Question

A piece of equipment costs$Rs6,00,000$ in a certain factory. If it depreciates in value$15\%$ the first year, $13.5\%$ the next year, $12\%$the third year, and so on. What will be its value at the end of$10$ years, all percentages applying to the original costs $?$

Answer 1

Finding the equipment cost:

The cost of equipment $=Rs6,00,000$

Step 1 - Finding the type of series

The value depreciates as

$15\%,13.5\%,12\%,--------$

Percentage of depreciation in the first year $=15\%$

Percentage of depreciation in the second year $=13.5\%$

Percentage difference $=$ $15\%-13.5\%$

$=-1.5\%$

Percentage of depreciation in the third year $=12\%-13.5\%$

$=-1.5\%$

Since, the percentage of depreciation is constant $=-1.5\%$

Therefore, the series is in arithmetic progression.

Step 2- Value of equipment after $10$years

Since the rate of depreciation is in arithmetic progression

So, the first term $\left(a\right)=15\%$

Common difference $\left(d\right)=-1.5\%$

Number of terms $\left(n\right)=10$

Total percentage depreciation in $10$ years

$s=\frac{n}{2}\left[2a+\left(n-1\right)\times d\right]$

$s=\frac{10}{2}\left[2\times 15+\left(10-1\right)\times \left(-1.5\right)\right]$

$s=5\left[30+9\times \left(-1.5\right)\right]$

$s=5\left[30-13.5\right]$

$s=5\times 16.5$

$s=82.5$

Therefore, percentage depreciation in $10$ years $=82.5\%$

Value of depreciation $=$$\left(\frac{100-\mathrm{depreciation}\%}{100}\right)\times$ cost of equipment

$$\begin{gathered} =\left(\frac{100-82.5}{100}\right)\times 6,00,000 \\ =\frac{17.5}{100}\times 6,00,000 \\ =17.5\times 6000 \\ =Rs105000 \end{gathered}$$

Hence, the value of the equipment after $10$ years is $Rs105000$.