210CHAPTER 16 Problem 16-1 1. 2. 3. 4. 5. C D D D B Problem 16-2 Cash paid for land and old building Removal of old building Payment to tenants of old building to vacate premises Architect fee Building permit Fee for title search Survey before construction Excavation Cost of new building constructed Assessment fee Cost of grading, leveling and landfill Driveways and walks Temporary quarters for construction crew Temporary building to house tools and materials Cost of construction changes Land 1,000,000 50,000 15,000 Building 200,000 30,000 10,000 20,000 100,000 6,000,000 5,000 45,000 _________ 1,145,000 40,000 80,000 60,000 50,000 6,560,000 Note: The cost of replacing windows is treated as expense. Problem 16-3 Land Land improvement Cost of land Legal fees Payment of mortgage Payment of taxes Cost of razing building Proceeds from sale of materials Grading and drainage Architect fee Payment to contractor Interest cost Driveway and parking lot Cost of trees, shrubs and other landscaping Cost of installing lights in parking lot Premium for insurance Building 2,000,000 10,000 50,000 20,000 30,000 ( 5,000) 15,000 200,000 8,000,000 300,000 _______ 2,120,000 25,000 8,525,000 40,000 55,000 5,000 _______ 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. 211 Problem 16-4 Office Land improvements Purchase price Materials Excavation Labor Remodeling Cash discounts Supervision Compensation insurance Clerical and other expenses Paving of streets Plans and specifications Legal cost - land 1,300,000 Factory building building Land 700,000 3,200,000 100,000 2,500,000 200,000 ( 60,000) 30,000 50,000 30,000 40,000 10,000 1,310,000 ________ 900,000 150,000 ________ 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 Payment for land Demolition of old building Total cost of land 50,000 1,000,000 100,000 1,150,000 Architect fee Payment to city hall Contract price Safety fence around construction site Safety inspection on building 30,000 Removal of safety fence Total cost of factory building 230,000 120,000 5,000,000 35,000 20,000 5,435,000 Problem 16-6 Purchase price Title clearance fee Cost of razing old building Scrap value of old building Total cost of land Construction cost of new building 3,000,000 50,000 100,000 ( 10,000) 3,140,000 8,000,000 212 Problem 16-7 Purchase price 4,000,000 Remodeling Salvage materials Grading, leveling and other permanent improvement Repairs Land Building 1,000,000 150,000 5,000) ( 50,000 ________ 1,050,000 10,000 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 Machinery Fair value Repairs Remodeling Invoice price Discount Base 1,500,000 _________ 1,500,000 Building 5,000,000 200,000 300,000 1,000,000 ( 20,000) _________ 50,000 5,500,000 1,030,000 The driveway and parking lot are charged to land improvements. Problem 16-9 Land Machinery Fair value 1,500,000 Repairs Special tax assessment Platform Remodeling Purchase price Discount Freight Installation 1,500,000 Building 4,000,000 200,000 30,000 70,000 400,000 ( _________ 1,530,000 _________ 4,600,000 800,000 40,000) 20,000 30,000 2,380,000 Problem 16-10 Purchase price Commission 2,000,000 100,000 Contract price 6,000,000 Plans, specification 000 900.000) 2.000.000 1. December 31 2.500.#621: Purchase price Commission Clearing cost Sale of timber and gravel Acquisition of land .000 500.000 4.000.000 350.500.000 The third tract of land should be presented as current asset because it was “classified as held for sale”.000 5.000 75. 1 1.000 1.000 650.000 and blueprint Architectural fee Cost of new building 100.000 Acquisition of land .000 9.000 ( 5.000.500.000 1.000.#622: Purchase price Cost of demolition New building: Construction cost Excavation fee Architectural design Building permit Improvements: Electrical work Construction extension (800.350.000 1.000 10.000. Jan.000 213 Problem 16-11 Land Leasehold Building improvements Machinery Balances.000. Problem 16-12 Land Land improvements Building Machinery Balances. Jan.000 6.800.000 20.000 .230.Legal fees Title guarantee Cost of razing old building Salvage value of materials Cost of land 50.000 15.000 60.000 ( 5.240.870. 1 1.000 x 1/2) Improvements on office space Purchase of new machine: Invoice price Freight Unloading charge Balances.000 Land acquired Issuance of share capital: 3.900.000 3.000 300.250.000 _________ _________ __ 30.000 150.000 40.000 250.000 7.000) 4.750.000 _________ 8.000 400.000 50. In the absence of a present obligation.000 Cost of transporting machine Installation cost Testing cost Safety rails and platform Water device Cost of adjustment Estimated dismantling cost Total cost of machine 30.000) 3.950.000) Discount (5% x 4.000 ________ _________ ( 500.000 20.200.000 400.000 65.000 Problem 16-17 1.000.000 10.000 250.000 150.000 3.000 80.250.000 ( 150. 31 4.000 30.000 Installation 100.000 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.000 60.000 Invoice cost 4.000 3.000 10.000 60.000 75. Materials Labor Installation 600.650.400.000 2.000 24/36 x 4.000 750.000 80.000) 50. Problem 16-16 Second hand market value Overhaul and repairs Installation Testing Hauling Safety device 2. Dec.000 Problem 16-15 Cost paid (896.000.000 – 96.12/36 x 4.000 1.500.000 3.000.000 Note that the estimated dismantling cost is capitalized because the company has a present obligation as required by contract.500.000 Transportation 40. the estimated dismantling cost is not capitalized.000 1.000) ( 200.000 ________ 6.000 40.000. 214 Problem 16-13 Invoice price Cash discount Freight Installation cost Testing cost Problem 16-14 3.000 50.000 New machinery New parking lot.400. street and sidewalk Machinery sold Balances.000) 800.000 Cash allowance ( 60.500.930.000 Trial run-salary of engineer 50.000.000 110.000) 1.000 .000.400. 000 7.600 Depreciation recorded Correct depreciation (1.000 150.000 88. Profit on construction Machinery 100.000 6.000 – 14.000 / 20) 400.000 2.000 128.600.600 40.000) 150.000 1.000 200. Depreciation – tools Tools (90.000 1.400.000 88. Machinery Accumulated depreciation Depreciation – machinery 90.000 1.600 Problem 16-18 Initial design fee Executive chairs and desks Storm windows and installation Installation of automatic door opening system Overhead crane Total capital expenditures 150.200.000 / 3 x 4/12) 10.000 405.000.Trial run Discount Overhead ( 30.000 350.600 40.000 10. Purchase discount Machinery 40.000 3.000 6.000 5.000 215 2.000 500. Tools Machinery 90.000 Problem 16-19 1.500.000 4.100.000 2.500.000 2.000 200.000 2.000 / 10 x 4/12) Overdepreciation 128.000 40.000) 6.000 40. Accumulated depreciation Loss on retirement of building Building Building Cash Depreciation (8.000 100. Machinery Factory overhead 150. Loss on retirement of old machinery Machinery (20.000 .200. Adjusting entries: 1. 000) 1.000 – 392.000 Building (9.000 – 2.000 216 Building Cash Depreciation (8.600 Building (9.100.000 1. Building (8.000 – 1.800.500.000 Book value 8.000. Building Cash 25 years 2.000 406.784) 8.960. beginning of current year 25 Add: Extension in life Revised useful life 10 5 15 Problem 16-21 .000 Original life 30 Less: Expired life Remaining useful life.000 / 30) 280.000) 9.568.400.000 392.000 / 15) Accumulated depreciation 7.000 Accumulated depreciation (1.400.000 / 280.500.000.800.900.400.000) b.000 – 400.000 Age of building (7.500.600 406.000 / 20) Accumulated depreciation 2.000) 9.000 + 2. Annual depreciation (8.000 260.900.000 3.900. Accumulated depreciation (1.960.132.000 Book value 2.000) 10.500.Accumulated depreciation 405.960.500.500.000.000.000 c.000 x 20%) Loss on retirement of building Building (2.000 Accumulated depreciation (1.000 Less: Accumulated depreciation Book value d.000 2.000.000 + 2.408.000 260.000 + 2.000) 1.500.132. Depreciation (3.000 x .540.000 1.500.000 2.500.000 Problem 16-20 a. 000 / 48) Accumulated depreciation 212.000 600. Building Cash 2.000 4.000 4.000 Cost Accumulated depreciation (900.000 – 2.000.500 217 Building (10.000 – 100.000 450.000.000 5. Machinery Cash 2.000.000 3.400.000.000 Accumulated depreciation (400.000 200. Depreciation Accumulated depreciation 3.000.000 + 3.000 / 5) Accumulated depreciation 660.000 / 6) Accumulated depreciation Cost Accumulated depreciation: 2005 2006 Book value Residual value Remaining depreciable cost – 1/1/2007 5.000 450.600.000.500.000 660.500.000 5. Building Cash Accumulated depreciation (2.000 450.000 + 600. Depreciation Accumulated depreciation 3.000 / 50 x 2) Loss on retirement of building Cash 10.000 300.000 100.500 212.000 5. Depreciation (3.500.500.000 .000 2.000 500.700.500.700.000 Problem 16-22 1.000) 300.000 3. Depreciation (3.500.000 4.000 600.000) 1.000 5.000 – 500.000 10.000 2.000 450.000 3.300.000.000 Book value – 1/1/2008 900.600.300.000 3.000 Book value – 1/1/2008 10. Machinery Cash 300.1.500.100.000) 11.800. Depreciation (10.000 200. 