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210CHAPTER 16
Problem 16-1
1. C2. D
3. D
4. D
5. B
Problem 16-2Land Building
Cash paid for land and old building 1,000,000
Removal of old building 50,000
Payment to tenants of old building to vacate premises 15,000
Architect fee 200,000
Building permit 30,000
Fee for title search 10,000Survey before construction 20,000
Excavation 100,000
Cost of new building constructed 6,000,000
Assessment fee 5,000
Cost of grading, leveling and landfill 45,000
Driveways and walks 40,000
Temporary quarters for construction crew 80,000
Temporary building to house tools and materials 60,000
Cost of construction changes _________ 50,000
1,145,000 6,560,000
Note: The cost of replacing windows is treated as expense.
Problem 16-3Land
Land Building improvement
Cost of land 2,000,000
Legal fees 10,000
Payment of mortgage 50,000
Payment of taxes 20,000
Cost of razing building 30,000
Proceeds from sale of materials ( 5,000)
Grading and drainage 15,000
Architect fee 200,000
Payment to contractor 8,000,000
Interest cost 300,000Driveway and parking lot 40,000
Cost of trees, shrubs and other landscaping 55,000
Cost of installing lights in parking lot 5,000
Premium for insurance _______ 25,000 _______
2,120,000 8,525,000 100,000
The payment for medical bills and the cost of open house party are outright expenses
because they are not a necessary cost of acquiring the land and building.
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211Problem 16-4
Office Factory Land
Land building building improvements
Purchase price 1,300,000 700,000Materials 3,200,000
Excavation 100,000
Labor 2,500,000
Remodeling 200,000
Cash discounts ( 60,000)
Supervision 30,000
Compensation insurance 50,000
Clerical and other expenses 30,000
Paving of streets 40,000
Plans and specifications 150,000
Legal cost - land 10,000 ________ ________ ______
1,310,000 900,000 6,000,000
40,000
1. The imputed interest on corporation’s own money is not capitalizable.
2. The payment of claim for injuries not covered by insurance and the legal cost of injury
claim are treated as expense.
3. Saving on construction is not recognized.
Problem 16-5
Taxes in arrears 50,000
Payment for land 1,000,000
Demolition of old building 100,000
Total cost of land 1,150,000
Architect fee 230,000
Payment to city hall 120,000
Contract price 5,000,000
Safety fence around construction site 35,000
Safety inspection on building 30,000
Removal of safety fence 20,000
Total cost of factory building 5,435,000
Problem 16-6
Purchase price 3,000,000
Title clearance fee 50,000Cost of razing old building 100,000
Scrap value of old building ( 10,000)
Total cost of land 3,140,000
Construction cost of new building 8,000,000
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212
Problem 16-7
Land BuildingPurchase price 1,000,000 4,000,000
Remodeling 150,000
Salvage materials ( 5,000)
Grading, leveling and other permanent improvement 50,000
Repairs ________ 10,000
1,050,000 4,155,000
The repairs are capitalized because they are necessary prior to the occupancy and intended
use of the building.
Problem 16-8
Land Building Machinery
Fair value 1,500,000 5,000,000
Repairs 200,000
Remodeling 300,000
Invoice price 1,000,000
Discount ( 20,000)
Base _________ _________ 50,000
1,500,000 5,500,000 1,030,000
The driveway and parking lot are charged to land improvements.
Problem 16-9
Land Building MachineryFair value 1,500,000 4,000,000 1,500,000
Repairs 200,000
Special tax assessment 30,000
Platform 70,000
Remodeling 400,000
Purchase price 800,000
Discount ( 40,000)
Freight 20,000
Installation _________ _________ 30,000
1,530,000 4,600,000 2,380,000
Problem 16-10
Purchase price 2,000,000 Contract price 6,000,000
Commission 100,000 Plans, specification
Legal fees 50,000 and blueprint 100,000
Title guarantee 10,000 Architectural fee 250,000
Cost of razing old building 75,000 Cost of new building 6,350,000
Salvage value of materials ( 5,000)
Cost of land 2,230,000
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213
Problem 16-11
LeaseholdLand Building improvements Machinery
Balances, Jan. 1 1,500,000 4,000,000 500,000 1,000,000
Acquisition of land - #621:
Purchase price 3,000,000
Commission 60,000
Clearing cost 15,000
Sale of t imber and gravel ( 5,000)
Acquisition of land - #622:
Purchase price 4,000,000
Cost of demolition 300,000
New building:
Construction cost 5,000,000
Excavation fee 50,000
Architectural design 150,000
Building permit 40,000
Improvements:
Electrical work 350,000
Construction extension
(800,000 x 1/2) 400,000
Improvements on office space 650,000
Purchase of new machine:
Invoice price 1,750,000
Freight 20,000
Unloading charge _________ _________ _________ __ 30,000Balances, December 31 8,870,000 9,240,000 1,900,000 2,800,000
The third tract of land should be presented as current asset because it was “classified as held
for sale”.
Problem 16-12Land
Land improvements Building Machinery
Balances, Jan. 1 3,500,000 900,000 7,000,000 1,500,000
Land acquired 1,250,000
Issuance of share capital:12/36 x 4,500,000 1,500,000
24/36 x 4,500,000 3,000,000
New machinery 3,400,000
New parking lot, street and
sidewalk 750,000
Machinery sold ________ ________ _________ ( 500,000)
Balances, Dec. 31 6,250,000 1,650,000 10,000,000 4,400,000
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The “assessed values” do not represent the fair values of the land and building but are used in
allocating the market value of the share capital.
214
Problem 16-13 Problem 16-14
Invoice price 3,000,000 Invoice cost 4,000,000
Cash discount ( 150,000) Discount (5% x 4,000,000) ( 200,000)
Freight 50,000 Transportation 40,000
Installation cost 30,000 Installation 100,000
Testing cost 20,000 Trial run-salary of engineer 50,000
2,950,000 Cash allowance ( 60,000)
3,930,000
Problem 16-15
Cost paid (896,000 – 96,000) 800,000
Cost of transporting machine 30,000
Installation cost 50,000
Testing cost 40,000
Safety rails and platform 60,000
Water device 80,000
Cost of adjustment 75,000
Estimated dismantling cost 65,000
Total cost of machine 1,200,000
Note that the estimated dismantling cost is capitalized because the company has a present
obligation as required by contract. In the absence of a present obligation, the estimated
dismantling cost is not capitalized.
Problem 16-16
Second hand market value 2,400,000
Overhaul and repairs 150,000
Installation 80,000
Testing 110,000
Hauling 10,000
Safety device 250,000
3,000,000Problem 16-17
1. Materials 600,000
Labor 400,000
Installation 60,000
Trial run 30,000
Discount ( 40,000)
Overhead 150,000
1,200,000
2. Adjusting entries:
1. Loss on retirement of old machinery 6,000
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Machinery (20,000 – 14,000) 6,000
2152. Purchase discount 40,000
Machinery 40,000
3. Machinery 150,000
Factory overhead 150,000
4. Profit on construction 100,000
Machinery 100,000
5. Tools 90,000
Machinery 90,000
6. Depreciation – tools 10,000
Tools (90,000 / 3 x 4/12) 10,000
7. Machinery 128,600
Accumulated depreciation 40,000
Depreciation – machinery 88,600
Depreciation recorded 128,600
Correct depreciation (1,200,000 / 10 x 4/12) 40,000
Overdepreciation 88,600
Problem 16-18
Initial design fee 150,000
Executive chairs and desks 200,000
Storm windows and installation 500,000
Installation of automatic door opening system 200,000
Overhead crane 350,000
Total capital expenditures 1,400,000
Problem 16-191. Accumulated depreciation 400,000
Loss on retirement of building 1,600,000
Building 2,000,000
Building 2,500,000
Cash 2,500,000
Depreciation (8,100,000 / 20) 405,000
Accumulated depreciation 405,000
Building (9,000,000 + 2,500,000 – 2,000,000) 9,500,000
Accumulated depreciation (1,800,000 – 400,000) 1,400,000
Book value 8,100,000
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2. Accumulated depreciation (1,960,000 x 20%) 392,000
Loss on retirement of building 1,568,000
Building (2,500,000 x .784) 1,960,000
216
Building 2,500,000Cash 2,500,000
Depreciation (8,132,000 / 20) 406,600
Accumulated depreciation 406,600
Building (9,000,000 – 1,960,000 + 2,500,000) 9,540,000
Accumulated depreciation (1,800,000 – 392,000) 1,408,000
Book value 8,132,000
Problem 16-20
a. Annual depreciation (8,400,000 / 30) 280,000
Age of building (7,000,000 / 280,000) 25 years
b. Building 2,500,000
Cash 2,500,000
c. Building (8,400,000 + 2,500,000) 10,900,000
Less: Accumulated depreciation 7,000,000
Book value 3,900,000
d. Depreciation (3,900,000 / 15) 260,000
Accumulated depreciation 260,000
Original life 30
Less: Expired life 25
Remaining useful life, beginning of current year 5
Add: Extension in life 10
Revised useful life 15
Problem 16-21
1. Building 10,500,000
