EXERCISE 112 (2025 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 doubledeclining 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 doubledeclining balance method.
(a)
The highest charge to income for Year 4 will be yielded by
the straightline 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 straightline 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 doubledeclining balance
method in this case.
EXERCISE 115
(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 

Allocated to 


Year 
Total 
2004 

2005 


1 
5/15 X $105,000 = $35,000 
$14,583* 

$20,417** 


2 
4/15 X $105,000 = $28,000 
______ 

11,667*** 




$14,583 

$32,084 


* 
$35,000 X 5/12 = $14,583 





** 
$35,000 X 7/12 = $20,417 





*** 
$28,000 X 5/12 = $11,667 




2005
Depreciation — SumoftheYears’Digits = $32,084
(d) 2004 40% X ($117,900) X 5/12 = $19,650
2005
40% X ($117,900 – $19,650) = $39,300
OR
1^{st}
full year (40% X $117,900) = $47,160
2^{nd}
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 116 (2030 minutes)
(a) 
2003 
Straightline 
$212,000 – $12,000 
= $25,000/year 
8 








3 months — Depreciation $6,250 =
($25,000 X 3/12) 
(b 
2003 
Output 
$212,000 – $12,000 
= $5.00/output unit 


40,000 











1,000 units X $5.00 = $5,000 


(c) 
2003 
Working hours 
$212,000 – $12,000 
= $10.00/hour 


20,000 











525 hours X $10.00 = $5,250 







(d) 
8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 OR 
n (n + 1) 
= 
8(9) 
= 36 

2 
2 




Allocated to 

Sumoftheyears’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 2^{nd} year of machine’s life plus 3/12 of 3^{rd}
year of machine’s life)
(e)
Doubledeclining 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
1^{st}
full year (25% X $212,000) = $53,000
2^{nd}
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 1116
(a) 
December 31, 2004 


Loss on Impairment................................................................ 
3,200,000 


Accumulated
Depreciation—Equipment.................. 

3,200,000 

Cost 
$9,000,000 



Accumulated
depreciation 
1,000,000 



Carrying
amount 
8,000,000 



Fair value 
4,800,000 



Loss
in impairment 
$3,200,000 


(b) 
December 31, 2005 


Depreciation Expense........................................................... 
1,200,000 



Accumulated
Depreciation—Equipment.................. 

1,200,000 


New carrying
amount 
$4,800,000 



Useful life 
4 years 



Depreciation
per year 
$1,200,000 

(c) No entry necessary. Restoration of any impairment loss is not
permitted.
EXERCISE 1117
(a) 
Loss on Impairment................................................................ 
3,220,000 



Accumulated
Depreciation—Equipment.................. 

3,220,000 


Cost 
$9,000,000 



Accumulated
depreciation 
1,000,000 



Carrying
amount 
8,000,000 



Less: Fair
value 
4,800,000 



Plus: Cost
of disposal 
20,000 



Loss
on impairment 
$3,220,000 

(a)
No entry necessary. Depreciation not taken on assets
intended to be sold.
(c) 
Accumulated Depreciation—Equipment............................ 
500,000 


Recovery
of Loss on Impairment............................ 

500,000 





Fair value 
$5,300,000 


Less: Cost of disposal 
20,000 
5,280,000 

Carrying amount 

4,780,000 

Recovery of impairment loss 

$
500,000 
EXERCISE 1123
(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 1124
(a) Asset turnover ratio:
$13,234 
= .96 times 
$14,212 +
$13,362 

2 

(a)
Rate of return on assets:
$76 
= .55% 
$14,212 +
$13,362 

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.

PROBLEM 113 

(a) 
Depreciation
Expense—Asset A 
2,900 



Accumulated Depreciation—Asset A 

2,900 


(5/55 X [$35,000 – $3,100]) 









Accumulated
Depreciation—Asset A 
26,100 



Asset A ($35,000 – $13,000) 

22,000 


Gain on Disposal of Plant Assets 

4,100 






(b) 
Depreciation
Expense—Asset B 
6,720 



Accumulated Depreciation—Asset B 

6,720 


([$51,000 – $3,000] ¸ 15,000 X 2,100) 








(c) 
Depreciation
Expense—Asset C 
6,000 



Accumulated Depreciation—Asset C 

6,000 


([$80,000 – $15,000 – $5,000] ¸ 10) 








(d) 
Asset E 
22,000 



Retained Earnings 

22,000 







Depreciation
Expense—Asset E 
4,400* 



Accumulated Depreciation—Asset E 

4,400 







*($22,000 X
.20) 










PROBLEM 118 


(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 



(1) Land 
$280,000 



$280,000 
(2)
Building 

$
84,000^{1} 
$100,000 
$21,600^{2} 
205,600 
(3)
Machinery 
_______ 
36,000^{1} 
_______ 
______ 
36,000 
Totals 
$280,000 
$120,000 
$100,000 
$21,600 
$521,600 
Supporting Calculations
^{1}Balance of
purchase price to be allocated. 


Total purchase price 
$400,000 

Less land appraisal 
280,000 

Balance
to be allocated 
$120,000 


Appraisal Values 



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 


1.00 

$120,000 
^{2}Capitalizable
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 



$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. 




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 


=
$10,280 



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 119 

(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
1110 

(1) 
$82,000 
Allocated in proportion to
appraised values 


(1/10 X $820,000). 



(2) 
$738,000 
Allocated in proportion to
appraised values 


(9/10 X $820,000). 



(3) 
Forty years 
Cost less salvage ($738,000 –
$40,000) divided by 


annual depreciation ($17,450). 



(4) 
$17,450 
Same as prior year since it is
straightline depreciation. 



(5) 
$91,000 
[Number of shares (2,500) times
fair value ($30)] 


plus demolition cost of existing building ($16,000). 



(6) 
None 
No depreciation before use. 



(7) 
$30,000 
Fair market value. 



(8) 
$4,500 
Cost ($30,000) times percentage
(1/10 X 150%). 



(9) 
$3,825 
Cost ($30,000) less prior
year’s depreciation ($4,500) 


equals $25,500. Multiply $25,500 times 15%. 



(10) 
$150,000 
Total cost ($164,900) less
repairs and maintenance 


($14,900). 



(11) 
$32,000 
Cost less salvage ($150,000 –
$6,000) times 8/36. 



(12) 
$9,333 
Cost less salvage ($150,000 –
$6,000) times 7/36 times 


onethird 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. 



(14) 
$2,600 
Cost
($52,000) divided by estimated life (20 years). 