Ch12Solutions

# Ch12Solutions - .1(a BV 3 = 100,000 –(40,000 24,000...

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Unformatted text preview: Chapter 12 12.1 (a) BV 3 = 100,000 – (40,000+24,000+14,000) = \$22,000 (b) Market value today and S = \$20,000, while BV 3 = \$22,000. (c) d 1 = 0.40; d 2 = 0.24; d 3 = 0.14 Percent written off is the sum = 0.78 or 78%. 12.2 (a) Depreciation = Rate(495,000) Year Depr BV 1 \$164,98 4 \$330,016 2 220,02 8 109,988 3 73,30 9 36,679 4 36,67 8 (b) BV 3 = \$36,679 while market value is much higher at \$150,000. The rates reduce the asset to zero salvage, not recognizing the \$150,000 salvage. 12.3 (a, b) Tax depreciation: D t = Rate(BV t-1 ) Book depreciation: D t = Rate(40,000) Tax Book Year Depr BV Depr BV 40,00 40,000 1 16,00 24,00 10,00 30,000 2 9,60 14,40 10,00 20,000 3 5,76 8,64 10,00 10,000 4 3,45 6 5,18 4 10,00 A spreadsheet solution with graphs follows. 12- 1 12.4 Productive life – Time the asset is actually expected to be in service. Tax recovery period – Time allow by tax laws to depreciate the asset’s value to salvage (or zero). Book recovery period – Time used on company accounting books for depreciation to salvage (or zero) Straight Line (SL) Depreciation 12.5 (a) d = 1/10 = 0.1 or 10% (b) S = 0.15(475,000) = \$71,250 D t = (550,000 – 71,250)/10 = \$47,875 per year (c) BV 5 = 550,000 – 5(47,875) = \$310,625 (d) BV 10 = 550,000 – 10(47,875) = \$71,250 = S 12.6 Spreadsheet solution uses SL depreciation. D t = (350,000 – 50,000)/5 = \$60,000 per year 12- 2 12.7 Book: D t = (150,000 – 25,000)/9.5 = \$13,158 per year Tax: D t = (150,000 – 0)5 = \$ 30,000 per year Spreadsheet solution uses SL depreciation for book and tax purposes to plot BV. 12.8 D t = 18,900 = (P – 0.25P)/8 0.75P = 8(18,900) P = \$201,600 12.9 DDB: D 3 = (2/5)(500,000)(3/5) 2 = \$72,000 150%DB: d = 1.5(1/5) = 0.3 D 3 = (0.3)(500,000)(0.7) 2 = \$73,500 SL: D 3 = (500,000-50,000)/5 = \$90,000 12.10 The DDB salvage amount is calculated using d = 0.4. BV 5 = B(1-d) t = 500,000(.6) 5 = \$38,880 Considerably less than the estimated S = \$50,000. For 150% DB, implied salvage is calculated using d = 0.3. BV 5 = 500,000(.7) 5 = \$84,035 Considerably more than the estimated S = \$50,000. 12.11 Use Equation [12.7] with d = 0.4. d 1 = 0.4(0.6) = 0.40 d 2 = 0.4(0.6) 1 = 0.24 d 3 = 0.4(0.6) 2 = 0.144 d 4 = 0.4(0.6) 3 = 0.0864 d 5 = 0.4(0.6) 4 = 0.05184 12- 3 12.12 Spreadsheet solution determines BV values and plots. Note the DDB depreciation in year 5 is \$14,800, so BV 5 = S = \$50,000 is maintained. 12.13 The d is the fixed percentage (in decimal form) by which the BV is reduced each year, while d max is a specific value of d for DDB, that is, double the SL rate of d = 1/n. Finally, the annual depreciation rate, d t is the fraction of P written off each year. In DB depreciation, d t and d are not equal, whereas in SL depreciation, they are equal....
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Ch12Solutions - .1(a BV 3 = 100,000 –(40,000 24,000...

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