CHAPTER 10
SOLUTIONS TO PROBLEMS: SET B
PROBLEM 10–1B
Item
Land
Buildings
1
2
3
($ 5,000)
$490,000
PROBLEM 10–2B
(a)
Year
Computation
Accumulated
Depreciation
12/31
MACHINE 1
MACHINE 2
2013
2012
2013
2014
$100,000 X 10% = $10,000
$100,000 X 10% = $10,000
$100,000 X 10% = $10,000
$ 10,000
20,000
30,000
(b)
Year
Depreciation Computation
Expense
MACHINE 2
(2)
(1)
2013
$180,000 X 25% X 8/12 = $30,000
$30,000
PROBLEM 10–3B
(a) (1) Purchase price ……………………………………………………….. $ 58,000
Sales tax ………………………………………………………………… 2,750
Shipping costs ……………………………………………………….. 100
Equipment …………………………………………………. 61,000
Cash …………………………………………………… 61,000
(2) Recorded cost ………………………………………………………… $ 61,000
Less: Salvage value ……………………………………………….. 5,000
(b) (1) Recorded cost ………………………………………………………… $120,000
Less: Salvage value ……………………………………………….. 10,000
(2)
Year
Book Value at
Beginning of
Year
DDB Rate
Annual
Depreciation
Expense
Accumulated
Depreciation
2015
2016
$120,000
60,000
*50%*
*50%*
$60,000
30,000
$60,000
90,000
PROBLEM 10-3B (Continued)
(3) Depreciation cost per unit = ($120,000 – $10,000)/25,000 units =
$4.40 per unit.
Annual Depreciation Expense
2015: $4.40 X 5,500 = $24,200
(c) The units-of-activity method reports the lowest amount of depreciation
expense the first year while the declining-balance 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
PROBLEM 10–4B
Year
Depreciation
Expense
Accumulated
Depreciation
2013
2014
$45,000(a)
45,000
$ 45,000
90,000
(a)
$300,000 – $30,000
6 years
= $45,000
PROBLEM 10–5B
(a) Apr. 1 Land…………………………………………. 1,200,000
Cash ………………………………….. 1,200,000
May 1 Depreciation Expense ……………….. 15,000
Accumulated Depreciation—
Equipment ……………………… 15,000
($450,000 X 1/10 X 4/12)
June 1 Cash ………………………………………… 1,000,000
Land ………………………………….. 340,000
Gain on Disposal of
Plant Assets …………………… 660,000
July 1 Equipment ………………………………… 1,500,000
Cash ………………………………….. 1,500,000
PROBLEM 10-5B (Continued)
Cost $300,000
Accum. depreciation—
(b) Dec. 31 Depreciation Expense ………………… 400,000
Accumulated Depreciation—
Buildings ………………………… 400,000
($20,000,000 X 1/50)
31 Depreciation Expense ………………… 3,000,000
Accumulated Depreciation—
(c) TORREALBA COMPANY
Partial Balance Sheet
December 31, 2016
Plant Assets*
Land …………………………………………….. $ 2,860,000
Buildings ………………………………………. $20,000,000
Less: Accumulated depreciation—
buildings ……………………………. 8,400,000 11,600,000
PROBLEM 10-5B (Continued)
Land
Apr. 1 1,200,000
Bal. 2,860,000
Bal. 2,000,000
June 1 340,000
Buildings
Bal. 20,000,000
Bal. 20,000,000
Accumulated Depreciation—Buildings
Dec. 31 adj. 400,000
Bal. 8,400,000
Bal. 8,000,000
Equipment
July 1 1,500,000
Dec. 31 300,000
Bal. 30,750,000
Bal. 30,000,000
May 1 450,000
Accumulated Depreciation—Equipment
May 1 195,000
Dec. 31 300,000
Dec. 31 30,000
Dec. 31 adj. 3,000,000
Bal. 6,550,000
Bal. 4,000,000
May 1 15,000
PROBLEM 10–6B
(a) Accumulated Depreciation—Equipment ……………… 26,000
Loss on Disposal of Plant Assets ………………………. 19,000
Equipment …………………………………………………. 45,000
PROBLEM 10–7B
(a) Jan. 2 Patents ………………………………………….. 36,000
Cash ……………………………………….. 36,000
Jan.– Research and Development
June Expense …………………………………….. 230,000
Cash ……………………………………….. 230,000
(b) Dec. 31 Amortization Expense …………………….. 14,000
Patents ……………………………………. 14,000
[($100,000 X 1/10) + ($36,000 X 1/9)]
(c) Intangible Assets
Patents ($136,000 cost – $24,000 amortization) (1) …………. $112,000
Copyrights ($360,000 cost – $31,500 amortization) (2) ……. 328,500
Total intangible assets …………………………………………… $440,500
(d) The intangible assets of the company consist of two patents and two
copyrights. One patent with a total cost of $136,000 is being amortized
in two segments ($100,000 over 10 years and $36,000 over 9 years); the
PROBLEM 10–8B
1. Research and Development Expense …………………. 110,000
Patents ………………………………………………………. 110,000
PROBLEM 10–9B
(a)
Auer Corp.
Marte Corp.
Asset turnover
$1,050,000
$1,000,000
= 1.05 times
$945,000
$1,050,000
= .90 times