Refer to figure.
To begin, we will determine the fuel consumption for both engines during the flight at FL360 and a Total Air Temperature (TAT) of -17°C, using the provided table. Given an aircraft mass of 65 000 kg, please consult the relevant section of the attached table:

The FF/ENG value provided is not the final figure. Please direct your attention to the accompanying note situated in the table:

"Increase/decrease fuel flow by 3% per 10°C above/below standard TAT."

Based on the given information, the TAT at FL360 is -17°C, which is 15°C higher than the standard TAT of -32°C.

Consequently, the final FF/ENG is calculated as follows:

• Final FF/ENG = 1163 kg/h + 4.5% of 1163 kg/h = 1215.335 kg/h
• The aircraft has two engines, therefore the total fuel flow is: 2 x 1215.335 kg/h = 2430.67 kg/h

Once more, we will make use of the attached table to establish the True Airspeed (TAS) at FL360. Kindly consult the relevant section of the attached table:

This, however, is not the final TAS value for the particular flight. Please refer to the accompanying note positioned above the table:

"Increase/decrease TAS by 1 knot for every 1°C above/below standard TAT."

According to the information provided, the TAT at FL360 is -17°C, which is 15°C higher than the standard TAT. Consequently, the final TAS is calculated as follows:

• Final TAS = 450 kt + 15 kt = 465 kt

The question states that the aircraft experiences a headwind component of 25 kt. Therefore, the ground speed will be GS = TAS - HWC; GS = 465 kt - 25 kt = 440 kt.

At a speed of 440 kt, the duration needed to cover a distance of 295 NM can be calculated as follows:

• Time = Distance/Speed = (295 NM)/(440 kt) = 0.6705 hours

Therefore, with a fuel flow rate of 2430.67 kg/h, we can determine the anticipated fuel consumption for the flight at FL360:

• Fuel consumption = (0.6705) x (2430.67 kg/h) = 1629.8 kg