In the 01 July 2016, Samsung Ltd acquired the machine A by $1,502,000. Estimated useful life is 10 years, estimated residual value is $2,000. Company uses straight-line method for depreciation. From 2019, Samsung Ltd measure the machine A by the revaluation model. At 30 June 2019, machine A was revalued to $ 1,400,000. After one year, at 30 June 2020, machine A was revalued to $1,000,000. At 30 June 2021, machine A was revalued to $1,500,000. Required: Prepare the journal entries from 01 July 2016 to 30 June 2021. (using both method for revaluation)

Answers

Answer 1

a) Samsung Ltd will record these Journal Entries from July 2016 to June 30, 2021, using the straight-line and revaluation methods of depreciation:

July 1, 2016 Debit Equipment $1,502,000

Credit Cash $1,502,000

To record the purchase of the machine A.

June 30, 2017 Debit Depreciation Expense $150,000

Credit Accumulated Depreciation $150,000

To record the depreciation expense for the first year.

June 30, 2018 Debit Depreciation Expense $150,000

Credit Accumulated Depreciation $150,000

To record the depreciation expense for the second year.

June 30, 2019 Debit Accumulated Depreciation $300,000

Credit Revaluation Reserve $300,000

To close the accumulated depreciation to Revaluation Reserve.

June 30, 2019 Debit Revaluation Reserve $198,000

Credit Revaluation Surplus $198,000

To record the Revaluation Surplus.

June 30, 2020 Debit Revaluation Expense $400,000

Credit Revaluation Reserve $400,000

To record the revaluation loss.

June 30, 2021 Debit Revaluation Reserve $500,000

Credit Revaluation Surplus $500,000

To record the revaluation surplus.

Data and Calculations:

Cost of machine on July 1, 2016 = $1,502,000

Estimated Residual Value = $2,000

Depreciable amount = $1,500,000

Estimated useful life = 10 years

Based on the straight-line method the depreciation expense for each year = $150,000 ($1,500,000/10).

Value of Machine at:

June 30, 2019 - $1,400,000

June 30, 2020 = $1,000,000

June 30, 2021 = $1,500,000

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[tex]\implies {\blue {\boxed {\boxed {\purple {\sf {C.\:None\:of\:the\:above.}}}}}}[/tex] ✔

[tex]\:\large{\boxed{\frak{ Step-by-step\:explanation: }}}[/tex]

A. [tex]x - ( - x) = -x[/tex]

Let us first solve for L. H. S.

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Hence, option B is ruled out too.

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