Pilot Prep

Gen Nav Question Series With Solutions
question 3

An aircraft at FL290 is required to commence descent when 50 NM from a VOR and to cross that VOR at FL80. Mean GS
during descent is 271 kt. What is the minimum rate of descent required ?
a) 1700 FT/MIN
b) 1900 FT/MIN
c) 1800 FT/MIN
d) 2000 FT/MIN

Since FL80 will be on VOR, an altitude of FL290 – FL80 = 210 = 21 000 feet must be lost,
If the speed is 271 kt per hour (60 minutes per hour) and 271 NM per hour, it covers a distance of 50 NM;
( 50 x 60 ) : 271 = 11.07011 ≈ travels in 11 minutes. In other words, 21,000 feet must be covered in 11 minutes, as a result;
21 000 : 11 =1909 ≈ 1900 FT/MIN.

Pilot Prep

Radio Navigation question

Gen Nav Question Series With Solutions
question 1
5 HR 20 MIN 20 SEC corresponds to a longitude difference of ;
a) 81°10'
b) 78°45'
c) 75°00'
d) 80°05'

Since the Earth completes one rotation around itself in 24 hours and there are 360 ​​meridians, the distance between two meridians is;
24 hours = 1440 minutes and 1440 : 360 = 4 minutes.
5 hours 20 minutes = 320 minutes = 320 : 4 = 80°
Since 20 seconds is 1/12 of 4 minutes (1 degree); 60: 12 = 5', resulting in difference; It is located at 80°05'