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22 Navigation Practice Questions & Answers

Every Navigation practice question from the Private Pilot Written Test Practice, with the correct answer and a short explanation.

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  1. 1. What distinguishes pilotage from dead reckoning as VFR navigation methods?

    • A.Pilotage uses precomputed heading, groundspeed and elapsed time; dead reckoning navigates by reference to visible landmarks.
    • B.Pilotage requires a current navigation database; dead reckoning requires a usable ground-based navaid signal.
    • C.Pilotage tracks VOR radials to a station; dead reckoning follows a sequence of GPS waypoints in order.
    • D.Pilotage navigates by reference to visible landmarks; dead reckoning uses precomputed heading, groundspeed and elapsed time.Answer

    Pilotage is navigation by visual reference to landmarks and charted features, while dead reckoning computes position from a heading, a groundspeed and the time flown since a known fix. Neither depends on radio or satellite equipment, and on a VFR cross-country the two are normally used together so that each cross-checks the other.

    Source: FAA-H-8083-25 Pilot's Handbook of Aeronautical Knowledge, Navigation chapter (pilotage and dead reckoning)Report a problem with this question

  2. 2. You are planning a low-altitude VFR cross-country over flat farmland crossed by several parallel highways. Which of these makes the most reliable visual checkpoint?

    • A.A single small pond about 12 NM to the right of the planned course line
    • B.A bend in a river where a railroad bridge crosses it, 2 NM from courseAnswer
    • C.A creek bed that fills only after heavy rain, directly beneath the course
    • D.One of several nearly identical grain silos beside the same straight highway

    A good checkpoint is unique, large enough to see from cruise altitude, and close to the course line; the intersection of two different linear features, such as a river and a railroad, satisfies all three because it cannot be confused with anything nearby. Small, repetitive or seasonal features invite misidentification, and a checkpoint far off course is hard to use for a position fix.

    Source: FAA-H-8083-25 Pilot's Handbook of Aeronautical Knowledge, Navigation chapter (selecting checkpoints)Report a problem with this question

  3. 3. On a sectional chart, a quadrangle bounded by the ticked latitude and longitude lines carries a bold blue Maximum Elevation Figure read as 3,800 ft. What does that figure represent?

    • A.The lowest altitude that may legally be flown across that quadrangle, 3,800 ft MSL
    • B.The height of the tallest obstacle above the ground there, which is 3,800 ft AGL
    • C.The floor of controlled airspace over that quadrangle, which is 3,800 ft MSL
    • D.The highest known terrain or obstacle elevation in that quadrangle, 3,800 ft MSLAnswer

    The Maximum Elevation Figure is the elevation above mean sea level of the highest feature in the quadrangle, terrain or obstacle, rounded up and increased by allowances for vertical error and for uncharted obstructions. It is charting information only: it is not field verified, it is not a minimum safe altitude, and it does not by itself guarantee any obstacle clearance margin.

    Source: FAA Aeronautical Chart Users' Guide, VFR charts — Maximum Elevation FiguresReport a problem with this question

  4. 4. A charted obstruction symbol shows 2,749 printed in bold with (1,320) in parentheses just below it. What is the approximate elevation of the ground at the base of that obstruction?

    • A.2,749 ft MSL
    • B.4,069 ft MSL
    • C.1,429 ft MSLAnswer
    • D.1,320 ft MSL

    On VFR charts the bold figure beside an obstruction is the elevation of its top above mean sea level and the number in parentheses is its height above the ground at its base. Subtracting the height above ground from the top elevation gives the ground elevation: 2,749 minus 1,320 equals 1,429 ft MSL.

    Source: FAA Aeronautical Chart Users' Guide, VFR charts — obstruction symbols and elevationsReport a problem with this question

  5. 5. A sectional shows an airport symbol printed in magenta with this data block: MAPLE RIDGE (7C9), AWOS-3 118.375, 412, *L, 54, and 122.7 followed by a circled C. Which statement about this airport is correct?