000 500.000 15.000 + 250.050.000 x 9/12) Accumulated depreciation Accumulated depreciation (2. Depreciation (900.000 Book value 4.650.000 3.000 400.100.000 600.000 – 750.000 4.000 .000 5.000 x 3/12) Accumulated depreciation Accumulated depreciation (480.000) 100.000.000 50. Depreciation (600.000 105.000 – 350.Residual value Remaining depreciable cost – 1/1/2008 500.000 x 5/12) Accumulated depreciation Delivery equipment – new Accumulated depreciation Cash (5.000) Loss on sale of office equipment Office equipment 3. Accumulated depreciation Office equipment 5.000 675.500.000 x 4/12) Accumulated depreciation 15.000 1.000 4.000) Fire loss Machinery 3.125.300.000) Delivery equipment – old Gain on exchange (750.000 3.000 1.000 250.000 1. Discount on bonds payable Machinery 500.375.000.000 Problem 16-23 1.000 + 15.200.000 350.700.000 Problem 16-24 1.000.000 Original cost Less: Accumulated depreciation to date (2.250.000 1. Depreciation (150.000) Loss on retirement of store equipment Store equipment 2.000 250.000 675.000 1.000 + 50.650.000 2.000 300. Depreciation (60.000 495.000) 2.400.000.000 218 Cash Accumulated depreciation (1.000 + 675.000 3.200.500.000 50. 000 670.000 390. Allowance for doubtful accounts Loss on exchange – accounts receivable 2. Interest expense Machinery (3.200.000 3.000.000 – 3.000.500.000) 840.500 Accumulated depreciation Depreciation 75.000 150.000 4.500.000 Depreciation for 12 months (600.000 / 10 x 9/12) Discount on bonds payable 37.000 540.000 300.000 37. Loss on exchange Machinery Cost per book Correct cost Trade in value Add: Cash paid Overstatement 700.610.000 1.000 Less: Depreciation recorded Overstatement 2.000 219 Correct depreciation for 9 months (3.000 390.000 3.500 75.000.000 150.000 75.000 / 5 x 9/12) 525.Interest expense (500.000 150.000 30.000 Adjusted 4.000 Depreciation for 9 months 600.000 .000 / 5) Overstatement 3.000 300.460.000 Trade in value Less: Book value Loss on exchange 4.000 Per book 5.000 60.000 Depreciation per book Correct depreciation (3.000 30.500.000 1.500.000 (390.000 4.000 x 5 years) Cost Less: Residual value Depreciable cost 800.000 390.000 2.000.000.000 150.350.000 / 9/12) Depreciable cost (800.000.000) Machinery Freight in Accumulated depreciation Depreciation 600.000 30. 200.000 x 5.000.600 / 6.000) Renovation Total cost of building 150.000 Per book Machinery Accounts receivable Treasury shares Machinery Should be Machinery Allowance for doubtful accounts (20% x 4.000) 3.000 2.400.200. Problem 16-25 Answer A Allocated cost of land (2.305.200.000 Total cost of land 100.000 Incidentally.000 200.000 The cost of treasury shares acquired for noncash consideration is usually measured by the recorded amount of the noncash asset surrendered (SFAS No.000) 2.000 3.300.000 500.300.000 840.000 3.000 / 6.000 x 5.400 / 6.000 x 250.000) Loss on accounts receivable Accounts receivable 4.000 100.600 / 6.000 3.300.300.000 Proceeds from sale of materials 4. the cost of the building is: Allocated cost (3.000 50.000 Property taxes (2.000 Problem 16-26 Answer A Purchase price Payments to tenants Demolition of old building Legal fees Title insurance 30.000.000 4.000 x 250.000) .500.000 4.000 Property taxes (3.500.000 4.000) Cost of survey 5.200.Treasury share 900.000 220 Treasury shares Machinery 3.000 4.200.200.000 60.950.000 ( 10. 18).200. Total cost of land 4.000 35. Problem 16-32 Answer A All costs are capitalized.000 .000 Problem 16-28 Answer D Acquisition price Option of building acquired Repairs Total cost 7.000 100.000 200.370.000.000 650.000 10.000 150.000 Problem 16-27 Answer D Land Building Purchase price of land Legal fees for contract Architect fee Demolition of old building Construction cost Total cost 600.000 3.000 50.000 5.000 300.000 7.500. Problem 16-33 Answer C Continuing and frequent repairs Repainting of the plant building Partial replacement of roof tiles Repair and maintenance expense 400.000 Problem 16-30 Answer A Problem 16-31 Answer A All expenditures are capitalized.580.000 80.000 20.000 221 Problem 16-29 Answer D Purchase price Shipping Installation Testing Total cost 250.000 _______ 670.000 500.000 3.700. 8. 6. 8. 7. 9. 3. 5. 2. 9. 4.Problem 16-34 Answer B Problem 16-35 Answer B 222 CHAPTER 17 Problem 17-1 1. 7. 2. 4. 5. 3. 10. A D B D D D D C C B Problem 17-3 Problem 17-2 1. 6. C A D D D B C B A A . 10. 000 120.000 155.000 600.000 480.000 x 4 29.000 35.000 Accumulated depreciation Book 635.000 120.000 x 10 265.000 2012 12.000 600.000 120.000 / 60.000 600.000 480.000 480.000 140.000 116.000 Accumulated Depreciation 136.000 x 10 155.000 x 4 30.000 240.000 120.000 499.000 x 10 35.000 .000 120.000 371.000 395.000 x 4 32.000 Depreciation Table – Service Hours Method Year Particular Acquisition cost 2008 14.000 = 10 223 Depreciation Table – Production Method Year 2008 2009 2010 2011 2012 Particular Acquisition cost 34.Depreciation Table – Straight Line Year Particular value Acquisition cost 2008 2009 2010 2011 2012 Depreciation 120.000 370.000 120.000 2010 10.000 120.000 x 10 365.000 360.000 35.000 130.000 100.000 364.000 x 4 Depreciation 136.000 275.000 = 4 Book value 635.000 100.000 600.000 264.000 x 10 495.000 / 150.000 271.000 Book value 635.000 128.000 155.000 x 4 25.000 Depreciation rate per hour = 600.000 270.000 2009 13.000 600.000 Depreciation rate per unit of output = 600.000 600.000 2011 11.000 110.000 Accumulated Depreciation depreciation 140.000 515. 000 SYD = 1 + 2 + 3 + 4 + 5 = 15 Depreciation Table – Double Declining Balance Year Particular Acquisition cost 2008 40% x 635.000 152.000 80.000 40. Straight line method: 2008 2009 27.000 254.500 55.000 360.000 Depreciation Accumulated depreciation 254.000 2012 1/15 x 600.296 Book value 635.160 2011 40% x 137.600 137.440 54.000 Rate per hour = ------------------.Depreciation Table – Sum of Years’ Digits Year Particular Acquisition cost 2008 5/15 x 600. Working hours method: 550.000 155.000 2010 3/15 x 600.000 275.000 560.= 11 50.000 Fixed rate = 100% / 5 = 20% x 2 = 40% Problem 17-4 a.000 381.840 552.000 35.000 480.000 600.000 228.000 2011 2/15 x 600.000 120.000 Book value 635.160 82.000 600.000 600.296 2012 82.000 435.600 2010 40% x 228.400 406.400 91.000 75.864 47.000 224 b.000 35.000 2009 40% x 381.000 Depreciation Accumulated depreciation 200.000 497.296 – 35.000 200.704 600.000 2009 4/15 x 600.000 hours .000 160. 4377) – 218.00 .000 2009 (570.196 450.000 April 1.2008 (3.000 38.5623 or .080.850 x .850 123. 1-June 30 July 1-Dec.4377 2008 (500.75) 2009 (22.000 Depreciation from April 1 to December 31.000 hours x 11) 33.000 2011 (500.000 x 20%) 102.600 Problem 17-5 Fixed rate = 1.000 Rate per unit = -------------------.000 x 8/36) 240.000 . 2010 (1.341.909 x .000 2009 (500.000 x 6/12) 50.75) 60.500 10 + 1 SYD = 10 (------------) = 55 2 2008 (10/55 x 550. 2008 (240.000 units x 2.080.000 95.000 e.75 200.000 x 7/36) 210. 2009 (1. Sum of years’ digits: 49.000 units 2008 (18.000 x 6/12) 50.000 units x 2.059 69. Sum of years’ digit April 1.000 45.4377) – 411.000 hours x 11) 2009 (5.000 55.= 2.000 2010 (500.000) 218.000 x 20% x 6/12) 57.000 2009 Jan.000 – 57..4377) . 2009 – March 31.000 c.895 x . 2008 – March 31.000 x 9/12) 225 180. Double declining balance: 2008 (570. Output method: 550. 31 (9/55 x 550.000 Problem 17-6 a.500 d.105 – 50. Service hours method: 960.50) 112.000 x 25%) 243.50) 225.000 x 9/12) 60. Sum of years’ digits: Sum of half years = 45 2008 (9/45 x 900.000 Depreciation rate per hour = ---------------------------.000 100.Depreciation for 2009: January 1 – March 31 (240.000 b.5 x 2 = 25% 2008 (1.200.000 x 3/6) 70.000 2009 January 1 – March 31 (9/45 x 900.500 217.000 x 5) 2009 (20.000) 160.200.000 hours x 112.000 x 25% x 9/12) 2009 (1.000 Problem 17-7 a.000 hours 2008 (1.000 25.000 hours x 112.000 x 25% x 6/12) 115.000 x 3/6) 90.000 x 3/12) April 1 – December 31 (210.50 8.000 October 1 – December 31 (7/45 x 900.000 . Double declining balance Fixed rate = 100 / 8 = 12.000 320.000 x 5) b.000 157.= 112.000 – 60.00 2008 (5.000 April 1 – September 30 (8/45 x 900.750 225. Rate per unit (900.000) 5.000 x 3/6) 90.000 Problem 17-8 a.000 / 180.500 2009 (2. Double declining balance: Fixed rate (100% / 8 x 2) 25% 2008 (920.500 b.000 – 225. 