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

I want to walk you through a set of worked answers from the General Navigation question bank. These are the solutions to the practice questions, and they show you exactly how to chain together the techniques you've learned. Let's go through them one by one, because each one demonstrates a different skill. First, question 8. This one is solved by scale drawing. You use the chart's scale to draw a line 480 nautical miles along the 110°E meridian, starting from the North Pole. Then you plot a second line, 300 nautical miles, in a direction of 154°Grid. The intersection of those two lines gives you your position. It's a pure plotting exercise. Question 10 is a departure question. A very similar worked example is given in Chapter 15, so if you struggled with this, go back and study that example carefully. Now, question 17 is a classic time, speed, and distance problem with a revised ground speed. Let's work through the arithmetic. The planned ground speed is 115 knots. For the first 74 nautical miles at 115 knots, the planned time is 74 divided by 115, which equals 38.6 minutes. For the full 250 nautical miles at 115 knots, the planned time is 250 divided by 115, which is 130.4 minutes. However, after covering the first 74 nautical miles, the actual elapsed time was 40.1 minutes, not the planned 38.6. So you revise your ground speed. The flight time for the remaining 250 nautical miles at the revised ground speed is calculated as 130.4 multiplied by 40.1, divided by 38.6. That comes to 135 minutes. So, 0900 UTC plus 135 minutes gives you an ETA of 1115 UTC. Question 18 is a simple proportion. ETA B and ETA C are not relevant. You fly 30 nautical miles in 17 minutes. The question asks how long it will take to fly the remaining 20 nautical miles at the same ground speed. It's a direct ratio. Question 20 is a plotting question, solved by measurement on the chart. Question 22 is a wind and TAS conversion problem. First, find the present TAS, apply the wind, and get the present ground speed, which is 230 knots. At 230 knots ground speed, 150 nautical miles takes 39 minutes. You need to arrive 5 minutes later, so your new time to go is 44 minutes. The new required ground speed is 150 nautical miles in 44 minutes, which is 205 knots. The wind should not change, so the new required TAS will be 240 knots. Then you convert that TAS to CAS. You can either use FL140 and minus 5°C in the airspeed window to convert the TAS to 190 knots CAS, or you can put the old TAS of 264 against the old IAS of 210, and then against 240 TAS, you will see 190. So 210 to 190 is a 20 knot reduction in IAS. Strictly speaking, it's CAS, but the PEC—the position error correction—should not change much in 20 knots. The answer is (d). Question 23 is another proportion. ETA Y is irrelevant. If it takes you 30 minutes to cover 30 nautical miles, it will take you a further 20 minutes to cover the remaining 20 nautical miles at the same ground speed. Question 24 is about pressure altitude. The QNH is 988 hPa. The datum for pressure altitude is 1013 hPa, which is 25 hPa greater. Using 27 feet per hPa, this equates to a distance of 675 feet. A greater static pressure occurs at a lower pressure level. Therefore, the 1013 hPa pressure level is below sea level by 675 feet. The airport is 1000 feet above sea level, which means it is 1675 feet above the 1013 hPa pressure level. Question 25 involves flying 2950 nautical miles north, which takes you to 4500N 17822W. Then you convert 314 kilometers to nautical miles and use the departure formula. Question 26 is about bearings. The true bearing from you TO the headland is 050°(T). However, you have to plot FROM the headland to the aircraft, because the headland is on your map, but the aircraft's position is not. You therefore plot the reciprocal of 155°(T). Question 41 is answered as per the reply to Q3. Question 42 is (d), and question 43 is (d). Now, let me pause here. These are the worked answers