CHAPTER 11
Depreciation, Impairments, and Depletion
SOLUTIONS TO B PROBLEMS
PROBLEM 11-1
(a) 1. Depreciable Base Computation:
Purchase price …………………………
$85,000
Less: Purchase discount (2%) …..
1,700
Installation ……………………………….
3,800
Less: Salvage value …………………
2. Sum-of-the-years’-digits for 2013
Machine Year
Total
Depreciation
2012
2013
1
8/36 X $86,400 =
$19,200
$12,800*
$ 6,400**
2
7/36 X $86,400 =
(b) An activity method.
PROBLEM 11-2
Depreciation Expense
2012
2013
(a)
Straight-line:
($89,000 – $5,000) ÷ 7 = $12,000/yr.
2012: $12,000 X 7/12
$7,000
2013: $12,000
$12,000
(b)
($89,000 – $5,000) ÷ 525,000 units = $.16/unit
2012: $.16 X 55,000
2013: $.16 X 48,000
(c)
Working hours:
($89,000 – $5,000) ÷ 42,000 hrs. = $2.00/hr.
2012: $2.00 X 6,000
12,000
2013: $2.00 X 5,500
11,000
(d)
Sum-of-the-years’-digits:
1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 or
n(n + 1)
=
7(8)
= 28
2
2
2012: 7/28 X $84,000 X 7/12
2013: 7/28 X $84,000 X 5/12 = $ 8,750
6/28 X $84,000 X 7/12 = 10,500
$19,250
(e)
Declining-balance:
Rate = 2/7
2012: 7/12 X 2/7 X $89,000
14,833
2013: 2/7 X ($89,000 – $14,833) = $21,191
2013: 5/12 X $10,595
($89,000 – $25,428) = 10,595
$21,190*
PROBLEM 11-3
(a)
Depreciation Expense ……………………………………………..
3,900
Accumulated Depreciation—Machinery (A)
(5/55 X [$46,000 – $3,100]) …………………………….
3,900
Accumulated Depreciation—Machinery (A) ………………
35,100
Machinery (A) ($46,000 – $13,000) ……………………
33,000
Gain on Disposal of Machinery ………………………..
2,100
(b)
Depreciation Expense ……………………………………………..
6,720
Accumulated Depreciation—Machinery (B)
([$51,000 – $3,000] ÷ 15,000 X 2,100) ……………..
(c)
Depreciation Expense ……………………………………………..
6,000
Accumulated Depreciation—Machinery (C)
([$80,000 – $15,000 – $5,000] ÷ 10) …………………
6,000
(d)
Machinery (E) ………………………………………………………...
28,000
Retained Earnings …………………………..……………..
28,000
Accumulated Depreciation—Machinery (E) ………
5,600
*($28,000 X .20)
Net
Income
As Adjusted
Retained
Acc. Dep.,
Per Company Books
Retained
Acc. Dep.
(a)
PROBLEM 11-4 (Continued)
3Book value of Truck #1 [$18,000 – ($18,000/5 X 4 yrs.)] =
$18,000 – $14,400 ……………………………………………………
= $3,600
Cash received on sale ……………………………………………………..
Loss on sale …………………………………………………………..
$ 100
$22,000/5
=
$24,000/5
=
$40,000/5
=
Total
5Book value of Truck #4 $24,000 – [($24,000/5 X 3 yrs.)] …….
= $9,600
Cash received ($700 + $2,500) ………………………………………….
= 3,200
Loss on disposal …………………………………………………….
$6,400
6Truck #2:
$22,000/5 X 1/2
=
$24,000/5 X 1/2
=
$40,000/5
$42,000/5 X 1/2
=
Total
7Truck #2:
(fully dep.)
=
$40,000/5
=
$42,000/5
=
Total
(b)
Compound journal entry December 31, 2013:
Accumulated Depreciation—Trucks ………………………
Trucks …………………………………………………………
Retained Earnings …………………………..…………..
Depreciation Expense …………………………………
PROBLEM 11-4 (Continued)
Summary of Adjustments:
Per
Books
As
Adjusted
Adjustment
Dr. or (Cr.)
Trucks
$152,000
$104,000
$(48,000)
Accumulated Depreciation
$129,150
$ 62,600
$ 66,550
Retained Earnings, 2010
Retained Earnings, 2012
Totals
$ 67,850
$ 63,300
$ (4,550)
Depreciation Expense, 2013
$ 30,400
$ 16,400
$(14,000)
PROBLEM 11-5
(a) Estimated depletion:
Estimated Depletion
Depletion
Base
Estimated
Yield
Per
Ton
1ST & 11th
Yrs.
Each of Yrs.
2-10 Incl.
$870,000*
120,000 tons
$7.25
$43,500**
$87,000***
Estimated depreciation:
Asset
Cost
Per ton
Mined
1st
Yr.
Yrs.
2–5
6th
Yr.
Yrs.
7–10
11th
Yr.
