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# ATPL theory questions

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# ATPL theory questions

2nd Apr 2019, 17:15

Join Date: Nov 2017
Location: Dubai
Posts: 32
4) The weight and balance sheet of an aircraft gives the following data: nose wheel weight 1000kg, left and right wheels 5000 kg each. The distance between the nose wheel and the main ones is 10 m. How many meters from the main landing wheels is located the centre of gravity?

a. 0.81m b.0.75m c. 0.91m d 9.1m

5) In windless conditions at 0830 UTC Pilot read radial = 315 degrees. Continuing flight with true track heading = 180 degrees at 0840 UTC he reads radial = 270 degrees. Assuming his ground speed is 240 kt, calculate distance to NDB at 0840 UTC:

a. 32NM b. 48NM c. 38NM d. 40NM

(Exams is in 2 days, so any help is appreciated!)

Last edited by asmith474; 2nd Apr 2019 at 18:19.
2nd Apr 2019, 18:50

Join Date: Nov 2000
Location: White Waltham, Prestwick & Calgary
Age: 67
Posts: 3,687
FL 370 for a PPL question?
An 11000 kg aircraft?
3rd Apr 2019, 13:45

Join Date: Nov 2017
Location: Dubai
Posts: 32
Originally Posted by 77Whiskey
1st : Station ALT 1000ft AMSL, So we'll subtract station height from 37,000(FL370) so 36,0000 (which converters to nearly 6NM) and the distance is 15 NM. We need to find the slant range. So formula of slant range is (S.R)^2 = (Height 1)^2+(Height 2)^2
(S.R)^2 = (15)^2 + (6)^2
(S.R)^2 = 225 + 36
(S.R)^2 = 261
S.R = √261
S.R = 16.1 NM, So the answer is A

2nd : You have the compass and magnetic heading 190° and 188° respectively. Remember Variation east magnetic least (-) so in order to get true track you need to go east(-) because variation is 3°E from the magnetic track. Therefore 188°-3° = 185°(T). The deviation was east(-) of magnetic by 2° So, -2°

3rd : I think it's a question of 1 in 60 rule but not sure
The 1st answer was correct but the 2nd was wrong according to the question bank results
3rd Apr 2019, 14:12

Join Date: Nov 2005
Location: Zulu Time Zone
Posts: 591
15 NM DME from the station. DME readout will be:
a.16nm b.15nm c18nm d.37nm
If the aircraft is at 15 DME (which the question states it is) the only possible correct answer is b. 15 DME! Perhaps find a better question bank.
3rd Apr 2019, 14:20

Join Date: Nov 2018
Location: New Delhi
Posts: 7
DME Distance is slant range, not the ground distance. Please don't confuse that.
3rd Apr 2019, 14:47

Join Date: Nov 2018
Location: New Delhi
Posts: 7
Originally Posted by asmith474
The 1st answer was correct but the 2nd was wrong according to the question bank results
My Bad, You have 188(M) and the variation is 3E. Draw Straight line and write 188 magnetic on top of that now variation is east means right hand side of magnetic (Assume straight line the center and left and right are west and east respectively). So add 3 to 188. The true track will be 191 and you already have the difference between 190 compass and 188 magnetic which is -2 degrees.

I suggest you to always draw on rough sheet
3rd Apr 2019, 15:42

Join Date: Nov 2005
Location: Zulu Time Zone
Posts: 591
Originally Posted by 77Whiskey
DME Distance is slant range, not the ground distance. Please don't confuse that.
I know it is slant range. But you didn't read the question which was in effect; 'if the aircraft is at 15 DME from the beacon, how many DME is the aircraft from the beacon?'

If you are having trouble understanding this try substituting the "DME" with "slant range" in the question and see if you get it:

The aircraft flies at FL370, 15 NM [slant range] from the station. [slant range] readout will be?
The answer is still 15 NM, which is the only possible correct answer.

Last edited by oggers; 3rd Apr 2019 at 15:54.
7th Apr 2019, 17:49

Join Date: Mar 2018
Location: Poland
Posts: 28
I am gonna go for the Operational Procedures exam tomorrow and while training in my 3x question banks I met a question of the kind "understand what the examiner thought about when creating it".

