Question
A sum of money is compound interest became doubles itself in 15 years. It will become eight times of itself in = ?
Answer: Option A
$$\eqalign{
& P{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{15}} = 2P \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{15}} = 2 \cr
& {\text{Let }}P{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = 8P \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = 8 = {2^3} = {\left\{ {{{\left( {1 + \frac{{\text{R}}}{{100}}} \right)}^{15}}} \right\}^3} \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{45}} \cr
& \Rightarrow n = 45 \cr
& \therefore {\text{Required time = 45 years}} \cr} $$
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$$\eqalign{
& P{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{15}} = 2P \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{15}} = 2 \cr
& {\text{Let }}P{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = 8P \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = 8 = {2^3} = {\left\{ {{{\left( {1 + \frac{{\text{R}}}{{100}}} \right)}^{15}}} \right\}^3} \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{45}} \cr
& \Rightarrow n = 45 \cr
& \therefore {\text{Required time = 45 years}} \cr} $$
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