Building
$36,000
$.30*
$1,800
$3,600
$3,600
$3,600
$1,800
Machinery (1/2)
0
0
$1,500
Total depreciation
(b) Depletion: $7.25 X 5,000 tons = $36,250
PROBLEM 11-6
(a)
Original cost
$550 X 3,000 =
$1,650,000
Deduct residual value of land
$200 X 3,000 =
600,000
Cost of logging road
150,000
$1,200,000
(b)
Inventory ………………………………………………………………..
240,000
Timber ……………………………………………………………
240,000
Depletion, 2012: 20% X 500,000 bd. ft. = 100,000 bd. ft.;
100,000 bd. ft. X $2.40 = $240,000
(c)
[$1,050,000 – ($1,050,000 X 20%)] ……………………
$ 840,000
Cost of salvaging timber …………………………………..
700,000
Less recovery ($3 X 400,000 bd. ft.) ……………………
$ 340,000
Loss of land value …………………………………………….
600,000
Loss of logging roads
[($150,000 – (20% X $150,000)] ………………….
120,000
Logging equipment …………………………………………..
300,000
Extraordinary loss due to the eruption
Loss of timber
PROBLEM 11-7
Cost per acre
$1,700
Land Cost
800
Timber Cost
$ 900
10,000 acres
Road Cost
Total Depletion Base
Estimated Depletion for 2012
$9,250,000
(540,000/6,750,000)
Depletion Expense for 2012
Instructors should note the changing depletion base in this problem.
2012
Computation of Depletion Base for 2012
Depreciation of Removable Equipment
Cost
$ 225,000
Salvage Value
(9,000)
Depreciable base
$ 216,000
Annual Depreciation using SL ($216,000/15)
$ 14,400
Depreciation Expense for 2012
$ 10,800
(9/12 X $14,400)
Depletion Base for 2013
Base for 2012
Less Depletion for 2012
Plus Seedling Planting Costs
Depletion Base for 2013
Depletion Base for 2013
Times
(774,000/6,450,000)
Depletion for 2013
Depreciation Expense for 2013
PROBLEM 11-7 (Continued)
2014
Depletion Base for 2014
Base for 2013
$ 8,630,000
Less: Depletion for 2013
1,035,600
Plus: Seedling Planting Costs
Depletion Base for 2014
$ 7,744,400
Depletion Base for 2014
Times
(650,000/6,500,000)
Depletion for 2014
Depreciation Expense for 2014
$ 14,400
PROBLEM 11-8
(a) The amounts to be recorded on the books of Darby Sporting Goods
Inc. as of December 31, 2012, for each of the properties acquired from
Encino Athletic Equipment Company are calculated as follows:
Cost Allocations to Acquired Properties
Appraisal
Value
Remaining
Purchase
Price
Allocations
Renovations
Capitalized
Interest
Total
(1) Land
$290,000
$290,000
$290,000
$521,000
Supporting Calculations
1Balance of purchase price to be allocated.
Total purchase price …………………………………………………..
$400,000
Machinery
PROBLEM 11-8 (Continued)
2Capitalizable interest.
Expenditures
Capitalization
Period
Weighted-Average
Accumulated Expenditures
Date
Amount
1/1
$ 50,000
12/12
$ 50,000
4/1
90,000
$500,000
$175,000
(b) Darby Sporting Goods Inc.’s 2013 depreciation expense, for book
purposes, for each of the properties acquired from Encino Athletic
Equipment Company is as follows:
1.
Land: No depreciation.
2013 depreciation expense
= Cost X Rate X 1/2 year
3.
Machinery: Depreciation rate
= 2.00 X 1/5 = .40
PROBLEM 11-8 (Continued)
(c) Arguments for the capitalization of interest costs include the following.
1. Diversity of practices among companies and industries called for
Arguments against the capitalization of interest include the following:
1. Interest capitalized in a period would tend to be offset by amorti–
PROBLEM 11-9
(a) Carrying value of asset: $10,000,000 – $2,500,000* = $7,500,000.
*($10,000,000 ÷ 8) X 2
(b) Depreciation Expense …………………………………….. 1,400,000**
Accumulated Depreciation—Equipment ….. 1,400,000
**($5,600,000 ÷ 4)
(c) No depreciation is recorded on impaired assets to be disposed of.
Recovery of impairment losses are recorded.
PROBLEM 11-10
(1)
$80,000
Allocated in proportion to appraised values
(1/10 X $800,000).
(2)
$720,000
Allocated in proportion to appraised values
(9/10 X $800,000).
(3)
Fifty years
Cost less salvage ($720,000 – $40,000) divided by
annual depreciation ($13,600).
(4)
$13,600
Same as prior year since it is straight-line 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)
$40,000
Fair value.
(8)
$6,000
Cost ($40,000) times percentage (1/10 X 150%).
(9)
$5,100
equals $34,000. Multiply $34,000 times 15%.
