Question
A finance company declares that, at a certain compound interest rate, a sum of money deposited by anyone will become 8 times in 3 years. If the same amount is deposited at the same compound rate of interest, then in how many years will it become 16 times ?
Answer: Option A
$$\eqalign{
& P{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^3} = 8P \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^3} = 8 \cr
& {\text{Let }}P{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = 16P \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = 16 = {2^4} = {\left( {{2^3}} \right)^{\frac{4}{3}}} \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = {\left( 8 \right)^{\frac{4}{3}}} \cr
& \Rightarrow {\left\{ {{{\left( {1 + \frac{{\text{R}}}{{100}}} \right)}^3}} \right\}^{\frac{4}{3}}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^4} \cr
& \Rightarrow n = 4 \cr
& \therefore {\text{ Required time 4 years}} \cr} $$
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$$\eqalign{
& P{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^3} = 8P \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^3} = 8 \cr
& {\text{Let }}P{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = 16P \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = 16 = {2^4} = {\left( {{2^3}} \right)^{\frac{4}{3}}} \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} = {\left( 8 \right)^{\frac{4}{3}}} \cr
& \Rightarrow {\left\{ {{{\left( {1 + \frac{{\text{R}}}{{100}}} \right)}^3}} \right\}^{\frac{4}{3}}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^4} \cr
& \Rightarrow n = 4 \cr
& \therefore {\text{ Required time 4 years}} \cr} $$
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