ATPL theory questions
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Have anyone done recently the exam in Radio Navigation? Do you know approximately how many of the 66 questions were about the PBN? Because Aviation Exam and other training databases got very few PBN questions and I heard they can be quite many on the exam so when training the questions in the data banks I do not know if I really know the whole subject or if I just remember the questions.
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KT - just in case they ask you which mosquito 
Dengue is another fever spread by infected mosquitoes that are active by day – usually the Aedes Aegypti.
With regard to PBN, probably about 5 in the RNAV exam. In the separate exam that you will have to take if you don't do it now, it's 25,
Dengue is another fever spread by infected mosquitoes that are active by day – usually the Aedes Aegypti.
With regard to PBN, probably about 5 in the RNAV exam. In the separate exam that you will have to take if you don't do it now, it's 25,
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@paco: Thanks, I did already the human performance exam, they did not ask at all about the mosquitoes on the real exam and most of the questions made sense, some were tricky but the result was 90 % so I can live with it (since I used most time learning General Navigation for that session).
So its about 5 questions of the 66 about PBN on the Radio Navigation exam? I am going to do the exam on 2nd April when next session start (If I get done with everything including June session, I got promised I will be done with all my licenses this year so I am learning everyday 14/7 rest is eating, sleeping or writing posts) and heard from a colleague who were not too lucky on the radio navigation exam that he got like 14-15 PBN questions from the 66 and that he got still RNAV (area navigation questions they were not replaced by PBN like Bristol ground school online suggested).
But maybe its the Polish exam where they added PBN questions but did not remove the previous area navigation that UK CAA removed according to Bristol. So I learned both and practice with both kind of questions. But using more time on one subject means less time used on the instruments, operational procedures and air law that I also plan to pass in April. So thats why I asked if anyone know if PBN replace the RNAV part of Radio navigation exam or if its just an addition using some of the RNAV question slots on the exam but not replacing it totally.
PS. From what I heard since September 2018 its not possible to do radio navigation exam without PBN so everyone doing the exam now will get it done with PBN.
So its about 5 questions of the 66 about PBN on the Radio Navigation exam? I am going to do the exam on 2nd April when next session start (If I get done with everything including June session, I got promised I will be done with all my licenses this year so I am learning everyday 14/7 rest is eating, sleeping or writing posts) and heard from a colleague who were not too lucky on the radio navigation exam that he got like 14-15 PBN questions from the 66 and that he got still RNAV (area navigation questions they were not replaced by PBN like Bristol ground school online suggested).
But maybe its the Polish exam where they added PBN questions but did not remove the previous area navigation that UK CAA removed according to Bristol. So I learned both and practice with both kind of questions. But using more time on one subject means less time used on the instruments, operational procedures and air law that I also plan to pass in April. So thats why I asked if anyone know if PBN replace the RNAV part of Radio navigation exam or if its just an addition using some of the RNAV question slots on the exam but not replacing it totally.
PS. From what I heard since September 2018 its not possible to do radio navigation exam without PBN so everyone doing the exam now will get it done with PBN.
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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.
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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
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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 Apr 2019 at 18:19.
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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
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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.
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I suggest you to always draw on rough sheet
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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 Apr 2019 at 15:54.
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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.
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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.
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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
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?
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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.
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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.