Cash 10,500,000
2. Depreciation 200,000Accumulated depreciation 200,000
3. Building 3,000,000
Cash 3,000,000
Accumulated depreciation (2,500,000 / 50 x 2) 100,000
Loss on retirement of building 2,400,000
Cash 2,500,000
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4. Depreciation (10,700,000 – 500,000 / 48) 212,500
Accumulated depreciation 212,500
217
Building (10,500,000 + 3,000,000 – 2,500,000) 11,000,000
Accumulated depreciation (400,000 – 100,000) 300,000
Book value – 1/1/2008 10,700,000
Problem 16-22
1. Machinery 5,000,000
Cash 5,000,000
2. Depreciation 450,000
Accumulated depreciation 450,000
3. Depreciation (3,600,000 / 6) 600,000
Accumulated depreciation 600,000
Cost 5,000,000
Accumulated depreciation:
2005 450,000
2006 450,000 900,000
Book value 4,100,000
Residual value 500,000
Remaining depreciable cost – 1/1/2007 3,600,000
4. Machinery 300,000
Cash 300,000
5. Depreciation (3,300,000 / 5) 660,000
Accumulated depreciation 660,000
Cost 5,300,000
Accumulated depreciation (900,000 + 600,000) 1,500,000
Book value – 1/1/2008 3,800,000
Residual value 500,000
Remaining depreciable cost – 1/1/2008 3,300,000
Problem 16-23
1. Depreciation (60,000 x 3/12) 15,000
Accumulated depreciation 15,000
Accumulated depreciation (480,000 + 15,000) 495,000
Loss on retirement of store equipment 105,000
Store equipment 600,000
2. Depreciation (150,000 x 4/12) 50,000
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Accumulated depreciation 50,000
218
Cash 100,000
Accumulated depreciation (1,050,000 + 50,000) 1,100,000
Loss on sale of office equipment 300,000
Office equipment 1,500,000
3. Depreciation (600,000 x 5/12) 250,000
Accumulated depreciation 250,000
Delivery equipment – new 5,000,000
Accumulated depreciation 2,650,000
Cash (5,000,000 – 750,000) 4,250,000
Delivery equipment – old 3,000,000Gain on exchange (750,000 – 350,000) 400,000
Original cost 3,000,000
Less: Accumulated depreciation to date (2,400,000 + 250,000) 2,650,000
Book value 350,000
4. Accumulated depreciation 1,200,000
Office equipment 1,200,000
5. Depreciation (900,000 x 9/12) 675,000
Accumulated depreciation 675,000
Accumulated depreciation (2,700,000 + 675,000) 3,375,000
Fire loss 1,125,000
Machinery 4,500,000
Problem 16-24
1. Discount on bonds payable 500,000
Machinery 500,000
Interest expense (500,000 / 10 x 9/12) 37,500
Discount on bonds payable 37,500
Accumulated depreciation 75,000Depreciation 75,000
Depreciation for 9 months 600,000
Depreciation for 12 months (600,000 / 9/12) 800,000
Depreciable cost (800,000 x 5 years) 4,000,000
Per book Adjusted
Cost 5,000,000 4,500,000
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Less: Residual value 1,000,000 1,000,000
Depreciable cost 4,000,000 3,500,000
219
Correct depreciation for 9 months (3,500,000 / 5 x 9/12) 525,000
Less: Depreciation recorded 600,000
Overstatement 75,000
2. Interest expense 300,000
Machinery (3,500,000 – 3,200,000) 300,000
Machinery 150,000
Freight in 150,000
Accumulated depreciation 30,000Depreciation 30,000
Depreciation per book 700,000
Correct depreciation (3,350,000 / 5) 670,000
Overstatement 30,000
3. Loss on exchange 390,000
Machinery 390,000
Cost per book 3,000,000
Correct cost
Trade in value 150,000Add: Cash paid 2,460,000 2,610,000
Overstatement 390,000
Trade in value 150,000
Less: Book value 540,000
Loss on exchange (390,000)
4. Allowance for doubtful accounts 840,000
Loss on exchange – accounts receivable 60,000
Treasury share 900,000
Per book
Machinery 4,200,000Accounts receivable 4,200,000
Treasury shares 4,200,000
Machinery 4,200,000
Should be
Machinery 3,300,000
Allowance for doubtful accounts (20% x 4,200,000) 840,000
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Loss on accounts receivable 60,000
Accounts receivable 4,200,000
220
Treasury shares 3,300,000
Machinery 3,300,000
The cost of treasury shares acquired for noncash consideration is usually
measured by the recorded amount of the noncash asset surrendered (SFAS No. 18).
Problem 16-25 Answer A
Allocated cost of land (2,400,000 / 6,000,000 x 5,500,000) 2,200,000
Property taxes (2,400 / 6,000 x 250,000) 100,000Cost of survey 5,000
Total cost of land 2,305,000
Incidentally, the cost of the building is:
Allocated cost (3,600 / 6,000 x 5,500,000) 3,300,000
Property taxes (3,600 / 6,000 x 250,000) 150,000
Renovation 500,000
Total cost of building 3,950,000
Problem 16-26 Answer A
Purchase price 4,000,000
Payments to tenants 200,000
Demolition of old building 100,000
Legal fees 50,000
Title insurance 30,000
Proceeds from sale of materials ( 10,000)
Total cost of land 4,370,000
Problem 16-27 Answer D
Land Building
Purchase price of land 600,000
Legal fees for contract 20,000Architect fee 80,000
Demolition of old building 50,000
Construction cost _______ 3,500,000
Total cost 670,000 3,580,000
Problem 16-28 Answer D
Acquisition price 7,000,000
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Option of building acquired 200,000
Repairs 500,000
Total cost 7,700,000
221
Problem 16-29 Answer D
Purchase price 250,000
Shipping 5,000
Installation 10,000
Testing 35,000
Total cost 300,000
Problem 16-30 Answer A
Problem 16-31 Answer A
All expenditures are capitalized.
Problem 16-32 Answer A
All costs are capitalized.
Problem 16-33 Answer C
Continuing and frequent repairs 400,000
Repainting of the plant building 100,000Partial replacement of roof tiles 150,000
Repair and maintenance expense 650,000
Problem 16-34 Answer B
Problem 16-35 Answer B
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222
CHAPTER 17
Problem 17-1 Problem 17-2
1. A 1. C
2. D 2. A
3. B 3. D
4. D 4. D
5. D 5. D
6. D 6. B
7. D 7. C
8. C 8. B
9. C 9. A
10. B 10. A
Problem 17-3
Depreciation Table – Straight Line
Accumulated
Year Particular Depreciation depreciation Book value
Acquisition cost 635,000
2008 120,000 120,000 515,000
2009 120,000 240,000 395,000
2010 120,000 360,000 275,000
2011 120,000 480,000 155,000
2012 120,000 600,000 35,000
600,000
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Depreciation Table – Service Hours Method
Accumulated
Year Particular Depreciation depreciation Book value
Acquisition cost 635,000
2008 14,000 x 10 140,000 140,000 495,000
2009 13,000 x 10 130,000 270,000 365,0002010 10,000 x 10 100,000 370,000 265,000
2011 11,000 x 10 110,000 480,000 155,000
2012 12,000 x 10 120,000 600,000 35,000
600,000
Depreciation rate per hour = 600,000 / 60,000 = 10
223
Depreciation Table – Production Method
Accumulated
Year Particular Depreciation Depreciation Book value
Acquisition cost 635,000
2008 34,000 x 4 136,000 136,000 499,000
2009 32,000 x 4 128,000 264,000 371,000
2010 25,000 x 4 100,000 364,000 271,000
2011 29,000 x 4 116,000 480,000 155,000
2012 30,000 x 4 120,000 600,000 35,000
600,000
Depreciation rate per unit of output = 600,000 / 150,000 = 4
Depreciation Table – Sum of Years’ Digits
AccumulatedYear Particular Depreciation depreciation Book value
Acquisition cost 635,000
2008 5/15 x 600,000 200,000 200,000 435,000
2009 4/15 x 600,000 160,000 360,000 275,000
2010 3/15 x 600,000 120,000 480,000 155,000
2011 2/15 x 600,000 80,000 560,000 75,000
2012 1/15 x 600,000 40,000 600,000 35,000
600,000
SYD = 1 + 2 + 3 + 4 + 5 = 15
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Depreciation Table – Double Declining Balance
Accumulated
Year Particular Depreciation depreciation Book value
Acquisition cost 635,000
2008 40% x 635,000 254,000 254,000 381,000
2009 40% x 381,000 152,400 406,400 228,6002010 40% x 228,600 91,440 497,840 137,160
2011 40% x 137,160 54,864 552,704 82,296
2012 82,296 – 35,000 47,296 600,000 35,000
600,000
Fixed rate = 100% / 5 = 20% x 2 = 40%
Problem 17-4
a. Straight line method:
2008 27,500
2009 55,000
224b. Working hours method:
550,000
Rate per hour = ------------------- = 11
50,000 hours
2008 (3,000 hours x 11) 33,000
2009 (5,000 hours x 11) 55,000
c. Output method:
550,000
Rate per unit = -------------------- = 2.75
200,000 units
2008 (18,000 units x 2.75) 49,500
2009 (22,000 units x 2.75) 60,500
d. Sum of years’ digits:
10 + 1
SYD = 10 (------------) = 55
2
2008 (10/55 x 550,000 x 6/12) 50,000
2009 Jan. 1-June 30 50,000
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July 1-Dec. 31 (9/55 x 550,000 x 6/12) 45,000
95,000
e. Double declining balance:
2008 (570,000 x 20% x 6/12) 57,000
2009 (570,000 – 57,000 x 20%) 102,600
Problem 17-5
Fixed rate = 1.00 - .5623 or .4377
2008 (500,000 x .4377) 218,850
2009 (500,000 – 218,850 x .4377) 123,059
2010 (500,000 - 341,909 x .4377) 69,196
2011 (500,000 – 411,105 – 50,000) 38,895
450,000
Problem 17-6
a. Sum of years’ digit
April 1, 2008 – March 31, 2009 (1,080,000 x 8/36) 240,000
April 1, 2009 – March 31, 2010 (1,080,000 x 7/36) 210,000
225
