The document provides information and instructions to calculate depreciation for equipment purchased by a company using different depreciation methods. It gives the cost of the equipment, estimated salvage value, service life, production and working hours estimates, and actual production and working hours for 2004 and 2005. The student is asked to calculate depreciation expense for 2004 and 2005 using straight-line, units-of-output, working hours, sum-of-years digits, and declining balance methods based on this information. Depreciation is calculated for each year and method as instructed and the answers are provided in a solution section.
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Accounting for Properties Plant & Equipment
1. ASSIGNMENT
ON
Accounting for Properties,
Plant, & Equipment(Book Exercise)
Course Title: Intermediate Accounting
Course No: AIS-208
Submitted To:
Mr. Mohammed Ali Arshad Chowdhury
Assistant Professor
Department of Accounting
University of Chittagong.
Submitted By:
Group No: 8
BBA, 4th
Semester
Session: 2015-2016
Department of Banking and Insurance
University of Chittagong.
Date of Submission: December 13, 2017
2. Group no: 08
List of the group members:
NO NAME ID
01 Afsana Mimi 16306098
02 Shakawat Hossain 16306099
03 Md.Obaidul Hoque 16306100
04 Md.Shakibul Hasan 16306101
05 Nusrat Jahan 16306102
06 Imrul Hasan Chowdhury 16306103
07 Md.Sohag Miah 16306104
08 Md.Sakibul Islam 16306105
09 Rakib Murad 16306106
10 Md.Robiul Hasan Reza 16306107
11 Nusrat Jahan 16306108
12 Moheuddin Masum Chowdhury 16306109
3. 10-1: (Acquisition Costs of Realty) The following expenditures and receipts are related to
land, land improvements, and buildings acquired for use in a business enterprise. The receipts
are enclosed in parentheses.
a) Money borrowed to pay building contractor (signed a note) $275,000
b) Payment for construction from note proceeds 275,000
c) Cost of land fill and clearing 8,000
d) Delinquent real estate taxes on property assumed by purchaser 7,000
e) Premium on 6- month insurance policy during construction 6,000
f) Refund on 1-month insurance premium because construction completed early (1,000)
g) Architect`s fee on building 22,000
h) Cost of real estate purchased as a plant site (land $200,000 and building $50,000) 250,000
i) Commission fee paid to real estate agency 9,000
j) Installation of fences around property 4,000
k) Cost of razing and removing building 11,000
l) Proceeds from salvage of demolished building (5,000)
m) Interest paid during construction on money borrowed for construction 13,000
n) Cost of parking lots and driveways 19,000
o) Cost of trees and shrubbery planted (permanent in nature) 14,000
p) Excavation costs for new building 3,000
Instructions
Identify each item by letter and list the items in columnar form, using the headings shown
below. All receipt amounts should be reported in parentheses. For any amounts entered in the
others Accounts column also indicate the account title.
Item - land Land Improvements Building Others Accounts
Solution:
Item Land Land Improvements Building Other
Accounts
a) ($275,000)
Notes
Acquisition and Disposition of Property, Plant, and Equipment
4. payable
b) $275,000
c) $8,000
d) 7,000
e) 6000
f) (1000)
g) 22,000
h) 250,000
i) 9,000
j) $4,000
k) 11,000
l) (5,000)
m) 13,000
n) 19,000
o) 14,000
p) 3,000
E10-9: (Capitalization of Interest) On July 31, 2004, Amsterdam company engaged Misk
toaling company to construct a special purpose piece of factory machinery, construction was
begun immediately and was completed on November 1, 2001. To help finance construction,
on July Amsterdam issued a $ 300000, 3 year, 12% note payable at Netherlands National
Bank, On Which interest is payable each July 31 $ 200000 of the proceeds of the note was
paid to Minsk on July 31. The remainder of the proceeds was temporarily invested in short
term marketable securities at 10% until November 1. On November 1 Amsterdam made a
final $ 100000 payment of Minsk. Other than the note to Netherlands, Amsterdam’s only
outstanding liability at December 31, 2001, is a $ 30000, 8%, 6 year note payable, dated
January 1, 1998. On which interest is payable each December 31.