So it is: Which of the following needs approval?

Aviation Exam says: anything on the flight deck (but what about the pilots sandwich its also on the flight deck for example or the time piece etc. itd. ) ?

Atpl questions says: child harness used in cabin

Hopefully I wont get such a question but who is correct here? What did the examiner think when making this question anyone have any idea? Because even the question banks are not sure about it.
8th Apr 2019, 06:19

Join Date: Nov 2000
Location: White Waltham, Prestwick & Calgary
Age: 67
Posts: 3,687
CAT.IDE.A.100 Instruments and equipment — general
(a) Instruments and equipment required by this Subpart shall be approved in accordance with the
applicable airworthiness requirements except for the following items:
(1) Spare fuses;
(2) Independent portable lights;
(3) An accurate time piece;
(4) Chart holder;
(5) First-aid kits;
(6) Emergency medical kit;
(7) Megaphones;
(8) Survival and signalling equipment;
(9) Sea anchors and equipment for mooring; and
(10) Child restraint devices.
8th Apr 2019, 13:52

Join Date: Apr 2002
Location: here today, there tomorrow
Age: 36
Posts: 264
Hi All,

I have seen that the same question has been posted earlier and it's still in the IR-OPS exam in South Africa. Wondering if someone can help with the below:

Total distance A to B 2000 nm TAS, Wind and endurance constant 5000kg fuel +500kg + reserve fuel which is assumed not to be used for the flight. PET from A 1200 nm What is the distance to the PNR/PSR?

How did they get to this?
Thanks and happy flying.

R
9th Apr 2019, 00:35

Join Date: Mar 2015
Location: The Loony Bin
Posts: 93
If you don't get the TAS/Wind etc... you can use the ETP equation to work out the necessary groundspeeds (Home and Out) etc... then use those figures to work out the PNR time etc...

ETP: https://www.skybrary.aero/index.php/FileTP.png

distance to ETP = (total dist * GS Home) / (GS Home + GS Out)
=> 1200 = 2000 * (H/(H+O))
=> 1200 / 2000 = H / (H+O)
=> 0.6 = H / (H+O)
=> H = 0.6 (H+O)

From here we can deduce that the value of H is 0.6 or 60% of the value of H+O... simply substitute a value for H to work out O...
ie. assume H = 1000
=> (1000+O) = 1000 / 0.60
=> 1000+O = ~1667
=> O = 1667 - 1000
=> O = 667

So, our Groundspeed home to A would be 1000 (tailwind)... and GS to B = 667 (hence why ETP is closer to B)

As per: https://www.skybrary.aero/index.php/...o_Return_(PNR)

To calculate the time to PNR, you need to know endurance (in hours), GS Home and GS Out.. Looks like we have elected to take 5000kg fuel... + contingency of 500kg... which is 5000/500 = 10% contingency...

Without an actual fuel burn or endurance figure, I would assume that the 5000kg fuel would have been taken based on our estimated travel time at calculated GS "out", so our basic endurance therefore equates to = 2000 dist / 667 GS out = 3 hrs... then we need to add the 10% due to contingency to get final "safe" endurance => 3.3 hours...

Time to PNR = (Endurance in Hours * GS Home) / (GS Home + GS Out)
=> 3.3 * 1000 / 1667
=> ~1.98

So... 1.98 hrs to PNR * 667 GS out => ~1320 nm

NOTE: I'm not sure that is the "correct" or most efficient way to go about getting from a stated distance, fuel load and PET to the PNR but it seems to work. Was there a fuel burn figure or endurance figure actually given?
10th Apr 2019, 15:07

Join Date: Apr 2002
Location: here today, there tomorrow
Age: 36
Posts: 264
Hi there RHSandLovingIt,

Much appreciate your way of solving it.

In a nutshell, its:

1- Find Distance to ETP
2 - Fond GSO and GSH
3- Time to PNR = ( Endurance in hrs * GS Home ) / GS Home + GS Out
4. Time to PNR * GS Out = Answer.

You are correct - I wonder what is the reason behind such questions but as one of the previous posters have said, some of the things that come up really make you scratch your head for the IR Ops exam.

Thanks once again.

R

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