Find the difference between compound interest and simple interest on ₹ 12000 and in $1\dfrac{1}{2}$ years at 10% compounded half-yearly.

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

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

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