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) Depreciation for 2009 87,500 187,500 Problem 17-9 Depreciable Life in Annual Assets Cost Salvage cost depreciation Machinery 310,000 10,000 300,000 5 Office equipment 110,000 10,000 100,000 10 Building 1,600,000 100,000 1,500,000 15 Delivery equipment 430,000 30,000 400,000 4 2,450,000 2,300,000 years 60,000 10,000 100,000 100,000 270,000 a. Composite rate = 270,000 / 2,450,000 = 11.02% b. Composite life = 2,300,000 / 270,000 = 8.52 years c. Depreciation Accumulated depreciation 270,000 270,000 Problem 17-10 Assets years Building Depreciable Salvage Cost depreciation 6,100,000 300,000 Machinery 2,550,000 500,000 Equipment 1,030,000 9,680,000 900,000 100,000 Life in Annual cost 6,000,000 50,000 2,500,000 30,000 1,000,000 9,500,000 20 5 10 100,000 a. Composite depreciation rate = 900,000 / 9,680,000 = 9.3% b. Average life = 9,500,000 / 900,000 = 10.56 years c. Depreciation Accumulated depreciation d. Cash 900,000 900,000 40,000 Accumulated depreciation Machinery 2,510,000 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 Cash Dec. 31 Depreciation (20% x 900,000) Accumulated depreciation 900,000 900,000 180,000 180,000 2004 Dec. 31 Depreciation Accumulated depreciation 180,000 2005 Dec. 31 Depreciation Accumulated depreciation 180,000 2006 Dec. 31 Depreciation Accumulated depreciation 180,000 Cash Accumulated depreciation Machinery (4 x 45,000) 2007 Dec. 31 Depreciation (720,000 x 20%) Accumulated depreciation Cash Accumulated depreciation Machinery (14 x 45,000) 2008 Dec. 31 Depreciation Accumulated depreciation 180,000 180,000 180,000 10,000 170,000 180,000 144,000 144,000 15,000 615,000 630,000 9,000 9,000 Remaining cost Less: Balance of accumulated depreciation Book value Less: Salvage proceeds Maximum depreciation Cash 90,000 79,000 11,000 2,000 9,000 2,000 Accumulated depreciation Machinery (4 x 45,000) 88,000 90,000 228 Problem 17-12 1. Old machinery overhauled (240,000 + 60,000) Accumulated depreciation 2005 (240,000 / 8) 2006 2007 Total Book value – January 1, 2008 300,000 30,000 30,000 30,000 90,000 210,000 Old machinery overhauled (210,000 / 7 years) Remaining cost of old machinery (1,152,000 – 240,000 / 8) New machinery (460,800 / 8 x 5/12) Total depreciation 2. Old machinery New machinery Cost of overhaul Total cost Accumulated depreciation: Balance – January 1 Depreciation for 2008 600,000 Book value – December 31, 2008 1,072,800 30,000 114,000 24,000 168,000 1,152,000 460,800 60,000 1,672,800 432,000 168,000 Problem 17-13 Main machine (7,500,000 / 10) First component – from January 1 to April 1, 2008 (1,200,000 / 6 x 3/12) 50,000 Second component – from April 1 to December 31, 2009 (2,000,000 – 400,000 / 4 x 9/12) Total depreciation for 2008 750,000 300,000 1,100,000 The second component is depreciated over the remaining life of the main machine. The original life is 10 years and 6 years already expired. Thus, the remaining life is 4 years. Problem 17-14 1. Tools Cash 40,000 2. Tools Cash 20,000 40,000 20,000 000 – 20.000 December 1 Electric meters Cash 200.000 250.000 1 Cash Depreciation Electric meters 15. Depreciation Tools 46.000 4.000 60.000) Cash 1 Electric meters Cash 230.000 185.000 1 Electric meters Cash 180.000 230. Cash Tools 4.000 135.000 December 1 Depreciation (200.000) Cash 185.000 120.000 46.000 229 Problem 17-15 Retirement method March July 1 Electric meters Cash 250.000 400.000 200.000 Cash (300 x 50) Depreciation Tools (300 x 200) 15.000 150.000 150.3.000 – 15.000 .000 46.000 45.000 400.000 Balance of tools account Less: Estimated cost on December 31 Depreciation 196.000 400.000 400.000 4.000 Problem 17-16 Retirement method 2008 Tools Cash 120.000 160.000 1 Cash Depreciation Electric meters 20.000 Replacement method March July 1 Depreciation (250. 000 120.000) Tools Problem 17-17 120.000 160.000 360.000 15.000 Cash Depreciation 49.000 Cash (700 x 70) Depreciation Tools 49.000 65.000 90.000 .000 .000 360.000 49.000 60.000 161.000 2009 Tools (200 x 400) Depreciation (700 x 400) Cash 80.000 111.350.000 120.000 360.2009 Tools Cash 360.000 15.000 280.000 65.000 360.000 161.000 15.000 Replacement method 2008 Tools (100 x 300) Depreciation (300 x 30) Cash 30.000 49.000 49.000 – 200.000) Tools 2009 Tools Cash Cash Tools Depreciation (511.000 160.000 Inventory method 2008 Tools Cash Cash Tools Depreciation (265.000 Cash Depreciation 15.000 230 500 x 200 200 x 300 Cost of tools retired 100. 000 – 108.000 Technically. Land (350.000 – 1. Beginning balance Acquisition (150.800.000 – 1.5) 50% (1.000 Total cost of land 875.000 / 5 years) 21.000 5.000) 800.500 x 8%) 468.000 + 450.000 + 25.000 / 15) 450.000 x 8%) 80.750 231 4.000 / 750.000 / 12 x 9/12) 12.000/750.500.050.000 = 1. 2.000 245.000 Depreciation – machinery 4. the land for undetermined use is an investment property.000 – 1. Depreciation – land improvements 192.000 .000 / 10) 30.000 Problem 17-19 548.440 225.644.000 143.1.000 x 7.250.000 + 45.000 x 50%) 228.250.500.440 New (600.000 / 10 400.000 x 1.000) 250.000 12.000 / 10 x 6/12) 3. 2.000. Fixed rate (100% / 3 x 1.000 x 1.000 110. Depreciation – leasehold improvements (216.000 Problem 17-18 1.000 / 10) (300.125.000 Depreciation – building 3.000 / 10 x 6/12 20. Old (7.000 – 60.000) 2.000 1.160. Depreciation of building (4.000 (60. Depreciation of land improvements (180.250.344.600 5.5%) 258.000 3.000 Land acquired (380. Depreciation of machinery and equipment (1. 672.000 19.000 4.000 x 27) Total cost 3.000 x 25) Variable (15.000 405.000 2.220 2.380.050.200.000 / 5 x 3) Gain 2.600.100.000 Problem 17-22 Answer B Accumulated depreciation – 12/31/2007 Add: Depreciation for 2008 Total 3.000 400.000 Straight line depreciation (1.000 7.000 50.1.000 300.200 x 10%) New building Direct cost Fixed (15.000 x 10% Total depreciation 467.000 / 10) New machinery Invoice cost Concrete embedding Wall demolition Rebuilding of wall Total cost 400.250.000 260.000 2. Old building (4.300.160.000 3.000 158.000 550.000 / 10) 105.000 356.220 Fixed rate (100 / 20 x 2) 10% 232 2.000 .000 4.000 Total depreciation 138.000 767.050.000.000 18.700.000 1.040.220.000.000 375.000 Problem 17-20 Answer A Cost of machinery (cash price) Less: Residual value Depreciable cost 1. Old machinery (1.000 / 10 x 6/12 20.000 Problem 17-21 Answer B Sales price Book value: Cost Accumulated depreciation (3. 000 800.000 790.000 2.000) 2008 (7/36 x 3.500.Less: Accumulated depreciation on property.000.000) Delivery cost Installation and testing Total cost Salvage value Depreciable cost 4.000 .600.000 / 200.000 4.000 310.000 ( Rate per unit (4.000 700.000 Book value.000) 20 Depreciation for 2008 (30.000 12.000 Depreciable cost 500.500. 4.500.000 16 years 233 Problem 17-24 Answer D Invoice price Cash discount (2% x 4.000 Salvage 50.800.000 40.000 B 200.000 is equal to 8/55.000.000 x 20) 600.000 45.000 4.000 / 45.000 90.000 4. plant and equipment retirements (squeeze) Accumulated depreciation – 12/31/2008 250.000) 1.000 20.600.000 C 40.000 180.000 Annual Life 20 15 5 25.500. 12/31/2008 Problem 17-26 Answer B The first three fractions are: 2006 2007 2008 10/55 9/55 8/55 Thus.000 Problem 17-25 Answer B Cost Accumulated depreciation 2007 (8/36 x 3.000.000 8.000) 80.000 Composite life = 720.000 720.000 Problem 17-23 Answer B Cost depreciation A 550.000.000 800. the 2008 depreciation of P240. 000) 240.000 .5% 25% 240.000 Problem 17-30 1. 2007 to March 31.000 (Answer A) 244.000) 1.000 240. March 31.000 1. 2008 depreciation (2/15 x 900.000 x 25%) 12.000) 180. 4.800.280.000) 300. Sales price Book value (2. 2008 (4/15 x 3.000.000 2007 (3/15 x 900.000) 1.000 – 1.800.700.560.000 / 8/55) Salvage Total cost 1.000.344.000 2006 (4/15 x 900.000 Gain (Answer D) (Answer A) 576. 1.456.000 x 25%) 320.000 2008 depreciation (1.000. The same is arrived at following the SYD as follows: SYD = 1 + 2 + 3 + 4 + 5 = 15 234 2005 (5/15 x 900.Depreciable cost (240.000) 800.000 Accumulated depreciation.700.000 Problem 17-27 Answer B April 1.000 Accordingly.000 – 2.000.000 1.000 Problem 17-29 Answer B Straight line rate (100% / 8 years) Fixed rate (12. 2007 (5/15 x 3.000 x 40% 2.000 50. 2006 to March 31.280. 2007 is recomputed following a certain method. the SYD is followed for 2008.5 x 2) 2007 depreciation (1.000) 120.000 Accumulated depreciation – 12/31/2007 720. 2008 1.000 Problem 17-28 Answer A The accumulated depreciation on December 31.650.000 – 320.800.000 x 2/15 (SYD) 3.000 April 1. 000 / 8) 192. December 31.400.000 – 3.000 ( 600.000 / 3) 600.000 235 Problem 17-34 Answer B Fixed rate (100% / 4 x 2) Cost Depreciation for 2007 (50% x 6.000.536.000 3.000) Book value – 1/1/2008 Residual value Maximum depreciation in 2008 Fixed rate in 2008 (100% / 2 x 2) 50% 6.000 Problem 17-33 Answer B Annual depreciation (1.000 3.000.200.Problem 17-31 Answer B Straight line rate (100% / 5 years) Fixed rate (20% x 2) 2006 depreciation (5.000) 2.200.000.160.000.000 / 10 x 3) Book value – 12/31/2007 7.000 Accumulated depreciation. 2008 20% 40% 3.040.260. 