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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


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.



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


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)


Construction cost of new building 8,000,000



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



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:


Commission 60,000

Clearing cost 15,000

Sale of t imber and gravel ( 5,000)

Acquisition of land - #622:


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:


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



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



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


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


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



2. Accumulated depreciation (1,960,000 x 20%) 392,000


Building (2,500,000 x .784) 1,960,000

216

Building 2,500,000Cash 2,500,000

Depreciation (8,132,000 / 20) 406,600


Building (9,000,000 – 1,960,000 + 2,500,000) 9,540,000

Accumulated depreciation (1,800,000 – 392,000) 1,408,000


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


d. Depreciation (3,900,000 / 15) 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


Cash 2,500,000



4. Depreciation (10,700,000 – 500,000 / 48) 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


3. Depreciation (3,600,000 / 6) 600,000


Cost 5,000,000

Accumulated depreciation:

2005 450,000

2006 450,000 900,000


Residual value 500,000

Remaining depreciable cost – 1/1/2007 3,600,000


Cash 300,000

5. Depreciation (3,300,000 / 5) 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 (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




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


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 (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



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


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


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



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


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



Payments to tenants 200,000


Legal fees 50,000

Title insurance 30,000

Proceeds from sale of materials ( 10,000)


Problem 16-27 Answer D

Land Building

Purchase price of land 600,000

Legal fees for contract 20,000Architect fee 80,000


Construction cost _______ 3,500,000

Total cost 670,000 3,580,000


Acquisition price 7,000,000



Option of building acquired 200,000

Repairs 500,000

Total cost 7,700,000

221


Purchase price 250,000

Shipping 5,000

Installation 10,000

Testing 35,000

Total cost 300,000



All expenditures are capitalized.


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




222

CHAPTER 17


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



Depreciation Table – Service Hours Method

Accumulated



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


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


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



Depreciation Table – Double Declining Balance

Accumulated



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



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



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




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


d. Cash 40,000


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


2004

Dec. 31 Depreciation 180,000


2005



2006



Cash 10,000


Machinery (4 x 45,000) 180,000



2007

Dec. 31 Depreciation (720,000 x 20%) 144,000


Cash 15,000

Accumulated depreciation 615,000Machinery (14 x 45,000) 630,000

2008



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


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


Accumulated depreciation:

Balance – January 1 432,000

Depreciation for 2008 168,000 600,000

Book value – December 31, 2008 1,072,800



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



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


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



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


3,000,000 x 10% 300,000


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



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


Sales price 2,300,000

Book value:



Cost 4,200,000

Accumulated depreciation (3,600,000 / 5 x 3) 2,160,000 2,040,000

Gain 260,000


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



Cash discount (2% x 4,500,000) ( 90,000)

Delivery cost 80,000

Installation and testing 310,000


Salvage value 800,000


Rate per unit (4,000,000 / 200,000) 20

Depreciation for 2008 (30,000 x 20) 600,000


Cost 4,000,000


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




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



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


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


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



3. Sales price 1,700,000

Book value (2,800,000 – 1,344,000) 1,456,000

Gain (Answer A) 244,000


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


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


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.



236

CHAPTER 18


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




Cash 4,000,000

4. Depletion 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



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


Cash 490,000

Depletion (600,000 x .75) 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


Cash 500,000

Depletion (700,000 x .44) 308,000Accumulated depletion 308,000


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



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


Cost of resource property 3,960,000



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



(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 – equipment 310,000



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


2010 Depletion (400,000 x 1.60) 640,000



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


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



Depreciation (Schedule A) 550,000

Schedule A – Computation of depreciation for third year

Cost of equipment 8,000,000


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


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



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



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


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


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




Depletion rate per unit (5,200,000 / 4,000,000) 1.30



Depletion for 2008 (500,000 x 1.30) 650,000


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



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.


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


Residual value ( 5,000,000)




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


Acquisition cost 26,400,000

Development cost 3,600,000

Estimated restoration cost 1,800,000


Less: Residual value 3,000,000


Rate per unit (28,800,000 / 1,200,000) 24

Depletion for 2008 (60,000 x 24) 1,440,000


Depletion rate per unit (9,200,000 / 4,000,000) 2.30


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



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




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


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



Rate per unit (7,500,000 / 1,500,000) 5.00

Depreciation for 2008 (150,000 x 5) 750,000





Development costs in 2007 300,000


Residual value 1,200,000


Rate in 2007 (8,100,000 / 2,000,000) 4.05

Depletion for 2007 (200,000 x 4.05) 810,000


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



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


Revaluation surplus 1,350,000

3. Depreciation (4,800,000 / 20) 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


Accumulated depreciation (40% x 6,000,000) 2,400,000


3. Depreciation (9,000,000 / 15) 600,000


247



Problem 19-5

Proportional approach

1. Building 3,000,000





2. Depreciation (8,000,000 / 40) or (6,000,000 / 30) 200,000


Gross replacement cost (6,000,000 / 75%) 8,000,000



Elimination approach


Building 1,250,000

Building (6,000,000 – 3,750,000) 2,250,000


2. Depreciation (6,000,000 / 30) 200,000


3. Revaluation surplus 75,000Retained earnings 75,000

Problem 19-6

1. Equipment 2,700,000



2. Depreciation (7,500,000 / 10) 750,000


3. Revaluation surplus (2,200,000 / 10) 220,000

Retained earnings 220,000

4. Cash 8,000,000


Equipment 9,200,000

Gain on sale of equipment 1,050,000

248

Revaluation surplus (2,200,000 - 220,000) 1,980,000


Problem 19-7

1. Building 10,000,000



2. Depreciation (13,000,000 / 5) 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



b. Correcting entry:

