All right, let's jump straight into these answers — Page 560, Lesson 554

All right, let's jump straight into these answers. We're working through the end-of-chapter questions, and I'll explain the reasoning behind each one as we go. Starting with question 44, the answer is (a). This is a glide slope question. A 5% glide slope means that for every 100 knots you travel forward horizontally, you descend 5 knots vertically. So, if your forward speed is 150 knots, 5% of that is 7.5 knots vertically. You then convert that vertical speed in knots to feet per minute, and you'll find option (a) is the nearest. Moving to question 45, the answer is (a). Here we have a distance of 65 nautical miles at 240 knots. That will take 16.25 minutes. You need to lose 25,000 feet in that same 16.25 minutes. So the required rate of descent, or ROD, is 25,000 divided by 16.25, which gives you 1,338 feet per minute. Question 46, the answer is (b). This one is about Jeppesen conventions. They differ slightly from the ICAO ones. The key to this is given in the introduction to the Jeppesen Student Pilots' Manual. Question 47, the answer is (c). Question 48, the answer is (b). This is about the Arctic Circle. The lowest latitude at which there is at least one day a year without a sunset, on Mid-summer Day, and one day a year without a sunrise, on Mid-winter Day, is the Arctic Circle, which is at 66½° North. Therefore, the Sun will rise and set every day at 62 North and 66 North. The higher of these two is 66 North. Question 49, the answer is (a). Here, pressure altitude plus the ISA deviation times 120 gives you 27,560. The CRP5 computer gives you 27,000. Question 50, the answer is (c). This is a sunrise calculation. You look up the Local Mean Time of Sunrise at 49 North on the 6th of December. There is a 3-minute change between the 4th and the 7th of December. Interpolating, the times at 50 North and 45 North 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 49 North on the 6th of December is 0738. Now, you set it out in a table. You have the LMT sunrise at Vancouver on the 6th of December at 07:38. Then you apply the arc/time for 123° 30' West. Since the longitude is West, UTC is best, so you add. That's plus 8 hours and 14 minutes. That gives you a UTC of 15:52 on the 6th of December. To calculate arc/time, you enter the degrees and minutes in DMS format into your calculator and divide by 15. Question 51, the answer is (b). Question 52, the answer is (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 touch-down at zero feet at the runway threshold. However, the height over the threshold is 50 feet, which has to be added. Question 53, the answer is (c). As always, use a table. You have the Standard Time at Queensland today at 06:00. The STD for longitude East is UTC least, so you subtract 10 hours to get UTC, which is yesterday at 20:00. Then, for Hawaii, the STD for longitude West is UTC best, so you add 10 hours to get the Standard Time at Hawaii, which is yesterday at 10:00. Question 54, the answer is (a). Question 55, the answer is (d). Question 56, the answer is (c). Question 57, the answer is (a). Question 58, the answer is (a). The first leg takes the aircraft down to the Equator at 170 West. The second leg takes it to the Equator at 143 West. The third leg takes it to 27 North, 143 West. You then work out the change of longitude of 1,620 nautical miles due west departure at 27 North, which is 30° 18' West, i.e., 173° 18' West.