FAC2601 May/June (Exam) Memo | Due 18 May 2026

, PLEASE USE THIS DOCUMENT AS A GUIDE ONLY

 QUESTION 1

A) Calculate the following amounts to be recorded by Standoff Ltd in respect of the Machine
Super A and the aircraft only, for the year ending 31 December 2025, in a columnar format

Machine Super A – Components (Straight-line)

Depreciation for 2024 (full year):


Component Cost (R) Residual (R) Depreciable (R) Life (yrs) Depr 2024 (R)

1 5,000,000 400,000 4,600,000 5 920,000

2 3,000,000 0 3,000,000 4 750,000

3 2,000,000 0 2,000,000 2 1,000,000

Total 10,000,000 2,670,000

Opening accumulated depreciation (1 Jan 2025) = R2,670,000

Additions in 2025:
 Replacement of part of Component 2 (partially): R1,100,000
 Replacement of Component 3: R2,100,000
 Day-to-day servicing (R45,000) is expensed, not capitalized.

But note: Replacement part for Comp 2 – original cost R1,000,000, accumulated depreciation?
Original Comp 2 cost R3,000,000, life 4 years, no residual. Annual depr R750,000.
From 1 Jan 2024 to 30 Sept 2025 = 1 year 9 months = 1.75 years.
Acc depr on original Comp 2 = 1.75 × R750,000 = R1,312,500.
Carrying amount of original Comp 2 at replacement date = R3,000,000 – R1,312,500 = R1,687,500.

But here, only a part of Comp 2 broke. IFRS: Derecognize the carrying amount of the replaced part.
Original cost of that part = R1,000,000.
Proportion of original Comp 2 cost: 1M / 3M = 1/3.
Acc depr on that part = 1/3 × R1,312,500 = R437,500.
Carrying amount of replaced part = R1,000,000 – R437,500 = R562,500 (disposal).

, Disposals at carrying amount (Machine Super A):

 Replaced part of Comp 2: R562,500

 Original Component 3 replaced 31 Dec 2025:
 Original Comp 3 cost R2,000,000, life 2 yrs, no residual, straight-line.
 Depr 2024 (full year) = R1,000,000.
 Depr 1 Jan – 31 Dec 2025 = R1,000,000.
 Total depr to disposal = R2,000,000. Carrying amount = R0.
 So disposal at R0.

Depreciation for 2025 (Machine Super A):

 Component 1: (5,000,000 – 400,000)/5 = 920,000

 Component 2: Original remaining part (R2,000,000 cost, life 4y, depr 500,000 for 2025) + New
part (R1,100,000, life remaining from 1 Oct 2025 to 31 Dec 2027? Wait, original Comp 2 life 4
years from Jan 2024, so end 2027. New part should match remaining life of Comp 2? Actually,
new part replaces broken part, so depreciate over remaining useful life of Comp 2 = from 1 Oct
2025 to 31 Dec 2027 = 2.25 years). Depr new part for 2025 (Oct–Dec = 3 months) =
R1,100,.25 yrs × 3/12 = R122,222. Depr original remaining Comp 2 for 2025 =
R500,000 (full year, since original 3M – 1M replaced = 2M cost, 4y life, annual depr 500k).
Total Comp 2 depr 2025 = 500,000 + 122,222 = R622,222.

 Component 3 new: Cost R2,100,000, life 2y, no residual, available 31 Dec 2025, so no
depreciation in 2025.

Total depr Machine Super A 2025 = 920,000 + 622,222 + 0 = R1,542,222

Aircraft (Units of production)

Depreciation for 2024:
 Engines: 18M / 10,000 hrs × 1,000 hrs = 1,800,000
 Airframe: 12M / 20,000 hrs × 1,000 = 600,000
 Furniture: 5M / 5,000 hrs × 1,000 = 1,000,000
 Inspection (original): 5M / 2,000 hrs × 1,000 = 2,500,000

Total depr 2024 = 5,900,000

Opening acc depr 1 Jan 2025 = R5,900,000

Additions in 2025:
Inspection on 30 June 2025: R4,000,000 capitalized as new component (since it’s a major
inspection). Useful life until next inspection: 2,000 hours, but done 6 months early, so remaining life
from 30 June 2025 = 2,000 hours. But actual hours flown after inspection: 500 hours in 2025
(July–Dec). Depreciation on this new inspection cost = 4M / 2,000 × 500 = R1,000,000.