The answer is (a) — Page 560, Lesson 554

This is the answer key for the chapter, so I’ll walk you through the reasoning behind each one, because that’s where the real learning is. Let’s start with question 44. The answer is (a). A 5% glide slope means that for every 100 knots you go forward horizontally, you go down 5 knots vertically. So if your forward speed is 150 knots, 5% of that is 7.5 knots vertically. You then convert that 7.5 knots into feet per minute, and you’ll find option (a) is the nearest. Question 45, answer (a). You have 65 nautical miles to cover at 240 knots. That takes 16.25 minutes. You need to lose 25,000 feet in that same 16.25 minutes. So the required rate of descent is 25,000 divided by 16.25, which gives you 1,338 feet per minute. Question 46, answer (b). The Jeppesen conventions differ slightly from the ICAO ones. The key is given in the introduction to the Jeppesen Student Pilots’ Manual. Question 47, answer (c). Question 48, answer (b). The lowest latitude at which there is at least one day a year without a sunset — that’s Mid-summer Day — and one day a year without a sunrise — Mid-winter Day — is the Arctic Circle, which is 66½°N. So the Sun will rise and set every day at 62N and 66N. The higher of those two is 66N, so that’s your answer. Question 49, answer (a). Pressure altitude plus the ISA deviation times 120 gives you 27,560. The CRP5 gives you 27,000. Question 50, answer (c). This one is a full worked example, so let’s take it step by step. You look up the LMT of sunrise at 49N on the 6th of December. There’s a 3-minute change between the 4th and the 7th of December. Interpolating, the times at 50N and 45N are 0742 and 0723 respectively. That’s a difference of 19 minutes. One-fifth of that is about 4 minutes, so the LMT of sunrise at 49N on the 6th of December is 0738. Then you set it out in a table. LMT sunrise at Vancouver is 6 December 07:38 LMT. The arc/time for 123°30’W — longitude west, UTC best — is plus 08 hours 14 minutes. That gives you UTC of 6 December 15:52. To calculate arc/time, you enter the degrees and minutes in DMS format into your calculator and divide by 15. Question 51, answer (b). Question 52, answer (a). This is just the 1 in 60 rule. A 3-degree glide path gives you 300 feet per nautical mile. 25 nautical miles at 300 feet per nautical mile is 7,500 feet. That would be for a touchdown at zero feet at the runway threshold. However, the height over the threshold is 50 feet, which has to be added. Question 53, answer (c). Again, use a table. Standard Time at Queensland is Today 06:00 ST. STD — longitude east, UTC least — minus 10 hours 00 minutes gives you UTC of Yesterday 20:00. Then STD — longitude west, UTC best — minus another 10 hours 00 minutes gives you Standard Time at Hawaii of Yesterday 10:00 ST. Question 54, answer (a). Question 55, answer (d). Question 56, answer (c). Question 57, answer (a). Question 58, answer (a). The first leg takes the aircraft down to the Equator at 170W. The second leg takes it to the Equator at 143W. The third leg takes it to 27N 143W. You then work out the change of longitude of 1,620 nautical miles due west departure at 27N, which is 30°18’W, i.e. 173°18’W. That’s the full set of answers for this chapter.