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 So, accumulated Depreciation at the end of 2010 = $14,400+$20,400 = $34,800
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This note was uploaded on 05/13/2010 for the course MECH 17657 taught by Professor Ravikant during the Spring '10 term at Indian Institute of Technology, Delhi.

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

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