CHAPTER 11 SOLUTIONS TO B EXERCISES E11-1B (15–20 minutes) (a) (b) (c) Straight-line method depreciation for each of Years through = $500,000 – $50,000 = $45,000 10 Sum-of-the-years’-digits = 10 X 11 = 55 10/55 X ($500,000 – $50,000) = $81,818 depreciation, Year 9/55 X ($500,000 – $50,000) = $73,636 depreciation, Year 8/55 X ($500,000 – $50,000) = $65,455 depreciation, Year Double-declining balance method depreciation rate = 100% X = 20.00% 10 $500,000 X 20.00% = $100,000 depreciation, Year ($500,000 – $100,000) X 20.00% = $ 80,000 depreciation, Year ($500,000 – $100,000 – $80,000) X 20.00% = $ 64,000 depreciation, Year E11-2B (20–25 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 = 40% Cost X 40% = $34,000 $34,000 ÷ 0.40 = $85,000 Cost of asset Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-1 E11-2B (Continued) (b) $85,000 cost [from (a)] – $75,000 total depreciation = $10,000 salvage value (c) The highest charge to income for Year will be yielded by the doubledeclining-balance method (d) The highest charge to income for Year will be yielded by the straight- line method (e) The method that produces the lowest book value at the end of Year would be the method that yields the highest accumulated depreciation at the end of Year 3, which is DDB Computations: St.-line = $85,000 – ($15,000 + $15,000 + $15,000) = $40,000 book value, end of Year S.Y.D = $85,000 – ($25,000 + $20,000 + $15,000) = $25,000 book value, end of Year D.D.B = $85,000 – ($34,000 + $20,400 + $12,240) = $18,360 book value, end of Year (f) The method that will yield the lowest gain (or highest loss) if the asset is sold at the end of Year is the method which will yield the highest book value at the end of Year 4, which is the double-declining balance method in this case E11-3B (15–20 minutes) (a) 20 (20 + 1) = 210 3/4 X 20/210 X ($355,500 – $30,000) = $23,250 for 2014 + 1/4 X 20/210 X ($355,500 – $30,000) = $ 7,750 3/4 X 19/210 X ($355,500 – $30,000) = 22,088 $29,838 for 2015 Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-2 E11-3B (Continued) (b) 100% = 5%; 5% X = 10% 20 3/4 X 10% X $355,500 = $26,663 for 2014 10% X ($355,500 – $26,663) = $32,884 for 2015 E11-4B (15–25 minutes) (a) $345,000 – $45,000 = $300,000; $300,000 ÷ 10 yrs = $30,000 (b) $300,000 ÷ 120,000 units = $2.50; 25,000 units X $2.50 = $62,500 (c) $300,000 ÷ 12,500 hours = $24.00 per hr.; 2,000 hrs X $24.00 = $48,000 (d) 10 + + + + + + + + + = 55 or n(n + 1) = 10(11) = 55 2 10 X $300,000 X 1/3 = $18,182 55 X $300,000 X 2/3 = 55 Total for 2015 (e) $345,000 X 20% X 1/3 = 32,727 $50,909 $23,000 [$345,000 – ($345,000 X 20%)] X 20% X 2/3 = 36,800 Total for 2015 $59,800 [May also be computed as 20% of ($345,000 – 2/3 of 20% of $345,000).] Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-3 E11-5B (20–25 minutes) (a) ($235,800 – $25,800) = $21,000/yr = $21,000 X 5/12 = $8,750 10 2014 depreciation—straight-line = $8,750 (b) ($235,800 – $25,800) = $5.00/hr 42,000 2014 depreciation—machine usage = 800 X $5.00 = $4,000 (c) Machine Year Total 10/55 X $210,000 = $38,182 9/55 X $210,000 = $34,364 Allocated to 2014 2015 $15,909* $22,273** 14,318*** $15,909 $36,591 * $38,182 X 5/12 = $15,909 ** $38,182 X 7/12 = $22,273 *** $34,364 X 5/12 = $14,318 2015 Depreciation—sum-of-the-years’-digits = $36,591 (d) 2014 40% X ($235,800) X 5/12 = $39,300 2015 40% X ($235,800 – $39,300) = $78,600 OR 1st full year (40% X $235,800) = $94,320 2nd full year [40% X ($235,800 – $94,320)] = $56,592 2014 depreciation = 5/12 X $94,320 = $39,300 2015 depreciation = 7/12 X $94,320 = $55,020 5/12 X $56,592 = 23,580 $78,600 Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-4 E11-6B (20–30 minutes) (a) 2014 $250,000 – $50,000 = $25,000/year Straight-line months—depreciation $6,250 = ($25,000 X 3/12) (b) 2014 $250,000 – $50,000 = $10.00/output unit 20,000 Output 1,500 units X $10.00 = $15,000 (c) 2014 Working hours $250,000 – $50,000 = $20.00/hour 10,000 900 hours X $20.00 = $18,000 (d) + + + + + + + = 36 or Sum-of-the-years’-digits Year 8/36 X $200,000 = 7/36 X $200,000 = 6/36 X $200,000 = Total $44,444 $38,889 $33,333 n (n + 1) = 2014 $11,111 $11,111 8(9) = 36 Allocated to 2015 $33,333 9,722 $43,055 2016 $29,167 8,333 $37,500 2016: $37,500 = (9/12 of 2nd year of machine’s life plus 3/12 of rd year of machine’s life) Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-5 E11-6B (Continued) (e) Double-declining balance 2005: 1/8 X = 25% 2014: 25% X $250,000 X 3/12 = $15,625 2015: 25% X ($250,000 – $15,625) = $58,594 OR 1st full year (25% X $250,000) = $62,500 2nd full year [25% X ($250,000 – $62,500)] = $46,875 2014 Depreciation 3/12 X $62,500 = $15,625 2015 Depreciation 9/12 X $62,500 = $46,875 3/12 X $46,875 = 11,719 $58,594 E11-7B (25–35 minutes) Methods of Depreciation Description A B C D Date Purchased 7/10/09 10/5/11 8/2/10 2/12/(g) Cost Salvage Life $216,000 (c) 210,000 148,000 $36,000 15,000 5,000 25,000 10 Accum Depr Method to 2014 2015 Depr (a) SYD SL DDB $105,000 46,667 (e) 29,600 (b) $ (d) (f) (h) Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-6 E11-7B (Continued) Machine A—Testing the methods (a) Straight-Line Method for 2011 $ 15,000 [($216,000 – $36,000) ÷ Straight-Line Method for 2012-14 6] X 1/2 $ 90,000 [($216,000 – $36,000) ÷ 6] X Total Straight Line $105,000 Double-Declining-Balance for 2011 Double-Declining-Balance for 2012 Double-Declining-Balance for 2013 Double-Declining-Balance for 2014 Total Double Declining Balance $ $ $ $ Sum-of-the-Years-Digits for 2011 $ 25,714 [($216,000 – $36,000) X Sum-of-the-Years-Digits for 2012 Sum-of-the-Years-Digits for 2013 Sum-of-the-Years-Digits for 2014 Total Sum-of-the-Years-Digits Method used must be (b) Using SL, 2015 Depreciation is 36,000 60,000 40,000 26,667 ($216,000 X 333 X 5) ($180,000 X 333) ($120,000 X 333 ($80,000 X 333) $162,667 6/21 X 5] $ 47,143 ($180,000 X 6/21 X 1/2) + ($180,000 X 5/21 X 1/2) $ 38,571 ($180,000 X 5/21 X 1/2) + ($180,000 X 4/21 X 1/2) $ 30,000 ($180,000 X 4/21 X 1/2) + ($180,000 X 3/21 X 1/2) $141,428 SL $30,000 ($216,000 – 36,000)/6 Machine B—Computation of the cost (c) Asset has been depreciated for 1/2 years using the SYD method Year depreciation is then equal to depreciable base X 5/15 X 1/2 and year depreciation is equal to the depreciable base X (5/15 X 1/2 + 4/15 X 1/2) Total depreciation equals depreciable base x (5/30 + 5/30 +4/30) $46,667 = depreciable base x 14/30, depreciable base = 100,000 Cost is $115,000 ($100,000 + $15,000] Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-7 E11-7B (Continued) (d) Using SYD, 2015 Depreciation is $23,333 Sum-of-the-Years-Digits for 2015 $23,333 ($100,000 X 4/15 X 1/2) + ($100,000 X 3/15 X 1/2) Machine C—Using the straight-line method of depreciation (e) 2012’s depreciation is 2013’s depreciation is 2014’s depreciation is Accumulated Depreciation December 31, 2014 (f) $10,250 $20,500 $20,500 ($210,000 – $5,000)/10 X 1/2) $205,000/10 $205,000/10 $51,250 Using SL, 2015 Depreciation is $20,500 Machine D—Computation of Year Purchased (g) First Half Year using SYD = $29,600 [$148,00 X 40% X 5] $29,600 Thus the asset must have been purchased on February 12, 2014 (h) Using DDB, 2015 Depreciation is $47,360 ($148,000 – 29,600) x 40% E11-8B (20–25 minutes) Old Machine October 1, 2010 Purchase Freight Installation Total cost Annual depreciation charge: ($69,000 – $4,000) ÷ 10 = $6,500 $68,000 300 700 $69,000 On October 1, 2013, debit the old machine for $7,000; the revised total cost is $76,000 ($69,000 + $7,000); thus the revised annual depreciation charge is: ($76,000 – $4,000 – $19,500) ÷ = $7,500 Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-8 E11-8B (Continued) Book value, old machine, October 1, 2016: [$76,000 – $19,500 – ($7,500 X 3)] = Fair value Loss on exchange Cost of removal Total loss $34,000 (22,000) 12,000 75 $12,075 (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 ($86,000 – $22,000) Fair value of old machine Installation cost Total cost of new machine $64,000 22,000 3,000 $89,000 Depreciation for the year beginning October 1, 2016 = ($89,000 – $10,000) ÷ 10 = $7,900 E11-9B (15–20 minutes) (a) Asset Cost Estimated Salvage A B C D E $121,500 100,800 108,000 57,000 70,500 $457,800 $16,500 14,400 10,800 4,500 7,500 $53,700 Depreciable Estimated Depreciation Cost Life per Year $105,000 86,400 97,200 52,500 63,000 $404,100 10 9 $10,500 9,600 10,800 7,500 10,500 $48,900 Composite life = $404,100 ÷ $48,900, or 8.26 years Composite rate = $48,900 ữ $457,800, or approximately 10.7% Copyright â 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-9 E11-9B (Continued) (b) Depreciation Expense—Plant Assets 48,900 Accumulated Depreciation—Plant Assets (c) Cash 14,400 Accumulated Depreciation—Plant Assets 42,600 Plant Assets 48,900 57,000 E11-10B (10–15 minutes) Sum-of-the-years’-digits = 8X9 = 36 Using Y to stand for the years of remaining life: Y/36 X ($860,000 – $140,000) = $120,000 Multiplying both sides by 36: $720,000 Y = $4,320,000 Y = $4,320,000 ÷ $720,000 Y=6 The year in which there are remaining years of life at the beginning of that given year is 2013 E11-11B (10–15 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 1/1/2015 [$100,000 – ($8,000 X 5)] = $60,000 Remaining useful life, years (9 years – years) Revised salvage value, $6,000 ($60,000 – $6,000) ÷ = $13,500 Depreciation Expense—Equipment .13,500 Accumulated Depreciation—Equipment 13,500 Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-10 E11-12B (20–25 minutes) (a) 1995–2001—($4,000,000 – $200,000) ÷ 40 = $95,000/yr (b) 2002–2015—Building ($4,000,000 – $200,000) ÷ 40 = Addition ($1,500,000 – $20,000) ÷ 32 = (c) No adjusting entry required (d) Revised annual depreciation Building Book value: ($4,000,000 – $1,995,000*) Salvage value $95,000/yr 46,250/yr $141,250/yr Remaining useful life Annual depreciation $2,005,000 200,000 1,805,000 29 years $ 62,241 *$95,000 X 21 years = $1,995,000 Addition Book value: ($1,500,000 – $647,500**) Salvage value $852,500 20,000 832,500 Remaining useful life ÷ 29 years Annual depreciation $ 28,707 **$46,250 X 14 years = $647,500 Annual depreciation expense—building ($62,241 + $28,707) $90,948 Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-11 E11-13B (15–20 minutes) (a) $3,000,000 ÷ 40 = $75,000 (b) Loss on Disposal of Plant Assets 100,000 Accumulated Depreciation—Building ($200,000 X 20/40) 100,000 Building 200,000 Building 500,000 Cash 500,000 Note: The most appropriate entry would be to remove the old roof and record a loss on disposal, because the cost of 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 Depreciation—Building 500,000 Cash 500,000 As indicated, this approach does not seem as appropriate as the first approach (c) No entry necessary (d) (Assuming the cost of old roof is removed:) Building ($3,000,000 – $200,000 + $500,000) Accumulated Depreciation ($75,000 X 20 – $100,000) Remaining useful life Depreciation—2015 ($1,900,000 ÷ 25) $3,300,000 1,400,000 1,900,000 25 years $ 76,000 OR (Assuming the cost of new roof is debited to accumulated