Question
At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years ?
Answer: Option C
$$\eqalign{
& {\text{A = P }}{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} \cr
& \Rightarrow 1348.32 = 1200{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow \frac{{134832}}{{120000}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow \frac{{231525}}{{200000}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow \frac{{2809}}{{2500}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow {\left( {\frac{{53}}{{50}}} \right)^2} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow \frac{{53}}{{50}} = 1 + \frac{{\text{R}}}{{100}} \cr
& \Rightarrow {\text{R}} = {\text{ 6% }} \cr} $$
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$$\eqalign{
& {\text{A = P }}{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} \cr
& \Rightarrow 1348.32 = 1200{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow \frac{{134832}}{{120000}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow \frac{{231525}}{{200000}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow \frac{{2809}}{{2500}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow {\left( {\frac{{53}}{{50}}} \right)^2} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr
& \Rightarrow \frac{{53}}{{50}} = 1 + \frac{{\text{R}}}{{100}} \cr
& \Rightarrow {\text{R}} = {\text{ 6% }} \cr} $$
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