    • A.The field has a part-time control tower and a longest runway of 5,400 ft.
    • B.The field has no operating control tower and a longest runway of 5,400 ft.Answer
    • C.The field has runway lights that burn all night and a 5,400 ft runway.
    • D.The field has no operating control tower and a longest runway of 540 ft.

    A magenta airport symbol and data block mean the airport has no operating control tower, while blue is used where a tower operates. In the data block the two-digit figure is the length of the longest active runway in hundreds of feet, so 54 means 5,400 ft; 412 is the field elevation in feet MSL, the asterisk before the L warns of lighting limitations to be looked up in the Chart Supplement, and the circled C marks the common traffic advisory frequency.

    Source: FAA Aeronautical Chart Users' Guide, VFR charts — airport symbols and airport data blocksReport a problem with this question

  6. 6. An airport on a sectional is enclosed by a segmented blue line, and a segmented blue box beside it contains the number 25. What does this depict?

    • A.Class D airspace from the surface up to and including 2,500 ft AGL
    • B.Class D airspace from the surface up to and including 2,500 ft MSLAnswer
    • C.Class D airspace beginning at 2,500 ft MSL and extending upward from there
    • D.Class E airspace beginning at 2,500 ft MSL above that airport

    Class D airspace is charted with a blue segmented line around the airport, and the number in the segmented box is the ceiling expressed in hundreds of feet above mean sea level, so 25 means 2,500 ft MSL. Class D normally extends upward from the surface, and a minus sign in front of the figure would mean the airspace goes up to but does not include that altitude.

    Source: AIM 3-2-5 (Class D airspace); FAA Aeronautical Chart Users' GuideReport a problem with this question

  7. 7. A wide magenta shaded (fuzzy) band encircles an airport on a sectional; the airport lies inside the band and no other airspace is charted there. What is the floor of controlled airspace directly over that airport?

    • A.14,500 ft above sea level
    • B.The surface of the airport
    • C.1,200 ft above the surface
    • D.700 ft above the surfaceAnswer

    The magenta vignette marks Class E airspace whose floor is 700 ft above the surface, lowered from the usual 1,200 ft to protect aircraft on instrument approaches into that airport. A Class E surface area would instead be outlined by a dashed magenta line, and where no lower floor is designated Class E begins at 1,200 ft above the surface.

    Source: AIM 3-2-6 (Class E airspace); FAA Aeronautical Chart Users' GuideReport a problem with this question

  8. 8. A Class B sector on a chart is outlined in solid blue and labeled with 100 above a line and 40 below it. What are the vertical limits of that sector?

    • A.From 4,000 ft MSL up to 10,000 ft MSLAnswer
    • B.From 4,000 ft AGL up to 10,000 ft AGL
    • C.From 4,000 ft AGL up to 10,000 ft MSL
    • D.From 400 ft MSL up to 1,000 ft MSL

    Class B sector altitudes are printed in solid blue with the last two zeros omitted and are always referenced to mean sea level, so 100 over 40 means a ceiling of 10,000 ft MSL and a floor of 4,000 ft MSL. Reading these figures as heights above the ground is a common error; a plus sign before a floor would mean the airspace starts upward from above that altitude.

    Source: AIM 3-2-3 (Class B airspace); FAA Aeronautical Chart Users' GuideReport a problem with this question

  9. 9. You lay your plotter along the line from a VORTAC's charted compass rose to your destination and read 062 degrees. The nearest isogonic line shows 7 degrees east variation. Which statement about that measured course is correct?

    • A.It is a magnetic course; the rose is oriented to magnetic north at that station.Answer
    • B.It is a true course; the 7 degree east variation must be subtracted to make it magnetic.
    • C.It is a true course; the 7 degree east variation must be added to make it magnetic.
    • D.It is a compass course; the deviation from the correction card is already applied to it.

    The compass rose printed around a VOR or VORTAC is aligned with magnetic north at that station, not with true north, so any course measured against it is already a magnetic course and no variation correction is applied. Variation is applied only to a course measured against the chart's meridians, which are true north references, and deviation is applied afterward to a heading, never to a course.

