The compound interest on ₹30000 at 10% per annum for n years is ₹9930. The value of n is :

3

In compound interest

A = P( 1+ \(\frac{r}{100}\))ⁿ

A = P + CI

30000 + 9930 = 30000( 1+ \(\frac{r}{100}\))ⁿ

39930 = 30000( 1+ \(\frac{10}{100}\))ⁿ

\(\frac{1331}{1000}\) =  \(\frac{11}{10}\))ⁿ

(\(\frac{11}{10}\))³ =  \(\frac{11}{10}\))ⁿ

on comparing ,

n = 3

The correct answer is option (3) : 3