When the present population (P) of a certain locality increases by r% per year, the population in n years will be :

1. $P\Big(1 + \dfrac{r}{100}\Big)^n$ - P

2. $P\Big(1 + \dfrac{r}{100}\Big)^n$ + P

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

4. $P\Big(1 + \dfrac{r}{100n}\Big)$

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

On a certain sum the rate of C.I. is x% per annum for the first two years and y% per annum for the next three years. Then the amount after 5 years is :

1. $\Big(1 + \dfrac{x}{100}\Big)\Big(1 + \dfrac{y}{100}\Big)$ times

2. $\Big(1 + \dfrac{x}{100}\Big)^3\Big(1 + \dfrac{y}{100}\Big)^2$ times

3. $\Big(1 + \dfrac{x}{100}\Big)^2\Big(1 + \dfrac{y}{100}\Big)^3$ times

4. $\Big(1 + \dfrac{x}{100}\Big)^2\Big(1 - \dfrac{y}{100}\Big)^3$ 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. ₹ $x \times \Big(\dfrac{80}{100}\Big)^2 \times \Big(\dfrac{75}{100}\Big)^2$