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TO SOLUTIONS EXERCISES EXERCISE 11-1 (1520 minutes) (a) Straight-line method depreciation for each of Years 1 through 3 = $518,000 $50,000 = $39,000 12 (b) Sum-of-the-Years-Digits = 12 X 13 2 = 78 12/78 X ($518,000 $50,000) = $72,000 11/78 X ($518,000 $50,000) = $66,000 depreciation Year 2 10/78 X ($518,000 $50,000) = $60,000 (c) depreciation Year 1 depreciation Year 3 Double-Declining-Balance method depreciation rate. 100% 12 X 2 = 16.67% $518,000 X 16.67% = $86,351 depreciation Year 1 ($518,000 $86,351) X 16.67% = $71,956 depreciation Year 2 ($518,000 $86,351 $71,956) X 16.67% = $59,961 depreciation Year 3 EXERCISE 11-2 (2025 minutes) (a) If there is any salvage value and the amount is unknown (as is the case here), the cost would have to be determined by looking at the data for the double-declining balance method. 100% = 20%; 20% X 2 = 40% 5 Cost X 40% = $20,000 $20,000 .40 = $50,000 Cost of asset EXERCISE 11-2 (Continued) (b) $50,000 cost [from (a)] $45,000 total depreciation = $5,000 salvage value. (c) The highest charge to income for Year 1 will be yielded by the double-declining-balance method. (d) The highest charge to income for Year 4 will be yielded by the straight- line method. (e) The method that produces the highest book value at the end of Year 3 would be the method that yields the lowest accumulated depreciation at the end of Year 3, which is the straight-line method. Computations: St.-line = $50,000 ($9,000 + $9,000 + $9,000) = $23,000 book value, end of Year 3. S.Y.D. = $50,000 ($15,000 + $12,000 + $9,000) = $14,000 book value, end of Year 3. D.D.B. = $50,000 ($20,000 + $12,000 + $7,200) = $10,800 book value, end of Year 3. (f) The method that will yield the highest gain (or lowest loss) if the asset is sold at the end of Year 3 is the method which will yield the lowest book value at the end of Year 3, which is the doubledeclining balance method in this case. EXERCISE 11-3 (1520 minutes) (a) 20 (20 + 1) = 210 2 3/4 X 20/210 X ($774,000 $60,000) = $51,000 for 2010 + 1/4 X 20/210 X ($774,000 $60,000) = 3/4 X 19/210 X ($774,000 $60,000) = $17,000 48,450 $65,450 for 2011 EXERCISE 11-3 (Continued) (b) 100% = 5%; 5% X 2 = 10% 20 3/4 X 10% X $774,000 = $58,050 for 2010 10% X ($774,000 $58,050) = $71,595 for 2011 EXERCISE 11-4 (1525 minutes) (a) $279,000 $15,000 = $264,000; $264,000 10 yrs. = $26,400 (b) $264,000 240,000 units = $1.10; 25,500 units X $1.10 = $28,050 (c) $264,000 25,000 hours = $10.56 per hr.; 2,650 hrs. X $10.56 = $27,984 (d) 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55 OR 10 X $264,000 X 1/3 = 55 $16,000 9 X $264,000 X 2/3 = 55 28,800 Total for 2011 (e) n(n + 1) 10(11) = = 55 2 2 $44,800 $279,000 X 20% X 1/3 = [$279,000 ($279,000 X 20%)] X 20% X 2/3 = Total for 2011 $18,600 29,760 $48,360 [May also be computed as 20% of ($279,000 2/3 of 20% of $279,000)] EXERCISE 11-5 (2025 minutes) (a) ($150,000 $24,000) = $25,200/yr. = $25,200 X 5/12 = $10,500 5 2010 DepreciationStraight line = $10,500 (b) ($150,000 $24,000) = $6.00/hr. 21,000 2010 DepreciationMachine Usage = 800 X $6.00 = $4,800 (c) Machine Year Allocated to 2010 2011 Total 1 2 5/15 X $126,000 = $42,000 4/15 X $126,000 = $33,600 $17,500* $17,500 $24,500** 14,000*** $38,500 * $42,000 X 5/12 = $17,500 ** $42,000 X 7/12 = $24,500 *** $33,600 X 5/12 = $14,000 2011 DepreciationSum-of-the-Years-Digits = $38,500 (d) 2010 40% X ($150,000) X 5/12 = $25,000 2011 40% X ($150,000 $25,000) = $50,000 OR 1st full year (40% X $150,000) = $60,000 2nd full year [40% X ($150,000 $60,000)] = $36,000 2010 Depreciation = 5/12 X $60,000 = $25,000 2011 Depreciation = 7/12 X $60,000 = 5/12 X $36,000 = $35,000 15,000 $50,000 