    Source: AIM 1-1-3 (VOR); FAA-H-8083-25 Pilot's Handbook of Aeronautical Knowledge, Navigation chapterReport a problem with this question

  10. 10. When you measure the true course of a long cross-country leg with a plotter, which meridian should be used for the measurement?

    • A.The meridian where the course line crosses an isogonic line
    • B.The meridian closest to the destination airport on the leg
    • C.The meridian closest to the midpoint of the entire course lineAnswer
    • D.The meridian closest to the departure airport on the leg

    Meridians converge toward the poles, so the angle between a straight course line and successive meridians changes along the route. Measuring at the meridian nearest the midpoint splits that difference and gives the best single average value for the true course of the leg.

    Source: FAA-H-8083-25 Pilot's Handbook of Aeronautical Knowledge, Navigation chapter (measuring true course with a plotter)Report a problem with this question

  11. 11. A leg has a true course of 095 degrees. The wind requires a 10 degree correction to the right. The isogonic line shows 6 degrees east variation and the compass correction card shows 3 degrees east deviation for this heading. What compass heading should be flown?

    • A.108 degrees
    • B.085 degrees
    • C.099 degrees
    • D.096 degreesAnswer

    The chain is true course plus wind correction equals true heading, then variation gives magnetic heading, then deviation gives compass heading. A right correction makes the true heading 105 degrees, east variation is subtracted for 099 degrees magnetic, and east deviation is likewise subtracted for a compass heading of 096 degrees.

    Source: FAA-H-8083-25 Pilot's Handbook of Aeronautical Knowledge, Navigation chapter (variation and deviation)Report a problem with this question

  12. 12. You plan level VFR cruising flight more than 3,000 ft above the surface. Your true course is 196 degrees and the chart shows 6 degrees east variation. Under 14 CFR 91.159, which cruising altitude is appropriate?

    • A.6,500 ft MSLAnswer
    • B.6,000 ft MSL
    • C.7,000 ft MSL
    • D.5,500 ft MSL

    The rule is based on magnetic course, not true course, so subtract the 6 degree east variation to get a magnetic course of 190 degrees. A magnetic course of 180 through 359 degrees requires an even thousand plus 500 ft, and the requirement applies only in level cruise more than 3,000 ft above the surface up to 18,000 ft MSL.

    Source: 14 CFR 91.159 (VFR cruising altitude or flight level)Report a problem with this question

  13. 13. On a day VFR flight the planned time en route to the first point of intended landing is 2 hours 24 minutes and the airplane burns 9.5 gallons per hour in cruise. Disregarding taxi and climb allowances, what is the least usable fuel that must be aboard at takeoff to comply with 14 CFR 91.151?

    • A.32.3 gallons
    • B.29.9 gallons
    • C.22.8 gallons
    • D.27.6 gallonsAnswer

    For day VFR the rule requires enough fuel to fly to the first point of intended landing plus 30 minutes at normal cruise consumption. That is 2.4 hours times 9.5 gallons per hour, or 22.8 gallons, plus 4.75 gallons for the half hour reserve, giving about 27.6 gallons; the 45 minute reserve figure of 29.9 gallons applies only at night.

    Source: 14 CFR 91.151(a)(1) (fuel requirements for flight in VFR conditions)Report a problem with this question

  14. 14. Your true airspeed is 115 kt, the true course is 270 degrees, and the forecast wind aloft is 240 degrees true at 20 kt. Which is closest to the wind correction angle and the resulting groundspeed?

    • A.10 degree correction to the left, groundspeed about 112 kt
    • B.5 degree correction to the left, groundspeed about 133 kt
    • C.5 degree correction to the left, groundspeed about 97 ktAnswer
    • D.5 degree correction to the right, groundspeed about 97 kt

    The wind is 30 degrees off the nose from the left, which resolves into about a 10 kt crosswind component and about a 17 kt headwind component. The crosswind against 115 kt of true airspeed calls for roughly a 5 degree crab into the wind, that is to the left, and the headwind reduces groundspeed to about 97 kt.

