Ch 11 Solutions

# Ch 11 Solutions - SOLUTIONS TO EXERCISES EXERCISE 11-1(1520...

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SOLUTIONS TO EXERCISES EXERCISE 11-1 (15–20 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 = 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 (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 2 = 40%

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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 double- declining balance method in this case. EXERCISE 11-3 (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

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\$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 (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

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