EXERCISE 11-6 (2030 minutes) (a) 2010 Straight-line $304,000 $16,000 = $36,000/year 8 3 monthsDepreciation ($36,000 X 3/12) = $9,000 (b) 2010 Output $304,000 $16,000 = $7.20/output unit 40,000 1,000 units X $7.20 = $7,200 (c) 2010 Working hours $304,000 $16,000 20,000 = $14.40/hour 525 hours X $14.40 = $7,560 (d) 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 OR n(n + 1) 2 = 8(9) 2 = 36 Allocated to Sum-of-the-years-digits Year 1 8/36 X $288,000 = 2 7/36 X $288,000 = 3 6/36 X $288,000 = Total $64,000 $56,000 $48,000 2010 $16,000 $16,000 2011 $48,000 14,000 $62,000 2012 $42,000 12,000 $54,000 2012: $54,000 = (9/12 of 2nd year of machines life plus 3/12 of 3rd year of machines life) (e) Double-declining-balance 2011: 1/8 X 2 = 25%. 2010: 25% X $304,000 X 3/12 = $19,000 2011: 25% X ($304,000 $19,000) = $71,250 OR 1st full year (25% X $304,000) = $76,000 EXERCISE 11-6 (Continued) 2nd full year [25% X ($304,000 $76,000)] = $57,000 2010 Depreciation 3/12 X $76,000 = $19,000 2011 Depreciation 9/12 X $76,000 = $57,000 3/12 X $57,000 = 14,250 $71,250 EXERCISE 11-7 (2535 minutes) Methods of Depreciation Date Description Purchased Accum. Depr. Cost Salvage Life Method to 2010 2011 Depr. A 2/12/09 $159,000 $16,000 10 (a) SYD $37,700 (b) $22,100 B 8/15/08 (c) 79,000 21,000 5 SL 29,000 (d) 11,600 C 7/21/07 88,000 28,500 8 DDB (e) 55,516 D (g) 10/12/09 219,000 69,000 5 SYD 70,000 Machine ATesting the methods (a) Straight-Line Method for 2009 (f) 3,984 (h) 35,000 $ 7,150 [($159,000 $16,000) 10] X 1/2 Straight-Line Method for 2010 Total Straight Line $14,300 $21,450 Double-Declining-Balance for 2009 Double-Declining-Balance for 2010 Total Double Declining Balance $15,900 ($159,000 X .2 X .5) $28,620 [($159,000 $15,900) X .2] $44,520 Sum-of-the-Years-Digits for 2009 $13,000 [($159,000 $16,000) X Sum-of-the-Years-Digits for 2010 10/55 X .5] $24,700 ($143,000 X 10/55 X 1/2) + ($143,000 X 9/55 X .5) Total Sum-of-the-Years-Digits Method used must be $37,700 SYD (b) Using SYD, 2011 Depreciation is $22,100 ($143,000 X 9/55 X 1/2) + ($143,000 X 8/55 X .5) EXERCISE 11-7 (Continued) Machine BComputation of the cost (c) Asset has been depreciated for 2 1/2 years using the straight-line method. Annual depreciation is then equal to $29,000 divided by 2 1/2 or $11,600. 11,600 times 5 plus the salvage value is equal to the cost. Cost is $79,000 [($11,600 X 5) + $21,000]. (d) Using SL, 2011 Depreciation is $11,600. Machine CUsing the double-declining-balance method of depreciation (e) 2007s depreciation is $11,000 ($88,000 X .25 X .5) 2008s depreciation is $19,250 ($88,000 $11,000) X .25 2009s depreciation is $14,438 ($88,000 $30,250) X .25 2010s depreciation is $10,828 ($88,000 $44,688) X .25 Accumulated Depreciation at 12/31/10 $55,516 (f) Using DDB, 2011 Depreciation is $8,121 ($88,000 $55,516) X 0.25 Machine DComputation of Year Purchased (g) First Half Year using SYD = $25,000 [($219,000 $69,000) X Second Year using SYD = 5/15 X .5] $45,000 ($150,000 X 5/15 X .5) + ($150,000 X 4/15 X .5) $70,000 Thus the asset must have been purchased on October 12, 2009 (h) Using SYD, 2011 Depreciation is $35,000 ($150,000 X 4/15 X .5) + ($150,000 X 3/15 X .5) EXERCISE 11-8 (2025 minutes) Old Machine June 1, 2008 Purchase ...................................... Freight .......................................... Installation ................................... Total cost ............................ $31,800 200 500 $32,500 Annual depreciation charge: ($32,500 $2,500) 10 = $3,000 On June 1, 2009, debit the old machine for $2,700; the revised total cost is $35,200 ($32,500 + $2,700); thus the revised annual depreciation charge is: ($35,200 $2,500 $3,000) 9 = $3,300. Book value, old machine, June 1, 2012: [$35,200 $3,000 ($3,300 X 3)] = .............................. Fair value ......................................................................... Loss on exchange .......................................................... Cost of removal............................................................... Total loss ................................................................ $22,300 (20,000) 2,300 75 $ 2,375 (Note to instructor: The above computation is done to determine whether there is a gain or loss from the exchange of the old machine with the new machine and to show how the cost of removal might be reported. Also, if a gain occurs, the gain is not deferred (1) because the exchange has commercial substance and (2) the cash paid exceeds 25% of the total value of the property received.) New Machine Basis of new machine Cash paid ($35,000 $20,000) Fair value of old machine Installation cost Total cost of new machine $15,000 20,000 1,500 $36,500 Depreciation for the year beginning June 1, 2012 = ($36,500 $4,000) 10 = $3,250. EXERCISE 11-9 (1520 minutes) (a) Asset A B C D E Cost Estimated Salvage Depreciable Cost $ 40,500 33,600 36,000 19,000 23,500 $152,600 $ 5,500 4,800 3,600 1,500 2,500 $17,900 Estimated Depreciation Life per Year $ 35,000 28,800 32,400 17,500 21,000 $134,700 10 9 8 7 6 $ 3,500 3,200 4,050 2,500 3,500 $16,750 Composite life = $134,700 $16,750, or 8.04 years Composite rate = $16,750 $152,600, or approximately 11.0% (b) (c) Depreciation ExpensePlant Assets ................ Accumulated DepreciationPlant Assets ...................................................... 16,750 Cash ..................................................................... Accumulated DepreciationPlant Assets ......... Plant Assets ................................................ 5,000 14,000 EXERCISE 11-10 (1015 minutes) Sum-of-the-years-digits = 8X9 2 = 36 Using Y to stand for the years of remaining life: Y/36 X ($502,000 $70,000) = $60,000 Multiplying both sides by 36: $432,000 X Y = $2,160,000 Y = $2,160,000 $432,000 Y=5 16,750 19,000 The year in which there are five remaining years of life at the beginning of that given year is 2010. EXERCISE 11-11 (1015 minutes) (a) No correcting entry is necessary because changes in estimate are handled in the current and prospective periods. (b) Revised annual charge Book value as of 1/1/2011 [$52,000 ($6,000 X 5)] = $22,000 Remaining useful life, 5 years (10 years 5 years) Revised salvage value, $4,500 ($22,000 $4,500) 5 = $3,500 Depreciation ExpenseEquipment ........................ Accumulated DepreciationEquipment........ 3,500 3,500 EXERCISE 11-12 (2025 minutes) (a) 19841993($1,900,000 $60,000) 40 = $46,000/yr. (b) 19942011Building ($1,900,000 $60,000) 40 = Addition ($470,000 $20,000) 30 = (c) No adjusting entry required. (d) Revised annual depreciation Building Book value: ($1,900,000 $1,288,000*) .............. Salvage value........................................................ $46,000/yr. 15,000/yr. $61,000/yr. Remaining useful life ........................................... Annual depreciation ............................................. $612,000 60,000 552,000 32 years $ 17,250 *$46,000 X 28 years = $1,288,000 EXERCISE 11-12 (Continued) Addition Book value: ($470,000 $270,000**) .................. Salvage value........................................................ Remaining useful life ........................................... Annual depreciation ............................................. $200,000 20,000 180,000 32 years $ 5,625 **$15,000 X 18 years = $270,000 Annual depreciation expensebuilding ($17,250 + $5,625) $22,875 EXERCISE (1520 11-13 minutes) (a) $2,400,000 40 = $60,000 (b) Loss on Disposal of Plant Assets ...................... Accumulated DepreciationBuilding ($180,000 X 20/40) ............................................. Building ....................................................... 