EXERCISE 11-2 (20-25 minutes)
(a)
If there is any salvage value and the amount is unknown (as
is the case here), the cost would have to be determined by looking at the data
for the double-declining balance method.
100% |
= 20%; 20% X
2 = 40% |
5 |
Cost X 40%
= $20,000
$20,000
¸ .40 = $50,000
Cost of asset
(b) $50,000
cost [from (a)] – $45,000 total depreciation = $5,000 salvage value.
(c) The highest charge to income for Year 1 will be yielded by
the double-declining balance method.
(a)
The highest charge to income for Year 4 will be yielded by
the straight-line method.
(b)
The method that produces the highest book value at the end
of Year 3 would be the method that yields the lowest accumulated depreciation
at the end of Year 3, which is the straight-line method.
Computations:
St.-line = $50,000 – ($9,000 + $9,000 + $9,000) = $23,000
book value, end of Year 3.
S.Y.D. = $50,000 – ($15,000 + $12,000 + $9,000) = $14,000
book value, end of Year 3.
D.D.B. = $50,000 – ($20,000 + $12,000 + $7,200) = $10,800
book value, end of Year 3.
(c)
The method that will yield the highest gain (or lowest loss)
if the asset is sold at the end of Year 3 is the method which will yield the
lowest book value at the end of Year 3, which is the double-declining balance
method in this case.
EXERCISE 11-5
(a) |
($117,900 – $12,900) |
= $21,000/yr. = $21,000 X 5/12 = $8,750 |
5 |
2004
Depreciation — Straight line = $8,750
(b) |
($117,900 – $12,900) |
= $5.00/hr. |
21,000 |
2004
Depreciation — Machine Usage = 800 X $5.00 = $4,000
(c) |
Machine |
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Allocated to |
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Year |
Total |
2004 |
|
2005 |
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1 |
5/15 X $105,000 = $35,000 |
$14,583* |
|
$20,417** |
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2 |
4/15 X $105,000 = $28,000 |
______ |
|
11,667*** |
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|
$14,583 |
|
$32,084 |
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* |
$35,000 X 5/12 = $14,583 |
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** |
$35,000 X 7/12 = $20,417 |
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*** |
$28,000 X 5/12 = $11,667 |
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2005
Depreciation — Sum-of-the-Years’-Digits = $32,084
(d) 2004 40% X ($117,900) X 5/12 = $19,650
2005
40% X ($117,900 – $19,650) = $39,300
OR
1st
full year (40% X $117,900) = $47,160
2nd
full year [40% X ($117,900 – $47,160)] = $28,296
2004 Depreciation = |
5/12 X $47,160 = |
$19,650 |
|
|
|
2005 Depreciation = |
7/12 X $47,160 = |
$27,510 |
|
5/12 X $28,296 = |
11,790 |
|
|
$39,300 |
EXERCISE 11-6 (20-30 minutes)
(a) |
2003 |
Straight-line |
$212,000 – $12,000 |
= $25,000/year |
8 |
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3 months — Depreciation $6,250 =
($25,000 X 3/12) |
(b |
2003 |
Output |
$212,000 – $12,000 |
= $5.00/output unit |
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40,000 |
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1,000 units X $5.00 = $5,000 |
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(c) |
2003 |
Working hours |
$212,000 – $12,000 |
= $10.00/hour |
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20,000 |
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525 hours X $10.00 = $5,250 |
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(d) |
8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 OR |
n (n + 1) |
= |
8(9) |
= 36 |
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2 |
2 |
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Allocated to |
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Sum-of-the-years’-digits |
Total |
2003 |
2004 |
2005 |
|
Year 1 |
8/36 X
$200,000 = |
$44,444 |
$11,111 |
$33,333 |
|
2 |
7/36 X $200,000 = |
$38,889 |
|
9,722 |
$29,167 |
3 |
6/36 X $200,000 = |
$33,333 |
_______ |
_______ |
8,333 |
|
|
|
$11,111 |
$43,055 |
$37,500 |
2005: $37,500
= (9/12 of 2nd year of machine’s life plus 3/12 of 3rd
year of machine’s life)
(e)
Double-declining balance 2004: 1/8 X 2 = 25%.
2003: 25% X $212,000 X 3/12 = $13,250
2004: 25% X ($212,000 – $13,250) = $49,688
OR
1st
full year (25% X $212,000) = $53,000
2nd
full year [25% X ($212,000 – $53,000)] = $39,750
2003
Depreciation 3/12 X $53,000 = $13,250
2004
Depreciation 9/12 X $53,000 = $39,750
3/12 X $39,750 = 9,938
$49,688
EXERCISE 11-16
(a) |
December 31, 2004 |
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Loss on Impairment................................................................ |
3,200,000 |
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Accumulated
Depreciation—Equipment.................. |
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3,200,000 |
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Cost |
$9,000,000 |
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Accumulated
depreciation |
1,000,000 |
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Carrying
amount |
8,000,000 |
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Fair value |
4,800,000 |
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Loss
in impairment |
$3,200,000 |
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(b) |
December 31, 2005 |
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Depreciation Expense........................................................... |
1,200,000 |
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Accumulated
Depreciation—Equipment.................. |
|
1,200,000 |
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New carrying
amount |
$4,800,000 |
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Useful life |
4 years |
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Depreciation
per year |
$1,200,000 |
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(c) No entry necessary. Restoration of any impairment loss is not
permitted.
