PracticeVideosBlogFlashcardsEspañol
All Practice Exams

100+ Free ATPL Performance Practice Questions

Pass your CASA Airline Transport Pilot Licence (Aeroplane) — Performance & Loading (APLA) exam on the first try — instant access, no signup required.

✓ No registration✓ No credit card✓ No hidden fees✓ Start practicing immediately
100+ Questions
100% Free

Loading practice questions...

Same family resources

Explore More Australian CASA Pilot Theory Exams

Continue into nearby exams from the same family. Each card keeps practice questions, study guides, flashcards, videos, and articles in one place.

CASA Aeronautical Radio Operator Certificate Exam (AROC, Australia)

Practice Questions

CASA ATPL(A) Aerodynamics & Aircraft Systems Exam (AASA, Australia)

Practice Questions

CASA ATPL(A) Air Law Exam (AALW, Australia)

Practice Questions

CASA ATPL(A) Flight Planning Exam (AFPA, Australia)

Practice Questions

CASA ATPL(A) Human Factors Exam (AHUF, Australia)

Practice Questions

CASA ATPL(A) Meteorology Exam (AMET, Australia)

Practice Questions
2026 Statistics

Key Facts: ATPL Performance Exam

100

Practice Questions

OpenExamPrep

40

Official Questions

CASA

70%

Pass Mark

CASA

2.5 hrs

Time Limit

CASA

The CASA ATPL Performance (APLA) exam is a 2.5-hour test requiring B727 flight performance graphs, %MAC weight/balance shifts, and takeoff safety speed derivations. Passing score is 70%. This prep features 100 questions.

