Computer equipment was acquired at the beginning of the year at a cost of $65,000 that has an estimated residual value of $3,800 and an estimated useful life of 8 years. (1)Determine the (a) depreciable cost for straight rate, (b) straight-line rate, and (c) annual depreciation for the straight line rate (d) double declining balance depreciation amount for 1st year (2) Journalize the sale of the equipment at the end of the 4th year for $20,000 cash, if using the straight line rate.

a) Depreciable cost = $61,200

b) Straight-line rate = 12.5%

c) Annual depreciation for the straight-line rate is:

$7,650.

d) The double-declining balance depreciation amount for the 1st year is:

$15,300.

e) Journal Entry of sale of equipment at the end of the 4th year for $20,000, using the straight-line rate:

Debit Sale of Equipment $65,000

Credit Equipment $65,000

To record the transfer of the equipment account.

Debit Accumulated Depreciation $30,600

Credit Sale of Equipment $30,600

To record the transfer of accumulated depreciation to the account.

Debit Cash Account $20,000

Credit Sale of Equipment $20,000

To record the cash received from the sale.

Debit Profit and Loss Account $14,400

Credit Sales of Equipment $14,400

To record the loss on sale of equipment.

Explanation:

a) Data and Calculations:

Cost of computer equipment = $65,000

Estimated residual value =             3,800

Depreciable value =                   $61,200

Estimated useful life = 8 years

Annual depreciation expense = $7,650 ($61,200/8) or $61,200 * 12.5%

Depreciable cost = $61,200 ($65,000 - $3,800)

Straight-line rate = 12.5% (100%/8)

Double-declining rate = 12.5% x 2 = 25%

Depreciation amount for the 1st year = $15,300 ($61,200 * 25%)

Book value of asset at the end of the 4th year, straight-line method:

Cost of equipment = $65,000

Accumulated Depreciation = $30,600 ($7,650 * 4)

Book value = $34,400

Sales proceed = $20,000

Loss on sale of asset = $14,400 ($34,400 - $20,000)

u = 30

continuing your method

= +

subtract from both sides

= - thus

= -

= ⇒ u = 30

answer; ///decision making under risk;