Depreciation from April 1 to December 31, 2008 (240,000 x 9/12) 180,000
Depreciation for 2009:
January 1 – March 31 (240,000 x 3/12) 60,000
April 1 – December 31 (210,000 x 9/12) 157,500
217,500
b. Double declining balance
Fixed rate = 100 / 8 = 12.5 x 2 = 25%
2008 (1,200,000 x 25% x 9/12) 225,000
2009 (1,200,000 – 225,000 x 25%) 243,750
Problem 17-7
a. Service hours method:
960,000 – 60,000
Depreciation rate per hour = ---------------------------- = 112.50
8,000 hours
2008 (1,000 hours x 112.50) 112,500
2009 (2,000 hours x 112.50) 225,000
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b. Sum of years’ digits:
Sum of half years = 45
2008 (9/45 x 900,000 x 3/6) 90,000
2009 January 1 – March 31 (9/45 x 900,000 x 3/6) 90,000
April 1 – September 30 (8/45 x 900,000) 160,000
October 1 – December 31 (7/45 x 900,000 x 3/6) 70,000
320,000
Problem 17-8
a. Rate per unit (900,000 / 180,000) 5.00
2008 (5,000 x 5) 25,000
2009 (20,000 x 5) 100,000
b. Double declining balance:
Fixed rate (100% / 8 x 2) 25%
2008 (920,000 x 25% x 6/12) 115,000
2009 (920,000 – 115,000 x 25%) 201,250
226
c. Sum of years’ digits:
July 1 – December 31, 2008 (900,000 x 8/36 x 6/12) 100,000
January 1 – June 30, 2009 (900,000 x 8/36 x 6/12) 100,000
July 1 – December 31, 2009 (900,000 x 7/36 x 6/12) 87,500
Depreciation for 2009 187,500
Problem 17-9Depreciable Life in Annual
Assets Cost Salvage cost years depreciation
Machinery 310,000 10,000 300,000 5 60,000
Office equipment 110,000 10,000 100,000 10 10,000
Building 1,600,000 100,000 1,500,000 15 100,000
Delivery equipment 430,000 30,000 400,000 4 100,000
2,450,000 2,300,000 270,000
a. Composite rate = 270,000 / 2,450,000 = 11.02%
b. Composite life = 2,300,000 / 270,000 = 8.52 years
c. Depreciation 270,000
Accumulated depreciation 270,000
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Problem 17-10
Depreciable Life in Annual
Assets Cost Salvage cost years depreciation
Building 6,100,000 100,000 6,000,000 20 300,000
Machinery 2,550,000 50,000 2,500,000 5 500,000Equipment 1,030,000 30,000 1,000,000 10 100,000
9,680,000 9,500,000 900,000
a. Composite depreciation rate = 900,000 / 9,680,000 = 9.3%
b. Average life = 9,500,000 / 900,000 = 10.56 years
c. Depreciation 900,000
Accumulated depreciation 900,000
d. Cash 40,000
Accumulated depreciation 2,510,000
Machinery 2,550,000
e. Depreciation 663,090
Accumulated depreciation (9,680,000 – 2,550,000 x 9.3%) 663,090
227
Problem 17-11
2003
Jan. 1 Machinery 900,000
Cash 900,000
Dec. 31 Depreciation (20% x 900,000) 180,000
Accumulated depreciation 180,000
2004
Dec. 31 Depreciation 180,000
Accumulated depreciation 180,000
2005
Dec. 31 Depreciation 180,000
Accumulated depreciation 180,000
2006
Dec. 31 Depreciation 180,000
Accumulated depreciation 180,000
Cash 10,000
Accumulated depreciation 170,000
Machinery (4 x 45,000) 180,000
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2007
Dec. 31 Depreciation (720,000 x 20%) 144,000
Accumulated depreciation 144,000
Cash 15,000
Accumulated depreciation 615,000Machinery (14 x 45,000) 630,000
2008
Dec. 31 Depreciation 9,000
Accumulated depreciation 9,000
Remaining cost 90,000
Less: Balance of accumulated depreciation 79,000
Book value 11,000
Less: Salvage proceeds 2,000
Maximum depreciation 9,000
Cash 2,000
Accumulated depreciation 88,000
Machinery (4 x 45,000) 90,000
228Problem 17-12
1. Old machinery overhauled (240,000 + 60,000) 300,000
Accumulated depreciation
2005 (240,000 / 8) 30,000
2006 30,000
2007 30,000
Total 90,000
Book value – January 1, 2008 210,000
Old machinery overhauled (210,000 / 7 years) 30,000
Remaining cost of old machinery (1,152,000 – 240,000 / 8) 114,000
New machinery (460,800 / 8 x 5/12) 24,000
Total depreciation 168,000
2. Old machinery 1,152,000New machinery 460,800
Cost of overhaul 60,000
Total cost 1,672,800
Accumulated depreciation:
Balance – January 1 432,000
Depreciation for 2008 168,000 600,000
Book value – December 31, 2008 1,072,800
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March 1 Depreciation (250,000 – 20,000) 230,000
Cash 230,000
July 1 Electric meters 400,000
Cash 400,000
December 1 Depreciation (200,000 – 15,000) 185,000Cash 185,000
Problem 17-16Retirement method
2008 Tools 120,000
Cash 120,000
Cash (300 x 50) 15,000
Depreciation 45,000
Tools (300 x 200) 60,000
2009 Tools 360,000
Cash 360,000
Cash (700 x 70) 49,000
Depreciation 111,000
Tools 160,000
230
500 x 200 100,000
200 x 300 60,000
Cost of tools retired 160,000
Replacement method
2008 Tools (100 x 300) 30,000
Depreciation (300 x 30) 90,000 Cash
120,000
Cash 15,000
Depreciation 15,000
2009 Tools (200 x 400) 80,000Depreciation (700 x 400) 280,000
Cash 360,000
Cash 49,000
Depreciation 49,000Inventory method
2008 Tools 120,000
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Cash 120,000
Cash 15,000
Tools 15,000
Depreciation (265,000 – 200,000) 65,000
Tools 65,000
2009 Tools 360,000
Cash 360,000
Cash 49,000
Tools 49,000
Depreciation (511,000 - 350,000) 161,000
Tools 161,000
Problem 17-17
1. Land (350,000 + 450,000) 800,000
Land acquired (380,000 + 25,000 + 45,000) 450,000
2. Depreciation of land improvements (180,000 / 15) 12,000
3. Depreciation of building (4,500,000 – 1,050,000 x 7.5%) 258,750
231
4. Depreciation of machinery and equipment
(1,160,000 – 60,000 / 10) 110,000
(300,000 / 10) 30,000
(60,000 / 10 x 6/12) 3,000
143,000
5. Fixed rate (100% / 3 x 1.5) 50%
(1,800,000 – 1,344,000 x 50%) 228,000
Problem 17-18
1. Beginning balance 875,000
Acquisition (150,000 / 750,000 x 1,250,000) 250,000
Total cost of land 1,125,000
Technically, the land for undetermined use is an investment property.
2. Old (7,500,000 – 1,644,500 x 8%) 468,440
New (600,000/750,000 x 1,250,000 = 1,000,000 x 8%) 80,000
Depreciation – building 548,440
3. 2,250,000 / 10 225,000
400,000 / 10 x 6/12 20,000
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Depreciation – machinery 245,000
4. Depreciation – leasehold improvements (216,000 – 108,000 / 5 years) 21,600
5. Depreciation – land improvements 192,000 / 12 x 9/12) 12,000
Problem 17-19
1. Old building (4,672,200 x 10%) 467,220
New building
Direct cost 2,220,000
Fixed (15,000 x 25) 375,000
Variable (15,000 x 27) 405,000
Total cost 3,000,000
3,000,000 x 10% 300,000
Total depreciation 767,220
Fixed rate (100 / 20 x 2) 10%
232
2. Old machinery (1,380,000 / 10) 138,000
New machinery
Invoice cost 356,000
Concrete embedding 18,000
Wall demolition 7,000
Rebuilding of wall 19,000
Total cost 400,000
400,000 / 10 x 6/12 20,000
Total depreciation 158,000
Problem 17-20 Answer A
Cost of machinery (cash price) 1,100,000
Less: Residual value 50,000
Depreciable cost 1,050,000
Straight line depreciation (1,050,000 / 10) 105,000
Problem 17-21 Answer B
Sales price 2,300,000
Book value:
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Cost 4,200,000
Accumulated depreciation (3,600,000 / 5 x 3) 2,160,000 2,040,000
Gain 260,000
Problem 17-22 Answer B
Accumulated depreciation – 12/31/2007 3,700,000
Add: Depreciation for 2008 550,000
Total 4,250,000
Less: Accumulated depreciation on property, plant and
equipment retirements (squeeze) 250,000
Accumulated depreciation – 12/31/2008 4,000,000
Problem 17-23 Answer BDepreciable Annual
Cost Salvage cost Life depreciation
A 550,00050,000 500,000 20 25,000B 200,00020,000 180,000 15 12,000
C 40,000 40,000 5 8,000
790,000 720,000 45,000
Composite life = 720,000 / 45,000 16 years
233
Problem 17-24 Answer D
Invoice price 4,500,000
Cash discount (2% x 4,500,000) ( 90,000)
Delivery cost 80,000
Installation and testing 310,000
Total cost 4,800,000
Salvage value 800,000
Depreciable cost 4,000,000
Rate per unit (4,000,000 / 200,000) 20
Depreciation for 2008 (30,000 x 20) 600,000
Problem 17-25 Answer B
Cost 4,000,000
Accumulated depreciation
2007 (8/36 x 3,600,000) 800,000
2008 (7/36 x 3,600,000) 700,000 1,500,000
Book value, 12/31/2008 2,500,000
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Problem 17-26 Answer B
The first three fractions are:
2006 10/55
2007 9/55
2008 8/55
Thus, the 2008 depreciation of P240,000 is equal to 8/55.
Depreciable cost (240,000 / 8/55) 1,650,000
Salvage 50,000
Total cost 1,700,000
Problem 17-27 Answer B
April 1, 2006 to March 31, 2007 (5/15 x 3,000,000) 1,000,000
April 1, 2007 to March 31, 2008 (4/15 x 3,000,000) 800,000
Accumulated depreciation, March 31, 2008 1,800,000
Problem 17-28 Answer A
The accumulated depreciation on December 31, 2007 is recomputed following a certain
method. The same is arrived at following the SYD as follows:
SYD = 1 + 2 + 3 + 4 + 5 = 15
234
2005 (5/15 x 900,000) 300,000
2006 (4/15 x 900,000) 240,000
2007 (3/15 x 900,000) 180,000
Accumulated depreciation – 12/31/2007 720,000
Accordingly, the SYD is followed for 2008.
