A sum of Rs. 12000 deposited at compound interest become double after 5 years. After 20 years it will become ?

1Rs. 9600
2Rs. 120000
3Rs. 124000
4Rs. 192000

Answer:- 1

Explanation:-

Solution:

\begin{aligned} 12000 \times {(1 + \frac{R}{100})}^{5} = 24000 \\ \Rightarrow {(1 + \frac{R}{100})}^{5} = 2 \\ ∴ {[{(1 + \frac{R}{100})}^{5}]}^{4} = 2^{4} = 16 \\ \Rightarrow {(1 + \frac{R}{100})}^{20} = 16 \\ \Rightarrow P {(1 + \frac{R}{100})}^{20} =  16P \\ \Rightarrow 12000 {(1 + \frac{R}{100})}^{20} = 16 \times 12000 \\ \Rightarrow 12000 {(1 + \frac{R}{100})}^{20} = 192000 \end{aligned}

$\eqalign{ & 12000 \times {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^5} = 24000 \cr & \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^5} = 2 \cr & \therefore {\left[ {{{\left( {1 + \frac{{\text{R}}}{{100}}} \right)}^5}} \right]^4} = {2^4} = 16 \cr & \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{20}} = 16 \cr & \Rightarrow {\text{P}}{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{20}}{\text{ = 16P}} \cr & \Rightarrow 12000{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{20}} = 16 \times 12000 \cr & \Rightarrow 12000{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{20}} = 192000 \cr}$