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Unformatted text preview: SOLUTIONS TO EXERCISES EXERCISE 111 (15–20 minutes) (a) Straightline method depreciation for each of Years 1 through 3 = $518,000 – $50,000 = $39,000 12 (b) SumoftheYears’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) DoubleDecliningBalance 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 112 (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 doubledeclining balance method. 100% = 20%; 20% X 2 = 40% 5 Cost X 40% = $20,000 $20,000 ÷ .40 = $50,000 Cost of asset EXERCISE 112 (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 doubledecliningbalance 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 straightline 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 113 (15–20 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) = $17,000 + 3/4 X 19/210 X ($774,000 – $60,000) = 48,450 $65,450 for 2011 EXERCISE 113 (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 114 (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 = 55 28,800 Total for 2011 $44,800 (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 115 (20–25 minutes) (a) ($150,000 – $24,000) = $25,200/yr. = $25,200 X 5/12 = $10,500= $25,200/yr....
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This note was uploaded on 07/28/2011 for the course ACCT 103 taught by Professor Huxhold during the Spring '11 term at San Diego.
 Spring '11
 Huxhold
 Cost Accounting, Depreciation

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