2007 Depreciation for 2008 – straight line (5.000 SYD for the remaining life of 7 years (1 + 2 + 3 + 4 + 5 + 6 + 7) 28 Depreciation for 2008 (5.400. Since there is a residual value of P600.000 or P2. December 31.000.000 100% This means that the computers should be fully depreciated in 2008.000 2.000 x 40%) 1.200.000 x 40%) 2.000 Problem 17-32 Answer A Cost – 1/1/2005 Accumulated depreciation – 12/31/2007 (7.000.000 5.000 2007 depreciation (3.000.000 minus the residual value of P600.000 3.000 x 7/28) 1. .000.040.000 Accumulated depreciation.000.200. the maximum depreciation for 2008 is equal to the book value of P3.000.200.000.800. 000 3. 2.000. Machinery Cash 4.000 3. B C C C D Problem 18-3 1.000. 3.000 4.236 CHAPTER 18 Problem 18-1 1. Ore property Cash 3. 5.000.000.000. Ore property Cash 5. Depletion 1. 3. 5.000 2.000 . 4. 4. D A A C A Problem 18-2 1.000.140. 2.000 4.000 5. 600.000 308.00 300.000 1.80 = 1.450.400.000 x 3.000 – 400.000 1.000.000 + 450.400.000 500.000 – 1.000 Less: Accumulated depletion Depletable cost Divide by estimated remaining output (2.000 + 490.000 1.000 Remaining depletable cost 1.100.75 500.000 x .000 400.000 450.Accumulated depreciation 1.000 x .000 x .000.000.000) 850.000 = 2.950.000 Divide by new estimated remaining output 2.000 490.40) Accumulated depletion 2009 Rock and gravel property Cash Depletion (600.000 5.000.000 237 Total cost (960.000 500.050.000 x 2.140.000 Problem 18-4 2008 Rock and gravel property Cash Depletion (1.000 600.600.000 308.000 = 3.000 .000) Revised depletion rate per ton 2010 Rock and gravel property Cash Depletion (700.140.00 = 600.000 400. Depreciation Accumulated depreciation 600.75) Accumulated depletion 960.450.44 .000 4.000 450.000 / 2.000.000 New depletion rate 400.000 = 7.000 490.500.80 300.000 1.000 7.000 / 2.000 .000 8.000) 1.000 960.44) Accumulated depletion Total cost Add: Additional development cost Total Less: Accumulated depletion (400.000. 000 3.000 Depreciation (25.400.000 960.000 / 4 years = 310.000 310.000 120.240.000) is shorter than the life of the building (8 years).000 200.000 Cost of resource property Less: Residual value 120. 238 Depreciation Accumulated depreciation (1.960.000 96.960.000 .400.000 x 32) 800.000 Depletable cost 3.= 8 120.000 800.000 5.000 x 8) Accumulated depreciation – building 200.000 x 8) Accumulated depreciation – building 3.240.000 1.000 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 x 32) Accumulated depletion 3.000 Depreciation rate per unit = ---------------.000) 310.200.000 310.000 Divide by estimated output Depletion rate per unit Depreciation (12.000 960.000 32 96.000 Depreciation Accumulated depreciation – equipment 310. 2009 Depletion Accumulated depletion (25.960.000 2.Problem 18-5 2008 Resource property Cash Building Equipment Cash Depletion (12.000 Problem 18-6 2008 Ore property Cash 5.840.000 384.000 384.000 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. 000 8.000 x 8) 1.560.000.400.000.000 x 2.000 2010 Depletion (400.120.Ore property Estimated liability for restoration cost Mine improvements Cash 2009 Depletion (600.000 3.250.000 1.000 = 1.000 x 8) 1.000 1.000) Depreciation rate (8.400.000 x 2.00 First year Depletion (200.000 2.370.000) 5.560.200.000) Less: Accumulated depreciation Book value (6.000.000 1.000 = 2.560.60) Accumulated depletion 2.000 1.000.000 x 5) Depreciation (250.275.000 770.275.00 8.000 x 5) Depreciation (200.000 Schedule A – Computation of depreciation for third year Cost of equipment 8.000 770.000.640.000 Second year Depletion (250.000 Third year Depletion Depreciation (Schedule A) none 550.000 / 2.000 5.60) Accumulated depletion Depreciation (600.770.000 640.640.400.000.370.000 / 1.000 450.600.000.80) Accumulated depreciation Cost (8.120.60) Mine improvements Cash Depreciation (400.000 + 770.000.000 1.000 .000.000 239 Problem 18-7 Depletion rate (5.000.000 x 1.000 8.000 2.000 640.000 2.000 Depletable cost Less: 2009 depletion Balance (3.000 1.80) 6.000 8.000 / 1.000 x 4) Accumulated depreciation 450.000 / 2. 850.000 550.000 2.000.900.000 Fourth year Depletion (100.000.000 x 20) 1. Cash (50.000 x 7) 700.000 Divide by remaining useful life in years (10 – 2) 8 Depreciation for third year 550.150.000) Depreciation for third year (100.000 3.600.000 7.000 Maximum dividend 1.Less: Accumulated depreciation Book value – beginning of third year 4.000.000 4.000 800.000 500.000.500.000 1.850.000 tons 450.000 240 2.000.000.000 2.500. Retained earnings Capital liquidated Dividends payable 1.800.500. Resource property Cash 3.000 .000 3.000 Original estimate of resource deposits Less: Extracted in first and second years Remaining output 1.100.00 Problem 18-8 1.000 Share premium 5. Retained earnings Accumulated depletion Total Less: Capital liquidated Depletion in ending inventory (5.000 x 110) Share capital (50.000 tons Depreciation rate per unit (3.000 / 550.000 x 5) Depreciation (Schedule B) 500.400.000 Schedule B – Computation of depreciation for fourth year Cost of equipment Less: Accumulated depreciation Book value – beginning of fourth year 8.000.000 4.000 200.800.000 2. Mining equipment Cash 800.000 2.000 3.000 100.000 Problem 18-9 1.000 3.000 x 100) 5.000 700. 268. Depreciation (90.000 Accumulated depletion (3.000 2.000 x 90.000 145.000.610.80) Accumulated depreciation .000 7.000 4. 2008 Sales Cost of sales Mining labor and other direct costs Depletion Depreciation Total production cost Less: Inventory.000 270.000 2.000 / 1.250.000 Depreciation rate (800.000 / 1.4.000 500. December 31 (5.000 72.000.000 x 29) Profit and loss 145.000 x . Depletion 270.268.768.000 1.000 . Mining and other direct cost Administrative expenses Cash 2.000 145.000 5.785. Cash (85.000 72.000 6.000 Gross income Administrative expenses Net income 4.80 8.000) = .465. Inventory.000 500.000.000 x 50) Sales 4.000 72.250.285.268.000 1.000) 270.000 Mining labor and other direct costs Depletion Depreciation Total production costs incurred Divide by number of units extracted Unit cost 2. December 31 2.000 29 241 Multinational Company Income Statement Year ended December 31.000 2.610.mining equipment 72.000 2.250.000 90.000 270. 000 6.000 3.000 145.000 5. Purchase price Road construction Improvements and development costs Total cost Residual value Depletable cost Depletion rate per unit (5.000 2.540.555.000 Maximum dividend Retained earnings Capital liquidated Dividends payable 5.327.000 3.800.000 242 Problem 18-10 1.000 270.000 800.30 Depletion for 2008 (500.000 750.200.000 3.785.000) .000 1.000.785.000 6.000 728.000) 5.285. 2008 Assets Current assets: Cash Inventory Noncurrent assets: Resource property Less: Accumulated depletion Mining equipment Less: Accumulated depreciation Total assets 3.000 1.000.000 / 4.000.000 ( 600.Multinational Company Statement of Financial Position December 31.000.000 5.000 1.000 255.000 270.000 ( 650.458.285.182.000 650.000 500.000 5.200.000 Equity Share capital Share premium Retained earnings Total equity Retained earnings Add: Accumulated depletion Total Less: Unrealized depletion in ending inventory (5.285.000 x 3) 15.730.200.000 x 1.000 72.000 1.30) Depletable cost Depletion in 2008 50.000 1.540.000) 1.000 1. 000) 1.000 2.90 900.000 x . 2009 Purchase price 28.800.90) 2. Cost of buildings Residual value Depreciable cost Depreciation rate per unit (1.000 Estimated restoration cost 2.000.500.000.30 Depreciation for 2009 (1.000 1.000 7.000 4.000 / 6.000) Depreciation for 2008 (500.000.800.850.000 / 6.000 .000.000. the output method is used in computing depreciation of mining equipment.30) 300.000 ( 225.000 3.000 Total cost 32.000 Development cost – 2009 1.45) .300.000 ( 200.000 Original estimated tons Additional estimate Total estimated tons Extracted in 2008 Remaining tons – 1/1/2009 6.000 x .000 243 Problem 18-11 2008 No depletion because there is no production.950.000 x . Depreciable cost Depreciation for 2008 Remaining depreciable cost Additional building in 2009 Total depreciable cost – 1/1/2009 1.000 5.000.575.000 / 4.Remaining depletable cost Development costs in 2009 Total depletable cost – 1/1/2009 4.000 ( 500.000.850.000 New depreciation rate per unit (1.000.000) New depletion rate per unit (5.000 In the absence of any statement to the contrary.000.000 .000) .000.45 225.000 1.000 375.950.000 Development cost – 2008 1.000) 1.500.800.000) Depletion for 2009 (1.000.000.500.550. 