22 Weight & Balance, Math & Physics Practice Questions & Answers
Every Weight & Balance, Math & Physics practice question from the FAA A&P Mechanic (General) Practice Test, with the correct answer and a short explanation.
Start practice test →1. On an aircraft weight and balance problem, where does a mechanic find the location of the datum for a particular aircraft?
- A.Measured from the nose of the aircraft by the mechanic
- B.In the pilot's information manual carried in the cockpit
- C.Stamped on the aircraft data plate
- D.In the Type Certificate Data Sheet or Aircraft Specifications✓ Answer
The datum is an imaginary vertical plane designated by the manufacturer, not chosen or measured by the mechanic; it may be at the nose, the firewall, the wing leading edge, or even a point ahead of the aircraft. Its exact location, along with the CG range and leveling means, is published in the TCDS or Aircraft Specifications. The data plate carries only identification data, and the pilot's manual is not the certificated source.
Source: FAA-H-8083-1 Ch. 2 (Datum); FAA-H-8083-30 Ch. 6Report a problem with this question
2. Which combination of items is included in the empty weight of an aircraft?
- A.Airframe, powerplant, installed equipment, and full usable fuel
- B.Airframe, powerplant, installed equipment, fixed ballast, and residual (unusable) fuel and oil✓ Answer
- C.Airframe and powerplant only, with all fluids and equipment excluded
- D.Airframe, powerplant, installed equipment, temporary ballast, and crew
Empty weight is the weight of the airframe, powerplant, all required and optional installed equipment, fixed (permanent) ballast, hydraulic fluid, and the residual or trapped fuel and oil that cannot be drained. Usable fuel, crew, payload, and temporary ballast are never part of empty weight because they are removable useful load. For most aircraft certificated after 1978 the full engine oil capacity is also included.
Source: FAA-H-8083-1 Ch. 2 (Empty Weight); FAA-H-8083-30 Ch. 6Report a problem with this question
3. An aircraft has a maximum allowable gross weight of 3,400 lb and an empty weight of 2,150 lb. What is its useful load?
- A.5,550 lb
- B.1,250 lb✓ Answer
- C.3,400 lb
- D.2,150 lb
Useful load is defined as maximum allowable gross weight minus empty weight, so 3,400 - 2,150 = 1,250 lb. It consists of usable fuel, fluids not counted in empty weight, crew, passengers, and baggage. A common error is subtracting from zero fuel weight or adding the two figures together.
Source: FAA-H-8083-1 Ch. 2 (Useful Load)Report a problem with this question
4. An airplane is leveled and weighed on three scales. The nosewheel scale reads 315 lb, which includes 15 lb of tare from a chock; the nosewheel arm is +20.0 in. The left main reads 852 lb and the right main reads 848 lb with no tare, both at an arm of +80.0 in. What is the empty-weight CG?
- A.+71.0 in aft of datum✓ Answer
- B.+74.5 in aft of datum
- C.+50.0 in aft of datum
- D.+70.6 in aft of datum
Tare must be subtracted from the scale reading before any moment is computed, so the nose reaction is 315 - 15 = 300 lb. Moments: 300 x 20.0 = 6,000 in-lb and (852 + 848) x 80.0 = 136,000 in-lb, giving 142,000 in-lb total over 2,000 lb total weight. CG = total moment / total weight = 142,000 / 2,000 = +71.0 in. Leaving the 15 lb of tare in the calculation yields the incorrect +70.6 in.
Source: FAA-H-8083-1 Ch. 3 (Tare Weight, EWCG)Report a problem with this question
5. An aircraft weighs 6,400 lb with its CG at station +78.0. The aft CG limit is +76.5. How much cargo must be moved from station +150.0 to station +30.0 to bring the CG exactly to the aft limit?
- A.320 lb
- B.120 lb
- C.80 lb✓ Answer
- D.64 lb
The weight-shift formula is weight shifted / total weight = change in CG / distance shifted, rearranged to weight to shift = (total weight x delta CG) / distance shifted. Delta CG is 78.0 - 76.5 = 1.5 in and the shift distance is 150.0 - 30.0 = 120 in, so (6,400 x 1.5) / 120 = 80 lb. Using a station number instead of the distance between stations is the classic error that produces 64 lb or 320 lb.
Source: FAA-H-8083-1 Ch. 4 (Shifting Weight)Report a problem with this question
6. A loaded airplane weighs 4,000 lb with the CG at station +38.0, but the aft CG limit is +37.0. Temporary ballast can be installed at station -13.0. How much ballast is required to bring the CG to the aft limit?
