5 years
Step-by-step explanation
Given
Cost price = $ 19,000
Depreciation yearly is % 25
What to find
Time to depreciate to $ 5, 800 or less
Step- by - Step Solution
After first year St
![\begin{gathered} 25\%\text{ }of\text{ 19,000} \\ \\ (25)/(100)*\text{ 19,000 = 4,750} \\ \\ 19,000\text{ - 4750 = 14, 250} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9zpqug9vrakf67djujikd3ccsfq6kar11t.png)
After the year the second year
![\begin{gathered} (25)/(100)\text{ }*\text{ 14, 250 = 3,562.5} \\ \\ 14,250\text{ - 3,562.5 =10, 687.5} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wndwodrtmmpcnnygixjjkjjvbm1sdpv2o3.png)
After Third year
![\begin{gathered} 25\%\text{ of 10,687.5} \\ \\ \frac{25}{100\text{ }}*\text{ 10, 687.5 = 2,671.875} \\ \\ 10,687.5\text{ - 2,671.875 = 8,015.625} \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wxg3t8r3h0z9nlo7gx41s2uxhwlygwx3tm.png)
After Fourth year
![\begin{gathered} 25\%\text{ of 8,015.625} \\ \\ (25)/(100)*\text{ 8,015.625 = 20003.906} \\ \\ 8\text{,015.625 - 20,003.906 = 6011.719} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/tzwnr6cxa5m14wyxp7fq2p62jm6qcazxnd.png)
After Fifth year
![\begin{gathered} 25\%\text{ of 6011.719} \\ \\ (25)/(100)*\text{ 6011.719 = 1502.930} \\ \\ 6011.719-1502.930\text{ = 4508.789} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/azpbbrbshwi7k0ppnconc2ae710twq6smn.png)
Therefore after 5 years the car be worth 5800. dollars or less Therefore after 5 years the car be worth 5800. dollars or less