160.15 Depletion in 2010 (3.000 Problem 18-13 Answer C Depletion rate per unit (9.000 Rate in 2009 (27.000 Estimated restoration cost 1.000 Less: Residual value 3.70 Depletion in 2009 (3.400.800.000) 3.440.100.600.800.800.000 Rate per unit (28.000 – 3.000.000.000 2010 Tons extracted in 2010 3.30 .000 Total estimated output – 1/1/2010 6.000 x 3.000 / 4.000) 2.000.000 – 8.70) 8.000.000) 24 Depletion for 2008 (60.000 Total cost 31.000) 20 2.000 New rate in 2010 (27.000.200.Residual value ( 5.000/6.000 Development cost 3.200.600.000 x 2.000.500.000.000) Depletable cost 27.100.000.15) 11.500.000 / 10.025.000) Problem 18-14 Answer C Rate per unit (46.000 x 24) 1.000 Tons remaining in 12/31/2010 2.000 / 2.500.800.000.000.000 Problem 18-12 Answer B Acquisition cost 26.000 Depletable cost 28.800.000 / 1. 225.000.000 / 10.000.350.000 = 8.000) 1.350.000.500.Depletion in cost of goods sold (240.000.000 = .000) Balance – 1/1/2008 31.56) 840.70 x 4.000.000.000.000 x .200.000 Production in 2008 New estimate – 12/31/2008 5.000 Remaining depletable cost – 1/1/2008 4.000 Less: Residual value 3.000) .000 244 Problem 18-15 Answer D Acquisition cost 10.225.000 Problem 18-17 Question 1 – Answer A Purchase price 14.000 New estimate – 1/1/2008 5.200.000) ( 1.000.000 x 20) 4.000 / 4.000 Problem 18-16 Answer B Depletable cost 33.800.000 Depletion for 2008 (31.25 x 200.000 / 7.000 New depletion rate (4.000 Depletable cost 12.56 Depletion for 2008 (1.000.000 Depletable cost 7.000 Less: Residual value 2.000 Depletion for 2007 (33.000 .000 225.650.000.800.000.000 / 5.000.350.000 Less: Accumulated depletion – 12/31/2007 (7.000 = 6 x 225.500.000.000) 2. 000 / 300.000) 4.000 Rate per unit (7.05) 810.500.300.100.000 Rate in 2007 (8.00 Depletion for 2008 (150.000 Question 2 – Answer C Production from July 1 to December 31.200.000 x 4.000 9.000.000 x 8) 1.000.000.000 Problem 18-18 Answer C Purchase price Development costs in 2007 Total cost Residual value Depletable cost 9.000 7.100.000 / 1.000 x 5) 750.500. the output method is used in computing depreciation.000 8.000 Production (25.000 Since the life of the mine is shorter than the life of the equipment.000.200.000) 8.000 300.100.000 Depletable cost Depletion in 2007 ( 810.000 / 1.000 x 12) tons Estimated life of mine (1.Depletion rate (12.05 Depletion for 2007 (200.000 500.000 x 6) 150.500. 2008 (25.500.000 x 6) 150.290.000 / 2.000) 5.000 tons Annual production (25.000 Development costs in 2008 135.000) Balance 7.000 .000) 5 years 300. 245 Equipment Less: Residual value Depreciable cost 8.500.000 8.000 1.00 Depreciation for 2008 (150. 000 246 CHAPTER 19 Problem 19-1 1.500. Depreciated replacement cost (7.760. Book value (4.000) 3. 7.425. B C A B A Problem 19-2 1.000) 4.760.50) 1. Revaluation surplus (5.000 Problem 19-3 .000 / 1.425. 5.700.200. 2.000 x 4.000 – 3.50 Depletion for 2008 (300. 3.000) 2.350.Depletable cost in 2008 7.000 Rate in 2008 (7.160.600. 10.000 – 4. 4. Appreciation (7.600. 8.000 – 900.000 4. C B D C C 6.200.000 3.650.000 x 80%) 5.000) 2. 9.500.000 2. 000 600.250.000 / 30) Elimination approach 1.000.000 / 360.000.000) years 2.000 / 75%) 8.800.000 Original life (3. Depreciation (4.000 240.000 1. Annual depreciation on cost (9.250.000 Problem 19-5 Proportional approach 1.000 Age of asset (3.1. Depreciation (8.000) years 2.400.000 / 15) Accumulated depreciation 10 6. Machinery Accumulated depreciation (40% x 6. Equipment Accumulated depreciation Revaluation surplus 3. Revaluation surplus Retained earnings (1. Revaluation surplus Retained earnings (3.000) Revaluation surplus 3.000.000 600.000 / 25) 360.600.350.000 .000 / 40) or (6.350. Building Accumulated depreciation Revaluation surplus 2.600.000. Accumulated depreciation Building 75.000.000 200.000 2.000 / 150.000 4.000 / 30) Accumulated depreciation 3.600.250.000.000 2.000.000 247 4.000 240. Annual depreciation on cost (750. Depreciation (9.000 240.000.000 / 20) Accumulated depreciation 20 1.000 1.800.000 750.000 Gross replacement cost (6.000 1.000 / 5) 150.000 200.000.000 Problem 19-4 1.000 / 15) 240. Revaluation surplus Retained earnings (2.000 450.000.250.000 90.000 75.000 3.000 3.000 / 15) 90. 000 600.000 750.000 Cost Appreciation 3.050. Revaluation surplus Retained earnings 75.000. Cash Accumulated depreciation Equipment Gain on sale of equipment 8.000) Revaluation surplus 2.000 2.000 4.000 9. Revaluation surplus (2.200.200.000.980.000.000 2.980.000 1.000) 1.000 4.220. Depreciation (7.000 1.000 / 5) Accumulated depreciation 2.600.000 2.000 2.200.000 2.000 / 10) Accumulated depreciation 750.000 / 30) Accumulated depreciation 200.200.000 4.000 200.000.000 2.000 75.000 .Building (6.000 Problem 19-8 Replacement Building Accumulated depreciation 400. Depreciation (13.600.400.000 1.000 / 10) Retained earnings 220.000 3.000 500.200.000 Problem 19-6 1.600.000 5.000 220.000 cost 2.250.000. Building Accumulated depreciation Revaluation surplus 10. Revaluation surplus Retained earnings (6.000 3.000.000.200.000 1. Equipment Accumulated depreciation Revaluation surplus 2.000 3.250.000 248 Revaluation surplus Retained earnings (2.000.000.000 6.000 2.000 Problem 19-7 1.000 / 5) 1.500.000.000 – 3.000 2.000.000.700.000.000 1.250.750. Depreciation (6.000 . 000 Accumulated depreciation on replacement cost (5.000 a.000. Correcting entry: Building Retained earnings Accumulated depreciation Revaluation surplus 1. “Should be entry: Building Accumulated depreciation Revaluation surplus 2.600. Retained earnings Revaluation surplus 400.000 45.000.600.000 Accumulated depreciation 25.000 1.000 Problem 19-9 1.000 2.000 Problem 19-10 Cost Appreciation 5.000 / 20) 80.000 / 20) Accumulated depreciation 200.000 .000.000 200.000.000. Accumulated depreciation Machinery 800.000 400.000.000 c.000 1.000 / 80%) 5.000 400.000. Revaluation surplus Retained earnings (1.000.600.000 x 20%) 600.000 1.000 249 d.000. Depreciation (4.000.000.000 b.000 x 20%) 1.000 Building 20.000 Land Replacement cost 10.000 80.000.000.000 800.000 Life of asset (100% / divided by 4%) 25 years Percent of accumulated depreciation (5 years / 25) 20% Gross replacement cost (4.000 400.000.000 5.Accumulated depreciation on cost (3.000. 440.000 _ 360.000 420. Depreciation Accumulated depreciation – building Accumulated depreciation – machinery Accumulated depreciation – equipment 5.000.000.000.000 2.000.000.000 Equipment 1.000.000 x 3/25) 2.000 360.000 4.200.600.000 5.000 20.000 1.000 3.000 / 2) Appreciation (2.000 a.000 __________ 15.000.000 3.000.000 Replacement cost 4.000 3. Land Building Machinery Equipment Accumulated depreciation – building Accumulated depreciation – machinery Accumulated depreciation – equipment Revaluation surplus 5.000 2.000.000 10.000 x 3/10) (4.000 5.000 / 22) Machinery: Cost (4.000.940.000 6.000 _________ 2.000 5.000 / 22) Appreciation (17.000.200.000.000 22.000.000 39.000.000.000.000 2.600.400.000 b.000 2.100.000.200.000 900.400.000.000 Accumulated depreciation (3.220.000 3.000 1.000 _________ 5.000 Machinery Accumulated depreciation (10.000 x 3/10) Cost Appreciation 3.(25.000 840.000.000 / 7) Total depreciation 1.000.000.000 x 3/5) (15.000.000 .000 120.000.000 1.000.600.000.100.000 25.000.000 / 7) Appreciation (840.000 17.220.000 9.000.000.000 420.000 1.400.800.000 6.000 x 3/25) (45.000 800.000 250 Building: Cost (22.000 1.000.000 / 2) Equipment: Cost (2.000 300.800.260.200.000 x 3/5) 3. 920.000.880.000 cost 7.000.200.000 29.200.000 Cost Appreciation 5.000 + 1.920.000.000.000 13.250.000 Replacement cost 8.000. Revaluation surplus Retained earnings (800.800.000.250.000.000.000.000 20.000 d.000 15.000 45.000.000 Less: Accumulated depreciation 20.000 74.750.000 The following disclosure should be made in the notes to financial statements: Replacement Land Building Machinery Equipment Total Accumulated depreciation Net carrying value Cost 5.c.000 1.000 53.200.000 Net carrying value 53.000.000 Equipment 4.000 1.880.000 1.000 2.200.000 .000 Building 45. plant and equipment (at revalued amounts): Land 10.000.000.000 8.000 12.320.000.000.000 3.000 + 120.000 Total 74.000 1.200.000.000 251 Problem 19-11 Answer B Building Accumulated depreciation 750.000 25.320.000 43.880.000 10.000 6.200.000 4.000) 1.000 Machinery 15.000 3.200.000.680.000. Property.000 2.000 cost 10.200.000.000 20.000 3.000 13.000.000 Schedule of Accumulated Depreciation Replacement Building Machinery Equipment Cost 4. 000 + 900. 2008 Net book value – December 31.500.000 / 3) Accumulated depreciation 900.Problem 19-12 Answer B Sound Revaluation value Book value surplus Land 5.000 1. 2008 Revaluation surplus 450.500.100.000 Problem 19-13 Answer D Fair value – December 31. D 9. 3.000) 7.500.000 252 3.000) 1. 15.000 3.000 Building (75% x 25.000 2.500.000 11.000.000 2. A C B A A 6.500. D 10. 2.500 Problem 19-14 Question 1 Question 2 Question 3 Answer A Answer B Answer B Problem 19-15 1.000 900.000 2.400.000) 4.000 2.100. 14.000.750.000.250. A A A D A Problem 19-16 1.000 900.500. A 7.000. Depreciation (1.000 3.000 18. Cost Accumulated depreciation (2.000. D 11.000 302.500. 