Question
The difference between compound interest and simple interest on an amount of Rs. 15,000 for 2 years is Rs. 96. What is the rate of interest per annum?
Answer: Option A
$$\left[ {15000 \times {{\left( {1 + \frac{R}{{100}}} \right)}^2} - 15000} \right]$$ $$ - $$ $$\left( {\frac{{15000 \times R \times 2}}{{100}}} \right)$$ $$ = 96$$
$$ \Rightarrow 15000\left[ {{{\left( {1 + \frac{R}{{100}}} \right)}^2} - 1 - \frac{{2R}}{{100}}} \right] = 96$$
$$ \Rightarrow 15000$$ $$\left[ {\frac{{{{\left( {100 + R} \right)}^2} - 10000 - \left( {200 \times R} \right)}}{{10000}}} \right]$$ $$ = 96$$
$$\eqalign{
& \Rightarrow {R^2} = {\frac{{96 \times 2}}{3}} = 64 \cr
& \Rightarrow R = 8 \cr
& \therefore {\text{Rate}} = 8\% \cr} $$
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$$\left[ {15000 \times {{\left( {1 + \frac{R}{{100}}} \right)}^2} - 15000} \right]$$ $$ - $$ $$\left( {\frac{{15000 \times R \times 2}}{{100}}} \right)$$ $$ = 96$$
$$ \Rightarrow 15000\left[ {{{\left( {1 + \frac{R}{{100}}} \right)}^2} - 1 - \frac{{2R}}{{100}}} \right] = 96$$
$$ \Rightarrow 15000$$ $$\left[ {\frac{{{{\left( {100 + R} \right)}^2} - 10000 - \left( {200 \times R} \right)}}{{10000}}} \right]$$ $$ = 96$$
$$\eqalign{
& \Rightarrow {R^2} = {\frac{{96 \times 2}}{3}} = 64 \cr
& \Rightarrow R = 8 \cr
& \therefore {\text{Rate}} = 8\% \cr} $$
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