Question
A sum of money doubles itself in 4 years compound interest. It will amount to 8 times itself at the same rate of interest in = ?
Answer: Option B
$$\eqalign{
& {\text{Let Principal = P}} \cr
& {\text{Rate = R% }} \cr
& {\text{T = 4 years}} \cr
& \therefore {\text{Amount = 2P}} \cr
& {\text{Case (I) 2P = P}}{\left( {1 + \frac{R}{{100}}} \right)^4} \cr
& 2 = {\left( {1 + \frac{R}{{100}}} \right)^4}.....(i) \cr
& {\text{Case (II) Let after t years it will be 8 times}} \cr
& {\text{8P = P}}{\left( {1 + \frac{R}{{100}}} \right)^t} \cr
& {\left( 2 \right)^3} = {\left( {1 + \frac{R}{{100}}} \right)^t}.....(ii) \cr
& {\text{By using equation (i) & equation (ii)}} \cr
& {\left( {1 + \frac{R}{{100}}} \right)^{12}} = {\left( {1 + \frac{R}{{100}}} \right)^t} \cr
& {\text{By comparing both sides,}} \cr
& {\text{t = 12 years}} \cr} $$
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$$\eqalign{
& {\text{Let Principal = P}} \cr
& {\text{Rate = R% }} \cr
& {\text{T = 4 years}} \cr
& \therefore {\text{Amount = 2P}} \cr
& {\text{Case (I) 2P = P}}{\left( {1 + \frac{R}{{100}}} \right)^4} \cr
& 2 = {\left( {1 + \frac{R}{{100}}} \right)^4}.....(i) \cr
& {\text{Case (II) Let after t years it will be 8 times}} \cr
& {\text{8P = P}}{\left( {1 + \frac{R}{{100}}} \right)^t} \cr
& {\left( 2 \right)^3} = {\left( {1 + \frac{R}{{100}}} \right)^t}.....(ii) \cr
& {\text{By using equation (i) & equation (ii)}} \cr
& {\left( {1 + \frac{R}{{100}}} \right)^{12}} = {\left( {1 + \frac{R}{{100}}} \right)^t} \cr
& {\text{By comparing both sides,}} \cr
& {\text{t = 12 years}} \cr} $$
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