- A.160.0 lb
- B.78.4 lb
- C.80.0 lb✓ Answer
- D.40.0 lb
Ballast weight = (loaded weight x distance the CG is out of limits) / (distance between the ballast arm and the affected limit). The CG is 1.0 in beyond the aft limit and the ballast arm is 37.0 - (-13.0) = 50 in from that limit, so 4,000 x 1.0 / 50 = 80 lb. Checking: new weight 4,080 lb and new moment (4,000 x 38.0) + (80 x -13.0) = 150,960 in-lb gives exactly +37.0. Dividing by the distance from the CG rather than from the limit gives the incorrect 78.4 lb, and note that moving the ballast even farther from the limit would reduce the weight required.
Source: FAA-H-8083-1 Ch. 4 (Ballast)Report a problem with this question
7. When performing an adverse-loaded CG check on a reciprocating-engine airplane whose engine is rated at 340 METO horsepower, what weight of fuel must be used as the minimum fuel?
- A.28.3 lb
- B.170 lb✓ Answer
- C.204 lb
- D.340 lb
For reciprocating-engine aircraft, minimum fuel is one-twelfth gallon per METO horsepower, which at 6 lb per gallon for avgas simplifies to METO horsepower divided by 2 expressed in pounds: 340 / 2 = 170 lb. The figure 28.3 is the quantity in gallons, not pounds. This rule does not apply to turbine aircraft, for which the manufacturer supplies the minimum fuel figure.
Source: FAA-H-8083-1 Ch. 4 (Minimum Fuel)Report a problem with this question
8. An airplane has a MAC of 80.0 in with the leading edge of the MAC (LEMAC) at station +120.0. If the loaded CG is at station +140.0, what is the CG expressed in percent MAC?
- A.175 percent MAC
- B.20 percent MAC
- C.58.3 percent MAC
- D.25 percent MAC✓ Answer
Percent MAC = ((CG - LEMAC) / MAC) x 100. Percent MAC is measured aft of the leading edge of the MAC, not from the datum, so the distance used is 140.0 - 120.0 = 20.0 in, and 20.0 / 80.0 x 100 = 25 percent. Dividing the raw station number by the MAC length yields the meaningless 175 percent.
Source: FAA-H-8083-1 Ch. 2 (Mean Aerodynamic Chord)Report a problem with this question
9. What is the principal effect of loading an airplane so that its center of gravity is aft of the rearward CG limit?
- A.Structural strength of the wing spar is reduced
- B.Longitudinal stability is reduced and recovery from a stall may be impossible✓ Answer
- C.The airplane becomes nose heavy and may not rotate for takeoff
- D.Stall speed increases and elevator control forces become excessively heavy
With the CG aft of limits the tail download needed for balance is reduced and the moment arm to the horizontal stabilizer becomes too short, so the airplane loses longitudinal (pitch) stability and may pitch up uncontrollably, making stall recovery impossible. Nose heaviness, high stall speed, and heavy elevator forces are the symptoms of a CG that is too far forward, which is the opposite condition.
Source: FAA-H-8083-1 Ch. 1 (Effects of Weight and Balance)Report a problem with this question
10. Evaluate the expression: the quantity (-5 + 2 cubed) multiplied by (-2), plus the quantity (3 - 3) multiplied by the square root of 64, with the entire result divided by 3.
- A.-2✓ Answer
- B.-6
- C.2
- D.-18
Order of operations requires parentheses first, then exponents and roots, then multiplication and division left to right, then addition and subtraction. Inside the parentheses, -5 + 8 = 3 and 3 - 3 = 0. Then 3 x (-2) = -6 and 0 x 8 = 0, so the numerator is -6, and -6 / 3 = -2. Answer -6 comes from skipping the final division and +2 from mishandling the negative sign.
Source: FAA-H-8083-30 Ch. 3 (Order of Operations)Report a problem with this question
11. Which is the correct expression of the number 0.00004631 in scientific notation?
- A.4.631 x 10 to the -4 power
- B.4.631 x 10 to the -5 power✓ Answer
- C.4.631 x 10 to the 5th power
- D.4.631 to the -5 power
Scientific notation writes a number as a coefficient between 1 and 10 multiplied by a power of ten. The decimal point must move five places to the right to turn 0.00004631 into 4.631, and moving the point to the right gives a negative exponent, so the value is 4.631 x 10 to the -5. A positive exponent would make the number larger than one, and raising 4.631 itself to a power is a different operation entirely.
