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Gleim Private Pilot FAA Knowledge Test 2017 Edition, 1st Printing

Update 2November 2016

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Study Unit 3 – Airports, Air Traffic Control, and Airspace

Page 72, Subunit 3.4, 5.c.1): This update provides an additional figure to clarify the text.

Figure 49. – Airport Diagram. Figure 49. – Airport Diagram. Figure 50. – Wind Cone to Wind Sock.

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Study Unit 4 – Federal Aviation Regulations

Page 158, Subunit 4.10, Question 144: This question was moved to Study Unit 9 from Study Unit 4 because Subunit 9.2 is a more appropriate location for the subject matter. Questions for Subunit 9.2 begin on page 312.

144 60. (Refer to Figure 78 on page 159.) What are the basic VFR weather minima required to takeoff from the Onawa, IA (K36) airport during the day?

A. 3 statute miles visibility, 500 feet below the clouds, 1,000 feet above the clouds and 2,000 feet horizontally from the clouds.

B. 0 statute miles, clear of clouds.

C. 1 statute mile, clear of clouds.

Answer (C) is correct. (AIM Para 3-1-4) DISCUSSION: Onawa, IA, (K36) airport is surrounded by Class G airspace. The VFR weather minima in Class G airspace below 1,200 feet AGL (regardless of MSL altitude) is 1 statute mile of visibility and clear of clouds. Answer (A) is incorrect. Class C (day and night), D (day and night), and E (day VFR only and less than 10,000 feet MSL) are all 3 statute miles of visibility, 500 feet below the clouds, 1,000 feet above the clouds, and 2,000 feet horizontally from the clouds. Answer (B) is incorrect. There is no VFR weather minima with visibility of 0 statute miles.

Study Unit 5 – Airplane Performance and Weight and Balance

Page 194, Subunit 5.3, Question 19: These edits were made to coincide with the FAA’s update of the referenced Figure 40.

19. (Refer to Figure 40 below.) Determine the approximate ground roll distance required for takeoff.

OAT . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100°F 38°CPressure altitude . . . . . . . . . . . . . . . . 2,000 ft Takeoff weight . . . . . . . . . . . . . . . . . . 2,750 lb Headwind component . . . . . . . . . . . . . . Calm

A. 1,150 feet.

B. 1,300 feet.

C. 1,800 feet.

Answer (A) is correct. (PHAK Chap 11) DISCUSSION: Begin on the left section of Fig. 40 at 100°F 38°C (see outside air temperature at the bottom). Move up vertically to the pressure altitude of 2,000 feet. Then proceed horizontally to the first reference line. Since takeoff weight is 2,750, move parallel to the closest guideline, to 2,750 pounds. Then proceed horizontally to the second reference line. Since the wind is calm, proceed again horizontally to the right-hand margin of the diagram (ignore the third reference line because there is no obstacle; i.e., ground roll is desired), which will be at 1,150 feet. Answer (B) is incorrect. This would be the ground roll distance required at maximum takeoff weight. Answer (C) is incorrect. This would be the total distance required to clear a 50-ft. obstacle.


20. (Refer to Figure 40 on page 194.) Determine the total distance required for takeoff to clear a 50-foot obstacle.

OAT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . StdPressure altitude . . . . . . . . . . . . . . . . Sea level Takeoff weight. . . . . . . . . . . . . . . . . . . . 2,700 lb Headwind component. . . . . . . . . . . . . . . . Calm

A. 1,000 feet.

B. 1,400 feet.

C. 1,700 feet.

Answer (B) is correct. (PHAK Chap 11) DISCUSSION: Begin in the left section of Fig. 40 by finding the intersection of the sea level pressure altitude and standard temperature (59°F 15°C) and proceed horizontally to the right to the first reference line. Then proceed parallel to the closest guideline, to 2,700 pounds. From there, proceed horizontally to the right to the third reference line. You skip the second reference line because the wind is calm. Then proceed upward, parallel to the closest guideline to the far right side. To clear the 50-ft. obstacle, you need a takeoff distance of about 1,400 feet. Answer (A) is incorrect. This would be the total distance required at 2,200 lb. takeoff weight. Answer (C) is incorrect. This would be the total distance required at maximum takeoff weight.