    Source: FAA-H-8083-25 Pilot's Handbook of Aeronautical Knowledge, Navigation chapter (the wind triangle)Report a problem with this question

  15. 15. You cross checkpoint A at 1412Z and checkpoint B, 35 NM farther along the course, at 1432Z. The next checkpoint lies 49 NM beyond B. If conditions stay the same, what is the estimated time over the next checkpoint?

    • A.1500ZAnswer
    • B.1508Z
    • C.1458Z
    • D.1452Z

    Timing a known distance gives the actual groundspeed: 35 NM in 20 minutes is 105 kt. Dividing 49 NM by 105 kt gives 0.47 hour, about 28 minutes, so 1432Z plus 28 minutes is 1500Z; revising the estimate this way is how a flight log is kept current in flight.

    Source: FAA-H-8083-25 Pilot's Handbook of Aeronautical Knowledge, Navigation chapter (groundspeed check and revised ETA)Report a problem with this question

  16. 16. You are cruising with an indicated altitude of 7,500 ft, the altimeter setting is 29.42 in Hg, and the outside air temperature is plus 11 degrees C. Using 1,000 ft per 1.00 in Hg and 120 ft per degree C of temperature deviation, what are the pressure altitude and the approximate density altitude?

    • A.Pressure altitude 8,000 ft; density altitude about 9,400 ftAnswer
    • B.Pressure altitude 8,000 ft; density altitude about 6,600 ft
    • C.Pressure altitude 7,500 ft; density altitude about 9,400 ft
    • D.Pressure altitude 7,000 ft; density altitude about 8,400 ft

    An altimeter setting below 29.92 puts pressure altitude above indicated altitude, and 29.92 minus 29.42 is 0.50 in Hg, so 7,500 plus 500 gives a pressure altitude of 8,000 ft. Standard temperature there is 15 minus 16, or minus 1 degree C, so the air is 12 degrees warmer than standard and 12 times 120 ft adds about 1,440 ft for a density altitude near 9,400 ft.

    Source: FAA-H-8083-25 Pilot's Handbook of Aeronautical Knowledge, chapters on flight instruments and aircraft performance (pressure and density altitude)Report a problem with this question

  17. 17. Your calibrated airspeed is 110 kt at a density altitude of 8,000 ft. Using the rule of thumb that true airspeed exceeds calibrated airspeed by about 2 percent per 1,000 ft, which airspeed belongs in the wind triangle?

    • A.110 kt - the calibrated airspeed is what is combined with the wind
    • B.128 kt - the true airspeed is what is combined with the wind vectorAnswer
    • C.95 kt - airspeed drops 2 percent per 1,000 ft for the wind triangle
    • D.118 kt - only the altitude above 4,000 ft counts in the 2 percent rule

    Air density falls with altitude, so the airplane moves through the air faster than the indicated or calibrated value suggests: 110 kt increased by 16 percent is about 128 kt true. The wind triangle combines true airspeed with the wind vector to produce groundspeed, and using calibrated airspeed there understates groundspeed and corrupts every time and fuel figure that follows.

    Source: FAA-H-8083-25 Pilot's Handbook of Aeronautical Knowledge (airspeed definitions and true airspeed)Report a problem with this question

  18. 18. En route you tune and identify a VOR, then rotate the OBS until the CDI centers. It centers on 118 degrees with a FROM indication. Which statement describes your position?

    • A.You are on the 118 degree radial, northwest of the station
    • B.You are on the 118 degree radial, southeast of the stationAnswer
    • C.You are on the 298 degree radial, northwest of the station
    • D.You are on a magnetic bearing of 118 degrees to the station

    A radial is a magnetic course outbound from the station, so when the CDI centers with a FROM flag the OBS reading is the radial the aircraft is on, and 118 degrees lies southeast of the VOR. The reciprocal, 298 degrees, would center the needle with a TO flag, and the indication does not depend on the aircraft's heading, which is why sketching the geometry is the reliable way to answer.

