11thA_Ch11HWSolutionsPlus

11thA_Ch11HWSolutionsPlus - CHAPTER 11 Depreciation,...

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CHAPTER 11 Depreciation, Impairments, and Depletion SOLUTIONS TO EXERCISES EXERCISE 11-1 (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 = 78 2 12/78 X ($518,000 – $50,000) = $72,000 depreciation Year 1 11/78 X ($518,000 – $50,000) = $66,000 depreciation Year 2 10/78 X ($518,000 – $50,000) = $60,000 depreciation Year 3 (c) Double-Declining-Balance method depreciation rate. 100% X 2 = 16.67% 12 $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 (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
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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 double-declining balance method in this case. EXERCISE 11-4 (15–25 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 n(n + 1) = 10(11) = 55 2 2 10 X $264,000 X 1/3 = $16,000 55 9 X $264,000 X 2/3 = 28,800 55 Total for 2011 $44,800
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(e) $279,000 X 20% X 1/3 = $18,600 [$279,000 – ($279,000 X 20%)] X 20% X 2/3 = 29,760 Total for 2011 $48,360 [May also be computed as 20% of ($279,000 – 2/3 of 20% of $279,000)] EXERCISE 11-6 (a) 2010 Straight-line $304,000 – $16,000 = $36,000/year 8 3 months—Depreciation ($36,000 X 3/12) = $9,000
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This note was uploaded on 04/14/2010 for the course TOBIN 615 taught by Professor Alan during the Spring '10 term at Adams State University.

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11thA_Ch11HWSolutionsPlus - CHAPTER 11 Depreciation,...

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