2008 depreciation (2/15 x 900,000) 120,000
Problem 17-29 Answer B
Straight line rate (100% / 8 years) 12.5%
Fixed rate (12.5 x 2) 25%2007 depreciation (1,280,000 x 25%) 320,000
2008 depreciation (1,280,000 – 320,000 x 25%) 240,000
Problem 17-30
1. 4,000,000 – 2,560,000 x 40% (Answer D) 576,000
2. 1,800,000 x 2/15 (SYD) (Answer A) 240,000
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3. Sales price 1,700,000
Book value (2,800,000 – 1,344,000) 1,456,000
Gain (Answer A) 244,000
Problem 17-31 Answer B
Straight line rate (100% / 5 years) 20%
Fixed rate (20% x 2) 40%
2006 depreciation (5,000,000 x 40%) 2,000,000
2007 depreciation (3,000,000 x 40%) 1,200,000
Accumulated depreciation, December 31, 2007 3,200,000
Depreciation for 2008 – straight line (5,000,000 – 3,200,000 / 3) 600,000
Accumulated depreciation, December 31, 2008 3,800,000
Problem 17-32 Answer A
Cost – 1/1/2005 7,200,000Accumulated depreciation – 12/31/2007 (7,200,000 / 10 x 3) 2,160,000
Book value – 12/31/2007 5,040,000
SYD for the remaining life of 7 years (1 + 2 + 3 + 4 + 5 + 6 + 7) 28
Depreciation for 2008 (5,040,000 x 7/28) 1,260,000
Problem 17-33 Answer B
Annual depreciation (1,536,000 / 8) 192,000
235Problem 17-34 Answer B
Fixed rate (100% / 4 x 2) 50%
Cost 6,000,000
Depreciation for 2007 (50% x 6,000,000) 3,000,000
Book value – 1/1/2008 3,000,000
Residual value ( 600,000)
Maximum depreciation in 2008 2,400,000
Fixed rate in 2008 (100% / 2 x 2) 100%
This means that the computers should be fully depreciated in 2008. Since there is a residual
value of P600,000, the maximum depreciation for 2008 is equal to the book value of
P3,000,000 minus the residual value of P600,000 or P2,400,000.
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236
CHAPTER 18
Problem 18-1 Problem 18-2
1. D 1. B
2. A 2. C
3. A 3. C
4. C 4. C
5. A 5. D
Problem 18-3
1. Ore property 5,000,000
Cash 5,000,000
2. Ore property 3,000,000
Cash 3,000,000
3. Machinery 4,000,000
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Cash 4,000,000
4. Depletion 1,140,000
Accumulated depreciation 1,140,000
8,000,000 – 400,000 = 7,600,000
7,600,000 / 2,000,000 = 3.80300,000 x 3.80 = 1,140,000
5. Depreciation 600,000
Accumulated depreciation 600,000
4,000,000 / 2,000,000 = 2.00
300,000 x 2.00 = 600,000
Problem 18-4
2008 Rock and gravel property 960,000
Cash 960,000
Depletion (1,000,000 x .40) 400,000
Accumulated depletion 400,000
2009 Rock and gravel property 490,000
Cash 490,000
Depletion (600,000 x .75) 450,000
Accumulated depletion 450,000
237
Total cost (960,000 + 490,000) 1,450,000
Less: Accumulated depletion 400,000
Depletable cost 1,050,000
Divide by estimated remaining output (2,400,000 – 1,000,000) 1,400,000
Revised depletion rate per ton .75
2010 Rock and gravel property 500,000
Cash 500,000
Depletion (700,000 x .44) 308,000Accumulated depletion 308,000
Total cost 1,450,000
Add: Additional development cost 500,000
Total 1,950,000
Less: Accumulated depletion (400,000 + 450,000) 850,000
Remaining depletable cost 1,100,000
Divide by new estimated remaining output 2,500,000
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New depletion rate .44
Problem 18-5
2008 Resource property 3,960,000
Cash 3,960,000
Building 960,000
Equipment 1,240,000
Cash 2,200,000
Depletion (12,000 x 32) 384,000
Accumulated depletion 384,000
Cost of resource property 3,960,000
Less: Residual value 120,000
Depletable cost 3,840,000
Divide by estimated output 120,000
Depletion rate per unit 32
Depreciation (12,000 x 8) 96,000
Accumulated depreciation – building 96,000
960,000
Depreciation rate per unit = ---------------- = 8
120,000
The output method is used in computing the depreciation of the building
because the life of the resource property (5 years or 120,000 / 24,000) is
shorter than the life of the building (8 years).
238
Depreciation 310,000
Accumulated depreciation 310,000
(1,240,000 / 4 years = 310,000)
The straight line method is used for the heavy equipment because the life of
4 years is shorter than the life of the resource property of 5 years.
2009 Depletion 800,000
Accumulated depletion (25,000 x 32) 800,000
Depreciation (25,000 x 8) 200,000
Accumulated depreciation – building 200,000
Depreciation 310,000
Accumulated depreciation – equipment 310,000
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Problem 18-6
2008 Ore property 5,400,000
Cash 5,400,000
Ore property 450,000Estimated liability for restoration cost 450,000
Mine improvements 8,000,000
Cash 8,000,000
2009 Depletion (600,000 x 2.60) 1,560,000
Accumulated depletion 1,560,000
Depreciation (600,000 x 4) 2,400,000
Accumulated depreciation 2,400,000
2010 Depletion (400,000 x 1.60) 640,000
Accumulated depletion 640,000
Depletable cost 5,200,000
Less: 2009 depletion 1,560,000
Balance (3,640,000 / 2,275,000 = 1.60) 3,640,000
Mine improvements 770,000
Cash 770,000
Depreciation (400,000 x 2.80) 1,120,000
Accumulated depreciation 1,120,000
Cost (8,000,000 + 770,000) 8,770,000Less: Accumulated depreciation 2,400,000
Book value (6,370,000 / 2,275,000 = 2.80) 6,370,000
239Problem 18-7
Depletion rate (5,000,000 / 1,000,000) 5.00
Depreciation rate (8,000,000 / 1,000,000) 8.00
First year
Depletion (200,000 x 5) 1,000,000Depreciation (200,000 x 8) 1,600,000
Second year
Depletion (250,000 x 5) 1,250,000
Depreciation (250,000 x 8) 2,000,000
Third year
Depletion none
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Depreciation (Schedule A) 550,000
Schedule A – Computation of depreciation for third year
Cost of equipment 8,000,000
Less: Accumulated depreciation 3,600,000
Book value – beginning of third year 4,400,000 Divideby remaining useful life in years (10 – 2) 8 Depreciation
for third year 550,000
Fourth year
Depletion (100,000 x 5) 500,000
Depreciation (Schedule B) 700,000
Schedule B – Computation of depreciation for fourth year
Cost of equipment 8,000,000
Less: Accumulated depreciation 4,150,000
Book value – beginning of fourth year 3,850,000
Original estimate of resource deposits 1,000,000 tons
Less: Extracted in first and second years 450,000
Remaining output 550,000 tons
Depreciation rate per unit (3,850,000 / 550,000) 7.00
Depreciation for third year (100,000 x 7) 700,000
Problem 18-8
1. Retained earnings 1,500,000
Accumulated depletion 2,500,000Total 4,000,000
Less: Capital liquidated 1,800,000
Depletion in ending inventory (5,000 x 20) 100,000 1,900,000
Maximum dividend 2,100,000
240
2. Retained earnings 1,800,000
Capital liquidated 200,000
Dividends payable 2,000,000
Problem 18-9
1. Cash (50,000 x 110) 5,500,000
Share capital (50,000 x 100) 5,000,000
Share premium 500,000
2. Resource property 3,000,000
Cash 3,000,000
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3. Mining equipment 800,000
Cash 800,000
4. Cash (85,000 x 50) 4,250,000
Sales 4,250,000
5. Mining and other direct cost 2,268,000
Administrative expenses 500,000
Cash 2,768,000
6. Depletion 270,000
Accumulated depletion (3,000,000 / 1,000,000 x 90,000) 270,000
7. Depreciation (90,000 x .80) 72,000
Accumulated depreciation - mining equipment 72,000
Depreciation rate (800,000 / 1,000,000) = .80
8. Inventory, December 31 (5,000 x 29) 145,000
Profit and loss 145,000
Mining labor and other direct costs 2,268,000
Depletion 270,000
Depreciation 72,000
Total production costs incurred 2,610,000
Divide by number of units extracted 90,000
Unit cost 29
241
Multinational CompanyIncome Statement
Year ended December 31, 2008
Sales 4,250,000
Cost of sales
Mining labor and other direct costs 2,268,000
Depletion 270,000
Depreciation 72,000
Total production cost 2,610,000
Less: Inventory, December 31 145,000 2,465,000
Gross income 1,785,000
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Administrative expenses 500,000
Net income 1,285,000
Multinational CompanyStatement of Financial Position
December 31, 2008
Assets
Current assets:
Cash 3,182,000
Inventory 145,000 3,327,000
Noncurrent assets:
Resource property 3,000,000
Less: Accumulated depletion 270,000 2,730,000
Mining equipment 800,000
Less: Accumulated depreciation 72,000 728,000 3,458,000
Total assets 6,785,000
Equity
Share capital 5,000,000
Share premium 500,000
Retained earnings 1,285,000
Total equity 6,785,000
Retained earnings 1,285,000
Add: Accumulated depletion 270,000
Total 1,555,000
Less: Unrealized depletion in ending inventory (5,000 x 3) 15,000
Maximum dividend 1,540,000
Retained earnings 1,285,000
Capital liquidated 255,000
Dividends payable 1,540,000
242
Problem 18-10
1. Purchase price 50,000Road construction 5,000,000
Improvements and development costs 750,000
Total cost 5,800,000
Residual value ( 600,000)
Depletable cost 5,200,000
Depletion rate per unit (5,200,000 / 4,000,000) 1.30
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Depletion for 2008 (500,000 x 1.30) 650,000
Depletable cost 5,200,000
Depletion in 2008 ( 650,000)
Remaining depletable cost 4,550,000
Development costs in 2009 1,300,000
Total depletable cost – 1/1/2009 5,850,000
Original estimated tons 4,000,000
Additional estimate 3,000,000
Total estimated tons 7,000,000
Extracted in 2008 ( 500,000)
Remaining tons – 1/1/2009 6,500,000
New depletion rate per unit (5,850,000 / 6,500,000) .90
Depletion for 2009 (1,000,000 x .90) 900,000
2. Cost of buildings 2,000,000
Residual value ( 200,000)
Depreciable cost 1,800,000
Depreciation rate per unit (1,800,000 / 4,000,000) .45
Depreciation for 2008 (500,000 x .45) 225,000
In the absence of any statement to the contrary, the output method is used in computing
depreciation of mining equipment.
