22 Performance & Weight and Balance Practice Questions & Answers

1. 1. An airport has a field elevation of 4,200 feet MSL. The automated weather station reports an altimeter setting of 29.70 in Hg, and the conversion table in the airplane's handbook shows a pressure altitude conversion factor of +205 feet for that setting. What is the pressure altitude at the field?

   • A.4,405 feet✓ Answer
   • B.4,420 feet
   • C.3,995 feet
   • D.4,200 feet

   Pressure altitude is the altitude indicated when the altimeter is set to 29.92 in Hg. When the reported setting is below 29.92 the conversion factor is ADDED to field elevation, so 4,200 + 205 = 4,405 feet. Estimating roughly at 1,000 feet per inch of mercury yields 4,420; subtracting the factor reverses the sign rule; using field elevation alone skips the correction entirely.

   Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 11 (Aircraft Performance) — pressure altitude conversionReport a problem with this question

2. 2. At one particular airport, which combination of conditions produces the HIGHEST density altitude?

   • A.A temperature well above standard, an altimeter setting below 29.92 in Hg, and high humidity✓ Answer
   • B.A temperature well below standard, an altimeter setting above 29.92 in Hg, and low humidity
   • C.A temperature well below standard, an altimeter setting below 29.92 in Hg, and high humidity
   • D.A temperature well above standard, an altimeter setting above 29.92 in Hg, and low humidity

   Density altitude rises whenever air density falls. High temperature expands the air, a low altimeter setting means low atmospheric pressure, and high humidity replaces heavier dry-air molecules with lighter water vapor. All three thin the air, so combining them gives the highest density altitude and the poorest performance.

   Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 11 — density altitudeReport a problem with this question

3. 3. An airplane departs a high-elevation airport on a hot afternoon, where the density altitude is far above the field elevation. Compared with a standard sea level day, what should the pilot expect and why?

   • A.The takeoff roll lengthens but the climb rate is unaffected, because only ground acceleration suffers
   • B.The takeoff roll shortens and the climb rate increases, because thin air reduces drag and speeds up acceleration
   • C.The takeoff roll and climb rate are unchanged, because the airspeed indicator compensates for density
   • D.The takeoff roll lengthens and the climb rate decreases, because thin air cuts engine power, propeller thrust and available lift✓ Answer

   Thin air contains fewer molecules per unit volume, so the engine draws in less air and produces less power, the propeller generates less thrust, and the wing must reach a higher true airspeed to produce the same lift. Acceleration is slower, the ground roll is longer, and the reduced excess power lowers the rate of climb.

   Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 11 — effects of density altitude on performanceReport a problem with this question

4. 4. A handbook chart gives density altitude at a pressure altitude of 6,000 feet as follows: OAT 0 °C — 5,900 ft; 10 °C — 7,150 ft; 20 °C — 8,400 ft; 30 °C — 9,650 ft. The airplane is at a pressure altitude of 6,000 feet with an outside air temperature of 25 °C. What is the approximate density altitude?

   • A.7,775 feet
   • B.8,400 feet
   • C.9,025 feet✓ Answer
   • D.9,650 feet

   25 °C lies exactly halfway between the 20 °C and 30 °C rows, so the value must be interpolated: 8,400 + (9,650 − 8,400) ÷ 2 = 8,400 + 625 = 9,025 feet. Rounding down or up to a charted row gives 8,400 or 9,650, and 7,775 comes from interpolating between the 10 °C and 20 °C rows instead.

   Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 11 — density altitude chartsReport a problem with this question

5. 5. An airplane's takeoff data, for a gross weight of 2,400 pounds on a paved, level, dry runway with zero wind, show: pressure altitude 2,000 ft at 20 °C — ground roll 1,000 ft, total distance to clear a 50-foot obstacle 1,800 ft; pressure altitude 4,000 ft at 20 °C — ground roll 1,270 ft, total distance to clear 50 feet 2,300 ft. A note states: decrease the distances by 10% for each 9 knots of headwind. Departing at 2,400 pounds from a paved runway at 4,000 feet pressure altitude and 20 °C with an 18-knot headwind, what total distance is required to clear a 50-foot obstacle?

