# ch11 - SOLUTIONS TO B EXERCISES E11-1B(15–20 minutes(a...

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Unformatted text preview: SOLUTIONS TO B EXERCISES E11-1B (15–20 minutes) (a) Straight-line method depreciation for each of Years 1 through 3 = \$500,000 – \$50,000 = \$45,000 10 (b) Sum-of-the-years’-digits = 10 X 11 = 55 2 10/55 X (\$500,000 – \$50,000) = \$81,818 depreciation, Year 1 9/55 X (\$500,000 – \$50,000) = \$73,636 depreciation, Year 2 8/55 X (\$500,000 – \$50,000) = \$65,455 depreciation, Year 3 (c) Double-declining balance method depreciation rate = 100% X 2 = 20.00% 10 \$500,000 X 20.00% = \$100,000 depreciation, Year 1 (\$500,000 – \$100,000) X 20.00% = \$ 80,000 depreciation, Year 2 (\$500,000 – \$100,000 – \$80,000) X 20.00% = \$ 64,000 depreciation, Year 3 11-1 E11-2B (15–20 minutes) (a) 20 (20 + 1) = 210 2 3/4 X 20/210 X (\$355,500 – \$30,000) = \$23,250 for 2007 1/4 X 20/210 X (\$355,500 – \$30,000) = \$ 7,750.00 + 3/4 X 19/210 X (\$355,500 – \$30,000) = 22,087.50 \$29,837.50 for 2008 (b) 100% = 5%; 5% X 2 = 10% 20 3/4 X 10% X \$355,500 = \$26,662.50 for 2007 10% X (\$355,500 – \$26,662.50) = \$32,883.75 for 2008 11-2 E11-3B (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 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55 or n(n + 1) = 10(11) = 55 2 2 10 X \$300,000 X 1/3 = \$18,182 55 9 X \$300,000 X 2/3 = 32,727 55 Total for 2008 \$50,909 (e) \$345,000 X 20% X 1/3 = \$23,000 [\$345,000 – (\$345,000 X 20%)] X 20% X 2/3 = 36,800 Total for 2008 \$59,800 [May also be computed as 20% of (\$345,000 – 2/3 of 20% of \$345,000).] 11-3 E11-4B (20–25 minutes) (a) (\$235,800 – \$25,800) = \$21,000/yr. = \$21,000 X 5/12 = \$8,750 10 2007 depreciation—straight-line = \$8,750 (b) (\$235,800 – \$25,800) = \$5.00/hr. 42,000 2007 depreciation—machine usage = 800 X \$5.00 = \$4,000 (c) Machine Allocated to Year Total 2007 2008 1 10/55 X \$210,000 = \$38,181.82 \$15,909* \$22,273** 2 9/55 X \$210,000 = \$34,363.64 14,318 *** \$15,909 \$36,591 * \$38,181.82 X 5/12 = \$15,909.09 ** \$38,181.82 X 7/12 = \$22,272.73 *** \$34,363.64 X 5/12 = \$14,318.18 2008 Depreciation—sum-of-the-years’-digits = \$36,591 (d) 2007 40% X (\$235,800) X 5/12 = \$39,300 2008 40% X (\$235,800 – \$39,300) = \$78,600 OR 1 st full year (40% X \$235,800) = \$94,320 2 nd full year [40% X (\$235,800 – \$94,320)] = \$56,592 2007 depreciation = 5/12 X \$94,320 = \$39,300 2008 depreciation = 7/12 X \$94,320 = \$55,020 5/12 X \$56,592 = 23,580 \$78,600 11-4 E11-5B (20–30 minutes) (a) 2006 Straight-line \$250,000 – \$50,000 = \$25,000/year 8 3 months—depreciation \$6,250 = (\$25,000 X 3/12) (b) 2006 Output \$250,000 – \$50,000 = \$10.00/output unit 20,000 1,500 units X \$10.00 = \$15,000 (c) 2006 Working hours \$250,000 – \$50,000 = \$20.00/hour 10,000 900 hours X \$20.00 = \$18,000 (d) 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 or n (n + 1) = 8(9) = 36 2 2 Allocated to Sum-of-the-years’-digits Total 2006 2007 2008 Year 1 8/36 X \$200,000 = \$44,444 \$11,111 \$33,333 2 7/36 X \$200,000 = \$38,889...
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## This note was uploaded on 05/10/2010 for the course ACC 440 BSB1658 taught by Professor Warwick during the Spring '09 term at University of Phoenix.

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ch11 - SOLUTIONS TO B EXERCISES E11-1B(15–20 minutes(a...

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