EXERCISE 11-17
(a) |
Loss on Impairment................................................................ |
3,220,000 |
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|
Accumulated
Depreciation—Equipment.................. |
|
3,220,000 |
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Cost |
$9,000,000 |
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Accumulated
depreciation |
1,000,000 |
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Carrying
amount |
8,000,000 |
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Less: Fair
value |
4,800,000 |
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Plus: Cost
of disposal |
20,000 |
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Loss
on impairment |
$3,220,000 |
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(a)
No entry necessary. Depreciation not taken on assets
intended to be sold.
(c) |
Accumulated Depreciation—Equipment............................ |
500,000 |
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Recovery
of Loss on Impairment............................ |
|
500,000 |
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Fair value |
$5,300,000 |
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Less: Cost of disposal |
20,000 |
5,280,000 |
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Carrying amount |
|
4,780,000 |
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Recovery of impairment loss |
|
$
500,000 |
EXERCISE 11-23
(a) |
$970,000 + $170,000 + $40,000* –
$100,000 |
= .09
depletion per unit |
12,000,000 |
*(Note
to instructor: The $40,000 should be depleted because it is a cost of the mine.
This cost is incurred to get the land back to its original value of $100,000.)
2,500,000
units extracted X $.09 = $225,000 depletion for 2004
(b) 2,100,000
units sold X $.09 = $189,000 charged to cost of goods sold for 2004
EXERCISE 11-24
(a) Asset turnover ratio:
$13,234 |
= .96 times |
$14,212 +
$13,362 |
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2 |
|
(a)
Rate of return on assets:
$76 |
= .55% |
$14,212 +
$13,362 |
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2 |
|
(b)
Profit margin on sales:
$76 |
= .57% |
$13,234 |
(c)
The asset turnover ratio times the profit margin on sales
provides the rate of return on assets computed for Eastman Kodak as follows:
Profit
margin on sales |
X |
Asset
Turnover |
|
Return on
Assets |
.57% |
X |
.96 |
= |
.55% |
Note the
answer .55% is the same as the rate of return on assets computed in (b) above.
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PROBLEM 11-3 |
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(a) |
Depreciation
Expense—Asset A |
2,900 |
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Accumulated Depreciation—Asset A |
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2,900 |
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(5/55 X [$35,000 – $3,100]) |
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Accumulated
Depreciation—Asset A |
26,100 |
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Asset A ($35,000 – $13,000) |
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22,000 |
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Gain on Disposal of Plant Assets |
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4,100 |
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(b) |
Depreciation
Expense—Asset B |
6,720 |
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Accumulated Depreciation—Asset B |
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6,720 |
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([$51,000 – $3,000] ¸ 15,000 X 2,100) |
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(c) |
Depreciation
Expense—Asset C |
6,000 |
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Accumulated Depreciation—Asset C |
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6,000 |
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([$80,000 – $15,000 – $5,000] ¸ 10) |
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(d) |
Asset E |
22,000 |
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Retained Earnings |
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22,000 |
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Depreciation
Expense—Asset E |
4,400* |
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Accumulated Depreciation—Asset E |
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4,400 |
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*($22,000 X
.20) |
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PROBLEM 11-8 |
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(a)
The amounts to be recorded on the books of Selig Sporting
Goods Inc. as of December 31, 2004, for each of the properties acquired from
Starks Athletic Equipment Company are calculated as follows:
Cost Allocations to Acquired Properties
|
|
Remaining Purchase Price
Allocations |
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|
(1) Land |
$280,000 |
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|
$280,000 |
(2)
Building |
|
$
84,0001 |
$100,000 |
$21,6002 |
205,600 |
(3)
Machinery |
_______ |
36,0001 |
_______ |
______ |
36,000 |
Totals |
$280,000 |
$120,000 |
$100,000 |
$21,600 |
$521,600 |
Supporting Calculations
1Balance of
purchase price to be allocated. |
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Total purchase price |
$400,000 |
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Less land appraisal |
280,000 |
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Balance
to be allocated |
$120,000 |
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Appraisal Values |
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Allocated Values |
|
Building |
$105,000 |
|
105/150 = |
.70 |
X
$120,000 |
$ 84,000 |
Machinery |
45,000 |
|
45/150 = |
.30 |
X $120,000 |
36,000 |
Totals |
$150,000 |
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1.00 |
|
$120,000 |
2Capitalizable
interest.