90,000 Building ................................................................ Cash ............................................................ 300,000 90,000 180,000 300,000 Note: The most appropriate entry would be to remove the old roof and record a loss on disposal, because the cost of the old roof is given. Another alternative would be to debit Accumulated Depreciation on the theory that the replacement extends the useful life of the building. The entry in this case would be as follows: Accumulated DepreciationBuilding ................ 300,000 Cash ............................................................ 300,000 As indicated, this approach does not seem as appropriate as the first approach. EXERCISE 11-13 (Continued) (c) No entry necessary. (d) (Assume the cost of the old roof is removed) Building ($2,400,000 $180,000 + $300,000) ........................ $2,520,000 Accumulated Depreciation ($60,000 X 20 $90,000) ........... 1,110,000 1,410,000 Remaining useful life ............................................................. 25 years Depreciation2011 ($1,410,000 25) ................................... $ 56,400 OR (Assume the cost of the new roof is debited to Accumulated Depreciation) Book value of the building prior to the replacement of roof $2,400,000 ($60,000 X 20) = ...................................... $1,200,000 Cost of new roof ..................................................................... 300,000 $1,500,000 Remaining useful life ............................................................. 25 years Depreciation2011 ($1,500,000 25) ................................... $ 60,000 EXERCISE 11-14 (2025 minutes) (a) (b) Repair Expense.................................................... Equipment ................................................... 500 The proper ending balance in the asset account is: January 1 balance ..................................... Add: New equipment: Purchases ............................................ $32,000 Freight .................................................. 700 Installation ........................................... 2,500 Less: Cost of equipment sold ................. December 31 balance ................................ 500 $133,000 35,200 (23,000) $145,200 (1) Straight-line: $145,200 10 = $14,520 EXERCISE 11-14 (Continued) (2) Sum-of-the-years-digits: 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55 OR n(n + 1) 2 = 10(11) = 55 2 For equipment purchased in 2009: $110,000 ($133,000 $23,000) of the cost of equipment purchased in 2009, is still on hand. 8/55 X $110,000 = $16,000 For equipment purchased in 2011: 10/55 X $35,200 = ..... 6,400 Total ........................................................................... $22,400 EXERCISE 11-15 (2535 minutes) (a) 2005 20062011 Incl. 2012 Total $ 3,400 18,250 0 9,125 $119,550 127,750 127,750 127,750 4,563 0 120,146 109,500 (1) $240,000 $21,000 = $219,000 (2) (3) (4) (5) (6) $219,000 12 = $18,250 per yr. ($50 per day) 133*/365 of $18,250 = $ 6,650 20062011 Include. (6 X $18,250) 68/365 of $18,250 = 0 18,250 9,125 4/12 of $18,250 6,083 20062011 Inc. 3/12 of $18,250 0 *(11 + 30 + 31 + 30 + 31) $109,500 109,500 109,500 109,500 109,500 109,500 (b) The most accurate distribution of cost is given by methods 1 and 5 if it is assumed that straight-line depreciation is satisfactory. Reasonable accuracy is normally given by 2, 3, or 4. The simplest of the applications are 6, 2, 3, 4, 5, and 1, in about that order. Methods 2, 3, and 4 combine reasonable accuracy with simplicity of application. EXERCISE 11-16 (1015 minutes) (a) December 31, 2010 Loss on Impairment ............................................. 