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Page 195, Subunit 5.3, Question 21: This edit was made to clarify the answer explanation.

21. (Refer to Figure 40 on page 194.) Determine the total distance required for takeoff to clear a 50-foot obstacle.

OAT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Std Pressure altitude. . . . . . . . . . . . . . . . . . . 4,000 ft Takeoff weight . . . . . . . . . . . . . . . . . . . . 2,800 lb Headwind component. . . . . . . . . . . . . . . . . Calm

A. 1,500 feet.

B. 1,750 feet.

C. 2,000 feet.

Answer (B) is correct. (PHAK Chap 11) DISCUSSION: The takeoff distance to clear a 50-ft. obstacle is required. Begin on the left side of the graph at standard temperature (as represented by the curved line labeled “ISA”). From the intersection of the standard temperature line and the 4,000-ft. pressure altitude, proceed horizontally to the right to the first reference line, and then move parallel to the closest guideline to 2,800 pounds. From there, proceed horizontally to the right to the third reference line (skip the second reference line because there is no wind), and move upward parallel to the closest guideline following equidistantly between the diagonal lines all the way to the far right. You are at 1,750 ft., which is the takeoff distance to clear a 50-ft. obstacle. Answer (A) is incorrect. This would be the total distance required with a 10-kt. headwind. Answer (C) is incorrect. This would be the total distance required at maximum takeoff weight.


22. (Refer to Figure 40 on page 194.) Determine the approximate ground roll distance required for takeoff.

OAT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90°F 32°CPressure altitude . . . . . . . . . . . . . . . . 2,000 ft Takeoff weight . . . . . . . . . . . . . . . . . . 2,500 lb Headwind component . . . . . . . . . . . . . . 20 kts

A. 650 feet.

B. 850 feet.

C. 1,000 feet.

Answer (A) is correct. (PHAK Chap 11) DISCUSSION: Begin with the intersection of the 2,000-ft. pressure altitude curve and 90°F 32°C in the left section of Fig. 40. Move horizontally to the right to the first reference line, and then parallel to the closest guideline to 2,500 pounds. Then move horizontally to the right to the second reference line, and then parallel to the closest guideline to the right to 20 knots. Then move horizontally to the right, directly to the right margin because there is no obstacle clearance. You should end up at about 650 ft., which is the required ground roll when there is no obstacle to clear. Answer (B) is incorrect. This would be the ground roll distance required if the wind were calm. Answer (C) is incorrect. This would be the ground roll distance required at maximum takeoff weight.

Page 197, Subunit 5.4, Question 27: These edits were made to update the measurements to coincide with the figure.

27. (Refer to Figure 35 on page 196.) Approximately what true airspeed should a pilot expect with 65 percent maximum continuous power at 9,500 feet with a temperature of 36°F below standard?

A. 163 KTS 178 MPH.

B. 161 KTS 181 MPH.

C. 158 KTS 183 MPH.

Answer (C) is correct. (PHAK Chap 11) DISCUSSION: The left part of the chart applies to 36°F below standard. At 8,000 ft., TAS is 157 KTS. At 10,000 ft., TAS is 160 KTS. At 9,500 ft., with a temperature 36°F below standard, the expected true airspeed is 75% above the 157 KTS at 8,000 ft. toward the 160 KTS at 10,000 ft., i.e., approximately 158 KTS. Refer to Figure 35 and locate the column for –36°F (ISA –20°C). Interpolation will be required to determine the TAS at 9,500 ft. At 8,000 ft., TAS is 181 MPH, and at 10,000 ft., TAS is 184 MPH; therefore, the difference is 3 MPH.

