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

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

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