Building 1,000,000

Retained earnings 1,000,000Accumulated depreciation 400,000


c. Depreciation (4,000,000 / 20) 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



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


(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


(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


Equipment 3,000,000 4,200,000 1,200,000


(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



b. Depreciation 5,220,000

Accumulated depreciation – building 1,800,000

Accumulated depreciation – machinery 3,000,000


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



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


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


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




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


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


Book value – January 1 2,400,000

Recoverable value 1,500,000

Impairment loss 900,000

2. Depreciation (1,500,000 / 3) 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



Problem 19-17

1. Impairment loss 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


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


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.



253

Problem 19-19

1. Net cash inflows PV factor Present value

2008 18,000,000 .930 16,740,000



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.


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








Accumulated depreciation (6,000,000 / 8) 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


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.



Problem 19-21


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


Accmulated depreciation 8,250,000

Buildings 25,000,000


Book value – 1/1/2008 18,250,000

Fair value – higher than value in use 10,000,000




Problem 19-23

1. Value in use (800,000 x 3.99) 3,192,000

2. Impairment loss 308,000


Machinery 5,000,000


Book value – 1/1/2008 3,500,000

Present value of cash flows – higher than fair value 3,192,000Impairment loss 308,000


Accumulated depreciation (3,192,000 / 5) 638,400

255

Problem 19-24



1. Total carrying amount 5,000,000



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


Goodwill 500,000


Inventory 300,000

Trademark 200,000

Problem 19-25

1. Carrying amount 16,000,000



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



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


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



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.


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


The loss is recorded as follows:



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



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



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



Cost – January 1, 2004 1,000,000

Accumulated depreciation, December 31, 2007 (900,000 / 10 x 4) 360,000


Depreciation for 2008 (640,000 – 40,000 / 4) 150,000



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


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



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



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


259


Carrying value 28,000,000

Decommissioning cost ( 8,000,000)



Adjusted carrying value 20,000,000

Fair value less cost to sell – higher (20,000,000 less 1,000,000) 19,000,000



Decommissioning cost ( 8,000,000)

Adjusted value in use 18,000,000


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




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


Patent 12,750

2010

Jan. 5 Legal expense 90,000

Cash 90,000


Patent 12,750

2011

Jan. 1 Patent 510,000

Cash 510,000


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



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



263Problem 20-9

1. Retained earnings 500,000

Patent 500,000

2. Patent 510,000


3. No adjustment.

4. Loss on damages 100,000

Legal expense 30,000

Accrued liabilities 130,000

5. Patent 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


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



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


Requirement b


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



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


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


Cash 400,000


Cash 100,000

5. Amortization of leasehold 100,000

Leasehold (1,000,000 / 10) 100,000



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


Cash 200,000




Cash 50,000

267

5. Amortization of leasehold 20,000Leasehold (100,000 / 5) 20,000



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



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


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



269

3. Average earnings 250,000Less: Normal earnings (10% x 1,700,000) 170,000


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%


Less: Net assets excluding goodwill 5,000,000

Goodwill 625,000

b. Average earnings 450,000


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



b. Goodwill (370,000 / 20%) 1,850,000

270

Problem 20-27

1. Share capital 2,000,000


Total shareholders’ equity 3,500,000

Less: Recorded goodwill 1,000,000


Average earnings (1,200,000 + 150,000 / 3) 450,000



Divide by 16%

Goodwill 1,250,000

2. Net assets before goodwill 2,500,000

Goodwill 1,250,000


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)


Goodwill 4,000,000

Problem 20-29

1. Value in use 60,000,000


Impairment loss (15,000,000)


Goodwill 5,000,000


Inventory 3,000,000


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



Trademark 6,000,000

Impairment loss (4,000,000)

271

Present value of cash flows from cash generating unit (9,000,000 x 8.51) 76,590,000


Impairment loss ( 3,410,000)


Trademark 4,000,000

Goodwill 3,410,000

Problem 20-31

1. Total carrying amount 5,000,000



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



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.

272

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


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


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




Cumulative earnings 550,000

Less: Gain on sale 50,000

Adjusted cumulative earnings 500,000

273

Average earnings (500,000 / 5) 100,000

Divide by capitalization rate 10%


Less: Net assets before goodwill 750,000

Goodwill 250,000


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



Less: Goodwill 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



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





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

274


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


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


Depreciation (3,600,000 / 6) 600,000


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


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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