SOLUTIONS TO PROBLEMS
PROBLEM 10-1A
Item
1
2
3
4
5
6
7
8
9
10
Land
($ 2,000)
Building
Other Accounts
$ 3,000
Property Taxes Expense
$600,000
25,000
125,000
15,000
Land Improvements
4,000
Land Improvements
10,000
( 21,000)
( (2,500)
($145,500)
0000,000
$635,000
PROBLEM 10-2A
(a)
(b)
Year
Computation
Cumulative
12/31
2003
2004
2005
2006
MACHINE 1
$70,000 X 10% = $7,000
$70,000 X 10% = $7,000
$70,000 X 10% = $7,000
$70,000 X 10% = $7,000
$ 7,000
14,000
21,000
28,000
2004
2005
2006
MACHINE 2
$80,000 X 25% = $20,000
$60,000 X 25% = $15,000
$45,000 X 25% = $11,250
$20,000
35,000
46,250
2006
MACHINE 3
1,000 X ($72,000 ÷ 24,000) = $3,000
$ 3,000
Year
Depreciation Computation
Expense
(1) 2004
MACHINE 2
$80,000 X 25% X 9/12 = $15,000
$15,000
(2) 2005
$65,000 X 25% = $16,250
$16,250
PROBLEM 10-3A
(a) (1) Purchase price .................................................................
Sales tax ...........................................................................
Shipping costs .................................................................
Insurance during shipping ..............................................
Installation and testing ....................................................
Total cost of machinery ...........................................
$38,500
2,200
175
75
50
$41,000
Machinery ...........................................................
Cash ............................................................
41,000
41,000
(2) Recorded cost ..................................................................
Less: Salvage value ........................................................
Depreciable cost ..............................................................
Years of useful life ...........................................................
Annual depreciation .................................................
Depreciation Expense ........................................
Accumulated Depreciation ........................
9,000
(b) (1) Recorded cost ..................................................................
Less: Salvage value ........................................................
Depreciable cost ..............................................................
Years of useful life ...........................................................
Annual depreciation .................................................
(2)
Year
2006
2007
2008
2009
Book Value at
Beginning of
Year
$100,000
50,000
25,000
12,500
DDB Rate
*50%*
*50%*
*50%*
*50%*
Annual
Depreciation
Expense
$50,000
25,000
12,500
4,500**
*100% ÷ 4-year useful life = 25% X 2 = 50%.
$41,000
5,000
$36,000
÷
4
$ 9,000
9,000
$100,000
8,000
$92,000
÷
4
$ 23,000
Accumulated
Depreciation
$50,000
75,000
87,500
92,000
**[($100,000 – $8,000) – $87,500] = $4,500.
PROBLEM 10-3A (Continued)
(3) Depreciable cost per unit = ($100,000 – $8,000)/25,000
units
=
$3.68 per unit.
Annual Depreciation Expense
2006:
2007:
2008:
2009:
$3.68 X 6,500 = $23,920
3.68 X 7,500 = 27,600
3.68 X 6,000 = 22,080
3.68 X 5,000 = 18,400
(c) The straight-line method reports the lowest amount of
depreciation expense the first year while the decliningbalance method reports the highest. In the fourth year,
the
declining-balance
method
reports the lowest amount of depreciation expense while
the straight-line method reports the highest.
These facts occur because the declining-balance method
is
an
accelerated depreciation method in which the largest
amount
of
depreciation is recognized in the early years of the
asset’s life. If the straight-line method is used, the same
amount
of
depreciation
expense is recognized each year. Therefore, in the early
years less depreciation expense will be recognized under
this method than under the declining-balance method
while more will be recognized in the later years.
The amount of depreciation expense recognized using the
units-of-activity method is dependent on production, so
this method could recognize more or less depreciation
expense than the other two methods in any year
depending on output.
No matter which of the three methods is used, the same
total amount of depreciation expense will be recognized
over the four-year period.
PROBLEM 10-4A
Year
2004
2005
2006
2007
2008
2009
2010
Depreciation
Expense
(b)
$9,000(a)
9,000
(b)
7,200(b)
7,200
7,200
(b)
8,700(c)
8,700
Accumulated
Depreciation
$ 9,000
18,000
25,200
32,400
39,600
48,300
57,000
(a)
$60,000 – $6,000
= $9,000
6 years
(b)
Book value – Salvage value $42,000 – $6,000
=
= $7,200
5 years
Remaining useful life
(c)
$20,400 – $3,000
= $8,700
2 years
PROBLEM 10-5A
(a) Apr. 1
May 1
1
Land .................................................
Cash..........................................
2,200,000
Depreciation Expense .....................
Accumulated Depreciation—
Equipment ............................
($750,000 X 1/10 X 4/12)
25,000
Cash .................................................
Accumulated Depreciation—
Equipment....................................
Equipment ................................
Gain on Disposal .....................
460,000
Cost
Accum. depreciation—
equipment
[($750,000 X 1/10 X 4) + $25,000]
Book value
Cash proceeds
Gain on disposal
June 1
July 1
Dec. 31
31
2,200,000
25,000
325,000
750,000
35,000
$750,000
325,000
0000,000
425,000
460,000
$ 35,000
Cash .................................................
Land ..........................................
Gain on Disposal .....................
1,800,000
Equipment........................................
