Free ATPL Questions and answers

Question ID: 627163

An aircraft diverts from the destination to the alternate and on arrival is instructed to hold by ATC.The speed is reduced to V minimum drag (Vmd) and the fuel flow is noted as 1170 kg/h.

The aircraft must be on the ground at the alternate with 30 minutes of Final Reserve Fuel remaining (based on fuel flow at Vmd).

If the total remaining usable fuel is 1225 kg, what is the remaining safe endurance/flight time?

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A

63 Minutes

B

41 Minutes

C

2 Minutes

D

32 minutes

Explanation

Reserve Fuel = 1170 x 30/60 = 585kg
1225 - 585 = 640kg

Endurance = 640/1170 = 0.547hrs = 32mins

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Question ID: 627145

Refer to DUBROVNIK - charts 19-1 and 19-2 from Jeppesen GSPRM

What is the elevation of the highest obstacle on the chart?

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A

5 017 ft AMSL

B

3000ft AGL

C

4000 ft AGL

D

1 907 ft AMSL

Explanation

The highest obstacle on the chart is highlighted by surrounding its elevation by a black box

The highest obstacle on the Dubrovnik 19-1 chart given in the question is a terrain highpoint with an elevation of 5 017 ft AMSL in the upper right of the chart.

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Question ID: 627130

Given:

CAS: 190 kt

Altitude: 9000 ft

Temperature: ISA -10° C

True Course TC: 350°

W/V: 320/40

Distance from departure to destination: 350 NM

Endurance: 3 hours
Actual time of departure: 1105 UTC

The distance from departure to Point of Equal Time PET is:

Answer in NM

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Answer:

Correct answer : 203

Explanation

PET = (D x H)/(O + H)

First we need to find TAS using CRP-5.
ISA Temp at 9,000ft is -13, since it is ISA - 10, Temp = -13.
Now use airspeed window with -3 and 9000ft.
Find CAS 190 on inner scale and read off TAS 214 on outer scale.

Now use either trig or CRP-5 to find Outbound and Homebound GS.
Outbound GS = 179kts
Homebound GS = 249kts

PET = (350 x 249)/(179+ 249) = 203NM

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Question ID: 627127

Given the following information, calculate the total fuel for engine start and taxi.

Start fuel: 20 litres
Taxi fuel flow: 150 litres per hour
Taxi time: 10 minutes
Fuel density: 0.82 kilograms per litre

Answer in KGs

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Answer:

Correct answer : 36.9

Explanation

Start fuel + taxi fuel = 20 l + (150:6) l = 45 l
45 l @ 0,82kg/l => 36,9 kg

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Question ID: 627102

Refer to NANTES - Charts 19-1, 19-2, 19-3 from Jeppesen GSPRM 2017.

What is the distance between reporting points E1 and E2?

Answer in NM (1NM tolerance each-way given)

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Answer:

Correct answer : 9

Explanation

Locate the two reporting points on 19-1.
Using either dividers or a piece of paper and the scale on the right hand side NM (or km on the left hand side), you should find a distance of 9NM or 16km.

Be careful which units the question asks for!

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Question ID: 627098

Refer to chart E LO 2 from Jeppesen GSPRM 2017.

A flight is planned along route T420, course 330°, FL380, inbound TRENT VOR/DME (TNT).

The descent should be initiated to pass waypoint ELVOS in FL200. Calculate the Top of Descent relating to TRENT for a ground speed of 200 kt and a rate of descent of 1 500 ft/min:

Answer in NM

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Answer:

Correct answer : 65

Explanation

Height to lose is 18,000 ft

at 1,500 per minute it will take 12 minutes to lose 18,000ft

In 12 minutes we will of covered a distance of 40NM with a ground speed of 200knots

Add 25 nm for the distance between the Trent and elvos to get the distance for the top of descent to Trent : 40NM + 25NM = 65NM

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Question ID: 626293

A flight has been diverted off the planned route due to the bad weather. ATC clears the flight to re-join the planned route at TUSKA. The pilot wishes to decide early, if there is enough fuel to safely complete the flight or if a fuel diversion is needed, so he/she checks the fuel at 18:06 UTC. Following the data, what is minimum fuel required at 18:06 UTC to complete the planned flight?

Minimum fuel at TUSKA to complete the flight: 144 litres

Average fuel flow: 45 kg/h

ETO TUSKA: 18:43 UTC

Fuel density: 0.72 kg/lt

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A

105 litres

B

183 litres

C

172 litres

D

164 litres

Explanation

First we need to find the volume of fuel required to fly from location at 18:06 to TUSKA at 18:43.
This is a total flight time of 37mins.
So total fuel burned = 45kg/hr x (37/60) = 27.75kg.

We then need to convert into litres, so this mass has a volume of
27.75kg ÷ 0.72kg/litres = 38.5litres

Since we want to arrive at TUSKA with a minimum of 144litres, we must add these two values to give the total fuel on board at the initial time of fuel check, 144 + 38.5 = 182.5 litres
Explanation Provided by Rachel Battrick

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Question ID: 626134

You refuel with 150 US Gallons. What is the mass in lb with an spesific gravity of 0.75?

