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Delicious77 [7]
3 years ago
11

Warren corporation purchased a truck at a cost of $60,000. It has an estimated useful life of five years and estimated residual

value of $5,000. At the beginning of year three, warren's managers concluded that the total useful life would be four years, rather than five years. There was no change in the estimated residual value. What is the amount of depreciation that warren should record for year 3 under the straight-line depreciation method?
Business
1 answer:
s344n2d4d5 [400]3 years ago
6 0

Answer:

What is the amount of depreciation that warren should record for year 3 under the straight-line depreciation method? $15500

Explanation:

Net Value Dep. year End Net value.

Year 1 55000 12000          43000

Year 2 43000 12000          31000

Year 3 31000 15500          15500

Year 4 15500 15500            0

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Answer:

Following are the solution to this question:

Explanation:

Assume that r_1  will be a 12-month for the spot rate:

\to 1.25 \% \times \frac{100}{2} \times 0.99 + \frac{(1.25\% \times \frac{100}{2}+100)}{(1+\frac{r_1}{2})^2}=98\\\\\to \frac{1.25}{100} \times \frac{100}{2} \times 0.99 + \frac{(\frac{1.25}{100} \times \frac{100}{2}+100)}{(1+\frac{r_1}{2})^2}=98\\\\\to \frac{1.25}{2} \times 0.99 + \frac{(\frac{1.25}{2} +100)}{(1+\frac{r_1}{2})^2}=98\\\\\to 0.61875 + \frac{( 0.625 +100)}{(\frac{2+r_1}{2})^2}=98\\\\\to 0.61875 + \frac{( 100.625)}{(\frac{2+r_1}{2})^2}=98\\\\\to 0.61875 + \frac{402.5}{(2+r_1)^2}=98\\\\

\to 0.61875 + \frac{402.5}{(2+r_1)^2}=98\\\\\to 0.61875 -98 = \frac{402.5}{(2+r_1)^2}\\\\\to -97.38125= \frac{402.5}{(2+r_1)^2}\\\\\to (2+r_1)^2= \frac{402.5}{ -97.38125}\\\\\to (2+r_1)^2= -4.13\\\\ \to r_1=3.304\%

Assume that r_2  will be a 18-month for the spot rate:

\to 1.5\% \times \frac{100}{2} \times 0.99+1.5\%  \times \frac{100}{2} \times \frac{1}{(1+ \frac{3.300\%}{2})^2}+\frac{(1.5\%  \times  \frac{100}{2}+100)}{(1+\frac{r_2}{2})^3}=97\\\\\to \frac{1.5}{100} \times \frac{100}{2} \times 0.99+\frac{1.5}{100}  \times \frac{100}{2} \times \frac{1}{(1+ \frac{\frac{3.300}{100}}{2})^2}+\frac{(\frac{1.5}{100}  \times  \frac{100}{2}+100)}{(1+\frac{r_2}{2})^3}=97\\\\

\to \frac{1.5}{2}  \times 0.99+\frac{1.5}{2}\times \frac{1}{(1+ \frac{\frac{3.300}{100}}{2})^2}+\frac{(\frac{1.5}{2} +100)}{(1+\frac{r_2}{2})^3}=97\\\\\to 0.7425+0.75 \times \frac{1}{(1+ \frac{\frac{3.300}{100}}{2})^2}+\frac{(0.75  +100)}{(1+\frac{r_2}{2})^3}=97\\\\\to 1.4925 \times \frac{1}{(1+0.0165)^2}+\frac{(100.75 )}{(1+\frac{r_2}{2})^3}=97\\\\\to 1.4925 \times \frac{1}{(1.033)}+\frac{(100.75 )}{(1+\frac{r_2}{2})^3}=97\\\\

\to 1.4925 \times 0.96+\frac{(100.75 )}{(1+\frac{r_2}{2})^3}=97\\\\\to 1.4328+\frac{(100.75 )}{(1+\frac{r_2}{2})^3}=97\\\\\to 1.4328-97= \frac{(100.75 )}{(1+\frac{r_2}{2})^3}\\\\\to -95.5672= \frac{(100.75 )}{(1+\frac{r_2}{2})^3}\\\\\to (1+\frac{r_2}{2})^3= -1.054\\\\\to r_2=3.577\%

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\to 1.25\% \times \frac{100}{2} \times 0.99+1.25\% \times \frac{100}{2} \times \frac{1}{(1+\frac{3.300\%}{2})^2}+1.25\%\times\frac{100}{2} \times \frac{1}{(1+\frac{3.577\%}{2})^3}+(1.25\% \times \frac{\frac{100}{2}+100}{(1+\frac{r_3}{2})^4})=96\\\\

to solve this we get r_3=3.335\%

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