Depreciable cost 1,800,000
Depreciation for 2008 ( 225,000)
Remaining depreciable cost 1,575,000Additional building in 2009 375,000
Total depreciable cost – 1/1/2009 1,950,000
New depreciation rate per unit (1,950,000 / 6,500,000) .30
Depreciation for 2009 (1,000,000 x .30) 300,000
243Problem 18-11
2008 No depletion because there is no production.
2009 Purchase price 28,000,000
Estimated restoration cost 2,000,000
Development cost – 2008 1,000,000
Development cost – 2009 1,000,000
Total cost 32,000,000
Residual value ( 5,000,000)
Depletable cost 27,000,000
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Rate in 2009 (27,000,000 / 10,000,000) 2.70
Depletion in 2009 (3,000,000 x 2.70) 8,100,000
2010 Tons extracted in 2010 3,500,000
Tons remaining in 12/31/2010 2,500,000Total estimated output – 1/1/2010 6,000,000
New rate in 2010 (27,000,000 – 8,100,000/6,000,000) 3.15
Depletion in 2010 (3,500,000 x 3.15) 11,025,000
Problem 18-12 Answer B
Acquisition cost 26,400,000
Development cost 3,600,000
Estimated restoration cost 1,800,000
Total cost 31,800,000
Less: Residual value 3,000,000
Depletable cost 28,800,000
Rate per unit (28,800,000 / 1,200,000) 24
Depletion for 2008 (60,000 x 24) 1,440,000
Problem 18-13 Answer C
Depletion rate per unit (9,200,000 / 4,000,000) 2.30
Problem 18-14 Answer C
Rate per unit (46,800,000 – 3,600,000 / 2,160,000) 20
Depletion in cost of goods sold (240,000 x 20) 4,800,000
244
Problem 18-15 Answer D
Acquisition cost 10,000,000
Less: Residual value 3,000,000Depletable cost 7,000,000
Less: Accumulated depletion – 12/31/2007
(7,000,000 / 10,000,000 = .70 x 4,000,000) 2,800,000
Remaining depletable cost – 1/1/2008 4,200,000
New depletion rate (4,200,000 / 7,500,000) .56
Depletion for 2008 (1,500,000 x .56) 840,000
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Problem 18-16 Answer B
Depletable cost
33,000,000
Depletion for 2007 (33,000,000 / 4,000,000 = 8.25 x 200,000) ( 1,650,000)
Balance – 1/1/2008 31,350,000
Production in 2008 225,000
New estimate – 12/31/2008 5,000,000
New estimate – 1/1/2008 5,225,000
Depletion for 2008 (31,350,000 / 5,225,000 = 6 x 225,000) 1,350,000
Problem 18-17
Question 1 – Answer A
Purchase price 14,000,000Less: Residual value 2,000,000
Depletable cost 12,000,000
Depletion rate (12,000,000 / 1,500,000) 8.00
Depletion for 2008 (150,000 x 8) 1,200,000
Production (25,000 x 6) 150,000
Question 2 – Answer C
Production from July 1 to December 31, 2008 (25,000 x 6) 150,000 tons
Annual production (25,000 x 12) 300,000 tonsEstimated life of mine (1,500,000 / 300,000) 5 years
Since the life of the mine is shorter than the life of the equipment, the output method is used
in computing depreciation.
245
Equipment 8,000,000
Less: Residual value 500,000
Depreciable cost 7,500,000
Rate per unit (7,500,000 / 1,500,000) 5.00
Depreciation for 2008 (150,000 x 5) 750,000
Problem 18-18 Answer C
Purchase price 9,000,000
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Development costs in 2007 300,000
Total cost 9,300,000
Residual value 1,200,000
Depletable cost 8,100,000
Rate in 2007 (8,100,000 / 2,000,000) 4.05
Depletion for 2007 (200,000 x 4.05) 810,000
Depletable cost 8,100,000
Depletion in 2007 ( 810,000)
Balance 7,290,000
Development costs in 2008 135,000
Depletable cost in 2008 7,425,000
Rate in 2008 (7,425,000 / 1,650,000) 4.50
Depletion for 2008 (300,000 x 4.50) 1,350,000
246
CHAPTER 19
Problem 19-1
1. C 6. B
2. B 7. C
3. D 8. A
4. C 9. B
5. C 10. A
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Problem 19-2
1. Appreciation (7,200,000 – 4,500,000) 2,700,000
2. Book value (4,500,000 – 900,000) 3,600,000
3. Depreciated replacement cost (7,200,000 x 80%) 5,760,000
4. Revaluation surplus (5,760,000 – 3,600,000) 2,160,000
Problem 19-3
1. Annual depreciation on cost (750,000 / 5) 150,000
Original life (3,000,000 / 150,000) 20 years
2. Equipment 1,800,000
Accumulated depreciation 450,000
Revaluation surplus 1,350,000
3. Depreciation (4,800,000 / 20) 240,000
Accumulated depreciation 240,000
4. Revaluation surplus 90,000
Retained earnings (1,350,000 / 15) 90,000
Problem 19-4
1. Annual depreciation on cost (9,000,000 / 25) 360,000
Age of asset (3,600,000 / 360,000) 10 years
2. Machinery 6,000,000
Accumulated depreciation (40% x 6,000,000) 2,400,000
Revaluation surplus 3,600,000
3. Depreciation (9,000,000 / 15) 600,000
Accumulated depreciation 600,000
247
4. Revaluation surplus 240,000
Retained earnings (3,600,000 / 15) 240,000
Problem 19-5
Proportional approach
1. Building 3,000,000
Accumulated depreciation 750,000
Revaluation surplus 2,250,000
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2. Depreciation (8,000,000 / 40) or (6,000,000 / 30) 200,000
Accumulated depreciation 200,000
Gross replacement cost (6,000,000 / 75%) 8,000,000
3. Revaluation surplus 75,000
Retained earnings (2,250,000 / 30) 75,000
Elimination approach
1. Accumulated depreciation 1,250,000
Building 1,250,000
Building (6,000,000 – 3,750,000) 2,250,000
Revaluation surplus 2,250,000
2. Depreciation (6,000,000 / 30) 200,000
Accumulated depreciation 200,000
3. Revaluation surplus 75,000Retained earnings 75,000
Problem 19-6
1. Equipment 2,700,000
Accumulated depreciation 500,000
Revaluation surplus 2,200,000
2. Depreciation (7,500,000 / 10) 750,000
Accumulated depreciation 750,000
3. Revaluation surplus (2,200,000 / 10) 220,000
Retained earnings 220,000
4. Cash 8,000,000
Accumulated depreciation 2,250,000
Equipment 9,200,000
Gain on sale of equipment 1,050,000
248
Revaluation surplus (2,200,000 - 220,000) 1,980,000
Retained earnings 1,980,000
Problem 19-7
1. Building 10,000,000
Accumulated depreciation 4,000,000
Revaluation surplus 6,000,000
2. Depreciation (13,000,000 / 5) 2,600,000
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Accumulated depreciation 2,600,000
3. Revaluation surplus 1,200,000
Retained earnings (6,000,000 / 5) 1,200,000
Problem 19-8
ReplacementCost cost Appreciation
Building 3,000,000 5,000,000 2,000,000
Accumulated depreciation 600,000 1,000,000 400,000
2,400,000 4,000,000 1,600,000
Accumulated depreciation on cost (3,000,000 x 20%) 600,000
Life of asset (100% / divided by 4%) 25 years
Percent of accumulated depreciation (5 years / 25) 20%
Gross replacement cost (4,000,000 / 80%) 5,000,000
Accumulated depreciation on replacement cost (5,000,000 x 20%) 1,000,000
a. “Should be entry:
Building 2,000,000
Accumulated depreciation 400,000
Revaluation surplus 1,600,000
b. Correcting entry:
Building 1,000,000
Retained earnings 1,000,000Accumulated depreciation 400,000
Revaluation surplus 1,600,000
c. Depreciation (4,000,000 / 20) 200,000
Accumulated depreciation 200,000
249
d. Revaluation surplus 80,000Retained earnings (1,600,000 / 20) 80,000
Problem 19-9
1. Accumulated depreciation 800,000
Machinery 800,000
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2. Retained earnings 400,000
Revaluation surplus 400,000
Problem 19-10Replacement
Cost cost Appreciation
Land 5,000,000 10,000,000 5,000,000
Building 25,000,000 45,000,000 20,000,000
Accumulated depreciation
(25,000,000 x 3/25) 3,000,000
(45,000,000 x 3/25) _________ 5,400,000 2,400,000
22,000,000 39,600,000 17,600,000
Machinery 10,000,000 15,000,000 5,000,000
Accumulated depreciation
(10,000,000 x 3/5) 6,000,000
(15,000,000 x 3/5) __________ 9,000,000 3,000,000
4,000,000 6,000,000 2,000,000
Replacement
Cost cost Appreciation
Equipment 3,000,000 4,200,000 1,200,000
Accumulated depreciation
(3,000,000 x 3/10) 900,000
(4,200,000 x 3/10) _________ 1,260,000 _ 360,000
2,100,000 2,940,000 840,000
a. Land 5,000,000
Building 20,000,000
Machinery 5,000,000
Equipment 1,200,000Accumulated depreciation – building 2,400,000
Accumulated depreciation – machinery 3,000,000
Accumulated depreciation – equipment 360,000
Revaluation surplus 25,440,000
b. Depreciation 5,220,000
Accumulated depreciation – building 1,800,000
Accumulated depreciation – machinery 3,000,000
Accumulated depreciation – equipment 420,000
250Building:
Cost (22,000,000 / 22) 1,000,000
Appreciation (17,600,000 / 22) 800,000 1,800,000
Machinery:
Cost (4,000,000 / 2) 2,000,000
Appreciation (2,000,000 / 2) 1,000,000 3,000,000
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Equipment:
Cost (2,100,000 / 7) 300,000
Appreciation (840,000 / 7) 120,000 420,000
Total depreciation 5,220,000
c. Revaluation surplus 1,920,000
Retained earnings (800,000 + 1,000,000 + 120,000) 1,920,000
d. Property, plant and equipment (at revalued amounts):
Land 10,000,000
Building 45,000,000
Machinery 15,000,000