depreciation:) Book value of building prior to the replacement of roof $3,000,000 – ($75,000 X 20) = Cost of new roof Remaining useful life Depreciation—2015 ($2,000,000 ÷ 25) $1,500,000 500,000 $2,000,000 25 years $ 80,000 Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-12 E11-14B (20–25 minutes) (a) (b) Repair Expense 1,250 Equipment The proper ending balance in the asset account is: January balance Add: New equipment: Purchases $81,000 Freight 500 Installation 3,000 Less: Cost of equipment sold December 31 balance 1,250 $212,000 84,500 (20,000) $276,500 (1) Straight-line: $276,500 ÷ 10 = $27,650 (2) Sum-of-the-years’-digits: 10 + + + + + + + + + = 55 OR n(n + 1) = 10(11) = 55 For equipment purchased in 2013: $192,000 ($212,000 – $20,000) of the cost of equipment purchased in 2013, is still on hand 8/55 X $192,000 = $27,927 For equipment purchased in 2015: 10/55 X $84,500 = 15,364 Total $43,291 Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-13 E11-15B (25–35 minutes) (a) 2010 $300,000 – $25,000 = $275,000 $275,000 ÷ 10 = $27,500 per yr 265*/365 of $27,500 = $19,966 2011–2015 Include (5 X $27,500) 166/365 of $27,500 = 27,500 13,750 8/12 of $27,500 18,333 2011–2015 Inc 6/12 of $27,500 27,500 2011–2015 Incl 2016 Total $137,500 137,500 137,500 137,500 $12,507 27,750 13,750 $169,973 165,000 165,000 165,000 13,750 169,583 165,000 137,500 137,500 *(20 + 31+ 30 + 31 + 31 + 30 + 31 + 30 + 31) (b) The most accurate distribution of cost is given by methods and if it is assumed that straight-line depreciation is satisfactory Reasonable accuracy is normally given by 2, 3, or The simplest of the applications are 6, 2, 3, 4, 5, and 1, in about that order Methods 2, 3, and combine reasonable accuracy with simplicity of application E11-16B (10–15 minutes) (a) December 31, 2014 Loss on Impairment 2,400,000 Accumulated Depreciation—Equipment Cost Accumulated depreciation Carrying amount Fair value Loss in impairment 2,400,000 $6,750,000 750,000 6,000,000 3,600,000 $2,400,000 Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-14 E11-16B (Continued) (b) December 31, 2015 Depreciation Expense 900,000 Accumulated Depreciation—Equipment New carrying amount Useful life Depreciation per year (c) 900,000 $3,600,000 years $ 900,000 No entry necessary Restoration of any impairment loss is not permitted E11-17B (15–20 minutes) (a) Loss on Impairment 2,415,000 Accumulated Depreciation—Equipment Cost Accumulated depreciation Carrying amount Less: Fair value Plus: Cost of disposal Loss on impairment 2,415,000 $6,750,000 750,000 6,000,000 3,600,000 15,000 $2,415,000 (b) No entry necessary Depreciation not taken on assets intended to be sold (c) Accumulated Depreciation—Equipment 375,000 Recovery of Loss on Impairment 375,000 Fair value $3,975,000 Less: Cost of disposal 15,000 $3,960,000 Carrying amount 3,585,000 Recovery of impairment loss $ 375,000 Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-15 E11-18B (15–20 minutes) (a) December 31, 2014 Loss on Impairment 300,000 Accumulated Depreciation—Equipment Cost Accumulated depreciation Carrying amount Fair value Loss in impairment 300,000 $1,200,000 300,000 900,000 600,000 $ 300,000 (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 E11-19B (15–20 minutes) (a) Depreciation expense: $90,000 = $3,000 per year 30 years Cost of timber sold: $2,000 – $500 = $1,500 $1,500 X 10,000 acres = $15,000,000 of value of timber ($15,000,000 ÷ 2,000,000 bd ft.) X 700,000 bd ft = $5,250,000 (b) Cost of timber sold: $15,000,000 – $5,250,000 = $9,750,000 $9,750,000 + $100,000 = $9,850,000 ($9,850,000 ÷ 4,000,000 bd ft.) X 900,000 bd ft = $2,216,250 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 Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-16 E11-20B (10–15 minutes) Cost per barrel of oil: Initial payment = $1,200,000 = 500,000 $2.40 Rental = $51,200 = 32,000 1.60 Premium, 5% of $90 = 4.50 Reconditioning of land = $265,000 = 500,000 0.53 Total cost per barrel $9.03 E11-21B (15–20 minutes) (a) $1,500 – $500 = $1,000 per acre for timber $1,000 X 8,000 acres X 1,000,000 bd ft = 5,000 bd ft X 8,000 acres $8,000,000 X 1,000,000 bd ft = $200,000 40,000,000 bd ft (b) (c) $80,000 X 1,000,000 bd ft = $2,000 40,000,000 bd ft Capitalize the cost of $35,000 ($10 X 3,500 trees) and adjust depletion the next time timber is harvested Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-17 E11-22B (15–20 minutes) Depletion base: $2,000,000 + $200,000 – $300,000 + $500,000 = $2,400,000 Depletion rate: $2,400,000 ÷ 50,000 = $48/ton (a) (b) (c) Per unit material cost: $48/ton 12/31/14 inventory: $48 X 5,000 tons = $240,000 Cost of goods sold 2014: $48 X 5,000 tons = $240,000 E11-23B (15–20 minutes) (a) $5,000,000 + $1,000,000 + $400,000* – $800,000 = $112 depletion per ton 50,000 *Note to instructor: The $400,000 should be depleted because it is an asset retirement obligation 30,000 tons extracted X $112 = $3,360,000 depletion for 2014 (b) 22,000 units sold X $112 = $2,464,000 charged to cost of goods sold for 2014 E11-24B (15–20 minutes) (a) Asset turnover ratio: $2,820 $1,436 + $1,484 (b) = 1.932 times Rate of return on assets: $138 $1,436 + $1,484 = 9.45% Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-18 E11-24B (Continued) (c) Profit margin on sales: $138 = 4.89% $2,820 (d) The asset turnover ratio times the profit margin on sales provides the rate of return on assets computed for Brinker International as follows: Profit margin on sales X Asset Turnover 4.89% X 1.932 = Return on Assets 9.45% Note the answer 9.45% is the same as the rate of return on assets computed in (b) above *E11-25B (20–25 minutes) (a) (b) 2015 Revenues $600,000 Operating expenses (excluding depreciation) 440,000 Depreciation [($54,000 – $6,000) ÷ 6] 8,000 Income before income taxes $152,000 2014 $600,000 440,000 8,000 $152,000 2015 Revenues $600,000 Operating expenses (excluding depreciation) 440,000 Depreciation* 10,800 Taxable income $149,200 2014 $600,000 440,000 17,280 $142,720 *2014 $54,000 X 0.20 = $10,800 2015 $54,000 X 0.32 = $17,280 (c) Book purposes ($54,000 – $6,000) Tax purposes (entire cost of asset) $48,000 $54,000 (d) Differences will occur for the following reasons: different depreciation methods half-year convention used for tax purposes estimated useful life and tax life different tax system ignores salvage value Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-19 *E11-26B (15–20 minutes) (a) (b) (c) ($21,000 – $1,000) X 1/8 X 6/12 = $1,250 depreciation expense for book purposes $21,000 X 1/5 X 1/2 = $2,100 depreciation for tax purposes $21,000 X 25% X 6/12 = $2,625 depreciation expense for book purposes $21,000 X 25% = $5,250 depreciation expense for tax purposes Differences will occur for the following reasons: half-year convention used for tax purposes estimated useful life and tax life different tax system ignores salvage value Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-20 ... 22,088 $29,838 for 2015 Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-2 E11-3B (Continued) (b) 100% = 5%;... 2/3 of 20% of $345,000).] Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-3 E11-5B (20–25 minutes) (a) ($235,800... $56,592 = 23,580 $78,600 Copyright © 2014 John Wiley & Sons, Inc. Kieso, Intermediate Accounting, 15/e, Exercise B Solutions (For Instructor Use Only) 11-4 E11-6B (20–30 minutes) (a) 2014