Instruction:
a. Calculate the interest revenue, weighted average accumulated expenditures, avoidable
interest and total interest cost to be capitalized during 2001. Round all computation to
the nearest dollar.
b. Prepare the Journal needed on the books of Amsterdam company at each of the tolling
dates:
i. July 31, 2004
5. ii. November 1, 2004
iii. December 31, 2004
Solution:
a) Computation of weighted Average Accumulated Expenditures are given here:
Expenditures × Current year
capitalization period
=
Weighted Average
Accumulated Expenditures
Date Amount
2004, July 31 200000 × 3/12 = 50000
November 1 100000 × 0/12 = 0
WAAE = 50000
Interest revenue:
[Investment on short term marketable securities 10%, $ 100000. Ended November 1]
= 100000 × 10% × 3/12
= 100000 × .10 × 3/12
= 2500
Weighted Average
Accumulated Expenditures
×
Interest Rate = Avoidable Interest
$ 50000 × 12% $ 6000
= $ 6000
Total Interest:
$ 300000, 3 year, 12% note
Payable = 300000 × 12% × 5/12 = 15000
$ 30000, 6 year, 8% note
Payable = 300000 × 8% = 2400
Total = 17400
Interest to be capitalized:
6. Capitalized interest = $ 6000
b) Preparing the Journal needed on the books of Amsterdam Company:
Journalizing the entries
Amsterdam Company
Date Particulars Ref. Debit Credit
2004
July, 1
Cash
Note Payable
300000
300000
Machinery
Trading Securities
Cash
200000
100000
300000
Nov. – 1 Cash
Interest Revenue
Trading Securities
102500
2500
100000
Machinery
Cash
100000
100000
Dec. – 31 Machinery
Interest Expense
Cash
Interest Payable
6000
11400
2400
15000
7. Depreciation, Impairments, and Depletion
E-11-3: Depreciation Computations__ SYD, DDB__ partial periods) Judds Company
purchased a new plants asset on April 1, 2004, at a cost of $711,000. It was estimated to have
a service life of 20 years and a salvage value of $60,000. Judd’s Accounting period is the
calendar year.
Instructions
a) Compute the depreciation for this asset for 2004 and 2005 using the sum-of-the-years-
digits method
b) Compute the depreciation for this asset for 2004 and 2005 using the double-declining
balance method.
SOLUTION:
a) Computation of depreciation for the asset for 2004-2005 using the sum- of-the-years-digits
method
1+2+......................+20
=n (n+1)/2
=20(20+1)/2
=210
1st year`s (full) depreciation:
= 20/210*651,000
=$62,000
Depreciation for 2004:
=62,000*9/12
=$46,500
Depreciation for 2005:
3 Months 62,000*3/12 = $ 15,500
9 Months 58,900*9/12 = $ 44,175
Total =$ 59,675
Here,
1st Year 2004 (April to December) = 9 Months
2005 (January to March) = 3 Months
8. 2nd Year (April 2005 to March 2006)
= 2005 (April to December) = 9 Months
= 2006 (January to March) = 3 Months
b) Computation of depreciation for the asset for 2004-2005 using the double-declining
balance method.
Rate of Depreciation:
= 100/20
= 5%
Applicable rate of depreciation:
= Normal Rate * 2
= 5 %*2
=10 %
Deprecation for 2004:
= (7, 11,000* 10%*9/12)
= $53,325
Depreciation for 2005:
= (7, 11,000-53,325)*10%
= $65,767.5
E11-4 :( Depreciation Computations)
Jon Secede Furnace Corp. purchased machinery for $315,000 on May 1,2014.It is estimated
that it will have useful life of 10 years, salvage value of $15,000,production of 240,000 units,
and working hours of 25,000.During 2015,Seceda Corp. uses the machinery for 2,650 hours,
and the machinery produces 25,500 units.
Instructions
From the information given, compute the depreciation charge for 2015 under each of the
following methods.
a) Straight-line
b) Units-of-output
c) Working hours
d) Sum-of-years-digits
e) Declining-balance.