3,600,000 Accumulated DepreciationEquipment.... 3,600,000 Note: The asset fails the recoverability test ($7,000,000 < $8,000,000) Cost ................................................ $9,000,000 Accumulated depreciation ............ 1,000,000 Carrying amount ............................ 8,000,000 Fair value........................................ 4,400,000 Loss on impairment ...................... $3,600,000 (b) December 31, 2011 Depreciation Expense .......................................... 1,100,000 Accumulated DepreciationEquipment.... 1,100,000 New carrying amount .................... $4,400,000 Useful life ....................................... 4 years Depreciation per year .................... $1,100,000 (c) No entry necessary. Restoration of any impairment loss is not permitted. EXERCISE 11-17 (1520 minutes) (a) Loss on Impairment ............................................. 3,620,000 Accumulated DepreciationEquipment.... Note: The asset fails the recoverability test ($7,000,000 < $8,000,000) Cost ................................................ $9,000,000 3,620,000 Accumulated depreciation ............ 1,000,000 Carrying amount ............................ 8,000,000 Less: Fair value ............................ 4,400,000 Plus: Cost of disposal .................. 20,000 Loss on impairment ...................... $3,620,000 EXERCISE 11-17 (Continued) (b) No entry necessary. Depreciation is not taken on assets intended to be sold. (c) Accumulated DepreciationEquipment ....... Recovery of Loss on Impairment ......... 700,000 700,000 Fair value......................................................... $5,100,000 Less: Cost of disposal ................................... 20,000 Carrying amount ............................................. Recovery of impairment loss ......................... 5,080,000 4,380,000* $ 700,000 *($9,000,000 $1,000,000 $3,620,000) EXERCISE 11-18 (1520 minutes) (a) December 31, 2010 Loss on Impairment ............................................. Accumulated DepreciationEquipment.... 220,000 220,000 Note: The asset fails the recoverability test ($300,000 < $500,000) Cost ................................................ Accumulated depreciation ............ Carrying amount ............................ Fair value........................................ Loss on impairment ...................... $900,000 400,000 500,000 280,000 $220,000 (b) It may be reported in the other expenses and losses section or it may be highlighted as an unusual item in a separate section. It is not reported as an extraordinary item. (c) No entry necessary. Restoration of any impairment loss is not permitted. (d) Management first had to determine whether there was an impairment. To evaluate this step, management does a recoverability test. The recoverability test estimates the future cash flows expected from use of that asset and its eventual disposition. If the sum of the expected future net cash flows (undiscounted) is less than the carrying amount of the asset, an impairment results. If the recoverability test indicates that an impairment has occurred, a loss is computed. The impairment loss is the amount by which the carrying amount of the asset exceeds its fair value. EXERCISE 11-19 (1520 minutes) (a) Depreciation Expense: $87,000 = $2,900 per year 30 years Cost of Timber Sold: $1,400 $400 = $1,000 $1,000 X 9,000 acres = $9,000,000 of value of timber ($9,000,000 3,000,000 bd. ft.) X 700,000 bd. ft. = $2,100,000 (b) Cost of Timber Sold: $9,000,000 $2,100,000 = $6,900,000 $6,900,000 + $100,000 = $7,000,000 ($7,000,000 5,000,000 bd. ft.) X 900,000 bd. ft. = $1,260,000 Note: The spraying costs as well as the costs to maintain the fire lanes and roads are expensed each period and are not part of the depletion base. EXERCISE 11-20 (1015 minutes) Cost per barrel of oil: Initial payment = Rental = $600,000 250,000 = $31,500 = 18,000 1.75 Premium, 5% of $65 = Reconditioning of land = $2.40 3.25 $30,000 = 250,000 .12 Total cost per barrel $7.52 EXERCISE 11-21 (1520 minutes) (a) $1,300 $300 = $1,000 per acre for timber $1,000 X 7,000 acres X 880,000 bd. ft. = 8,000 bd. ft. X 7,000 acres $7,000,000 56,000,000 bd. ft. (b) (c) X 880,000 bd. ft. = $110,000. $84,000 X 880,000 bd. ft. = $1,320. 56,000,000 bd. ft. Jonas should capitalize the cost of $70,000 ($20 X 3,500 trees) and adjust the depletion the next time the timber is harvested. EXERCISE 11-22 (1520 minutes) Depletion base: $1,250,000 + $90,000 $100,000 + $200,000 = $1,440,000 Depletion rate: $1,440,000 60,000 = $24/ton (a) (b) (c) Per unit mineral cost: $24/ton 12/31/10 inventory: $24 X 6,000 tons = $144,000 Cost of goods sold 2010: $24 X 24,000 tons = $576,000 EXERCISE 11-23 (1520 minutes) (a) $850,000 + $170,000 + $40,000* $100,000 = .08 depletion per unit 12,000,000 *Note to instructor: The $40,000 should be depleted because it is an asset retirement obligation. 2,500,000 units extracted X $.08 = $200,000 depletion for 2010 (b) 2,200,000 units sold X $.08 = $176,000 charged to cost of goods sold for 2010 EXERCISE 11-24 (1520 minutes) (a) Asset turnover ratio: $10,301 = .736 times $13,659 + $14,320 2 (b) Rate of return on assets: $676 = 4.83% $13,659 + $14,320 2 (c) Profit margin on sales: $676 = 6.56% $10,301 (d) The asset turnover ratio times the profit margin on sales provides the rate of return on assets computed for Eastman Kodak as follows: Profit margin on sales X Asset Turnover 6.56% X .736 Return on Assets = 4.83% Note the answer 4.83% is the same as the rate of return on assets computed in (b) above. *EXERCISE 11-25 (2025 minutes) 2010 (a) Revenues ............................................................. $200,000 Operating expenses (excluding depreciation)... 130,000 Depreciation [($41,000 $6,000) 7].................. 5,000 Income before income taxes............................... $ 65,000 2010 (b) Revenues ............................................................. $200,000 Operating expenses (excluding depreciation)... 130,000 Depreciation* ....................................................... 8,200 Taxable income.................................................... $ 61,800 *2010 2011 2011 $200,000 130,000 5,000 $ 65,000 2011 $200,000 130,000 13,120 $ 56,880 $41,000 X .20 = $8,200 $41,000 X .32 = $13,120 (c) Book purposes ($41,000 $6,000) Tax purposes (entire cost of asset) $35,000 $41,000 (d) Differences will occur for the following reasons: 1. different depreciation methods. 2. half-year convention used for tax purposes. 3. estimated useful life and tax life different. 4. tax system ignores salvage value. *EXERCISE 11-26 (1520 minutes) (a) (1) ($36,000 $3,000) X 1/10 X 10/12 = $2,750 depreciation expense for book purposes. (2) $36,000 X 1/5 X 1/2 = $3,600 depreciation for tax purposes. *EXERCISE 11-26 (Continued) (b) $36,000 X 20% X 10/12 = $6,000 depreciation expense for book purposes. (2) (c) (1) $36,000 X 20% = $7,200 depreciation expense for tax purposes. Differences will occur for the following reasons: 1. half-year convention used for tax purposes. 2. estimated useful life and tax life different. 3. tax system ignores salvage value. ... View Full Document

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