ATPL theory questions
Joined: Nov 2017
Posts: 58
Likes: 0
From: Portugal
Some PPL navigation questions that I dont understand
1)DME station is situated at an altitude of 1000 ft AMSL. QNH = 1013.25 hPa. The aircraft flies at FL370, 15 NM DME from the station. DME readout will be:
a.16nm b.15nm c18nm d.37nm
2)Variation = 3° E, Magnetic track = 188 °, Compass heading = 190 °. True track and Compass deviation values are respectively:
a. 194°, +4. b. 189°, -3. c. 185°, -2. d. 191°, -2.
3)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. 32 NM. b. 48 NM. c. 38 NM. d. 40 NM.
1)DME station is situated at an altitude of 1000 ft AMSL. QNH = 1013.25 hPa. The aircraft flies at FL370, 15 NM DME from the station. DME readout will be:
a.16nm b.15nm c18nm d.37nm
2)Variation = 3° E, Magnetic track = 188 °, Compass heading = 190 °. True track and Compass deviation values are respectively:
a. 194°, +4. b. 189°, -3. c. 185°, -2. d. 191°, -2.
3)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. 32 NM. b. 48 NM. c. 38 NM. d. 40 NM.
Joined: Nov 2018
Posts: 7
Likes: 0
From: New Delhi
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
(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
Joined: Nov 2017
Posts: 58
Likes: 0
From: Portugal
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!)
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 April 2019 at 18:19.
Joined: Nov 2017
Posts: 58
Likes: 0
From: Portugal
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
(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
Joined: Nov 2005
Posts: 729
Likes: 1
From: Zulu Time Zone
15 NM DME from the station. DME readout will be:
a.16nm b.15nm c18nm d.37nm
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.
Joined: Nov 2018
Posts: 7
Likes: 0
From: New Delhi
I suggest you to always draw on rough sheet
Joined: Nov 2005
Posts: 729
Likes: 1
From: Zulu Time Zone
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?
Last edited by oggers; 3rd April 2019 at 15:54.
Joined: Mar 2018
Posts: 38
Likes: 0
From: Poland
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.
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.
Joined: Nov 2000
Posts: 4,330
Likes: 74
From: White Waltham, Prestwick & Calgary
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.
(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.
Joined: Apr 2002
Posts: 264
Likes: 0
From: here today, there tomorrow
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?
Answer is 1320nm.
How did they get to this?
Thanks and happy flying.
R
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?
Answer is 1320nm.
How did they get to this?
Thanks and happy flying.
R
Joined: Mar 2015
Posts: 114
Likes: 0
From: The Loony Bin
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/File
TP.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?
ETP: https://www.skybrary.aero/index.php/File
TP.pngdistance 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?
Joined: Apr 2002
Posts: 264
Likes: 0
From: here today, there tomorrow
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.
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.
Joined: Apr 2019
Posts: 1
Likes: 0
From: adelaide
Help with question.
G'day
Can anyone help in regards to a question I don't understand.
The question is, Is it possible for TAS to be greater than ISA at mean sea level.
The book has yes as the correct answer but does not explain why.
Can anyone explain why this is the case.
Thanks.
Can anyone help in regards to a question I don't understand.
The question is, Is it possible for TAS to be greater than ISA at mean sea level.
The book has yes as the correct answer but does not explain why.
Can anyone explain why this is the case.
Thanks.
Joined: Dec 2006
Aviation Qualifications: CPL
Posts: 433
Likes: 0
From: Hamburg
Neglecting position and instrument error (IAS/CAS) and compressibility (CAS/EAS), the ratio of IAS and TAS is determined by density. If actual air density is below ISA MSL density, TAS will be greater than IAS.
Joined: Nov 2017
Posts: 26
Likes: 1
From: Milan
Meteo warm occlusion
Hello everybody, I’d like to ask your help regarding this meterology question. Honestly I can’t figure out the answer.
during the summer a warm occlusion is approaching the east coast of the united kingdom. what weather pattern would you expect in the north sea?
A) medium level 3 to 8 oktas clouds, isolated showers
B) high level CI clouds
C)TSs, CBs, showers
D)Low level ST clouds
correct D.
Honestly I choose A, as my idea was embedded CBs can give showers while the characteristics of a warm front are cloudy sky.
I found good also B, as a typical sign of warm front approaching could be CIs.
Any help to understand this question?
thanks
during the summer a warm occlusion is approaching the east coast of the united kingdom. what weather pattern would you expect in the north sea?
A) medium level 3 to 8 oktas clouds, isolated showers
B) high level CI clouds
C)TSs, CBs, showers
D)Low level ST clouds
correct D.
Honestly I choose A, as my idea was embedded CBs can give showers while the characteristics of a warm front are cloudy sky.
I found good also B, as a typical sign of warm front approaching could be CIs.
Any help to understand this question?
thanks
Joined: May 2019
Posts: 2
Likes: 0
From: ireland
help with mass and balance question
I'm studying for my upcoming flight performance and planning exam and I'm doing practice questions on PPLtutor and a question has come up that I just can't get my head around it.
Q: Before refuelling, an aeroplane has a weight of 1850 lb and the total moments are 153000 in.lb aft of the datum. Fuel weight equalling 520 lb with an effective arm of 9 ft aft of datum is then loaded. what is the new GC of the aeroplane.
I'd calculate as follows by making a table and the formula mass x arm = moment:
Aeroplane weight: 1850 82.70 1530000
Fuel Weight: 520 9 4680.
Total 2370 157680
New GC: 157680/2370 = 66.53.
However, this is not the answer.I looked at the solution this is what the answer was:
Aeroplane weight: 1850 82.70 1530000
Fuel Weight: 520 108 56160.
Total 2370 209160
New GC: 209160/2370 = 88.25.
Can anyone explain where the value of 108 Came from for the fuel weight ARM?
Q: Before refuelling, an aeroplane has a weight of 1850 lb and the total moments are 153000 in.lb aft of the datum. Fuel weight equalling 520 lb with an effective arm of 9 ft aft of datum is then loaded. what is the new GC of the aeroplane.
I'd calculate as follows by making a table and the formula mass x arm = moment:
Aeroplane weight: 1850 82.70 1530000
Fuel Weight: 520 9 4680.
Total 2370 157680
New GC: 157680/2370 = 66.53.
However, this is not the answer.I looked at the solution this is what the answer was:
Aeroplane weight: 1850 82.70 1530000
Fuel Weight: 520 108 56160.
Total 2370 209160
New GC: 209160/2370 = 88.25.
Can anyone explain where the value of 108 Came from for the fuel weight ARM?
