So your true heading is 100. VOR radials are always magnetic, so you are… — Page 552, Lesson 550

This is the answers section of a General Navigation question bank, so I’m going to walk you through the reasoning behind each answer, because that’s where the real learning lives. We’ll go in order, and I’ll explain the technique behind each one. Question 8: This is a scale-drawing problem. You take the chart’s scale and draw a line 480 NM along the 110°E meridian, starting from the North Pole. Then you plot a second line, 300 NM long, in a direction of 154°Grid. The intersection of those two lines gives you your position. The key here is that you’re working in grid navigation, so the direction is measured relative to grid north, not true north. Question 9: Answer is (d). No working shown, but it’s a standard plotting or calculation result. Question 10: Answer is (c). This is a departure question — that’s the east-west distance between two meridians at a given latitude. A very similar worked example is given in Chapter 15, so if you’re unsure, go back and study that example, because the method is identical. Question 11: (c). Question 12: (a). Question 13: (d). Question 14: (a). Question 15: (b). These are single-letter answers with no working shown, so they’re straightforward calculation or plotting results. Now Question 17, this is a good one — it’s a revised ground speed and ETA problem. You plan to fly 74 NM at a planned ground speed of 115 knots. That gives you a planned time of 38.6 minutes. Then you plan 250 NM at the same 115 knots, giving 130.4 minutes. But after you’ve actually flown the first 74 NM, your actual elapsed time is 40.1 minutes — slower than planned. So your revised ground speed is slower. To find the new flight time for the remaining 250 NM, you take the planned time for 250 NM, 130.4 minutes, and multiply it by the ratio of actual time to planned time for the first leg: 40.1 divided by 38.6. That comes to 135 minutes. Add that to your departure time of 0900 UTC, and you get 1115 UTC. That’s your revised ETA. Question 18: Answer (c). The trick here is that ETA B and ETA C are irrelevant — you ignore them. You’ve flown 30 NM in 17 minutes. At the same ground speed, how long for the remaining 20 NM? It’s a simple proportion: 30 NM in 17 minutes means 20 NM takes two-thirds of that, which is about 11.3 minutes. Question 19: (a). Question 20: (c). This is a plotting question — you solve it by measuring directly on the chart. Question 22: This is a detailed one, so let’s take it step by step. First, find your present TAS, apply the wind, and get your present ground speed, which is 230 knots. At that speed, 150 NM takes 39 minutes. But you need to arrive 5 minutes later than planned, so your new time to go is 44 minutes. The new required ground speed is therefore 150 NM in 44 minutes, which works out to 205 knots ground speed. The wind doesn’t change, so the new required TAS is 240 knots. Now you convert that TAS to CAS. You can either use FL140 and -5°C in the airspeed window of your CRP-5 to convert 240 TAS to 190 knots CAS, or you can use the ratio method: put the old TAS of 264 against the old IAS of 210, then read against 240 TAS, and you’ll see 190. So your IAS drops from 210 to 190 — a 20-knot reduction. Strictly speaking it’s CAS, but the position error correction shouldn’t change much over 20 knots. Answer is (d). Question 23: (c). Same logic as Q18 — ETA Y is irrelevant. If you cover 30 NM in 30 minutes, you cover the remaining 20 NM in a further 20 minutes at the same ground speed. Question 24: This is a pressure altitude question. Your QNH is 988 hPa. The datum for pressure altitude is 1013 hPa, which is 25 hPa greater than your QNH. Using the standard 27 feet per hPa, that equates to 675 feet. Now think about this: a greater static pressure occurs at a lower pressure level. So the 1013 hPa pressure level is actually below sea level by 675 feet. Your airport is 1000 feet above sea level, which means it’s 1675 feet above the 1013 hPa pressure level. That’s your pressure altitude. Answer is (b). Question 25: (b). The aircraft flies 2950 NM north, which takes it to 45°N, 178°22’W. Then you convert 314 km to nautical miles, and use the departure formula — that’s the formula that relates the change in longitude to the distance along a parallel of latitude. Question 26: (d). This is a bearing-reciprocal problem. The true bearing from you TO the headland is 050°(T). But you have to plot FROM the headland to the aircraft, because the headland is on your map and the aircraft’s position is not. So you plot the reciprocal of 050°(T), which is 230°(T). That’s the bearing you draw on the chart. Question 41: (c). This is answered the same way as Question 3 — same technique, same reasoning. Question 42: (d). Question 43: (d). So the big takeaways from this set: for revised ETA problems, always work with the ratio of actual to planned time. For pressure altitude, remember the 1013 hPa datum and the 27 feet per hPa rule. And for bearings, always ask yourself whether you’re plotting TO or FROM the known point — that determines whether you use the bearing or its reciprocal.