10,000 ft. – 8,000 ft. = 2,000 ft. 9,500 ft. – 8,000 ft. = 1,500 ft.1,500 ft. ÷ 2,000 ft. = .75

.75 × 3 MPH = 2.25 MPH181 MPH + 2.25 MPH = 183.25 MPH

Answer (A) is incorrect. The expected TAS at 6,000 ft. is 163 KTS 178 MPH. Answer (B) is incorrect. The expected TAS at 8,000 ft. is 161 KTS 181 MPH.


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Study Unit 7 – Aviation Weather

Page 245, Subunit 7.8, Question 33: This edit clarifies the answer explanation.

33. Which conditions result in the formation of frost?

A. The temperature of the collecting surface is at or below freezing when small droplets of moisture fall on the surface.

B. The temperature of the collecting surface is at or below the dewpoint of the adjacent air and the dewpoint is below freezing.

C. The temperature of the surrounding air is at or below freezing when small drops of moisture fall on the collecting surface.

Answer (B) is correct. (AvW Chap 5) DISCUSSION: Frost forms when both the collecting surface is below the dew point of the adjacent air and the dew point is below freezing. Frost is the direct sublimation deposition of water vapor to ice crystals. Answer (A) is incorrect. If small droplets of water fall on the collecting surface, which is at or below freezing, ice (not frost) will form. Answer (C) is incorrect. If small droplets of water fall while the surrounding air is at or below freezing, ice (not frost) will form.

Study Unit 8 – Aviation Weather Services

Page 265, Subunit 8.4, Question 25: These edits clarify and correct the answer choice and explanation.

25. The section of the Area Forecast entitled “VFR CLDS/WX” contains a general description of

A. cloudiness and weather significant to flight operations broken down by states or other geographical areas covering at least 350 square miles.

B. area and route briefings, as well as airspace procedures and special announcements.

C. clouds and weather which cover an area greater than 3,000 square miles and is significant to VFR flight operations.

Answer (C) is the best answer. (AWS Sect 7) DISCUSSION: The VFR CLDS/WX section contains a 12-hr. specific forecast plus a 6-hour categorical outlook section. The specific forecast section gives a general description of clouds and weather that covers an area greater than 3,000 square miles and is significant to VFR flight operations. Answer (A) is incorrect. While tThe VFR CLDS/WX section may will be broken down by states or well-known geographical areas, and the area of coverage will always be greater than 3,000 square miles, not 350 square miles. Answer (B) is incorrect. TIBS (Telephone Information Briefing Service) is a continuous telephone recording of area and route briefings, as well as airspace procedures and special announcements. It is designed to be a preliminary briefing tool and is available 24 hours a day.

Study Unit 9 – Navigation: Charts and Publications

Page 302, Subunit 9.1, Question 1: These edits reflect the FAA’s update of the referenced Figure 22.

1. (Refer to Figure 22 on page 303.) (Refer to area 3.) Determine the approximate latitude and longitude of Shoshone County Airport.

A. 47°02'N – 116°11'W.

B. 47°33'N – 116°11'W.

C. 47°32'N – 116°41'W.

Answer (B) is correct. (PHAK Chap 16) DISCUSSION: Shoshone County Airport is below 3, just west of the 116° line of longitude (find the 116° line in the 8,000 MSL northeast of Shoshone). There are 60 min. between the 116° line and the 117° line, depicted in 1-min. tick marks. Shoshone is 11 tick marks (11 min.) past the 116° line. Find the labeled 48° latitude line just northeast of the 116° line. The latitude and longitude lines are presented each 30 min. Since lines of latitude are also divided into 1 in. tick marks, the airport is three tick marks above the 47°30' line or 47°33'N. The correct latitude and longitude is thus 47°33'N – 116°11'W. Answer (A) is incorrect. Shoshone Airport is just north of the 47°30' line of latitude (not the 47°00' line). Answer (C) is incorrect. Shoshone Airport is 11 ticks past the 116°00' line of longitude (not the 116°30' line).


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Page 318, Subunit 9.2, Question 23: This edit clarifies the answer choice to better mirror the wording of the Legend.