Equipment 4,200,000
Total 74,200,000
Less: Accumulated depreciation 20,880,000
Net carrying value 53,320,000
The following disclosure should be made in the notes to financial statements:
Replacement
Cost cost
Land 5,000,000 10,000,000
Building 25,000,000 45,000,000
Machinery 10,000,000 15,000,000
Equipment 3,000,000 4,200,000
Total 43,000,000 74,200,000
Accumulated depreciation 13,200,000 20,880,000
Net carrying value 29,800,000 53,320,000
Schedule of Accumulated Depreciation
ReplacementCost cost
Building 4,000,000 7,200,000
Machinery 8,000,000 12,000,000
Equipment 1,200,000 1,680,000
13,200,000 20,880,000
251Problem 19-11 Answer B
Replacement
Cost cost Appreciation
Building 5,000,000 8,000,000 3,000,000
Accumulated depreciation 1,250,000 2,000,000 750,000
3,750,000 6,000,000 2,250,000
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Problem 19-12 Answer B
Sound Book Revaluation
value value surplus
Land 5,000,000 2,000,0003,000,000
Building (75% x 25,000,000) 18,750,000 11,250,0007,500,000
Machinery (50% x 5,000,000) 2,500,000 1,500,000 1,000,00011,500,000
Problem 19-13 Answer D
Fair value – December 31, 2008 450,000
Net book value – December 31, 2008 302,500
Revaluation surplus 142,500
Problem 19-14
Question 1 Answer A
Question 2 Answer BQuestion 3 Answer B
Problem 19-15
1. A 6. A 11. A
2. C 7. A 12. A
3. B 8. D 13. A
4. A 9. D 14. D
5. A 10. D 15. A
Problem 19-16
1. Impairment loss 900,000Accumulated depreciation 900,000
Cost 4,500,000
Accumulated depreciation 2,100,000
Book value – January 1 2,400,000
Recoverable value 1,500,000
Impairment loss 900,000
2. Depreciation (1,500,000 / 3) 500,000
Accumulated depreciation 500,000
252
3. Cost 4,500,000
Accumulated depreciation (2,100,000 + 900,000 + 500,000) 3,500,000
Book value – December 31 1,000,000
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Problem 19-17
1. Impairment loss 1,125,000
Accumulated depreciation 1,125,000
Cost – January 1 2,500,000
Accumulated depreciation (2,500,000 – 500,000 / 8 x 2) 500,000Book value – January 1 2,000,000
Recoverable value 875,000
Impairment loss 1,125,000
2. Depreciation 375,000
Accumulated depreciation (875,000 – 125,000 / 2) 375,000
3. Cost 2,500,000
Accumulated depreciation (500,000 + 1,125,000 + 375,000) 2,000,000
Book value – December 31 500,000
Problem 19-18
1. Offer price
25,000,000
Cost of dismantling and removal assumed by the bidder 5,000,000
Fair value less cost to sell 30,000,000
Present value of future cash flows 33,000,000
Less: Estimated liability 5,000,000
Value in use 28,000,000
Carrying amount 39,000,000
Less: Estimated liability 5,000,000
Adjusted carrying amount 34,000,000Recoverable amount – fair value less cost to sell, being the higher amount 30,000,000
Impairment loss 4,000,000
PAS 36, paragraph 78, provides that the fair value less cost to sell is equal to the estimated
selling price plus the estimated liability assumed by the buyer.
The standard further provides that to perform a meaningful comparison between the carrying
amount and recoverable amount, the estimated liability assumed by the buyer is deducted
in determining both the value in use and carrying amount of the asset.
2. Impairment loss 4,000,000
Accumulated depreciation 4,000,000
253
Problem 19-19
1. Net cash inflows PV factor Present value
2008 18,000,000 .930 16,740,000
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2009 15,000,000 .857 12,855,000
2010 15,000,000 .794 11,910,000
2011 12,000,000 .735 8,820,000
60,000,000
Total value in use 50,325,000
2. The recoverable amount is the value in use of P50,325,000 because this is higher than thefair value less cost to sell of P48,000,000.
3. Impairment loss 14,675,000
Accumulated depreciation (65,000,000 – 50,325,000) 14,675,000
4. Depreciation 12,581,250
Accumulated depreciation (50,325,000 / 4) 12,581,250
Problem 19-20
1. Depreciation 1,000,000
Accumulated depreciation (10,000,000 / 10) 1,000,000
2. Depreciation 1,000,000
Accumulated depreciation 1,000,000
3. Impairment loss 2,000,000
Accumulated depreciation 2,000,000
4. Depreciation 750,000
Accumulated depreciation (6,000,000 / 8) 750,000
5. Accumulated depreciation 1,750,000
Gain on impairment recovery 1,750,000
Cost – 1/1/2006 10,000,000
Accumulated depreciation (10,000,000 / 10 x 2) 2,000,000
Book value – 12/31/2007 8,000,000
Impairment loss – 2007 2,000,000
Adjusted book value – 12/31/2007 6,000,000
Depreciation – 2008 (6,000,000 / 8) 750,000
Book value – 12/31/2008 5,250,000
Cost – 1/1/2006 10,000,000
Accumulated depreciation (10,000,000 / 10 x 3) 3,000,000
Book value – 12/31/2008 (assuming no impairment) 7,000,000
Recorded book value 5,250,000
Gain on reversal of impairment 1,750,000
254
The fair value or recoverable value of P7,500,000 cannot exceed the “book value” that would
have been determined assuming no impairment is recognized.
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Problem 19-21
1. Impairment loss 5,000,000
Accumulated depreciation (35,000,000 – 30,000,000) 5,000,000
2. Depreciation 6,000,000Accumulated depreciation (30,000,000 / 5) 6,000,000
Observe that the undiscounted net cash flows from the asset amount to P37,500,000 for 5
years. This amount is more than the book value of the machinery. Under American
Standard, no impairment loss should be recognized in this case. However, under the
PAS 36, if the recoverable amount is less than carrying amount, an impairment loss is
recognized, regardless of the amount of undiscounted cash flows whether less than or
more than the carrying amount. PAS 36 has totally rejected the concept of
undiscounted cash flows for impairment purposes.
Problem 19-22
1. Value in use (1,500,000 x 5.65) 8,475,000
2. Impairment loss 8,250,000
Accmulated depreciation 8,250,000
Buildings 25,000,000
Accumulated depreciation (22,500,000 / 20 x 6) 6,750,000
Book value – 1/1/2008 18,250,000
Fair value – higher than value in use 10,000,000
Impairment loss 8,250,000
3. Depreciation 1,000,000
Accumulated depreciation (10,000,000 / 10) 1,000,000
Problem 19-23
1. Value in use (800,000 x 3.99) 3,192,000
2. Impairment loss 308,000
Accumulated depreciation 308,000
Machinery 5,000,000
Accumulated depreciation 1,500,000
Book value – 1/1/2008 3,500,000
Present value of cash flows – higher than fair value 3,192,000Impairment loss 308,000
3. Depreciation 638,400
Accumulated depreciation (3,192,000 / 5) 638,400
255
Problem 19-24
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1. Total carrying amount 5,000,000
Value in use 3,600,000
Impairment loss 1,400,000
2. Impairment loss allocated to goodwill 500,000
Impairment loss allocated to the other assets 900,0001,400,000
When an impairment loss is recognized for a cash generating unit, the loss is
allocated to the assets of the unit in the following order:
a. First, to the goodwill, if any.
b. Then, to all other assets of the unit prorata based on their carrying amount.
Carrying amount Fraction Loss
Building 2,000,000 20/45 400,000
Inventory 1,500,000 15/45 300,000
Trademark 1,000,000 10/45 200,000
4,500,000 900,000
3. Impairment loss 1,400,000
Goodwill 500,000
Accumulated depreciation – building 400,000
Inventory 300,000
Trademark 200,000
Problem 19-25
1. Carrying amount 16,000,000
Value in use 11,000,000
Impairment loss 5,000,000
2. Allocation of impairment loss
Building (8/16 x 5,000,000) 2,500,000
Equipment (4/16 x 5,000,000) 1,250,000
Inventory (4/16 x 5,000,000) 1,250,000
5,000,000
Observe that after allocating the P2,500,000 loss to the building, the carrying amount of the
building would be P5,500,000 which is lower than its fair value of P6,500,000.
Accordingly, only P1,500,000 loss is allocated to the building and the balance of P1,000,000
is reallocated to the equipment and inventory prorata.
256
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Building Equipment Inventory
Allocated loss 2,500,000 1,250,000 1,250,000
Reallocated loss (1,000,000)
(4/8 x 1,000,000) 500,000
(4/8 x 1,000,000) _________ _________ 500,000
Impairment loss 1,500,000 1,750,000 1,750,000
3. Impairment loss 5,000,000
Accumulated depreciation – building 1,500,000
Accumulated depreciation – equipment 1,750,000
Inventory 1,750,000
Problem 19-26
All Unimart’s stores are in different locations and probably have different customer profile. So
although Smart is managed at the corporate level, Smart generates cash inflows that are
largely independent from those of the other Unimart’s stores. Therefore, it is likely that Smart in
itself is a cash generating unit.