12.000.500 142.000 Machinery (50% x 5.000 + 500. 5.500. Impairment loss Accumulated depreciation Cost Accumulated depreciation Book value – January 1 Recoverable value Impairment loss 2. 13.000 500.000 1.000 4.500.000 . 4.000 11. A 8.000 500. 000 3.125.000 253 Problem 19-19 .000 2.125.000.000 Impairment loss 4.000 28. 2. being the higher amount 30.000 Problem 19-17 1.000.000 + 1.000 Adjusted carrying amount 34.500.000.000 Carrying amount 39. Impairment loss Accumulated depreciation Cost – January 1 Accumulated depreciation (2.000 – 125. the estimated liability assumed by the buyer is deducted in determining both the value in use and carrying amount of the asset.000. Impairment loss Accumulated depreciation 4.000 Less: Estimated liability 5.000. The standard further provides that to perform a meaningful comparison between the carrying amount and recoverable amount.000 375.500.000 875.Book value – December 31 1.000.000 1.000.000.000) Book value – December 31 1.125.000 + 375. Depreciation Accumulated depreciation (875.000 Book value – January 1 Recoverable value Impairment loss 2.000 / 2) 375.000 1.000 – 500.000 Cost of dismantling and removal assumed by the bidder Fair value less cost to sell Present value of future cash flows 33.000 500.000 2. provides that the fair value less cost to sell is equal to the estimated selling price plus the estimated liability assumed by the buyer.000 / 8 x 2) 500.000 5.000.000 Problem 19-18 1. Offer price 25.000.000 2.000. Cost Accumulated depreciation (500.000. paragraph 78.000 Recoverable amount – fair value less cost to sell.000 30.500.000.000 2.000 Less: Estimated liability Value in use 5.000.000 4.000 PAS 36.000.125.000. 000.000 8.000 5.000 4.000 / 10 x 2) Book value – 12/31/2007 Impairment loss – 2007 Adjusted book value – 12/31/2007 Depreciation – 2008 (6.000 .000 .000 5.000.000 2.325.000 Cost – 1/1/2006 Accumulated depreciation (10.740.000 6.000 .000 – 50.910. Impairment loss Accumulated depreciation (65.675.000 Book value – 12/31/2008 10.750.000 3.000.000) 14.581. Depreciation Accumulated depreciation (10.000.000.000 / 8) 5.000. Accumulated depreciation Gain on impairment recovery 1.000 4.000. Impairment loss Accumulated depreciation 2.000 1.000 Problem 19-20 1.000 PV factor Present value .930 15.000 1.000.820.000 because this is higher than the fair value less cost to sell of P48.000 / 8) 750.000 / 10 x 3) Book value – 12/31/2008 (assuming no impairment) 7.250 12.000.000 Cost – 1/1/2006 Accumulated depreciation (10.000.000 2010 11.000.250.750.000 2.000. 3.000.000.000.000.000 2011 8.000.855.675.000.250.000 750.325. Depreciation Accumulated depreciation 1.000 3.000 2.000 / 10) 1.1. Depreciation Accumulated depreciation (50.000.000.735 60.000 2.581.000.000 Net cash inflows 18.325.250 14. The recoverable amount is the value in use of P50.000 2009 12.000 750.000.000 2.000 Total value in use 50. Depreciation Accumulated depreciation (6.000 1.000.000.000.794 12.857 15.325.000. 2008 16.000.000 .000 / 4) 12.000 Recorded book value 10. Impairment loss Accmulated depreciation 8.000.65) 2.500.500. regardless of the amount of undiscounted cash flows whether less than or more than the carrying amount.192. Value in use (1.000.000 10.000 8.000.000 – 30. Impairment loss Accumulated depreciation Machinery Accumulated depreciation Book value – 1/1/2008 Present value of cash flows – higher than fair value 3. PAS 36 has totally rejected the concept of undiscounted cash flows for impairment purposes. Value in use (800.000 Problem 19-23 1.000 8.500.500.000.000 3.000. an impairment loss is recognized.000 1.750.000.000.000.000 x 5.000 18.000 254 The fair value or recoverable value of P7.000.000) 5.000 . under the PAS 36.000 / 10) 25.000 6. if the recoverable amount is less than carrying amount. However.000 / 20 x 6) 6.000 5.000. Depreciation Accumulated depreciation (10.000 for 5 years.000 Observe that the undiscounted net cash flows from the asset amount to P37. no impairment loss should be recognized in this case. Depreciation Accumulated depreciation (30.000 x 3.99) 2.000 2. Impairment loss Accumulated depreciation (35.000 308.000 3.000 8.000.000 1.000 Book value – 1/1/2008 Fair value – higher than value in use Impairment loss 3.250.250.192.500.000 5.250.000 308.250. Problem 19-22 1.000 cannot exceed the “book value” that would have been determined assuming no impairment is recognized.000 1. Under American Standard.475.000 / 5) 6.000. This amount is more than the book value of the machinery. Problem 19-21 1.000.500.000 Buildings Accumulated depreciation (22.Gain on reversal of impairment 1.750. 000 3.000 which is lower than its fair value of P6. Impairment loss Goodwill Accumulated depreciation – building Inventory Trademark Fraction 20/45 15/45 10/45 Loss 400.000 / 5) 638.400 255 Problem 19-24 1.500.400.000) 2.000 Observe that after allocating the P2.000.000 1. the carrying amount of the building would be P5.000 900.000 loss to the building. to all other assets of the unit prorata based on their carrying amount.000 Equipment (4/16 x 5.000 1.000 300.000 200. Depreciation Accumulated depreciation (3. Then.000 5.Impairment loss 308.500.000 When an impairment loss is recognized for a cash generating unit.000) 1. Total carrying amount Value in use Impairment loss 5.000 900.000.000) Inventory (4/16 x 5.000 5.000 1.000 Problem 19-25 1.000.000 500.250.400.000.400. First. if any. Impairment loss allocated to goodwill Impairment loss allocated to the other assets 500.000 2.000.000 3.000 1.000 1.000.000.000.000 1.500.000 4.000 2. b. Allocation of impairment loss Building (8/16 x 5.000 200.000 400.500.500.192.500.000.400 638.250.000 11.600. Building Inventory Trademark Carrying amount 2. . to the goodwill.000. Carrying amount Value in use Impairment loss 16.000.000 300. the loss is allocated to the assets of the unit in the following order: a.000 3. 000 Problem 19-26 All Unimart’s stores are in different locations and probably have different customer profile.250. Therefore. It is unlikely that cash inflows from B and C can be determined individually. Smart generates cash inflows that are largely independent from those of the other Unimart’s stores.250. Problem 19-28 Case 1 1. together . Problem 19-27 It is likely that the recoverable amount of an individual magazine title can be assessed. the individual magazine titles generate cash inflows that are largely independent from one another and therefore. Impairment loss Accumulated depreciation – building Accumulated depreciation – equipment Inventory Building 2. A is separate cash generating unit because there is an active market for A’s products.000 is reallocated to the equipment and inventory prorata. cash inflows from B and C depend on the allocation of production across two countries.500.000 _________ 1.000 5.000) (4/8 x 1.000. In addition. B and C.000.000. 256 Allocated loss Reallocated loss (4/8 x 1.000 1.000.000) Equipment 1.000 1.000.Accordingly.000 Impairment loss 3. cash inflows from direct sales and advertising are identifiable for each title. Although there is an active market for the products of B and C.000 1. Therefore.750. So although Smart is managed at the corporate level.750.000 500.000) 500. decisions to abandon titles are made on an individual basis.750. 2. it is likely that Smart in itself is a cash generating unit.500. Accordingly.500.000 loss is allocated to the building and the balance of P1.000 _________ 1.000 (1.500.750. each magazine title is a separate cash generating unit. Even though the level of advertising income for a title is influenced to a certain extent by the other titles in the customer segment.000 Inventory 1. only P1.000 1. b. 2007 Recoverable value Impairment loss 300. Problem 19-30 Answer C Cost. 2007 (100. Therefore.000 600. January 1.000 + 300.000 40. and therefore. since there is no active market for A’s product. B and C. Thus.000 . The building is not held for investment.000 300. December 31. should be treated as the largest single cash generating unit.should be treated as a cash generating unit. the cash generating unit is Litmus Company as a whole. As a consequence.000 500.000 The loss is recorded as follows: Impairment loss Accumulated depreciation Cost Accumulated depreciation (300. 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. January 1.000 / 5) Book value. Problem 19-29 The primary purpose of the building is to serve as a corporate asset supporting Litmus Company’s manufacturing operations.000 300. 2008 300. as a whole.000 Depreciation for 2008 (200. A. 2005 800. December 31. In this case. December 31. it is not appropriate to determine the value in use of the building based on the cash inflows of related rent. 