​A sum becomes ₹8640 after two years and ₹12,441.6 after four years at the same compound interest. Find the sum. ​

Correct option is B

Given:

Amount after 2 years = ₹8640

Amount after 4 years = ₹12,441.6

Interest is compounded annually.

Formula Used:

​$$\frac{A_2}{A_1} = (1 + r)^{t_2 - t_1}$$​​

Then use:

A = $$P(1 + r)^t$$​​

​$$\Rightarrow P = \frac{A}{(1 + r)^t}$$​​

Solution:

​$$\frac{12441.6}{8640} = \frac{A_4}{A_2} = (1 + r)^2$$​​

​$$\Rightarrow (1 + r)^2 = \frac{12441.6}{8640} = 1.44$$​​

​$$\Rightarrow 1 + r = \sqrt{1.44} = 1.2$$​​

​$$\Rightarrow r = 0.2 = 20\%$$​​

Using amount after 2 years to find principal:

​$$8640 = P(1.2)^2 \\ \ \\8640 = P \times 1.44$$​​

​$$\Rightarrow P = \frac{8640}{1.44} = 6000$$​​

Thus, the correct option is (b) ₹6000

Alternate Method:

Since in 2 years amount becomes $$\times\frac{12441.6}{8640}$$ times of itself.

=> P $$\times\frac{12441.6}{8640}$$ = 8640

​=> P = $$\frac{8640 \times 8640}{12441.6}$$ 

​=> P = Rs 6000