An asset dri g was purchased and pla ed in servi by a pell leum oduction compan
ID: 2515392 • Letter: A
Question
An asset dri g was purchased and pla ed in servi by a pell leum oduction compan s sl basis is $70,000 and it has an es nal d M ? $10,000 at the end o an es na ed useful l of 1 1 yuais o npule lhe depleual on amount in the third yea' and the BV at the end of the sixth year of life by each of these methods a. The SL method b. The 200% DB method with switchover to SL. c. The GDS d. The ADS Clic the con to view the partial listing of depreciable assets used in business Click the icon to view the GDS Recovery Rates ) a. Using the SI method the deprecistion amount in the third year is s 5455 (Round to the nearest dollar) ising the Sl method the BV at the end o the sixth year ot lite i3r2r3 (Round to the nearest dolar) b. Ising the 200% 1)B method the depreciatan amount in me third year is si- (Round to the nearest dollarExplanation / Answer
Depreciation is the value of asset used during a period. It has several methods for computing. These are as below:
a.
SL method of depreciation = (Cost – Salvage value) / Number of years of useful life
= (70,000 – 10,000) / 11
= 60,000 / 11
= $5,455.
The above depreciation is charged in each year. Therefore, the amount of depreciation in the third year is $5,455 (Answer).
Book value (BV) after 6th year = Cost – Total depreciation up to 6th year
= 70,000 – ($5,455 × 6)
= 70,000 – 32,730
= $37,270 (Answer)
b.
200% Double declining (DB) method:
Rate of depreciation = (1 / Life years) × 200
= (1 / 11) × 200
= 18.18 %
Depreciation in the 1st year = Cost × Rate of depreciation = $70,000 × 18.18% = $12,726
BV at the end of 1st year = Cost – Depreciation in the 1st year = 70,000 – 12,726 = 57,274
Depreciation in the 2nd year = BV of 1st year × 18.18% = 57,274 × 18.18% = 10,412
BV at the end of 2nd year = BV at the end of 1st year - Depreciation in the 2nd year = 57,274 – 10,412 = 46,862
Depreciation in the 3rd year = BV of 2nd year × 18.18% = 46,862 × 18.18% = 8,520 (Answer)
BV at the end of 3rd year = BV at the end of 2nd year - Depreciation in the 3rd year = 46,862 – 8,520 = 38,342
Depreciation in the 4th year = BV of 3rd year × 18.18% = 38,342 × 18.18% = 6,971
BV at the end of 4th year = BV at the end of 3rd year - Depreciation in the 4th year = 38,342 – 6,971 = 31,371
Depreciation in the 5th year = BV of 4th year × 18.18% = 31,371 × 18.18% = 5,703
BV at the end of 5th year = BV at the end of 4th year - Depreciation in the 5th year = 31,371 – 5,703 = 25,668
Depreciation in the 6th year = BV of 5th year × 18.18% = 25,668 × 18.18% = 4,666
BV at the end of 6th year = BV at the end of 5th year - Depreciation in the 6th year = 25,668 – 4,666 = 21,002 (Answer)
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