Source: FAA-H-8083-30 Ch. 3 (Scientific Notation)Report a problem with this question
12. A sheet-metal doubler is shaped as a trapezoid with parallel sides of 14 in and 10 in and a perpendicular height of 6 in. What is its area?
- A.144 square inches
- B.84 square inches
- C.60 square inches
- D.72 square inches✓ Answer
The area of a trapezoid is one-half the sum of the two parallel sides multiplied by the height: 0.5 x (14 + 10) x 6 = 0.5 x 24 x 6 = 72 square inches. Multiplying only one parallel side by the height (84 or 60) treats the shape as a rectangle, and omitting the one-half factor gives 144.
Source: FAA-H-8083-30 Ch. 3 (Area of a Trapezoid)Report a problem with this question
13. A six-cylinder engine has a cylinder bore of 5.5 in and a stroke of 4.5 in. What is the total piston displacement of the engine?
- A.148.5 cubic inches
- B.106.9 cubic inches
- C.2,565.9 cubic inches
- D.641.5 cubic inches✓ Answer
Piston displacement is the volume of a cylinder, piston area times stroke, and total displacement is that value times the number of cylinders. Piston area = 0.7854 x diameter squared = 0.7854 x 30.25 = 23.76 square inches; 23.76 x 4.5 = 106.9 cubic inches per cylinder; 106.9 x 6 = 641.5 cubic inches. Using the bore as the radius instead of the diameter quadruples the answer to about 2,566 cubic inches.
Source: FAA-H-8083-30 Ch. 3 (Volume of a Cylinder)Report a problem with this question
14. The total volume of an engine cylinder with the piston at bottom dead center is 96 cubic inches, and the volume with the piston at top dead center is 12 cubic inches. What is the compression ratio?
- A.8 to 1✓ Answer
- B.12 to 1
- C.7 to 1
- D.9.6 to 1
A ratio compares two quantities of the same units, and compression ratio is the total cylinder volume with the piston at bottom dead center divided by the volume with the piston at top dead center: 96 / 12 = 8, expressed as 8 to 1. Subtracting the two volumes first and then dividing (84 / 12 = 7) is a common error because it uses swept volume rather than total volume.
Source: FAA-H-8083-30 Ch. 3 (Ratio, Compression Ratio)Report a problem with this question
15. A 36-tooth drive gear turning at 1,200 rpm meshes with a 20-tooth driven gear. At what speed does the driven gear turn?
- A.2,400 rpm
- B.667 rpm
- C.2,160 rpm✓ Answer
- D.1,200 rpm
Gear speeds are inversely proportional to tooth counts, so the proportion is driven speed / drive speed = drive teeth / driven teeth. Solving, driven speed = 1,200 x 36 / 20 = 2,160 rpm. The smaller gear must turn faster to keep the number of teeth passing the mesh point equal; inverting the proportion gives the incorrect 667 rpm.
Source: FAA-H-8083-30 Ch. 3 (Ratio and Proportion)Report a problem with this question
16. A hoist raises a 1,650 lb engine a vertical distance of 20 ft in 30 seconds. How much horsepower is being developed?
- A.0.5 horsepower
- B.4 horsepower
- C.2 horsepower✓ Answer
- D.1 horsepower
Work equals force times distance: 1,650 x 20 = 33,000 ft-lb. Power is work divided by time, so 33,000 ft-lb in 30 seconds equals 1,100 ft-lb per second. Since one horsepower is defined as 550 ft-lb per second (or 33,000 ft-lb per minute), 1,100 / 550 = 2 horsepower.
Source: FAA-H-8083-30 Ch. 5 (Work, Power, Horsepower)Report a problem with this question
17. Which statement correctly describes a third-class lever, such as a retractable landing gear actuated by a hydraulic cylinder?
- A.The effort is applied between the fulcrum and the resistance, and mechanical advantage is always less than 1✓ Answer
- B.The resistance is between the fulcrum and the effort, and mechanical advantage is always greater than 1
- C.Mechanical advantage is always exactly 2 because the effort arm is twice the resistance arm
- D.The fulcrum is between the effort and the resistance, and mechanical advantage is always 1
In a third-class lever the effort is applied between the fulcrum and the resistance, so the effort arm is always shorter than the resistance arm. Since mechanical advantage equals effort arm divided by resistance arm, it is always less than one, meaning more force is required than the load, but the load moves farther and faster. A fulcrum between effort and resistance describes a first-class lever, and resistance in the middle describes a second-class lever.