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. together. 257 Case 2 a.000 200.000 160.000) Recoverable value. 2008 200.000 800. Maximus Company.000 x 3) Book value.000 Accumulated depreciation. 000 x 5) 1.350.000 Problem 19-34 Answer C Depreciation for 2008 (10% x 2.050.12/31/08 (200.000) Cost – 1/2/2004 Accumulated depreciation .000 1.000 650.000 / 10 x 4) 360.000 1.000 .Problem 19-31 Answer B From August 31.000 Problem 19-32 Answer B Cost – January 1.000.600.000. 2008 1.000 Book value.000. December 31.000 1.000 490.000 Book value. December 31.000 640.000 258 Cost Accumulated depreciation – 5/31/2008 (3.350.400.000 x 33/60) Book value – 5/31/2008 Fair value Impairment loss 3. 2007 Depreciation for 2008 (640.000 Book value-12/31/2008 Estimated cost of disposal Impairment loss Problem 19-35 Answer C 200.000 1.000 / 4) 150.000 Book value. 2005 to May 31.715. 2007 (900. the remaining life of the machine is 27 months.000.000 – 500.485.350.000 365.000.000 Problem 19-33 Answer C Book value.000 50.000. 2008 is a period of 33 months. 2004 Accumulated depreciation. December 31. 60 months original life minus 33. Depreciation for the month of June 2008 (1.000 1.000 1.000 Depreciation for 2008 (1. Thus.200. 1/1/2008 2.000 / 27 months) 50.000 2. 12/31/2008 Sales price-recoverable value Impairment loss 2.000 / 4) 400.000 – 40.200. 000.000 Decommissioning cost ( 8.000 Value in use 26.000.000 Impairment loss 200.000 7.000.000 Fair value less cost to sell – higher (20.000 Decommissioning cost ( 8.000.Cost Accumulated depreciation – 1/1/2008 (2.000) Adjusted carrying value 20.000 / 10 x 2.000 Impairment loss 1.600.5) Book value – 1/1/2008 Fair value Impairment loss 2.000 259 Problem 19-37 Answer C Carrying value 28.000.000 1.000.000 – 100.000 5.500.000 Book value – 8/31/2008 1.100.000 Fair value 1.800.000 475.000.000 less 1.000) Adjusted value in use 18.600.000 .000 Problem 19-36 Answer C Cost – 12/31/2004 2.000) 19.400.000.000) Carrying value – 12/31/2008 Carrying value – 12/31/2008 (assuming no impairment) Reversal of impairment loss 7.000.000.000.000 1.525.000 Problem 19-38 Answer C Carrying value – 12/31/2007 Depreciation for 2008 (20%) (1.000 Accumulated depreciation – 8/31/2008 (2.700.000 925.000 / 96 months x 44) 1.000 600.200.000.400.000. 750 2009 Dec. 31 Amortization of patent Patent (255.000 255. 3. 8. D A D D B Problem 20-2 1. 10. 31 Amortization of patent Patent 12. 2.000 / 20) 255. 4.750 90. D 9. 9. 3. 5. 5 Legal expense Cash 90. B 7. D Problem 20-3 2008 Jan. A A C D D 6. 4.000 12. D C D A D 6. B 8.000 .000 12. 5. 7. D 10.750 2010 Jan.260 CHAPTER 20 Problem 20-1 1. 1 Patent Cash Dec.750 12. 2. 000 / 15) 45.000 6.000 Cash Amortization of patent Patent 540.000 / 17) 12.000 261 Amortization of patent (720.000 On related patent On competing patent (540.000 / 16) Patent 2012 Patent 45.000 600.000 81.750 42.Dec. 31 Amortization of patent Patent 2011 Jan.000 36.000 510.000 6.750 On original cost On competing patent (510.000 81.000 540.000 42.000 250.000 Cash 720. 31 Amortization of patent Patent 12. 1 Patent Cash Dec.000 600.000 81.000 45.000 42.000 2011 Patent 720.750 Problem 20-4 2008 Research and development expense Cash 510.000 Cash Amortization of patent Patent (60.750 12.000 Problem 20-5 2008 Research and development expense Cash 2009 Patent 250.000 510.750 510.750 30.000 / 10) 2010 Patent Cash 60.000 .000 60. Acquisition cost Amortization for 2005.712.000 43.000 / 10) 900.000 43. 2006 and 2007 (476. Amortization of patent Patent (7.600 43.000 ( 816.600 43. 2009 Amortization of patent (654.800 566.000 654.140.000 90.000 660. Amortization of patent Patent (5.000 816.000 (1.000) 4.000 7.800 262 Problem 20-6 1.000) 5.000 476.896.600 43.600 87.000 6. Patent Cash 7.712.000 .600 566.800 654.140.200 566. Patent Cash 2.000 Problem 20-7 1.428. Amortization of patent Patent (900.000 / 15) 476.000 600.Original cost New patent Total cost Less: Amortization for 2008 Balance – January 1.140.000 90.000 3.000 x 3) Carrying amount – 1/1/2008 Amortization for 2008 Carrying amount – 12/31/2008 7.600 43.140.000 2.000 / 15) Patent 2011 Amortization of patent Patent Patent written off Patent Balance – 1/1/2010 Less: Amortization 2010 2011 Unamortized cost 60.000 / 7) 816.000 4.000 900. 000 – 450.000 30.000 500. Patent Retained earnings 510.000 3. Amortization of patent Patent 50. 4.050.000 / 5) Z (3.000 1. Retained earnings Patent 500.500 25. Patent written off Patent 540.500 24.000) Carrying amount – 12/31/2008 900.500 Problem 20-10 2008 Copyright Cash 285.000 Cost Amortization for 2005.500 130.000.500 Amortization per book (500.000 263 Problem 20-9 1. Patent Retained earnings 24.000 / 8) Y (2. Legal expenses Cash 3.000 400. 2007 and 2008 (90.000 540.000 500.000 / 20) Overamortization 6. No adjustment.000 .000 x 4) (360. Loss on damages Legal expense Accrued liabilities 100.000 2.000 Problem 20-8 1.000 510.000 450.000 6.000 / 6) 150.050.000 450.200. 2006.000 5.000 25.050.200.000 24. Amortization of patent Patent 6.000 285.000 X (1. Patent Cash 2.000.000) Correct amortization for 2007 (510.500 25.200.000 540.000 1.000 1.3. 000 .000 / 5) Book value 2.000) Amortization of patent Copyright Patent Problem 20-13 Books of Franchisee 400.000 60.000 60.000 / 15 x 2.000 Book value 2.000 240.000 240.000 / 10 x 2) 100.000 Book value Copyright 2 Less: Amortization from 7/1/2005 to 12/31/2007 (360.000 = 3 per copy 50.000 264 Problem 20-12 1.000 / 20 x 4) 80.000 320.000 x 3 = 150.000 150.000 400.Amortization of copyright Copyright 150.000 Cost of copyright Less: Amortization (300.000 / 95.000 50.000 Copyright 1 Less: Amortization from 1/1/2004 to 12/31/2007 (400.000 285.000 x 3) 90.000 44. Copyright Patent Retained earnings 620.020.000 60.000 60. Amortization of copyright (20. Copyright Retained earnings 240.000 90.000 44.000 500. Amortization of copyright Copyright 300.5) Book value Patent Less: Amortization for 2006 and 2007 (500.000 1.000 + 24.000 300.000 50.000 Problem 20-11 1.000 2009 Amortization of copyright Copyright (30.000 400.000 360. 210. Cash Sales 25.210.000 x 3.79) 3.250.000 5.000 / 20) 300.000.000.000) Discount on note payable Cash Note payable 6.000 / 4) Interest expense (15.000.000.750.790.000 15.000 3. Note payable Cash 4. Interest expense (10% x 3.000.000 Problem 20-14 Books of Franchisee 1.000 Implied interest 2.000.000.000 1.000 679.000) Discount on note payable Problem 20-16 5.000 5.000 2.000 5.000 25.000 1.790.790.000 6.000.000.000 Note payable Present value of note (1.000.000 x 10%) Cash 20.250.000 300. There is no amortization because the franchise is for an indefinite period. Franchise fee expense Cash (25. Franchise Cash 6.000 1.000 .000.000 1.000 4. Amortization of franchise Franchise (6.000.000.000 1.000 3.1.000 / 10) 3.000 3.000 265 3.000 x 5%) 1.000 679.000.000.000 379.500.000 + 3. Amortization of franchise Franchise (6. Franchise (3. Franchise Cash Note payable 2.000 1.000 379.000.000. Note payable (15.000.790.790.250. Problem 20-15 Books of Franchisee 1. 000 100. Leasehold improvement – building Cash 2.000 5.000 / 10) Accumulated depreciation 5.000 x 12) Cash 3.000 100.000 500.Requirement a 1.000 5.000 600.000. Depreciation (5.000 1.000 400. Rent expense (50.000 100.500.000.000.000 4. Leasehold Cash 1.000 4.000.000 1.000 2. Rent expense (150.000 2.000.000 3.000 2. Leasehold improvement Cash 500.000 600. Rent expense Prepaid rent Cash 1.000 x 12) Cash 1.000 / 5) 400. Depreciation (500.400.000 Problem 20-17 1.000 .000 5.000 500.000 1.000 400. Leasehold improvement Cash 400.800.000 5.500.000 Problem 20-18 1.000 / 5) Accumulated depreciation 500.000.200. Amortization of leasehold Leasehold (2.000 3.000 Requirement b Accumulated depreciation Loss on leasehold cancelation Leasehold improvement – building 2.000.200.000 2.000 266 2.800. Leasehold Cash 2.000. Amortization of leasehold 100. Leasehold improvement Cash 100.000. Amortization of leasehold Leasehold (100.000 60.000 3. Rent expense Cash 600.000 / 4 40.000 – 240. Trademark (800. Leasehold improvement Cash 200.000 2.000 600.000 2.000 100.500 Problem 20-20 1.000 400.000 267 5.000 50. Depreciation Accumulated depreciation 100.Leasehold (1.000 100.920.000 / 5 20.000 12. Leasehold improvement Cash 50.000 280.000. Amortization of patent Accumulated amortization (1.000 4.000 3.000 4.000 / 10 40.500 52.000 60.000 .000 / 5 50.000 Problem 20-19 1.000 Problem 20-21 40.000 800. Depreciation Accumulated depreciation 52.500 20.000 / 6) 280.000 200.000 / 5) 20.000 200.000 60.500 200.000 50.000 6.000 52. Royalty expense Cash 50.000 / 10) 6. Amortization of noncompetition agreement Accumulated amortization (200.000 / 5) 40. Leasehold Cash 100.000 x 3/4) Noncompetition agreement Cash 600. 