Problem 19-27
It is likely that the recoverable amount of an individual magazine title can be assessed. Even
though the level of advertising income for a title is influenced to a certain extent by the other
titles in the customer segment, cash inflows from direct sales and advertising are identifiable
for each title. In addition, decisions to abandon titles are made on an individual basis.
Accordingly, the individual magazine titles generate cash inflows that are largely
independent from one another and therefore, each magazine title is a separate cash
generating unit.
Problem 19-28
Case 1
1. A is separate cash generating unit because there is an active market for A’s products.
2. Although there is an active market for the products of B and C, cash inflows from B and
C depend on the allocation of production across two countries. It is unlikely that cash
inflows from B and C can be determined individually. Therefore, B and C, together
should be treated as a cash generating unit.
257
Case 2
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a. A cannot be treated as a separate cash generating unit because its cash inflows depend
on the sales of the final product by B and C, since there is no active market for A’s
product.
b. As a consequence, A, B and C, together, and therefore, Maximus Company, as a whole,
should be treated as the largest single cash generating unit.
Problem 19-29
The primary purpose of the building is to serve as a corporate asset supporting Litmus
Company’s manufacturing operations. Therefore, the building in itself cannot be considered
to generate cash inflows that are largely independent of the cash inflows from the entity as a
whole. In this case, the cash generating unit is Litmus Company as a whole.
The building is not held for investment. Thus, it is not appropriate to determine the value in use
of the building based on the cash inflows of related rent.
Problem 19-30 Answer C
Cost, January 1, 2005 800,000
Accumulated depreciation, December 31, 2007 (100,000 x 3) 300,000
Book value, December 31, 2007 500,000
Recoverable value 200,000
Impairment loss 300,000
The loss is recorded as follows:
Impairment loss 300,000
Accumulated depreciation 300,000
Cost 800,000
Accumulated depreciation (300,000 + 300,000) 600,000
Recoverable value, January 1, 2008 200,000
Depreciation for 2008 (200,000 / 5) 40,000
Book value, December 31, 2008 160,000
Problem 19-31 Answer B
From August 31, 2005 to May 31, 2008 is a period of 33 months. Thus, the remaining life of the
machine is 27 months, 60 months original life minus 33.
Depreciation for the month of June 2008 (1,350,000 / 27 months) 50,000
258
Cost 3,200,000
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Accumulated depreciation – 5/31/2008 (3,200,000 – 500,000 x 33/60) 1,485,000
Book value – 5/31/2008 1,715,000
Fair value 1,350,000
Impairment loss 365,000
Problem 19-32 Answer B
Cost – January 1, 2004 1,000,000
Accumulated depreciation, December 31, 2007 (900,000 / 10 x 4) 360,000
Book value, December 31, 2007 640,000
Depreciation for 2008 (640,000 – 40,000 / 4) 150,000
Book value, December 31, 2008 490,000
Problem 19-33 Answer C
Book value, 1/1/2008 2,400,000
Depreciation for 2008 (1,600,000 / 4) 400,000
Book value, 12/31/2008 2,000,000
Sales price-recoverable value 650,000Impairment loss 1,350,000
Problem 19-34 Answer C
Depreciation for 2008 (10% x 2,000,000) 200,000
Cost – 1/2/2004 2,000,000
Accumulated depreciation - 12/31/08 (200,000 x 5) 1,000,000
Book value-12/31/2008 1,000,000
Estimated cost of disposal 50,000
Impairment loss 1,050,000
Problem 19-35 Answer C
Cost 2,000,000
Accumulated depreciation – 1/1/2008 (2,000,000 – 100,000 / 10 x 2.5) 475,000
Book value – 1/1/2008 1,525,000
Fair value 600,000
Impairment loss 925,000
Problem 19-36 Answer C
Cost – 12/31/2004 2,800,000
Accumulated depreciation – 8/31/2008 (2,400,000 / 96 months x 44) 1,100,000
Book value – 8/31/2008 1,700,000Fair value 1,500,000
Impairment loss 200,000
259
Problem 19-37 Answer C
Carrying value 28,000,000
Decommissioning cost ( 8,000,000)
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Adjusted carrying value 20,000,000
Fair value less cost to sell – higher (20,000,000 less 1,000,000) 19,000,000
Impairment loss 1,000,000
Value in use 26,000,000
Decommissioning cost ( 8,000,000)
Adjusted value in use 18,000,000
Problem 19-38 Answer C
Carrying value – 12/31/2007 7,000,000
Depreciation for 2008 (20%) (1,400,000)
Carrying value – 12/31/2008 5,600,000
Carrying value – 12/31/2008 (assuming no impairment) 7,200,000
Reversal of impairment loss 1,600,000
260CHAPTER 20
Problem 20-1 Problem 20-2
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1. D 6. D 1. A 6. B
2. C 7. A 2. A 7. B
3. D 8. D 3. C 8. D
4. A 9. D 4. D 9. D
5. D 10. B 5. D 10. D
Problem 20-3
2008
Jan. 1 Patent 255,000
Cash 255,000
Dec. 31 Amortization of patent 12,750
Patent (255,000 / 20) 12,750
2009
Dec. 31 Amortization of patent 12,750
Patent 12,750
2010
Jan. 5 Legal expense 90,000
Cash 90,000
Dec. 31 Amortization of patent 12,750
Patent 12,750
2011
Jan. 1 Patent 510,000
Cash 510,000
Dec. 31 Amortization of patent 42,750
Patent 42,750
On original cost 12,750
On competing patent (510,000 / 17) 30,000
42,750
Problem 20-4
2008 Research and development expense 510,000
Cash 510,000
2011 Patent 720,000
Cash 720,000
261
Amortization of patent (720,000 / 16) 45,000
Patent 45,000
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2012 Patent 540,000
Cash 540,000
Amortization of patent 81,000
Patent 81,000
On related patent 45,000
On competing patent (540,000 / 15) 36,000
81,000
Problem 20-5
2008 Research and development expense 250,000
Cash 250,000
2009 Patent 60,000
Cash 60,000
Amortization of patent 6,000
Patent (60,000 / 10) 6,000
2010 Patent 600,000
Cash 600,000
Original cost 60,000
New patent 600,000
Total cost 660,000
Less: Amortization for 2008 6,000
Balance – January 1, 2009 654,000
Amortization of patent (654,000 / 15) 43,600Patent 43,600
2011 Amortization of patent 43,600
Patent 43,600
Patent written off 566,800
Patent 566,800
Balance – 1/1/2010 654,000
Less: Amortization
2010 43,600
2011 43,600 87,200
Unamortized cost 566,800
262
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263Problem 20-9
1. Retained earnings 500,000
Patent 500,000
2. Patent 510,000
Retained earnings 510,000
3. No adjustment.
4. Loss on damages 100,000
Legal expense 30,000
Accrued liabilities 130,000
5. Patent 24,500
Retained earnings 24,500
Amortization per book (500,000 – 450,000) 50,000Correct amortization for 2007 (510,000 / 20) 25,500
Overamortization 24,500
6. Amortization of patent 25,500
Patent 25,500
Problem 20-10
2008 Copyright 285,000
Cash 285,000
Amortization of copyright 150,000
Copyright 150,000
285,000 / 95,000 = 3 per copy
50,000 x 3 = 150,000
2009 Amortization of copyright 90,000
Copyright (30,000 x 3) 90,000
Problem 20-11
1. Copyright 240,000
Retained earnings 240,000
Cost of copyright 300,000
Less: Amortization (300,000 / 5) 60,000
Book value 240,000
2. Amortization of copyright 60,000
Copyright 60,000
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265
3. There is no amortization because the franchise is for an indefinite period.
Problem 20-15
Books of Franchisee
1. Franchise (3,000,000 + 3,790,000) 6,790,000
Discount on note payable 1,210,000
Cash 3,000,000
Note payable 5,000,000
Note payable 5,000,000
Present value of note (1,000,000 x 3.79) 3,790,000
Implied interest 1,210,000
2. Amortization of franchise 679,000Franchise (6,790,000 / 10) 679,000
3. Note payable 1,000,000
Cash 1,000,000
4. Interest expense (10% x 3,790,000) 379,000
Discount on note payable 379,000
Problem 20-16
Requirement a
1. Leasehold improvement – building 5,000,000
Cash 5,000,000
2. Rent expense (50,000 x 12) 600,000
Cash 600,000
3. Depreciation (5,000,000 / 10) 500,000
Accumulated depreciation 500,000
Requirement b
Accumulated depreciation 2,500,000
Loss on leasehold cancelation 2,500,000Leasehold improvement – building 5,000,000
Problem 20-17
1. Rent expense 1,200,000
Prepaid rent 1,200,000
Cash 2,400,000
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266
2. Leasehold 2,000,000
Cash 2,000,000
3. Leasehold improvement 500,000
Cash 500,000
4. Amortization of leasehold 400,000
Leasehold (2,000,000 / 5) 400,000
5. Depreciation (500,000 / 5) 100,000
Accumulated depreciation 100,000
Problem 20-18
1. Leasehold 1,000,000Cash 1,000,000
2. Rent expense (150,000 x 12) 1,800,000
Cash 1,800,000
3. Leasehold improvement 400,000
Cash 400,000
4. Leasehold improvement 100,000
Cash 100,000
5. Amortization of leasehold 100,000
Leasehold (1,000,000 / 10) 100,000
6. Depreciation 60,000
Accumulated depreciation 60,000
400,000 / 10 40,000
100,000 / 5 20,000
60,000
Problem 20-19
1. Rent expense 600,000
Cash 600,000
2. Leasehold 100,000
Cash 100,000
3. Leasehold improvement 200,000
Cash 200,000
4. Leasehold improvement 50,000
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Cash 50,000
267
5. Amortization of leasehold 20,000Leasehold (100,000 / 5) 20,000
6. Depreciation 52,500
Accumulated depreciation 52,500
200,000 / 5 40,000
50,000 / 4 12,500
52,500
Problem 20-20
1. Amortization of patent 280,000
Accumulated amortization (1,920,000 – 240,000 / 6) 280,000
2. Trademark (800,000 x 3/4) 600,000
Noncompetition agreement 200,000
Cash 800,000
3. Amortization of noncompetition agreement 40,000
Accumulated amortization (200,000 / 5) 40,000
4. Royalty expense 50,000
Cash 50,000
Problem 20-21
1. Acquisition cost 7,500,000
Net assets acquired (4,600,000)
Goodwill 2,900,000
2. Cash 50,000
Accounts receivable 800,000
Inventory 1,350,000
Property, plant and equipment 4,300,000
Goodwill 2,900,000
Accounts payable 900,000
Note payable – bank 1,000,000
Cash 7,500,000
Problem 20-22
1. Acquisition cost 6,000,000
Net assets acquired at fair value (3,300,000)
Goodwill 2,700,000
Total assets at fair value 5,300,000
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Total liabilities 2,000,000
Net assets acquired at fair value 3,300,000
268
2. Cash 50,000
Accounts receivable 500,000Inventory 1,500,000
Patent 250,000
Property, plant and equipment 3,000,000
Goodwill 2,700,000
Accounts payable 2,000,000
Cash 6,000,000
Problem 20-23
1. Cash 1,000,000
Inventory 500,000
In-process R and D 5,000,000Total assets 6,500,000
Total liabilities 3,000,000
Net assets 3,500,000
Acquisition cost 8,000,000
Net assets acquired at fair value (3,500,000)
Goodwill 4,500,000
The goodwill includes the fair value of the assembled workforce of P1,200,000.