Sample ATPL Performance Practice Questions

Try these sample questions to test your ATPL Performance exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1An aircraft has a total mass of 68,000 kg and the Center of Gravity (CG) is located at balance station 865 inches. If the Leading Edge of the Mean Aerodynamic Chord (LEMAC) is at 840 inches and the MAC is 180 inches long, what is the CG position expressed in percent of Mean Aerodynamic Chord (%MAC)?
A.15.22% MAC
B.12.50% MAC
C.11.11% MAC
D.13.89% MAC
Explanation: To calculate the CG in %MAC, use the formula: CG (%MAC) = ((CG - LEMAC) / MAC) * 100. Substituting the values: ((865 - 840) / 180) * 100 = (25 / 180) * 100 = 13.89% MAC. Other options represent errors in applying the formula or math errors.
2An airliner's CG is located at 25% MAC. If the LEMAC is at 620.0 inches and the MAC is 140.0 inches, calculate the CG location in inches from the datum.
A.665.0 inches
B.635.0 inches
C.655.0 inches
D.645.0 inches
Explanation: To convert %MAC to inches from the datum, use: CG (inches) = LEMAC + (%MAC * MAC / 100). Thus, CG = 620.0 + (25 * 140.0 / 100) = 620.0 + 35.0 = 655.0 inches. Other options are incorrect mathematical derivations.
3A cargo compartment has a maximum floor load intensity limit of 750 kg/m². A crate weighing 600 kg is loaded on a pallet. What is the minimum base area of the crate required to avoid exceeding the floor limit?
A.0.60 m²
B.0.45 m²
C.1.25 m²
D.0.80 m²
Explanation: Floor load intensity is weight divided by area (Load = Weight / Area). Therefore, Area = Weight / Limit = 600 kg / 750 kg/m² = 0.80 m². A base area smaller than this would exceed the maximum limit, while larger areas are acceptable but not the minimum required.
4An aircraft has a takeoff weight of 75,000 kg and the CG is currently at 18% MAC. The crew moves a load of 450 kg from the forward cargo hold (FS 420) to the aft cargo hold (FS 980). If the MAC is 170 inches, what is the new CG position in %MAC?
A.21.15% MAC
B.18.82% MAC
C.19.97% MAC
D.20.54% MAC
Explanation: Use the weight shift formula: Delta CG (inches) = (Weight Shifted * Shift Distance) / Total Aircraft Weight. Distance = 980 - 420 = 560 inches. Delta CG = (450 * 560) / 75,000 = 3.36 inches. Convert to %MAC: Delta %MAC = (3.36 / 170) * 100 = 1.976%. Since cargo moved aft, CG moves aft: 18% + 1.976% = 19.976% MAC (rounds to 19.97%).
5What is the aerodynamic consequence of loading an aircraft such that the Center of Gravity (CG) is at the forward limit?
A.Decreased longitudinal stability, lower stall speed, and decreased fuel consumption.
B.Increased longitudinal stability, lower stall speed, and decreased trim drag.
C.Decreased longitudinal stability, higher stall speed, and increased elevator authority at low speeds.
D.Increased longitudinal stability, higher stall speed, and increased fuel consumption due to higher trim drag.
Explanation: A forward CG increases the tail-down force required to balance the nose-heavy pitching moment. This increases the effective weight of the aircraft, resulting in a higher stall speed and greater induced drag (trim drag), leading to higher fuel consumption. The longer arm between the CG and the tail increases longitudinal stability but reduces elevator authority.
6What is the primary operational risk associated with an aircraft loaded beyond its Maximum Zero Fuel Weight (MZFW) but within its Maximum Takeoff Weight (MTOW)?
A.Inability to climb out of ground effect due to excessive induced drag.
B.Exceeding the maximum structural landing weight upon arrival at the destination.
C.Exceeding the certified tire speed limit during the takeoff roll.
D.Excessive structural bending moments at the wing root during flight, risking structural failure.
Explanation: MZFW is established to limit the upward bending moment on the wing root. Fuel in the wing tanks provides bending relief (weight counteracting lift). If payload replaces fuel such that MZFW is exceeded, the wing root experiences excessive structural stress under flight loads.
7An aircraft has a Basic Empty Weight of 41,500 kg, a Dry Operating Weight (DOW) of 43,000 kg, and a Maximum Zero Fuel Weight of 58,500 kg. If the Maximum Takeoff Weight is 74,400 kg and the Maximum Landing Weight is 62,000 kg, what is the maximum allowable payload for a flight with a fuel load of 18,000 kg?
A.17,000 kg
B.15,500 kg
C.13,400 kg
D.11,400 kg
Explanation: Let's check the limits: 1) Payload limited by MZFW: MZFW - DOW = 58,500 - 43,000 = 15,500 kg. 2) Payload limited by MTOW: MTOW - DOW - Fuel = 74,400 - 43,000 - 18,000 = 13,400 kg. The most restrictive structural limit is the takeoff weight limit, which allows a maximum payload of 13,400 kg. Thus, maximum payload is 13,400 kg.
8How does the consumption of fuel from a center wing tank typically affect the CG position on a sweep-back wing aircraft where the center tank is forward of the main wing tanks?
A.The CG moves forward as fuel is burned from the center tank.
B.The CG moves laterally, requiring fuel cross-feed to maintain balance.
C.The CG moves aft as fuel is burned from the center tank.
D.The CG remains stationary because the center tank lies exactly on the mean aerodynamic chord.
Explanation: Because the center tank is located forward of the main wing tanks due to the swept-back design of the wings, burning fuel from the center tank removes weight from a forward arm, causing the CG to shift aft.
9Which of the following describes the effect of an aft Center of Gravity (CG) location on aircraft performance and handling?
A.Reduced longitudinal stability, lighter control forces, lower stall speed, and decreased drag leading to lower fuel flow.
B.Reduced longitudinal stability, heavier control forces, lower stall speed, and increased drag.
C.Increased longitudinal stability, heavier control forces, higher stall speed, and increased drag.
D.Increased longitudinal stability, lighter control forces, higher stall speed, and decreased drag.
Explanation: An aft CG reduces the tail-down force required to balance the aircraft, which decreases the effective weight (aerodynamic download). This results in a lower stall speed and less induced drag (lower fuel flow). The shorter static margin reduces longitudinal stability and results in lighter pitch control forces, making the aircraft more sensitive to pitch inputs.
10A cargo compartment pallet location has a load limit of 680 kg. A heavy machinery part is mounted on a wooden crate measuring 1.2 m by 0.8 m. If the part weighs 620 kg, what is the average floor load in kg/m², and does it exceed a floor load limit of 600 kg/m²?
A.645.8 kg/m²; it does not exceed the floor load limit because the total weight is under 680 kg.
B.775.0 kg/m²; it exceeds the floor load limit.
C.645.8 kg/m²; it exceeds the floor load limit.
D.516.7 kg/m²; it does not exceed the floor load limit.
Explanation: The area of the crate base is 1.2 m * 0.8 m = 0.96 m². The average floor load is 620 kg / 0.96 m² = 645.83 kg/m². Since the floor load limit is 600 kg/m², this exceeds the limit, regardless of the fact that the total weight (620 kg) is less than the pallet load limit of 680 kg.