900.000 3.500.000.000 500.000 6.500.000.000 1.000 5.300.000 2.000 4.000 6.700.000 3.000) 2.300.500.000 500.000 1.000.000.000 268 2.000) 4. The assembled workforce is not accounted for separately as an asset.200.000 5.000 500.000 800. plant and equipment Goodwill Accounts payable Cash 50.000 5. plant and equipment Goodwill Accounts payable Note payable – bank Cash 7.500.000 (3.700.000 7.500.600. Acquisition cost Net assets acquired at fair value Goodwill Total assets at fair value Total liabilities Net assets acquired at fair value 6.000.000. Cash Accounts receivable Inventory Patent Property.000) 2.500.000 250. Cash Inventory In-process R and D Total assets Total liabilities Net assets Acquisition cost Net assets acquired at fair value Goodwill 1.000 Problem 20-23 1.000 2.000.000 8. Cash Accounts receivable Inventory Property. Cash Inventory In process R and D Goodwill 1.000 The goodwill includes the fair value of the assembled workforce of P1. Acquisition cost Net assets acquired Goodwill 2.000 1.000.500.000 Problem 20-22 1.000 (3.000 2.000. 2.000.000 50.000 4.000 .000 3.000 900.350.300.1.000.000 2.000.500.900.000.000 (4.000 3.300. 600. Average future earnings Divide by Net assets including goodwill Less: Net assets excluding goodwill Goodwill b.000 450.000 400. Excess earnings Multiply by Goodwill 80.000) 170.000 Earnings for goodwill computation a.000 5. Average earnings Divide by Net assets including goodwill Less: Net assets before goodwill Goodwill 2.000 5.000 625.000 114.000 269 3.000 Problem 20-25 Average earnings or prior years (1.65 452.000 / 5 years) 100.500.625.000 550.000 / 3) Increase in average earnings (10% x 500. Average earnings Less: Normal earnings (10% x 1.000.000 .000 15% 760.000 450.000 250.000 Excess earnings Divide by Goodwill 250.000 4.000 Problem 20-24 1.700.000 80.000 450.700.000 5 400.000 800.000.000) Total Less: Patent amortization (500.000 50. Average earnings Less: Normal earnings (8% x 1.000.000 8.000) 136. Average earnings Less: Normal earnings (8% x 5.Accounts payable Notes payable Cash 2.000 50.000 Excess earnings Multiply by Goodwill 250.500.000 10% 2.000 400.000 1.700.000 8% 5.000) Average excess earnings 500. 000 x 10%) 480.Divide by Goodwill c.000. Share capital Retained earnings Total shareholders’ equity 3.17) 10% 500.000 2.000.000.000 200.000) 100.000 4.000) 38.750.000 Average earnings (1.000 1.000 b.000.500.000.500.000 1.000 370.000 850.000.000 / 20%) 1.500. Value in use Net assets including goodwill at carrying amount Impairment loss ( 4.000 / 3) Less: Normal earnings (10% x 2.000 Less: Recorded goodwill Net assets before goodwill 2.480.000 + 150.000 200. Average earnings Expected increase (1.000 2.000 Goodwill (370.800.250.000 42.000 Excess earnings Divide by Goodwill 450.000 x 3.850.000 Excess earnings 750.000 Total Less: Normal earnings (4.500.000 Problem 20-28 1.800.000) 250.000 1.000 Shareholders’ equity per book Less: Recorded goodwill Net assets before goodwill 5.200.000 – 900.000 270 Problem 20-27 1.000. Goodwill (370.000 x 4) 1. Net assets before goodwill Goodwill Purchase price 2.000 16% 1.000 3.500. Goodwill (50.000 .000 158.500 Problem 20-26 a.250. 000 30.000.000 5.000) Inventory (15/45 x 270.000 90.2.000.000) Trademark (5/45 x 270.000) 271 Present value of cash flows from cash generating unit (9.51) 76.000) 2.000.000.410. Impairment loss Trademark Goodwill 7.000. Total carrying amount Value in use Impairment loss 2.000 2.230.000 3.590.000.000 / 10%) 2. is allocated to the other noncash assets on a prorata basis.000 3. Impairment loss Goodwill Accumulated depreciation – building (25/45 x 270.000) Problem 20-32 5.000 770.000.000 Net assets including goodwill at carrying amount 80.000.000 75.000.000 . Impairment loss Goodwill 4.000. Present value of indefinite cash flows (200. Problem 20-30 1.000 4. after deducting the loss applicable to goodwill.000.000.000 15. Value in use Net assets including goodwill at carrying amount Impairment loss (15.000.000 The remaining impairment loss of P10.000.000 x 8.000 Trademark Impairment loss 6.000.410.000 Problem 20-29 1.000 5.000 (4.000.000.000 4.000.000 770.000 500.000 Problem 20-31 1.000 Impairment loss ( 3.000) 2.410.000 150. Impairment loss Goodwill Accounts receivable Inventory Accumulated depreciation 60.000.000 4. 000 3.300. The costs incurred from the time of technological feasibility to the time when product costs are incurred should be capitalized as computer software cost.000 1.000 3.400.000 5.000 1.300.000 2.000.400.000 800.000.500.000 / 4) Total expense in 2009 1.500. Completion of detail program design Cost incurred for coding and testing to establish technological feasibility Total costs charged as expense 2.000 Cash 100.500.625.000 1.000 2.000 Cash 11/15/2009 Patent 350.000 1.000 500.000.500.500.000 Patent 350.12/31/2008 1/1/2009 7/1/2009 11/1/2009 R and D expense Cash 2.000 Cash 12/31/2009 Patent 800.000 1.000 100.200.200. Product costs which are associated wit inventory items are: Duplication of computer software and training materials Packaging product Total inventory 2.000 is capitalized as software cost to be subsequently amortized.000 R and D expense Cash 500.000 900. Cost of producing the software program in 2009 Amortization of software cost (2.000 625.000 2.000 __500.500.000 2.000 Problem 20-33 1. 2. Designing and planning Code development Testing Total R and D expense in 2008 1.900.000.000 R and D expense Cash 1.000 .000 Problem 20-34 1.000.000 The cost of producing the product master of P2.000 1. 272 Other coding costs after establishment of technological feasibility Other testing costs after establishment of technological feasibility Costs of producing product masters for training materials Total costs to be capitalized 3.500. 000 285.000 500.000 10% 750.000 Amortization for 2008 (4.000 / 15 x 3) 71.400 Book value – 12/31/2007 Amortization for 2008 (285.000 2.Problem 20-35 Answer C Cost Accumulated amortization from 2005 to 2007 (357.000.27 353.000.600 40.000 4.000.000.000 Problem 20-37 Answer C Cumulative earnings Less: Gain on sale Adjusted cumulative earnings 550.000 / 15 x 5) Book value – 1/1/2008 6.800.000 / 5) 800.000 Less: Net assets before goodwill Goodwill 100.000 4.800 Problem 20-36 Answer C Cost 1/1/2003 Accumulated depreciation – 12/31/2007 (6.000 273 Average earnings (500.600 / 7) Book value – 12/31/2008 357.500.000 Problem 20-38 Answer C Net assets Multiply by excess rate (16% minus 10%) Excess earnings Multiply by present value factor Goodwill 1.000 / 5) Divide by capitalization rate Net assets including goodwill 1.000 50.000.000 6% 108.000 Net assets before goodwill 5.000 .800 244.000.000 3.160 Problem 20-39 Answer D Purchase price Less: Goodwill 500.000 250.000. 000.000 274 Problem 20-43 Answer A Design costs Legal fees of registering trademark Registration fee with Patent Office Total cost of trademark 1.000 500.000 / 15) Book value 12 years 8 20 2 18 15 540.000 150.910.000.000 20% 500.000 100.000 (1.000.000 (2.700.000.000 Problem 20-41 Answer A Problem 20-42 Answer C Downpayment Present value of annual payment for 4 years (1.500.500.000 Problem 20-44 Answer B Original lease Extension Total life Less: Years expired (2006 and 2007) Remaining life years Life of improvement (shorter) years Leasehold improvement Less: Depreciation for 2008 (540.000 .000 450.000 504.000 x 10%) Excess or superior earnings Divide by capitalization rate Goodwill 550.000 2.000) 4.000.91) Cost of franchise 2.000 x 2.000 Acquisition cost Net assets at fair value Goodwill 5.000 50.000.000 500.Estimated annual earnings (squeeze) Less: Normal earnings (4.000) 1.000 Problem 20-40 Answer C Accounts receivable Inventory Equipment Short-term payable Net assets at fair value 2.000.910.000.000 1.000 36.000 4. 600.000.000 500.000 1.000 215.000.000 / 5) 400.000 Impairment loss 2.000 2.000.000 x 3.000 -_ _ Problem 20-50 Answer B Carrying amount of net assets Value in use (8.235.5) 12.000 Impairment loss – applicable to goodwill 16.000. jigs.000 150.000 520.000 Book value – 12/31/2008 Value in use (500.000.47) 1. molds and dies 170.735.000 200.000 x 1.000 Laboratory research Total research and development expense 135.000.000.000 250.600.000 .Problem 20-45 Answer D Depreciation (3.000 Problem 20-46 Answer C Depreciation of equipment Materials used Compensation costs of personnel Outside consulting fees Indirect costs allocated 135.000 Problem 20-48 Answer D All costs are charged to R and D expense.000 4.000 / 6) 600. 275 Problem 20-49 Answer A Trademark Value in use (120.000 / 6%) Impairment loss 3.000.000.000 Patent Amortization for 2008 (2.000 Problem 20-47 Answer A Modification to the formulation of a chemical product Design of tools.000 1.000 1.