    Source: AIM 1-1-3 (VOR); FAA-H-8083-25 Pilot's Handbook of Aeronautical Knowledge, Navigation chapterReport a problem with this question

  19. 19. What positively confirms that the VOR you are navigating by is the intended station and that it is usable?

    • A.The off/warning flag is out of sight and the TO-FROM flag reads steady
    • B.The frequency selected matches the one printed in the chart's navaid box
    • C.The CDI centers with a TO indication when the OBS is set to the course
    • D.The Morse code or recorded voice identification of the station is heardAnswer

    The aural identifier is the only positive identification of a VOR, because the identification is removed or replaced with a T-E-S-T coded signal whenever the facility is undergoing maintenance and is unreliable. A centered needle, a matching frequency or a flag that is out of view can all occur while the received signal is not the station you believe it to be or is not usable for navigation.

    Source: AIM 1-1-3 (VOR identification and unusable facilities)Report a problem with this question

  20. 20. You are navigating with a VFR panel-mounted GPS receiver that has no RAIM capability. What is the practical consequence for your VFR navigation?

    • A.The receiver must have a current database before it may be used under VFR
    • B.The receiver will annunciate a loss of integrity and revert to dead reckoning
    • C.The receiver can display an erroneous position without any warning to the pilotAnswer
    • D.The receiver stops computing a position with fewer than five satellites

    Receiver autonomous integrity monitoring needs at least five satellites, or four plus baro-aiding, to detect a faulty signal, and a unit without it simply cannot tell the pilot that the position it shows has gone bad. For that reason a satellite position used under VFR must be cross-checked against pilotage, dead reckoning or ground-based navaids; there is no regulatory requirement that the database be current for VFR, but an outdated one should not be trusted near critical airspace.

    Source: AIM 1-1-17 (Global Positioning System, RAIM and VFR use of GPS)Report a problem with this question

  21. 21. You decide to divert. On the sectional you draw a straight line from your present position to the alternate and it spans 21 minutes of latitude on the nearby scale. Your groundspeed on the new heading will be about 90 kt. What distance and time en route should you plan?

    • A.21 NM and about 23 minutes
    • B.21 NM and about 14 minutesAnswer
    • C.21 SM and about 14 minutes
    • D.24 NM and about 16 minutes

    One minute of latitude equals one nautical mile, so the latitude scale along a meridian is the quickest in-flight ruler and 21 minutes is 21 NM. At 90 kt that leg takes 21 divided by 90 of an hour, about 14 minutes; the recommended technique is to turn toward the alternate on an estimated heading first and refine the computation once the airplane is going the right way.

    Source: FAA-H-8083-25 Pilot's Handbook of Aeronautical Knowledge, Navigation chapter (diversion; latitude scale distance measurement)Report a problem with this question

  22. 22. You become uncertain of your position over unfamiliar terrain in good visibility. Why is climbing recommended as the first of the four C's?

    • A.Climbing widens the area of visible landmarks and improves radio and radar receptionAnswer
    • B.Climbing reduces fuel flow enough to extend endurance while position is resolved
    • C.Climbing is required before the transponder may be set to the 7700 emergency code
    • D.Climbing places the airplane in controlled airspace where radar service is given

    Radio and radar coverage are line of sight, so altitude brings more ground features into view, extends VHF range to reach a facility that can help, makes the airplane visible to radar sooner, and increases gliding distance if the engine quits. Once established higher the pilot communicates, confesses the situation and complies with the instructions received.

    Source: FAA-H-8083-25 Pilot's Handbook of Aeronautical Knowledge, Navigation chapter (lost procedures)Report a problem with this question

Practice questions based on the aeronautical knowledge areas codified at 14 CFR 61.105(b) and on FAA handbooks (the Pilot's Handbook of Aeronautical Knowledge, the Airplane Flying Handbook) and the Aeronautical Information Manual. This site is not affiliated with or endorsed by the FAA. Taking the real knowledge test requires an endorsement from an authorized instructor or evidence of completing a ground training course. Charts, weather products and aircraft performance data are republished on a cycle — always fly and test from current official sources. About FAA airman testing →