About the ATPL Performance Exam

The CASA ATPL Performance & Loading Exam (APLA) is a mandatory subject for the Airline Transport Pilot Licence in Australia. It tests advanced performance and weight/balance calculations for large jet transports, covering loading index systems, Center of Gravity (CG) limits, Mean Aerodynamic Chord (%MAC) conversions, takeoff limits (field length, tire speed, brake energy, obstacle clearance), jet climb segments (first, second, third, final), cruise performance, and wet/contaminated runway landing limits using the Boeing 727 Performance Manual.

Assessment

Closed-book computer-based exam administered at approved ASPEQ centers. Candidates are permitted to use a navigation computer, approved calculators, and reference manuals (such as Boeing 727 Performance Manual).

Time Limit

2.5 hours

Passing Score

70%

Exam Fee

Approx. $150 - $250 AUD (plus test center provider fees) (CASA / ASPEQ Exam Delivery)

ATPL Performance Exam Content Outline

25%

Weight & Balance / Loading

Large aircraft loading charts, weight limitations, CG limits, Mean Aerodynamic Chord (%MAC), and index units

20%

Takeoff Performance Limits

Balanced field lengths, tire speed limits, brake energy limits, obstacle clearance, V1/VR/V2 speed schedules

20%

Climb & Cruise Performance

Jet engine climb segments (1st, 2nd, 3rd, final), net climb paths, cruise floors, and drag curves

20%

Landing Performance Limits

Landing field length limits, quick turnaround charts, brake cooling calculations, and wet/contaminated runways

15%

Dry Operating Index Calculations

Operating index shifts, cargo/passenger layout index adjustments, basic empty weights, and payload calculations

How to Pass the ATPL Performance Exam

What You Need to Know

  • Passing score: 70%
  • Assessment: Closed-book computer-based exam administered at approved ASPEQ centers. Candidates are permitted to use a navigation computer, approved calculators, and reference manuals (such as Boeing 727 Performance Manual).
  • Time limit: 2.5 hours
  • Exam fee: Approx. $150 - $250 AUD (plus test center provider fees)

Keys to Passing

  • Complete 500+ practice questions
  • Score 80%+ consistently before scheduling
  • Focus on highest-weighted sections
  • Use our AI tutor for tough concepts

ATPL Performance Study Tips from Top Performers

1Master the %MAC formula: CG (%MAC) = ((CG Distance from Datum - Leading Edge of MAC) / Length of MAC) * 100
2Understand takeoff segments: the critical second segment climb extends from gear retraction to a minimum height of 400 feet above the runway, requiring a specific minimum gradient of climb (e.g. 2.4% for twin-jets)
3Learn how brake energy limits change: high altitude airport elevation and high ambient temperatures reduce the maximum speed at which a rejected takeoff can be executed without blowing fuse plugs

Frequently Asked Questions

What is the passing score for the ATPL APLA exam?

The passing score is 70%.

What is %MAC and how is it used?

Mean Aerodynamic Chord (MAC) percentage is the standard method for expressing CG location on large transport category aircraft, calculated relative to the leading edge of the wing's mean aerodynamic chord.