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# solutions P10 - Solution to P10 1A Item 1 2 3 4 5 6 7 8 9...

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Solution to P10 – 1A Item: Land: Building: Other Account: 1 \$4,000 2 \$700,000 3 \$5,000 * 4 100,000 45,000 5 35,000 6 10,000 7 2,000 8 14,000 ** 9 15,000 10 (3,500) \$154,000 \$745,000 \$19,000 * Property Tax Expenses ** Land Improvements Solution to P10 – 2A Depreciation for Bus 1 Depreciation as per Straight Line Method = (Cost of the asset – salvage value) / life of the asset Yearly Depreciation = (96,000 – 6,000) / 5 = \$18,000 per year Accumulated Depreciation on 31 Dec 2010 = 18000 x 3 = \$54,000 Depreciation for Bus 2 Formula for Double Declining Balance = depreciable base * (2 * 100% / useful life in years) For DDB, you initially ignore salvage value. Useful life: 4 yrs, so straight-line rate is 25%, and DDB rate is 50% Cost \$120,000 2008 Depreciation = 120,000 x 50% = \$60,000 Net Book Value at end 2008 - \$60,000 2009 Depreciation = 60,000 x 50% = \$30,000 Net Book Value at end 2009 - \$30,000 2010 Depreciation = 30,000 x 50% = \$15,000 So Accumulated Depreciation at end of 2010 = \$(60,000+30,000+15,000) = \$105,000

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Depreciation for Bus 3 Unit of Production method = [(Cost of the asset – salvage value)/total no. of production] x no. of units produced in a particular period Depreciable Cost = \$80,000 - \$8,000 = \$72,000 Therefore Depreciation cost per mile = \$72,000/120,000 = \$0.6 per mile Depreciation for the year 2009 = 24,000 x 0.6 = \$14,400 Depreciation for the year 2010 = 34,000 x 0.6 = \$20,400
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