The assembled workforce is not accounted for separately as an asset.
2. Cash 1,000,000
Inventory 500,000
In process R and D 5,000,000
Goodwill 4,500,000
Accounts payable 2,600,000
Notes payable 400,000
Cash 8,000,000
Problem 20-24
1. Average earnings 250,000
Divide by 10%
Net assets including goodwill 2,500,000
Less: Net assets before goodwill 1,700,000Goodwill 800,000
2. Average earnings 250,000
Less: Normal earnings (8% x 1,700,000) 136,000
Excess earnings 114,000
Divide by 15%
Goodwill 760,000
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3. Average earnings 250,000Less: Normal earnings (10% x 1,700,000) 170,000
Excess earnings 80,000
Multiply by 5
Goodwill 400,000
4. Excess earnings 80,000
Multiply by 5.65
Goodwill 452,000
Problem 20-25
Average earnings or prior years (1,500,000 / 3) 500,000Increase in average earnings (10% x 500,000) 50,000
Total 550,000
Less: Patent amortization (500,000 / 5 years) 100,000
Earnings for goodwill computation 450,000
a. Average future earnings 450,000
Divide by 8%
Net assets including goodwill 5,625,000
Less: Net assets excluding goodwill 5,000,000
Goodwill 625,000
b. Average earnings 450,000
Less: Normal earnings (8% x 5,000,000) 400,000
Average excess earnings 50,000
Divide by 10%
Goodwill 500,000
c. Goodwill (50,000 x 3.17) 158,500
Problem 20-26
a. Average earnings 750,000
Expected increase (1,000,000 – 900,000) 100,000
Total 850,000
Less: Normal earnings (4,800,000 x 10%) 480,000Excess earnings 370,000
Goodwill (370,000 x 4) 1,480,000
Shareholders’ equity per book 5,000,000
Less: Recorded goodwill 200,000
Net assets before goodwill 4,800,000
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b. Goodwill (370,000 / 20%) 1,850,000
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Problem 20-27
1. Share capital 2,000,000
Retained earnings 1,500,000
Total shareholders’ equity 3,500,000
Less: Recorded goodwill 1,000,000
Net assets before goodwill 2,500,000
Average earnings (1,200,000 + 150,000 / 3) 450,000
Less: Normal earnings (10% x 2,500,000) 250,000
Excess earnings 200,000
Divide by 16%
Goodwill 1,250,000
2. Net assets before goodwill 2,500,000
Goodwill 1,250,000
Purchase price 3,750,000
Problem 20-28
1. Value in use 38,000,000
Net assets including goodwill at carrying amount 42,000,000
Impairment loss ( 4,000,000)
2. Impairment loss 4,000,000
Goodwill 4,000,000
Problem 20-29
1. Value in use 60,000,000
Net assets including goodwill at carrying amount 75,000,000
Impairment loss (15,000,000)
2. Impairment loss 15,000,000
Goodwill 5,000,000
Accounts receivable 2,000,000
Inventory 3,000,000
Accumulated depreciation 5,000,000
The remaining impairment loss of P10,000,000, after deducting the loss
applicable to goodwill, is allocated to the other noncash assets on a
prorata basis.
Problem 20-30
1. Present value of indefinite cash flows (200,000 / 10%) 2,000,000
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Trademark 6,000,000
Impairment loss (4,000,000)
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Present value of cash flows from cash generating unit (9,000,000 x 8.51) 76,590,000
Net assets including goodwill at carrying amount 80,000,000
Impairment loss ( 3,410,000)
2. Impairment loss 7,410,000
Trademark 4,000,000
Goodwill 3,410,000
Problem 20-31
1. Total carrying amount 5,000,000
Value in use 4,230,000
Impairment loss 770,000
2. Impairment loss 770,000
Goodwill 500,000
Accumulated depreciation – building (25/45 x 270,000) 150,000
Inventory (15/45 x 270,000) 90,000
Trademark (5/45 x 270,000) 30,000
Problem 20-32
12/31/2008 R and D expense 2,500,000
Cash 2,500,000
1/1/2009 R and D expense 1,200,000
Cash 1,200,000
7/1/2009 R and D expense 500,000
Cash 500,000
11/1/2009 Patent 350,000
Cash 350,000
11/15/2009 Patent 800,000
Cash 800,000
12/31/2009 Patent 100,000
Cash 100,000
Problem 20-33
1. Product costs which are associated wit inventory items are:
Duplication of computer software and training materials 2,500,000
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Packaging product 900,000
Total inventory 3,400,000
2. The costs incurred from the time of technological feasibility to the time when
product costs are incurred should be capitalized as computer software cost.
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Other coding costs after establishment of technological feasibility 2,400,000
Other testing costs after establishment of technological feasibility 2,000,000
Costs of producing product masters for training materials 1,500,000
Total costs to be capitalized 5,900,000
3. Completion of detail program design 1,300,000
Cost incurred for coding and testing to establish technological feasibility 1,000,000
Total costs charged as expense 2,300,000
Problem 20-34
1. Designing and planning 1,000,000
Code development 1,500,000
Testing __500,000
Total R and D expense in 2008 3,000,000
The cost of producing the product master of P2,500,000 is capitalized as
software cost to be subsequently amortized.
2. Cost of producing the software program in 2009 1,000,000
Amortization of software cost (2,500,000 / 4) 625,000
Total expense in 2009 1,625,000
Problem 20-35 Answer C
Cost 357,000
Accumulated amortization from 2005 to 2007 (357,000 / 15 x 3) 71,400
Book value – 12/31/2007 285,600
Amortization for 2008 (285,600 / 7) 40,800
Book value – 12/31/2008 244,800
Problem 20-36 Answer C
Cost 1/1/2003 6,000,000Accumulated depreciation – 12/31/2007 (6,000,000 / 15 x 5) 2,000,000
Book value – 1/1/2008 4,000,000
Amortization for 2008 (4,000,000 / 5) 800,000
Problem 20-37 Answer C
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Cumulative earnings 550,000
Less: Gain on sale 50,000
Adjusted cumulative earnings 500,000
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Average earnings (500,000 / 5) 100,000
Divide by capitalization rate 10%
Net assets including goodwill 1,000,000
Less: Net assets before goodwill 750,000
Goodwill 250,000
Problem 20-38 Answer C
Net assets 1,800,000
Multiply by excess rate (16% minus 10%) 6%Excess earnings 108,000
Multiply by present value factor 3.27
Goodwill 353,160
Problem 20-39 Answer D
Purchase price 5,000,000
Less: Goodwill 500,000
Net assets before goodwill 4,500,000
Estimated annual earnings (squeeze) 550,000
Less: Normal earnings (4,500,000 x 10%) 450,000
Excess or superior earnings 100,000
Divide by capitalization rate 20%
Goodwill 500,000
Problem 20-40 Answer C
Accounts receivable 2,000,000
Inventory 500,000
Equipment 500,000
Short-term payable (2,000,000)
Net assets at fair value 1,000,000
Acquisition cost 5,000,000Net assets at fair value (1,000,000)
Goodwill 4,000,000
Problem 20-41 Answer A
Problem 20-42 Answer C
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Downpayment 2,000,000
Present value of annual payment for 4 years (1,000,000 x 2.91) 2,910,000
Cost of franchise 4,910,000
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Problem 20-43 Answer A
Design costs 1,500,000
Legal fees of registering trademark 150,000
Registration fee with Patent Office 50,000
Total cost of trademark 1,700,000
Problem 20-44 Answer B
Original lease 12 years
Extension 8Total life 20
Less: Years expired (2006 and 2007) 2
Remaining life 18 years
Life of improvement (shorter) 15 years
Leasehold improvement 540,000
Less: Depreciation for 2008 (540,000 / 15) 36,000
Book value 504,000
Problem 20-45 Answer D
Depreciation (3,600,000 / 6) 600,000
Problem 20-46 Answer C
Depreciation of equipment 135,000
Materials used 200,000
Compensation costs of personnel 500,000
Outside consulting fees 150,000
Indirect costs allocated 250,000
1,235,000
Problem 20-47 Answer A
Modification to the formulation of a chemical product 135,000
Design of tools, jigs, molds and dies 170,000
Laboratory research 215,000
Total research and development expense 520,000
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