23. (Refer to Figure 24 below, and Legend 1 on page 319.) (Refer to area 1.) For information about the parachute jumping at Caddo Mills Airport, refer to

A. notes on the border of the chart.

B. the Airport/Facility Directory section of the Chart Supplement.

C. the Notices to Airmen (NOTAM) publication.

Answer (B) is correct. (ACL) DISCUSSION: The miniature parachute near the Caddo Mills Airport (at 1 on Fig. 24) indicates a parachute jumping area. In Legend 1, the symbol for a parachute jumping area instructs you to see the Airport/Facility Directory section of the Chart Supplement for more information. Note, as of March 31, 2016, the Airport/Facility Directory book has been retitled to Chart Supplement. Answer (A) is incorrect. The sectional chart legend identifies symbols only. Answer (C) is incorrect. NOTAMs are issued only for hazards to flight.

Page 340, Subunit 9.2, Question 54: This edit corrects the airport code.

54. (Refer to Figure 75 on page 341.) The airspace surrounding the Gila Bend AF AUX Airport (GBN GXF) (area 6) is classified as Class

A. B.

B. C.

C. D.

Answer (C) is correct. (AIM Chap 3) DISCUSSION: The GBN GXF airport is surrounded by a dashed blue line, which indicates it is within Class D airspace. Answer (A) is incorrect. Class B airspace is surrounded by a solid blue line. Answer (B) is incorrect. Class C airspace is surrounded by a solid magenta line.


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Page 342, Subunit 9.2, Question 56: This edit corrects the image.

56. (Refer to Figure 22 23 below, and Legend 1 on page 343.) (Refer to area 3.) For information about glider operations at Ridgeland Airport, refer to

A. notes on the border of the chart.

B. the Chart Supplement.

C. the Notices to Airmen (NOTAM) publication.

Answer (B) is correct. (ACL) DISCUSSION: The miniature glider near the Ridgeland Airport (at 3 on Fig. 22 23) indicates a glider operations area. The Chart Supplement will have information on the glider operations at Ridgeland Airport. Answer (A) is incorrect. The sectional chart legend identifies symbols only. Answer (C) is incorrect. NOTAMs are issued only for hazards to flight.

Figure 22 23. – Sectional Chart Excerpt.NOTE: Chart is not to scale and should not be used for navigation. Use associated scale.


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Page 348, Subunit 9.3, Question 63: This edit clarifies and corrects the answer explanation.

63. (Refer to Figure 24 on page 349.) Which public use airports depicted are indicated as having fuel?

A. Commerce (area 6) and Rockwall (area 1).

B. Rockwall (area 1) and Sulphur Springs (area 5).

C. Commerce (area 6) and Sulphur Springs (area 5).

Answer (B) is correct. (ACL) DISCUSSION: On Fig. 24, the requirement is to identify the airports having fuel available. Airports having fuel available are designated by small squares extending from the top, bottom, and both sides of the airport symbol. Only Rockwall (area 1) and Sulphur Springs (area 5) have such symbols. Answer (A) is incorrect. Commerce does not indicate it has fuel. Answer (C) is incorrect. Commerce does not indicate it has fuel.

Study Unit 10 – Navigation Systems


6. (Refer to Figure 28 below, and Figure 20 on page 385.) The VOR is tuned to Elizabeth City VOR/DME (area 3 in Figure 20), and the aircraft is positioned over Shawboro, a small town 3 NM west of Currituck County Regional (ONX). Which VOR indication is correct?

A. 5

B. 9 2

C. 2 8

Answer (C A) is correct. (PHAK Chap 16) DISCUSSION: See Fig. 20, northeast of 3 along the compass rose. Shawboro is northeast of the Elizabeth City VOR on the 030° radial; zoom in to see the tiny black circle located to the lower left of the “S” in Shawboro; that corresponds to the town of Shawboro. To be over it, the needle should be centered with either an OBS setting of 210° and a TO indication or with an OBS setting of 030° and a FROM indication. VOR 2 5 matches the latter description. Answer (A B) is incorrect. VOR 5 2 indicates that the aircraft is southwest, not northeast, of Elizabeth City VOR. Answer (B C) is incorrect. VOR 9 8 indicates that the aircraft is southwest, not northeast, of Elizabeth City VOR.