   • A.2,300 feet
   • B.2,070 feet
   • C.1,840 feet✓ Answer
   • D.1,610 feet

   An 18-knot headwind is two full 9-knot increments, so the note removes 20% of the charted figure: 2,300 − 460 = 1,840 feet. Crediting only one increment gives 2,070, ignoring the wind leaves 2,300, and crediting three increments gives 1,610. The wind credit is applied to the charted value for the correct pressure altitude and temperature line.

   Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 11 — takeoff distance charts and correction notesReport a problem with this question

6. 6. A handbook chart shows that an airplane needs 1,100 feet of ground roll at its present weight and at the airport's density altitude of 6,500 feet. The available runway is 1,800 feet of dry grass, the afternoon is hot, and the engine has nearly reached its overhaul time. What is the soundest decision?

   • A.Depart now but abort the takeoff at mid-field, since dry grass adds only a negligible roll
   • B.Delay until the cooler evening or offload weight, because the chart assumes a new engine and a dry paved runway✓ Answer
   • C.Depart as planned, since the runway available is over 60 percent longer than the charted ground roll
   • D.Depart as planned, because published chart figures already contain a built-in safety margin

   Performance charts are flight-tested with a new airplane, a properly leaned engine and precise technique on a smooth, level, dry, paved runway. A worn engine, ordinary piloting and a grass surface all add distance, and typical grass corrections alone run 15–20%, so the 700-foot apparent margin can disappear. The pilot must add a personal margin rather than treat the chart number as achievable.

   Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 11 — chart assumptions and runway surface effectsReport a problem with this question

7. 7. A landing distance table for 1,600 pounds, full flaps, power off, a hard-surface runway and zero wind shows: sea level at 59 °F — ground roll 445 ft, total to clear a 50-foot obstacle 1,075 ft; 2,500 ft at 50 °F — 470 ft and 1,135 ft; 5,000 ft at 41 °F — 495 ft and 1,195 ft. A note states: for a dry grass runway, increase both the ground roll and the total distance by 20% of the 'total to clear 50 ft' figure. Landing at a pressure altitude of 5,000 feet at 41 °F with no wind on dry grass, what ground roll should be expected?

   • A.734 feet✓ Answer
   • B.1,434 feet
   • C.495 feet
   • D.594 feet

   The note is deliberately worded around the total-to-clear-50-ft figure, not the ground roll: 20% of 1,195 is 239 feet, which is added to the charted ground roll, giving 495 + 239 = 734 feet. Taking 20% of the 495-foot ground roll produces the classic wrong answer of 594; 1,434 is the corrected total distance; 495 ignores the note.

   Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 11 — landing distance tables and correction notesReport a problem with this question

8. 8. A landing distance table for 1,600 pounds, full flaps, power off, a hard-surface runway and zero wind shows: sea level at 59 °F — ground roll 445 ft, total to clear a 50-foot obstacle 1,075 ft; 2,500 ft at 50 °F — 470 ft and 1,135 ft; 5,000 ft at 41 °F — 495 ft and 1,195 ft. A note states: decrease the distances by 10% for each 4 knots of headwind. Landing at sea level at 59 °F on a hard-surface runway with an 8-knot headwind, what total distance is required to clear a 50-foot obstacle?

   • A.1,075 feet
   • B.968 feet
   • C.860 feet✓ Answer
   • D.645 feet

   An 8-knot headwind is two complete 4-knot increments, so 20% is removed from the charted total: 1,075 − 215 = 860 feet. Crediting a single increment gives 968, crediting three gives 645, and ignoring the wind leaves 1,075. The credit is given only for whole increments named in the note, not pro-rated per knot.

   Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 11 — landing distance tables and correction notesReport a problem with this question

9. 9. The tower reports the wind as 260° at 20 knots and the airplane will depart from Runway 22. A crosswind component graph shows that for a 40° angle between the wind and the runway, the crosswind component is about 64% of the wind velocity and the headwind component about 77%. What is the approximate crosswind component?

   • A.15 knots
   • B.10 knots
   • C.20 knots
   • D.13 knots✓ Answer

   Runway 22 has a magnetic heading of about 220°, so the angle between the 260° wind and the runway is 40°. The crosswind component is 0.64 × 20 = 12.8, or about 13 knots; 0.77 × 20 ≈ 15 knots is the headwind component, not the crosswind. Assuming half the wind gives 10 knots, and treating the whole wind as crosswind gives 20 knots.

   Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 11 (crosswind components) and Chapter 14 (runway numbering by magnetic heading)Report a problem with this question

10. 10. An airplane is landing on Runway 09 and the tower reports the wind as 180° at 15 knots. What are the headwind and crosswind components?

    • A.Headwind 0 knots, crosswind 15 knots✓ Answer
    • B.Headwind 13 knots, crosswind 7 knots
    • C.Headwind 11 knots, crosswind 11 knots
    • D.Headwind 15 knots, crosswind 0 knots

    Runway 09 has a magnetic heading of about 090°, and a wind from 180° is exactly 90° from that heading, so the entire 15 knots acts as crosswind and there is no headwind component. Both the runway number and the tower-reported wind are referenced to magnetic north, so no variation correction is applied before comparing them.

    Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 11 (crosswind components) and Chapter 14 (runway numbering by magnetic heading)Report a problem with this question

11. 11. What does an airplane's published 'maximum demonstrated crosswind component' represent?

    • A.The strongest crosswind permitted for student pilots without an endorsement
    • B.The strongest crosswind permitted by regulation for any pilot flying that type
    • C.The strongest crosswind at which the airplane can still be controlled on landing
    • D.The strongest crosswind actually flown during certification testing, not an operating limit✓ Answer

    The figure records the strongest crosswind a company test pilot actually handled during certification flight testing. It is information, not a certificated limitation, so it neither guarantees controllability at that value for an average pilot nor legally forbids operation above it; the pilot in command must judge the conditions and personal capability.

    Source: FAA Airplane Flying Handbook, Chapter 9 — crosswind approach and landingReport a problem with this question

12. 12. A cruise power setting table for 65% power at 2,450 RPM shows, at a pressure altitude of 6,000 feet: ISA −20 °C — 21.0 in Hg, 11.5 gal/hr, TAS 155 knots; standard day — 21.5 in Hg, 11.5 gal/hr, TAS 160 knots; ISA +20 °C — 22.0 in Hg, 11.5 gal/hr, TAS 165 knots. The airplane is cruising at 6,000 feet pressure altitude with an outside air temperature of +23 °C. Which block applies and what true airspeed should be expected?

    • A.The ISA −20 °C block; expect a true airspeed of about 155 knots
    • B.The ISA +20 °C block; expect a true airspeed of about 155 knots
    • C.The standard-day block; expect a true airspeed of about 160 knots
    • D.The ISA +20 °C block; expect a true airspeed of about 165 knots✓ Answer

    Standard temperature is 15 °C at sea level and falls about 2 °C per 1,000 feet, so at 6,000 feet it is 15 − 12 = +3 °C. An outside air temperature of +23 °C is therefore 20 °C above standard, which selects the ISA +20 °C block and a true airspeed of 165 knots. Entering the table by raw temperature instead of by deviation from standard is the usual error.

    Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 11 — cruise power setting tables and the standard atmosphere lapse rateReport a problem with this question

13. 13. In airplane weight and balance work, which statement correctly describes the datum, the arm and the moment?

    • A.The datum is always the wing leading edge, the arm is measured in feet, and the moment is weight plus arm
    • B.The datum is an arbitrary reference plane, the arm is the distance in inches from it, and the moment is weight times arm✓ Answer
    • C.The datum is an arbitrary reference plane, the arm is measured in pounds, and the moment is the arm divided by the weight
    • D.The datum is always the engine firewall, the arm is the distance in inches from it, and the moment is weight times arm

    The datum is an arbitrary vertical reference plane chosen by the manufacturer, which may lie at the nose, at the firewall, or even ahead of the airplane. An arm is the horizontal distance in inches from that plane, positive aft and negative forward, and a moment is weight multiplied by arm, expressed in pound-inches. Center of gravity is then total moment divided by total weight.

    Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 10 — Weight and Balance terminologyReport a problem with this question

14. 14. An airplane has a maximum certificated gross weight of 2,550 pounds and a basic empty weight of 1,750 pounds, which already includes full engine oil and unusable fuel. The pilot and passengers weigh 500 pounds and the baggage weighs 60 pounds. Usable fuel capacity is 48 gallons of avgas at 6 pounds per gallon. What is the maximum fuel that may be loaded without exceeding gross weight?