Dates of
loans in 2004 |
|
|
|
Periods
Outstanding |
|
Interest |
1/1 |
|
$ 50,000 |
X |
12/12 |
X .12 |
$ 6,000 |
4/1 |
|
130,000 |
X |
9/12 |
X .12 |
11,700 |
10/1 |
|
130,000 |
X |
3/12 |
X .12 |
3,900 |
12/31 |
|
190,000 |
X |
|
X .12 |
_______ |
Totals |
|
$500,000 |
|
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|
$21,600 |
Note t: If the interest is
allocated between the building and the machinery, $15,120 ($21,600 X 105/150)
would be allocated to the building and $6,480 ($21,600 X 45/150) would be
allocated to the machinery.
(b)
Selig Sporting Goods Inc.’s 2004 depreciation expense, for
book purposes, for each of the properties acquired from Starks Athletic
Equipment Company is as follows:
1. |
Land:
No depreciation. |
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|
|
2. |
Building:
Depreciation rate |
= 1.50 X
1/15 = .10 |
|
2004 depreciation expense |
= Cost X
Rate X 1/2 year |
|
|
= $205,600 X
.10 X 1/2 |
|
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=
$10,280 |
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3. |
Machinery:
Depreciation rate |
= 2.00 X 1/5
= .40 |
|
2004 depreciation expense |
= Cost X
Rate X 1/2 |
|
|
= $36,000 X
.40 X 1/2 |
|
|
=
$7,200 |
(c)
Arguments for the capitalization of interest costs include
the following.
(1)
Diversity of practices among companies and industries called
for standardization in practices.
(2)
Total interest costs should be allocated to enterprise
assets and operations, just as material, labor, and overhead costs are
allocated. That is, under the concept of historical costs, all costs incurred
to bring an asset to the condition and location necessary for its intended use
should be reflected as a cost of that asset.
Arguments against the
capitalization of interest include the following:
(1) Interest capitalized in a period would tend to be offset by
amortization of interest capitalized in prior periods.
(2) Interest cost is a cost of financing, not of construction.
|
PROBLEM 11-9 |
|
(a)
Carrying value of asset: $8,000,000 – $2,000,000 =
$6,000,000.
Future cash flows ($5,300,000) < Carrying value
($6,000,000)
Impairment entry:
Loss on
Impairment.. 1,600,000*.....
Accumulated Depreciation 1,600,000
*$6,000,000 – $4,400,000
(b) Depreciation Expense....... 1,100,000**
Accumulated
Depreciation 1,100,000
**(4,400,000 ¸ 4)
(c) No depreciation is recorded on impaired assets to be disposed
of. Recovery of impairment losses are recorded.
Accumulated
Depreciation........... 200,000
Recovery of Impairment Loss 200,000
|
PROBLEM
11-10 |
|
(1) |
$82,000 |
Allocated in proportion to
appraised values |
|
|
(1/10 X $820,000). |
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(2) |
$738,000 |
Allocated in proportion to
appraised values |
|
|
(9/10 X $820,000). |
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(3) |
Forty years |
Cost less salvage ($738,000 –
$40,000) divided by |
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annual depreciation ($17,450). |
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(4) |
$17,450 |
Same as prior year since it is
straight-line depreciation. |
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(5) |
$91,000 |
[Number of shares (2,500) times
fair value ($30)] |
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plus demolition cost of existing building ($16,000). |
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(6) |
None |
No depreciation before use. |
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(7) |
$30,000 |
Fair market value. |
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(8) |
$4,500 |
Cost ($30,000) times percentage
(1/10 X 150%). |
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(9) |
$3,825 |
Cost ($30,000) less prior
year’s depreciation ($4,500) |
|
|
equals $25,500. Multiply $25,500 times 15%. |
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(10) |
$150,000 |
Total cost ($164,900) less
repairs and maintenance |
|
|
($14,900). |
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|
(11) |
$32,000 |
Cost less salvage ($150,000 –
$6,000) times 8/36. |
|
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|
(12) |
$9,333 |
Cost less salvage ($150,000 –
$6,000) times 7/36 times |
|
|
one-third of a year. |
(13) |
$52,000 |
Annual payment ($6,000) times present
value of annuity due at 8% for 11 years (7.710) plus down payment ($5,740).
This can be found in an annuity due table since the payments are at the
beginning of each year. Alternatively, to convert from an ordinary annuity to
an annuity due factor, proceed as follows: For eleven payments use the
present value of an ordinary annuity for 11 years (7.139) times 1.08.
Multiply this factor (7.710) times $6,000 annual payment to obtain $46,260,
and then add the $5,740 down payment. |
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|
(14) |
$2,600 |
Cost
($52,000) divided by estimated life (20 years). |