Cash..........................................
2,400,000
Depreciation Expense .....................
Accumulated Depreciation—
Equipment ............................
($500,000 X 1/10)
50,000
Accumulated Depreciation—
Equipment....................................
300,000
1,500,000
2,400,000
50,000
500,000
Equipment ................................
500,000
PROBLEM 10-5A (Continued)
Cost
Accum. depreciation—
equipment
$500,000
500,000
($500,000 X 1/10 X 10)
Book value
(b) Dec. 31
31
$
0
Depreciation Expense .....................
Accumulated Depreciation—
Buildings ..............................
($26,500,000 X 1/50)
530,000
Depreciation Expense .....................
Accumulated Depreciation—
Equipment ............................
3,995,000
530,000
3,995,000
($38,750,000* X 1/10)
$3,875,000
[($2,400,000 X 1/10) X 6/12]
120,000
$3,995,000
*($40,000,000 – $750,000 – $500,000)
(c)
WALTON COMPANY
Partial Balance Sheet
December 31, 2007
Plant Assets*
Land .....................................................
Buildings..............................................
Less: Accumulated depreciation—
buildings ...............................
Equipment ...........................................
Less: Accumulated depreciation—
equipment .............................
Total plant assets ........................
*See T-accounts which follow.
$ 4,900,000
$26,500,000
12,630,000
41,150,000
13,870,000
8,245,000
32,905,000
$51,675,000
PROBLEM 10-5A (Continued)
Bal.
Apr. 1
Bal.
Land
3,000,000 June 1
2,200,000
4,900,000
Bal.
Bal.
Buildings
26,500,000
26,500,000
Bal.
July 1
Bal.
Equipment
40,000,000 May 1
2,400,000 Dec. 31
41,150,000
300,000
750,000
500,000
Accumulated Depreciation—Buildings
Bal.
12,100,000
Dec. 31 adj.
530,000
Bal.
12,630,000
Accumulated Depreciation—Equipment
May 1
325,000 Bal.
5,000,000
Dec. 31
500,000 May 1
25,000
Dec. 31
50,000
Dec. 31 adj. 3,995,000
Bal.
8,245,000
PROBLEM 10-6A
(a) Accumulated Depreciation—Delivery
Equipment ..............................................................
Loss on Disposal .......................................................
Delivery Equipment ............................................
22,000
28,000
50,000
(b) Cash............................................................................
Accumulated Depreciation—Delivery
Equipment ..............................................................
Gain on Disposal ................................................
Delivery Equipment ............................................
31,000
(c) Cash............................................................................
Accumulated Depreciation—Delivery
Equipment ..............................................................
Loss on Disposal .......................................................
Delivery Equipment ............................................
18,000
22,000
3,000
50,000
22,000
10,000
50,000
PROBLEM 10-7A
(a) Jan. 2
Jan.June
Patents ...................................................
Cash................................................
Research and Development
Expense .............................................
Cash................................................
Sept. 1
Oct. 1
(b) Dec. 31
36,000
36,000
140,000
140,000
Advertising Expense .............................
Cash................................................
75,000
Copyright ...............................................
Cash................................................
80,000
Amortization Expense—Patents ..........
Patents ...........................................
10,000
75,000
80,000
10,000
[($60,000 X 1/10) + ($36,000 X 1/9)]
31
Amortization Expense—Copyright ......
Copyright........................................
4,000
4,000
[($36,000 X 1/10) + ($80,000 X 1/50 X 3/12)]
(c) Intangible Assets
Patents ($96,000 cost – $16,000 amortization) (1) ...............
Copyright ($116,000 cost – $18,400 amortization) (2) .........
Total intangible assets...................................................
(1) Cost ($60,000 + $36,000); amortization ($6,000 +
$10,000).
(2) Cost ($36,000 + $80,000); amortization ($14,400 +
$4,000).
(d) The intangible assets of the company consist of two patents
and two copyrights. One patent with a total cost of $96,000
is being amortized in two segments ($60,000 over 10 years
$ 80,000
97,600
$177,600
and $36,000 over 9 years); the other patent was obtained at
no recordable cost. A copyright with a cost of $36,000 is
being amortized over 10 years; the other copyright with a
cost of $80,000 is being amortized over 50 years.
PROBLEM 10-8A
1.
2.
Research and Development Expense ......................
Patents ................................................................
95,000
Patents .......................................................................
Amortization Expense—Patents .......................
[$6,100 – ($27,000 X 1/20)]
4,750
Goodwill .....................................................................
Amortization Expense—Goodwill .....................
800
95,000
Note: Goodwill should not be amortized because it has an
indefinite life unlike Patents.
4,750
800
PROBLEM 10-9A
(a)
Dirks Corp.
Hewes Corp.
(1) Asset turnover
ratio
$1,200,000
= .60 times
$2,000,000
$1,140,000
= .76 times
$1,500,000
(2) Return on assets
ratio
$400,000
= 20%
$2,000,000
$450,000
$1,500,000
(b)
Based on the asset turnover ratio, Hewes Corp. is more effective in u sing
assets to generate sales. Its asset turnover ratio is 27% higher than Dirks’ asset
turnover ratio. In addition, Hewes’ return on assets
ratio of 30% is 50% higher than Dirks’ 20% return.
= 30%