Source: FAA-H-8083-30 Ch. 5 (Simple Machines, Levers)Report a problem with this question
18. In a hydraulic system, a force of 90 lb is applied to an input piston having an area of 1.5 square inches. What force is developed at an output piston with an area of 12 square inches?
- A.60 lb
- B.1,080 lb
- C.720 lb✓ Answer
- D.11.25 lb
Pascal's law states that pressure applied to a confined fluid acts undiminished and equally in all directions, so the system pressure is the same at both pistons. Pressure = force / area = 90 / 1.5 = 60 psi, and force at the output = pressure x area = 60 x 12 = 720 lb. Multiplying the input force by the output area alone (1,080 lb) ignores the input piston area and is the classic error.
Source: FAA-H-8083-30 Ch. 5 (Pascal's Law)Report a problem with this question
19. According to Bernoulli's principle, what happens to the air flowing through the restricted throat of a venturi?
- A.Velocity decreases and static pressure increases
- B.Velocity increases and static pressure decreases✓ Answer
- C.Both velocity and static pressure decrease
- D.Both velocity and static pressure increase
The same mass of air must pass every station in the tube each second, so where the cross-section is reduced the air must speed up. Bernoulli's principle states that the total energy in the stream is constant, so the increase in velocity (kinetic energy) must be paid for by a drop in static pressure. In the diverging section downstream, velocity falls again and static pressure is recovered.
Source: FAA-H-8083-30 Ch. 5 (Bernoulli's Principle)Report a problem with this question
20. An aircraft tire is inflated to 45 psig at 60 degrees F. If the tire is then heated to 120 degrees F and its volume does not change, what is the new gauge pressure?
- A.66.6 psig
- B.90.0 psig
- C.50.2 psig
- D.51.9 psig✓ Answer
Gas law problems must be worked in absolute pressure and absolute temperature. Convert: 45 + 14.7 = 59.7 psia; 60 + 460 = 520 R and 120 + 460 = 580 R. At constant volume the general gas law reduces to P1/T1 = P2/T2, so P2 = 59.7 x 580 / 520 = 66.6 psia. Converting back to gauge, 66.6 - 14.7 = 51.9 psig. Working in gauge pressure gives 50.2, and working in degrees F gives the wildly wrong 90.0.
Source: FAA-H-8083-30 Ch. 5 (General Gas Law, Absolute Pressure)Report a problem with this question
21. A temperature of -40 degrees F is equal to how many degrees Rankine?
- A.-40 degrees R
- B.233 degrees R
- C.420 degrees R✓ Answer
- D.500 degrees R
The Rankine scale is the absolute scale that uses Fahrenheit-sized degrees, and its zero point is 460 degrees below zero F, so degrees Rankine = degrees F + 460. Adding a negative number is the same as subtracting, so -40 + 460 = 420 degrees R. Adding 460 as if the reading were positive gives 500, and 233 is the Kelvin equivalent, which belongs to the Celsius-based absolute scale.
Source: FAA-H-8083-30 Ch. 5 (Temperature Scales)Report a problem with this question
22. Which of the five basic stresses is actually a combination of two other stresses acting at the same time?
- A.Torsion
- B.Tension
- C.Bending✓ Answer
- D.Shear
When a member is bent, the material on the outside of the curve is stretched while the material on the inside is squeezed, so bending is a combination of tension and compression occurring simultaneously in the same part. Tension pulls a part apart, compression crushes it, shear tends to slide adjacent layers past each other (the stress a rivet resists), and torsion is a twisting stress produced by torque.
Source: FAA-H-8083-30 Ch. 5 (Stress and Strain)Report a problem with this question
Practice questions based on 14 CFR Parts 43 and 65 and the FAA Aviation Maintenance Technician Handbook—General (FAA-H-8083-30). This site is not affiliated with or endorsed by the Federal Aviation Administration. This bank covers the GENERAL written test only; the Airframe and Powerplant written tests and the oral and practical exams are separate. Torque values, servicing quantities, corrosion rework limits, and inspection criteria always come from the manufacturer's current maintenance data and the applicable airworthiness directives — never from a practice test. Confirm current eligibility and testing requirements with the FAA before you test. About the A&P mechanic certificate →