    • A.48 gallons
    • B.32 gallons
    • C.44 gallons
    • D.40 gallons✓ Answer

    Useful load is maximum gross weight minus basic empty weight: 2,550 − 1,750 = 800 pounds. Removing 500 pounds of occupants and 60 pounds of baggage leaves 240 pounds for fuel, and 240 ÷ 6 = 40 gallons. Converting at 7.5 pounds per gallon, which is the weight of oil rather than avgas, gives the wrong answer of 32 gallons, and full tanks of 48 gallons would weigh 288 pounds and exceed gross weight by 48 pounds.

    Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 10 — standard weights and useful loadReport a problem with this question

15. 15. An airplane's loading data give a basic empty weight of 2,015 pounds with a moment/100 of 1,554, and these arms: front seats 85, rear seats 121, baggage 140, main fuel tanks 75. Avgas weighs 6 pounds per gallon and maximum takeoff weight is 2,950 pounds. The moment limits versus weight table shows: 2,930 lb — minimum 2,405, maximum 2,483; 2,940 lb — minimum 2,412, maximum 2,489; 2,950 lb — minimum 2,419, maximum 2,495. The airplane is loaded with 200 pounds in the front seats, 400 pounds in the rear seats, 85 pounds of baggage and 40 gallons of fuel. What is the loading condition?

    • A.Over the maximum takeoff weight, though the moment is within limits
    • B.Within the weight limit, but the moment is below the minimum for that weight
    • C.Within the weight limit, but the moment exceeds the maximum for that weight✓ Answer
    • D.Within both the maximum weight and the moment limits for that weight

    Fuel weighs 40 × 6 = 240 pounds, so total weight is 2,015 + 200 + 400 + 85 + 240 = 2,940 pounds, which is under the 2,950-pound maximum. The moments/100 total 1,554 + 170 + 484 + 119 + 180 = 2,507, above the 2,489 maximum listed for 2,940 pounds, so the CG is aft of the aft limit. Confirming that weight is legal and stopping there is the classic error.

    Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 10 — moment index and loading table methodReport a problem with this question

16. 16. An airplane is loaded as follows: empty weight 1,400 pounds at an arm of 38.0 inches; front seats 340 pounds at 37.0 inches; rear seats 170 pounds at 73.0 inches; baggage 40 pounds at 95.0 inches; and 38 gallons of avgas at an arm of 48.0 inches, at 6 pounds per gallon. Where is the center of gravity?

    • A.42.7 inches aft of datum✓ Answer
    • B.58.2 inches aft of datum
    • C.51.1 inches aft of datum
    • D.42.2 inches aft of datum

    Fuel weighs 38 × 6 = 228 pounds, so total weight is 2,178 pounds and total moment is 53,200 + 12,580 + 12,410 + 3,800 + 10,944 = 92,934 pound-inches; the CG is 92,934 ÷ 2,178 = 42.7 inches. Leaving fuel in gallons gives 42.2, omitting the airplane's own weight and moment gives 51.1, and simply averaging the five arms gives 58.2.

    Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 10 — computation method (CG = total moment ÷ total weight)Report a problem with this question

17. 17. An airplane weighs 2,800 pounds with its center of gravity at 89.5 inches aft of datum, and the aft CG limit is 88.5 inches. The pilot moves a 60-pound bag from the rear baggage compartment at an arm of 150 inches to the front baggage compartment at an arm of 100 inches. Where is the new center of gravity?

    • A.90.6 inches aft of datum
    • B.87.4 inches aft of datum
    • C.88.4 inches aft of datum✓ Answer
    • D.89.5 inches aft of datum

    The CG shift equals weight moved times distance moved divided by total weight: (60 × 50) ÷ 2,800 = 1.07 inches. Because the bag moved forward, the CG moves forward as well: 89.5 − 1.07 = 88.4 inches, which is just inside the 88.5-inch aft limit. Using the destination arm of 100 inches instead of the 50-inch distance gives 87.4, and adding rather than subtracting gives 90.6.

    Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 10 — weight-shift formulaReport a problem with this question

18. 18. An airplane is loaded near its aft CG limit for takeoff, and its fuel tanks are at an arm of 75 inches, well forward of the loaded center of gravity. What happens to the center of gravity during the flight, and what does that mean for the pilot?

