The population of a town decreases by 10% in a particular year and then increases by 15% in the next year. The population at the end of two years is :

1. $\Big(1 + \dfrac{10}{100}\Big)^3\Big(1 + \dfrac{15}{100}\Big)$ times

2. $\Big(1 - \dfrac{10}{100}\Big)\Big(1 + \dfrac{15}{100}\Big)$ times

3. $\Big(1 - \dfrac{10}{100}\Big)\Big(1 - \dfrac{15}{100}\Big)$ times

4. $\Big(1 + \dfrac{10}{100}\Big)\Big(1 - \dfrac{15}{100}\Big)$ times

When the cost of a machine decreases by r% per year; the cost of machine in 3 years is :

1. $\Big(1 + \dfrac{r}{100}\Big)^3$ times

2. $\Big(1 - \dfrac{r}{100}\Big)^3$ times

3. $\Big(1 - \dfrac{r}{100}\Big)^2$ times

4. $\Big(1 + \dfrac{r}{100}\Big)^2$ times

The cost (₹ x) of a machine increases by 20% in the first two years and then decreases by 25% in the next two years. Then the cost of machine becomes :

1. ₹ $x \times \dfrac{120}{100} \times \dfrac{75}{100}$

2. ₹ $x \times \Big(\dfrac{120}{100}\Big)^2 \times \dfrac{75}{100}$

3. ₹ $x \times \Big(\dfrac{120}{100}\Big)^2 \times \Big(\dfrac{75}{100}\Big)^2$

4. ₹ $x \times \Big(\dfrac{80}{100}\Big)^2 \times \Big(\dfrac{75}{100}\Big)^2$

The cost of a machine is supposed to depreciate each year by 12% of its value at the beginning of the year. If the machine is valued at ₹ 44,000 at the beginning of 2008, find its value :

(i) at the end of 2009.

(ii) at the beginning of 2007.