9. Solution:
We are given,
Date of acquisition: May1, 2014
Purchase price: $315,000
Useful life: 10 years
Salvage value: $15,000
Working hours: 25,000
Estimated production: 240,000 units
During 2015 production: 25,500 units
During 2015 working hours: 2,650
a) Straight line method:
Depreciation per year= cost-salvage value ÷ useful life
= ($315,000-$15,000) ÷ 10 years
= $30,000
b) Units-of-output method:
Per unit depreciation= cost-salvage value ÷ estimated production
= $315,000-$15,000/240,000 units
=1.25
Depreciation for 2015 = production × depreciation per unit
= 25,500×1.25
= $31,875
10. c) Working hour method:
Depreciation per hour= (cost-savage value) ÷ working hours
= ($315,000 - $15,000) ÷25,000
= $12
Depreciation for 2015 = actual usage × depreciation per hour
= 2,650×12
= $31,800
d) Sum-of-years-digits method:
Sum of the estimated useful years=1+2+3+4+5+6+7+8+9+10
=n (n+1) ÷2
=10(10+1) ÷2
=55
1st year (full) depreciation = 10/55×$300,000
= $54,545
Depreciation for 2014(8 months) = $54,545×8/12
= $36,363
Depreciation for 2015 (4 month) = 9/55×$300,000×4/12
= $16,363
Total depreciation for 2015 = ($36,363+$16363)
= $52,726
e) Double declining balance method:
Rate of depreciation= 100÷estimated life
=100÷10
=10%
11. Applicable rate of depreciation = normal rate×2
=10%×2
=20%
Depreciation for 2014 (8 month) = $315,000×20%×8/12
=$ 42,000
Depreciation for 2015 (4 month) = ($315,000-$42,000) ×20%×4/12
= 18,200
Total depreciation for 2015 = ($42,000+$18,200)
= $60,200
P11-2 (Depreciation for Partial Periods - SL, Act. SYD, and DDB)
The cost of equipment purchased by Boris Becker, Inc., on June 1, 2004 is $67,000. It’s
estimated that the machine will have a $4,000 salvage value at the end of the service life. Its
service life is estimated at 7 years; its total working hours are estimated at 42,000 and its total
production is estimated at 525,000 units. During 2004 the machine was operated 6,000 hours
and produced 55,000 unit. During 2005 the machine was operated 5,500 hours and produced
48,000 units.
Instructions:
Compute depreciation expense on the machine for the year ending December 31, 2004, and
the year ending December 31, 2005, using the following methods.
(a) Straight-line.
(b)Units-of-output.
(c) Working hours.
(d)Sum-of-the-years’-digits.
(e) Declining balance (twice the straight-line rate)
12. Answer to the Q no. (a)
Straight-line Method:
Depreciation per year =
Cost − Salvage value
Useful life
=
$67,000−$4000
7
= $9,000
Depreciation for 2004 = ($9,000 × 7/12)
= $5,250
Depreciation for 2005 = $9,000
Answer to the Question no. (b)
Units-of-output Method:
Per unit depreciation =
Cost − Salvage value
Estimated production
=
$67,000 −$4,000
525,000
unit
= $.12 unit
Depreciation for 2004 = Actual production × Depreciation per unit
= (55,000 units × $.12 per unit)
= 6,600 units
Depreciation for 2005 = Actual production × Depreciation per unit
= (48,000 units × $.12 per unit)
= 5,760 units
13. Answer to the Question no. (c)
Working hours Method:
Per hour depreciation =
Cost − Salvage value
Estimated working hour
=
$67,000 −$4,000
42,000
hours
= $1.5 hours
Depreciation of 2004 = (Actual working hour × Depreciation per hour)
= (6,000 × $1.5)
= 9,000
Depreciation 0f 2005 = (Actual working hour × Depreciation per hour)
= (5,500 × $1.5)
= $8,250
Answer to the Question no. (d)
Sum-of-the-years’-digits Method:
Sum of the estimated useful years:-
= (1+2+3+4+5+6+7) or
n(n+1)
2
=
7(7+1)
2
= 28
Depreciation for 2004 (7 month) = (7/28 ×$63,000 × 7/12)
= $9,188
Depreciation for 2005 = (7/28 ×$63,000 × 5/12) + (6/28 × $63,000 ×7/12)
= $14,438
14. Answer to the Question no. (e)
Declining balance (twice the straight-line rate)
First we have to find out rate,
=
100
7
= 14.285%
Applicable rate of depreciation = (Normal rate × 2)
= (14.285% × 2)
= 28.57%
Depreciation of 2004 = ($67,000 × 28.57% × 7/12)
= $11,166
Depreciation of 2005 = ($67,000 – $11,166) × 28.57%
= $15,952
[Reference: Intermediate Accounting – Eleventh edition by Kieso, Weygandt, and
Warfield)