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A

697 lb

B

827 lb

C

937.5 lb

1025.5 lb

Explanation

150 USG X 3.78 = 567 Litres

567 X 0.75 = 425 (To get the weight from specific gravity)

425 X 2.2 (to get kg in to LBS)

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Question ID: 626033

Given:

Distance from departure to destination:
190 NM
Safe Endurance:
2.4 hrs
True Track:
120°
W/V:
030/40
TAS:
130 kts

What is the distance of the PSR from the departure point?

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A

95 NM

B

73 NM

C

149 NM

D

44 NM

Explanation

PSR = EOH/(O+H), E =Endurance, O = Outbound GS, H = Homebound GS

First we need to calculate the wind. The easiest method of doing this is using trigonometry...
HW/TW = wind speed x cos(difference between wind direction and track)
= 40 x cos(120 - 30) = 0
Thus, TAS = O = H = 130kts

Now we can use the original equation...
PSR = (2.4 x 130 x 130)/(130+ 130) = 156NM
Closest answer is 149NM

Explanation Provided by Rachel Battrick

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Question ID: 625671

What is the maximum extra fuel that may be uplifted?

MZFM: 170,000 kg
MTOM: 223,000 kg
MLM: 182,000 kg
Planned ZFM: 165,767 kg
TOF: 51,616 kg
Trip Fuel: 45854 kg

Fuel Penalty:
Take Off: 282.7/1000 kg
Landing: 394/1000 kg

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A

2396 kg

B

4328 kg

C

4988 kg

D

5617 kg

Explanation

To calculate uplifts in fuel, we need to use the 'LEAST OF THREE' method...

1. MTOM - ZFM
2. MLM - ZFM + Trip Fuel
3. Tank Capacity - Taxi Fuel

1. 223,000 - 165,767 = 57,233kg
2. 182,000 - 165,767 + 45,854 = 62,087kg
3. Tank Capacity not provided

The least of these three options is 1. using MTOM, so this is the value we shall take forward and consider in our calculations.

Since we plan to take off with TOF = 51,616kg, the total uplift in fuel is 57,233 - 51,616 = 5,617kg

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Question ID: 625341

An aircraft operating as Commercial Air Transport is flying from Paris to Barcelona. Using the following information, determine the Contingency Fuel required.

Trip Fuel: 4560 kg
Fuel flow at 1500 feet above destination aerodrome: 2304 kg/h
Fuel penalty when flying 2000 feet lower than optimum altitude: 232 kg

Answer in KG

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Answer:

Correct answer : 228

Explanation

5 % of trip fuel or 5 minutes of holding.

Select the largest!

228 is the highest and is therefore the answer!

*ignore the penalty as that seems to have been loaded already!

Explanation Provided by atpl2airline.com

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Question ID: 625336

Given the following information, what is the expected leg time?

Leg track: 090°
Distance: 100 NM
TAS: 350 kt
Wind: 030°/100 kt

Answer in Minutes

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Answer:

Correct answer : 21

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Question ID: 625322

Given: Course A to B: 088°T
Distance: 1250 NM
Mean TAS: 330 kt
Mean W/V: 340°/60 kt

The time from A to the PET between A and B is

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A

2 hours 02 minutes

B

1 hour 43 minutes

C

1 hour 39 minutes

D

1 hour 54 minutes

Explanation

PET = DH/(O+H), D = Distance, O = Outbound GS, H = Homebound GS

First we need to calculate the wind, the easiest way to do this is using trigonometry...
HW/TW Component = wind speed x cos(difference in wind direction and track)
=60 x cos(340-88) = -18.5kts
Since this is a NEGATIVE value, this is a TAILWIND. (If the value was positive, it would be a headwind).

We can now use this wind along with the TAS to calculate our O and H GS...
O = 330 + 18.5 = 348.5kts
H = 330 - 18.5 = 3115.5kts

We can now use these values in the original equation to find the following...
PET = (1250 x 311.5)/(348.5 + 311.5) = 590NM

To find the time taken to get to PET, we use this distance and our outbound GS since this is how fast we will be travelling on our way to the PET.

speed = distance/time => time = distance/speed = 590/348.5 = 1hr 42mins

Explanation Provided by Rachel Battrick

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Question ID: 625321

On a VFR flight, the traffic circuit altitude has to be established 5 NM before reaching the destination aerodrome. The descent must begin no later than how many nautical miles? ground distance from the airfield?

Rate of descent: 500 ft/min
Ground speed during descent: 130 kt
Initial altitude: 4500 ft AMSL
Traffic circuit altitude: 2500 ft AMSL

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A

14 NM

B

9 NM

C

20 NM

D

25 NM

Explanation

Total descent height = 4500 - 2500 = 2000ft

Time taken to descend = 2000ft/500fpm = 4min

speed = distance/time => distance = speed x time = 130kts x (4min/60) = 8.6NM

We then need to account for the 5NM before reaching destination provided in the question,
8.6 + 5 = 13.6NM (approx 14NM)

Explanation Provided by Rachel Battrick

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Question ID: 622381

3643 Lbs of fuel; Specific gravity of 0.84

How many US Gallons does that equal

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Answer:

Correct answer : 520

Explanation

Start with 3643.
3643÷2.205 = 1652 LBS into KGS
1652 ÷ 0.84 = 1966 KGS to Liters (SG)
1966 ÷ 3.785 = 519/521 Liters to USG

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