The value of an article decreased for two years at the rate of 10% per year and then in the third year it increased by 10%. Find the original value of the article, if its value at the end of 3 years is ₹ 40,095.

Let original value be ₹ x.

Given, it decreases by 10% for two years. So,

For depreciation :

Value after n years = Present value $\times \Big(1 - \dfrac{r}{100}\Big)^n$

$\text{Value after 2 years} = x \times \Big(1 - \dfrac{10}{100}\Big)^2 \\[1em] = x \times \Big(\dfrac{90}{100}\Big)^2\\[1em] = x \times \Big(\dfrac{9}{10}\Big)^2 \\[1em] = x \times \dfrac{81}{100} \\[1em] = \dfrac{81x}{100}.$

Given,

In the third year value is increased by 10%.

For growth :

Value after n year = Present value $\times \Big(1 + \dfrac{r}{100}\Big)^n$

$\text{Value after 1 year} = \dfrac{81x}{100} \times \Big(1 + \dfrac{10}{100}\Big)^1 \\[1em] = \dfrac{81x}{100} \times \dfrac{110}{100} \\[1em] = \dfrac{891x}{1000}.$

Given,

At the end of 3 years value is ₹ 40095.

$\therefore \dfrac{891x}{1000} = 40095 \\[1em] \Rightarrow x = \dfrac{40095 \times 1000}{891} \\[1em] \Rightarrow x = ₹ 45000.$

Hence, original value of article = ₹ 45000.