Page 386, Subunit 10.2, Question 9: This edit reflects the FAA’s update of the referenced Figure 28.

9. (Refer to Figure 28 above, and Figure 24 on page 387.) The VOR is tuned to Bonham VORTAC (area 3 in Figure 24) and the aircraft is positioned over the town of Sulphur Springs (area 5 in Figure 24). Which VOR indication is correct?

A. 1

B. 8

C. 7

Answer (C) is correct. (PHAK Chap 16) DISCUSSION: The town of Sulphur Springs (south-southwest of 5) is on the 120° radial of Bonham VORTAC. Illustration 7 in Fig. 28 shows the VOR receiver tuned to the 210° radial, which is perpendicular to (90° away from) the 120° radial. This places the aircraft in the zone of ambiguity, which results in neither a TO nor a FROM indication and an unstable CDI, which can be deflected left or right. Answer (A) is incorrect. With indication 1, the aircraft would have to be west north of Sulphur Springs. Answer (B) is incorrect. It shows the aircraft on the 030° radial, which is well to the north of Sulphur Springs.


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Study Unit 11 – Cross-Country Flight Planning

Page 420, Subunit 11.6, Questions 31-33: These edits reflect the FAA’s update of the referenced Figure 22.

31. (Refer to Figure 22 on page 421.) What is the magnetic heading for a flight from Priest River Airport (area 1) to Shoshone County Airport (area 3)? The wind is from 030° at 12 knots and the true airspeed is 95 knots.

A. 121°.

B. 143°.

C. 136°.

Answer (A) is correct. (PHAK Chap 16) DISCUSSION: On Fig. 22, begin by computing the true course from Priest River Airport (upper left corner) to Shoshone County Airport (just below 3) by laying a flight plotter between the two airports. The grommet should coincide with the meridian (vertical line with cross-hatchings). Note the 143° true course on the edge of the protractor. Next, find the magnetic variation that is given by the dashed line marked 14°E (rounded to 15°E), slanting in a northeasterly fashion just south of Carlin Bay Private to the east of Shoshone County Airport. Subtract the 15°E variation from TC to obtain a magnetic course of 128°. Since the wind is given true, reduce the true wind direction of 030° by the magnetic variation of 15°E to a magnetic wind direction of 15°. Now use the wind side of your computer. Turning the inner circle to 15° under the true index, mark 12 kt. above the grommet. Set the magnetic course of 128° under the true index. Slide the grid so the pencil mark is on 95 kt. TAS. Note that the pencil mark is 7° left of the center line, requiring you to adjust the magnetic course to a 121° magnetic heading (128° – 7°). Subtract left, add right. That is, if you are on an easterly flight and the wind is from the north, you will want to correct to the left. Answer (B) is incorrect. This is the true course, not the magnetic heading. Answer (C) is incorrect. This would be the magnetic heading if the wind was from 215° at 19 kt., not 030° at 12 kt.

32. (Refer to Figure 22 on page 421.) Determine the magnetic heading for a flight from St. Maries Airport (area 4) to Priest River Airport (area 1). The wind is from 340° at 10 knots and the true airspeed is 90 knots.

A. 330°.

B. 325°.

C. 345°.

Answer (A) is correct. (PHAK Chap 16) DISCUSSION:

1. This flight is from St. Maries (just below 4) to Priest River (upper left corner) on Fig. 22.

2. TC is 345°. 3. MC = 345° – 15°E variation (14° 30E rounded up) = 330°. 4. Wind magnetic = 340° – 15° (14° 30E rounded up) = 325°. 5. Mark 10 kt. up when 325° under true index. 6. Put MC 330° under true index.7. Slide grid so pencil mark is on 90 kt. TAS. 8. Note that the pencil mark is 1° left. 9. Subtract 1° from 330° MC for 329° MH.

