On January 1, 20X2, X Company made modifications to an asset used in its manufacturing operations in order to extend its useful life. The asset had an original cost of $30,000 and accumulated depreciation of $12,000, with a remaining useful life of 5 years. As a result of the modification, which cost $6,000, the remaining useful life was increased to 8 years. If X Company uses straight-line depreciation, in 20X2, X Company:

Will recognize repairs and maintenance expense of $6,000 and depreciation expense of $2,250.

Will recognize repairs and maintenance expense of $6,000 and depreciation expense of $3,600.

Will capitalize the cost of the modification and will recognize depreciation expense of $3,000.

Will capitalize the cost of the modification and will recognize depreciation expense of $4,800.

Question

Business

Jazlyn Shelton

29 July, 10:25

On January 1, 20X2, X Company made modifications to an asset used in its manufacturing operations in order to extend its useful life. The asset had an original cost of $30,000 and accumulated depreciation of $12,000, with a remaining useful life of 5 years. As a result of the modification, which cost $6,000, the remaining useful life was increased to 8 years. If X Company uses straight-line depreciation, in 20X2, X Company:

Will recognize repairs and maintenance expense of $6,000 and depreciation expense of $2,250.

Will recognize repairs and maintenance expense of $6,000 and depreciation expense of $3,600.

Will capitalize the cost of the modification and will recognize depreciation expense of $3,000.

Will capitalize the cost of the modification and will recognize depreciation expense of $4,800.

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Answers (1)

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Philip · Answer 1

Will capitalize the cost of the modification and will recognize depreciation expense of $3,000.

Explanation:

Provided information, we have the following details.

Original value of asset = $30,000

Accumulated depreciation = $12,000

Therefore carrying value of asset = $30,000 - $12,000 = $18,000

Remaining life of asset = 5 years

When an additional modification is done, cost of modification = $6,000

Revised life of machine due to modification = 8 years

Thus, the modification cost will also be capitalized.

Depreciation each year = ($18,000 + $6,000) / 8 = $3,000 each year.

Therefore, correct option is

c. Will capitalize cost of repairs and will charge depreciation of $3,000 each year