(10)
$168,000
Total cost ($182,900) less repairs and maintenance
($14,900).
(11)
$36,000
Cost less salvage ($168,000 – $6,000) times 8/36.
(12)
$10,500
Cost less salvage ($168,000 – $6,000) times 7/36 times
one-third of a year.
PROBLEM 11-10 (Continued)
(13)
$52,000
obtain $46,260, and then add the $5,740 down payment.
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
(14)
$2,600
Cost ($52,000) divided by estimated life (20 years).
PROBLEM 11-11
(a)
1.
Straight-line Method:
$90,000 – $6,000
= $16,800 a year
5 years
2.
Activity Method:
$90,000 – $6,000
= $.84 per hour
100,000 hours
Year
2010
20,000 hrs. X $.84 =
$16,800
2011
25,000 hrs. X $.84 =
21,000
2012
15,000 hrs. X $.84 =
2013
30,000 hrs. X $.84 =
2014
10,000 hrs. X $.84 =
3. Sum-of-the-Years’-Digits: 5 + 4 + 3 + 2 + 1 = 15
Year
2010
5/15 X ($90,000 – $6,000) =
$28,000
2011
4/15 X $84,000 =
22,400
2012
3/15 X $84,000 =
2013
2/15 X $84,000 =
11,200
2014
1/15 X $84,000 =
4. Double-Declining-Balance Method: Each year is 20% of its total
life. Double the rate to 40%.
Year
2010
40% X $90,000 =
$36,000
2011
40% X ($90,000 – $36,000) =
2012
40% X ($90,000 – $57,600) =
2013
40% X ($90,000 – $70,560) =
2014
Enough to reduce to salvage =
(b) 1. Straight-line Method:
Year
2010
$90,000 – $6,000
X 9/12 =
$12,600
5 years
2011
Full year
16,800
2012
Full year
2013
Full year
16,800
2015
Full year X 3/12 year =
PROBLEM 11-11 (Continued)
2. Sum-of-the-Years’-Digits:
2010
(5/15 X $84,000) X 9/12 =
$21,000
2011
(5/15 X $84,000) X 3/12 =
$ 7,000
(4/15 X $84,000) X 9/12 =
16,800
23,800
2012
(4/15 X $84,000) X 3/12 =
(3/15 X $84,000) X 9/12 =
12,600
18,200
2013
(3/15 X $84,000) X 3/12 =
(2/15 X $84,000) X 9/12 =
12,600
2014
(2/15 X $84,000) X 3/12 =
(1/15 X $84,000) X 9/12 =
2015
(1/15 X $84,000) X 3/12 =
3. Double-Declining Balance Method:
Year
Cost
Accum.
Depr. at
beg. of
year
Book
Value at
beg. of
year
Depr.
Expense
2010
$90,000
—
$90,000
$27,000 (1)
2011
25,200 (2)
2014
5,443 (5)
8,165
(1) $90,000 X 40% X 9/12
(2) ($90,000 – $27,000) X 40%
*PROBLEM 11-12
(a) The straight-line method would provide the highest total net income
Computations of depreciation expense and accumulated depreciation under
various assumptions:
(1) Straight-line:
$1,260,000 – $60,000
= $240,000
5 years
Year
Depreciation
Expense
Accumulated
Depreciation
2011
$240,000
$ 240,000
2012
$ 480,000
2013
$ 720,000
$720,000
(2) Double-declining-balance:
Year
Depreciation
Expense
Accumulated
Depreciation
2011
$504,000
(40% X $1,260,000)
$ 504,000
2012
(40% X $756,000)
$ 806,400
2013
(40% X $453,600)
$ 987,840
$987,840
(3) Sum-of-the-years’-digits:
Year
Depreciation
Expense
Accumulated
Depreciation
2011
$400,000
(5/15 X $1,200,000)
$ 400,000
2012
(4/15 X $1,200,000)
$ 720,000
2013
(3/15 X $1,200,000)
$ 960,000
$960,000
*PROBLEM 11-12 (Continued)
(4) Units-of-output:
Year
Depreciation
Expense
Accumulated
Depreciation
2011
$288,000
($24* X 12,000)
$288,000
2013
($24 X 10,000)
$792,000
$792,000
*$1,200,000 ÷ 50,000 (total units) = $24 per unit
(b) General MACRS method:
Total Cost
MACRS
Rates (%)*
Annual
Depreciation
Accumulated
Depreciation
2011
$1,260,000
X
14.29
=
$180,054
$180,054
2012
X
=
2013
X
=
*Taken from the MACRS rates schedule.
Optional straight-line method:
Total Cost
Depreciation
Rate
Annual
Depreciation
Accumulated
Depreciation
2011
$1,260,000
X
(1/7 X 1/2)
=
$ 90,000
$ 90,000
2012
X
=
$270,000
2013
X
=
$450,000
The general MACRS method would have higher depreciation expense
($709,002) than that of the optional straight-line method ($450,000) for the