    • A.It moves forward as fuel burns, so a takeoff check within limits also covers the landing
    • B.It stays where it is, because fuel is drawn symmetrically from both wing tanks
    • C.It moves aft as fuel burns, so the weight and balance for the landing condition must also be checked✓ Answer
    • D.It moves aft as fuel burns, but only when the airplane is above maximum landing weight

    Burning fuel removes weight acting at the tanks' arm. When that arm is forward of the current CG, removing the weight shifts the balance point aft, exactly as removing a forward load would. An airplane legally loaded at takeoff can therefore drift beyond the aft limit before landing, which is why the landing condition must be computed as well.

    Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 10 — effect of fuel burn on center of gravityReport a problem with this question

19. 19. An airplane is loaded so that its center of gravity is at or ahead of the forward limit. What effects should the pilot expect?

    • A.Lighter elevator control forces, a higher stall speed and reduced longitudinal stability
    • B.Heavier elevator control forces, a higher stall speed and a longer takeoff roll✓ Answer
    • C.Lighter elevator control forces, a lower stall speed and a shorter takeoff roll
    • D.Heavier elevator control forces, a lower stall speed and a higher cruise speed

    A forward CG lengthens the arm between the CG and the tail, so the horizontal stabilizer must produce more download to balance the airplane. The wing therefore carries the aircraft weight plus that download, raising the stall speed, and heavier elevator forces delay rotation and lengthen the takeoff roll. In the extreme the elevator may lack the authority to flare for landing.

    Source: FAA Pilot's Handbook of Aeronautical Knowledge, Chapter 10 — effects of a forward center of gravityReport a problem with this question

20. 20. Under 14 CFR 91.103, which preflight information must the pilot in command obtain before EVERY flight, including a short local flight that stays in the traffic pattern?

    • A.Any known air traffic control delays affecting the airports of intended use
    • B.Alternatives available if the planned flight cannot be completed as intended
    • C.Runway lengths at airports of intended use, with takeoff and landing distance data✓ Answer
    • D.Weather reports and forecasts for the route and the destination airport

    Section 91.103 requires familiarity with all available information concerning the flight, then lists two categories. Weather, fuel requirements, alternatives and known ATC delays are required for IFR flights and for flights not in the vicinity of an airport. Runway lengths at airports of intended use, together with takeoff and landing distance data, are required for any flight, local pattern work included.

    Source: 14 CFR 91.103 (Preflight action)Report a problem with this question

21. 21. Under 14 CFR 91.151, how much fuel must an airplane carry to begin a flight in VFR conditions?

    • A.Enough to reach the first point of intended landing plus 5 gallons of reserve by day and 10 gallons at night
    • B.Enough to reach the first point of intended landing plus 30 minutes at cruise by day and 45 minutes at night✓ Answer
    • C.Enough to reach the first point of intended landing plus 45 minutes at cruise by day and 30 minutes at night
    • D.Enough to reach the first point of intended landing plus one hour at normal cruising speed by day or night

    The rule expresses the reserve in time flown at normal cruising speed rather than in gallons, because the gallons needed depend on the airplane and power setting. For airplanes the reserve is 30 minutes by day and 45 minutes at night, and the whole computation must consider wind and forecast weather conditions along the route.

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

22. 22. A daytime VFR cross-country leg is 245 nautical miles, the planned groundspeed is 105 knots, and the airplane burns 9.2 gallons per hour in cruise. Applying the reserve required by 14 CFR 91.151, what is the minimum usable fuel that must be on board at takeoff?

    • A.21.5 gallons
    • B.30.7 gallons
    • C.28.4 gallons
    • D.26.1 gallons✓ Answer

    Time en route is 245 ÷ 105 = 2.33 hours, so the leg needs 2.33 × 9.2 = 21.5 gallons. The daytime VFR reserve of 30 minutes at cruise adds 0.5 × 9.2 = 4.6 gallons, giving 26.1 gallons minimum. Stopping at 21.5 omits the reserve, 28.4 applies the 45-minute night reserve, and 30.7 applies a full hour.

    Source: 14 CFR 91.151(a) (Fuel requirements for flight in VFR conditions)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 →