A magnetic heading of 330° is the best answer of the choices given. Answer (B) is incorrect. This would be the magnetic heading if the wind was from 300° at 14 kt., not 340° at 10 kt. Answer (C) is incorrect. This is the approximate true course, not magnetic heading.


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33. (Refer to Figure 22 on page 421.) Determine the magnetic heading for a flight from Sandpoint Airport (area 1) to St. Maries Airport (area 4). The wind is from 215° at 25 knots and the true airspeed is 125 knots.

A. 352°.

B. 172°.

C. 166°.

Answer (B) is correct. (PHAK Chap 16) DISCUSSION:

1. This flight is from Sandpoint Airport (above 1), to St. Maries (below 4) on Fig. 22

2. TC = 181°. 3. MC = 181° – 15°E (14° 30E rounded up) variation = 166°. 4. Wind magnetic = 215° – 15° (14° 30E rounded up) = 200°. 5. Mark up 25 kt. with 200° under true index. 6. Put MC 166° under true index. 7. Slide grid so pencil mark is on 125 kt. TAS. 8. Note that the pencil mark is 6° right. 9. Add 6° to 166° MC for 172° MH.

Answer (A) is incorrect. This would be the magnetic heading for a flight from St. Maries Airport to Sandpoint Airport, not from Sandpoint to St. Maries, with the wind from 145°, not 215°, at 25 kt. Answer (C) is incorrect. This is the magnetic course, not the magnetic heading.


37. (Refer to Figure 23 on page 425.) Determine the magnetic heading for a flight from Allendale County Airport (area 1) to Claxton-Evans County Airport (area 2). The wind is from 090° at 16 knots and the true airspeed is 90 knots. Magnetic variation is 7°W.

A. 230°.

B. 212 213°.

C. 209 210°.

Answer (C) is correct. (PHAK Chap 16) DISCUSSION:

1. This flight is from Allendale County (above 1) to Claxton-Evans County Airport (left of 2) on Fig. 23. Variation is shown on Fig. 23 as 6 7°W.

2. TC = 212°. 3. MC = 212° TC + 6 7°W variation = 218 219°. 4. Wind magnetic = 090° + 6 7°W variation = 096 097°. 5. Mark up 16 kt. With 096 097° under true index. 6. Place MC 218 219° under true index. 7. Move wind mark to 90 kt. TAS arc. 8. Note that the pencil mark is 9° left. 9. Subtract 9° from 218 219° MC for 209 210° MH.

Answer (A) is incorrect. This would be the approximate magnetic heading if the wind was out of 330° at 23 kt., not 090° at 16 kt. Answer (B) is incorrect. This is the true heading, not the magnetic heading.


38. (Refer to Figure 58 below, and Figure 23 on page 425.) Determine the compass heading for a flight from Claxton-Evans County Airport (area 2) to Hampton Varnville Airport (area 1). The wind is from 280° at 8 knots, and the true airspeed is 85 knots. Magnetic variation is 7°W.

A. 033°.

B. 042 044°.

C. 038°.

Answer (B) is the best answer. (PHAK Chap 16) DISCUSSION:

1. This flight is from Claxton-Evans (left of 2) to Hampton Varnville (right of 1) on Fig. 23.

2. TC = 045°. 3. MC = 045° TC + 6 7°W variation = 051 052°. 4. Wind magnetic = 280° + 6 7°W variation = 286 287°. 5. Mark up 8 kt. With 286 287° under true index. 6. Place MC 051 052° under true index. 7. Move wind mark to 85 kt. TAS arc. 8. Note that the pencil mark is 4° left. 9. Subtract 4° from 051 052° MC for 047 048° MH.

10. Subtract 4° compass variation (obtained from Fig. 58) from 047 048° to find the compass heading of 043 044°.

After determining a compass heading of 043°, 042° is the best answer available. Answer (A) is incorrect. This would be the approximate compass heading if the wind were out of 295° at 22 kt., not 280° at 8 kt. Answer (C) is incorrect. This would be the approximate compass heading if the wind were out of 295° at 12 knots.


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