce speed and form algebraic equations to solve for x. Is the answer 75nm/25 mins from start?
Hope this has helped. If I am wrong, someone please correct me, for my own benefit atleast.
1) To accelerate in a climb, without increasing the power setting, the only way will be to lower the nose. This will lead to a decrease in ROC and also angle of climb. Hence answer A.
2) A/c is HDG 325. NDB is on relative bearing 330. Hence the QDM to NDB is 295 i.e. radial 115 from NDB. To track inbound on 280 means intercept radial 100 inbound to NDB. For this we have to turn right. To intercept at angle 50 would mean heading 280+50 = 330.
3) This is a sum of 'doubling the angle at the bow'. Basically the angle to the NDB is increased fromo 45 to 90 degrees. This forms a 45-45-90 isoceles triangle, so the distance travelled to double the angle (150 x (10/60) = 25 Nm) is the same as the distance from beacon when it is abeam. So the answer is 25Nm. Difficult to explain here without a diagram, but I hope you get the idea.
4) I find the easiest way to do this is with algebraic equation because it is fool proof and I am comfortable with them, but there are simpler ways. In this case you would assume x to the distance you would reduce speed and form algebraic equations to solve for x. Is the answer 75nm/25 mins from start?
Hope this has helped. If I am wrong, someone please correct me, for my own benefit atleast.
Join Date: Jul 2011
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I have just got a mail from Capt Upendranath for the 29th July exam.....Started to get really nervous now....Best of luck to everyone appearing for both the July exams.
Join Date: Sep 2009
Location: VT - WTF
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Hey guys can you please list all the email addresses of the CaE guys.
Requests emails of the following.
Sandra Pinto
Capt. Upendranath
Anyone else from who any indigo Exam related email has been forwarded to you.
Thanks guys.
Requests emails of the following.
Sandra Pinto
Capt. Upendranath
Anyone else from who any indigo Exam related email has been forwarded to you.
Thanks guys.
Join Date: Jul 2011
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@ vivekh
thanks...
Hello people can u please help me with these:
1. Given True track= 352 degrees, variation=11 W, deviation=-5 and drift- 10 S.
what would be the compass heading????
2. when acknowledging mode and code,pilot needs to:
a.acknowledge mode only. b. mode and code both
3.the procedure for an aircraft that suffers comm failure during an IFR flt in VFR condition is:
a. proceed direct to its destination
b. maintain last cleared level and speed for 20 mins and then continue in accordance wid flt plan
c. continue to fly in vmc, land at nearest aerodrome, report its arrival.
4. At medium latitude, coriolis force experienced is:
a.7.5 degrees to the rgt
b. 15 degrees to left
c. 7.5 degrees to lft
d. 10.5 degrees to rgt
thanks.
Hello people can u please help me with these:
1. Given True track= 352 degrees, variation=11 W, deviation=-5 and drift- 10 S.
what would be the compass heading????
2. when acknowledging mode and code,pilot needs to:
a.acknowledge mode only. b. mode and code both
3.the procedure for an aircraft that suffers comm failure during an IFR flt in VFR condition is:
a. proceed direct to its destination
b. maintain last cleared level and speed for 20 mins and then continue in accordance wid flt plan
c. continue to fly in vmc, land at nearest aerodrome, report its arrival.
4. At medium latitude, coriolis force experienced is:
a.7.5 degrees to the rgt
b. 15 degrees to left
c. 7.5 degrees to lft
d. 10.5 degrees to rgt
thanks.
Join Date: Apr 2009
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@flyhigh27
1) Simple CDMVT sum. Compass hdg in nil wind comes to 008deg. So for compensating 10deg Starboard drift, we need a compass heading of 358deg.
2) Mode & Code both
3) Radio failure of IFR flight in VFR conditions - land at nearest aerodrome and report nearest ATC unit.
4) If in the question you are talking about the Northern hemisphere, then the answer is 10.5deg right. Med lats ~ 45deg. So Coriolis force = 15sin45 ~ 10.5deg. Since it is northern hemisphere, it will be to the right.
1) Simple CDMVT sum. Compass hdg in nil wind comes to 008deg. So for compensating 10deg Starboard drift, we need a compass heading of 358deg.
2) Mode & Code both
3) Radio failure of IFR flight in VFR conditions - land at nearest aerodrome and report nearest ATC unit.
4) If in the question you are talking about the Northern hemisphere, then the answer is 10.5deg right. Med lats ~ 45deg. So Coriolis force = 15sin45 ~ 10.5deg. Since it is northern hemisphere, it will be to the right.
Join Date: Oct 2009
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Q.on a particular take off you can accept upto 10 kts tailwind.the runway qdm is 047 the variation is 17E and the ATIS gives the wind direction as 210.what is the max wind strength you can accept?
a 18kts
b 11kts
c 8kts
d 4kts
the answer given is b 11kts...but i cant figur how to work this question out
a 18kts
b 11kts
c 8kts
d 4kts
the answer given is b 11kts...but i cant figur how to work this question out
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@ vivekh
for question 1 above (CDMVT)
Given True track= 352 degrees, variation=11 W, deviation=-5 and drift- 10 S.
what would be the compass heading????
TRUE TRACK is given
drift is diff b/w hdg and track.
therefore, hdg(T)= 342
then applying variation and deviation,
compass hdg= 348
I am not sure, I did it this way, jst wanna confirm is it right or not.
Given True track= 352 degrees, variation=11 W, deviation=-5 and drift- 10 S.
what would be the compass heading????
TRUE TRACK is given
drift is diff b/w hdg and track.
therefore, hdg(T)= 342
then applying variation and deviation,
compass hdg= 348
I am not sure, I did it this way, jst wanna confirm is it right or not.