Question
The compound interest on Rs. 30000 at 7% per annum is Rs. 4347. The period (in years) is = ?
Answer: Option A
$$\eqalign{
& {\text{Amount = Rs}}{\text{. }}\left( {30000 - 4347} \right) \cr
& {\text{Amount = Rs}}{\text{. }} 34347 \cr
& {\text{Let the time be }}n{\text{ years}} \cr
& {\text{Then,}} \cr
& {\text{30000}}{\left( {1 + \frac{7}{{100}}} \right)^n} = 34347 \cr
& \Leftrightarrow {\left( {\frac{{107}}{{100}}} \right)^n} = \frac{{34347}}{{30000}} \cr
& \Leftrightarrow {\left( {\frac{{107}}{{100}}} \right)^n} = \frac{{11449}}{{10000}} = {\left( {\frac{{107}}{{100}}} \right)^2} \cr
& \therefore n = {\text{ 2 years}} \cr} $$
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$$\eqalign{
& {\text{Amount = Rs}}{\text{. }}\left( {30000 - 4347} \right) \cr
& {\text{Amount = Rs}}{\text{. }} 34347 \cr
& {\text{Let the time be }}n{\text{ years}} \cr
& {\text{Then,}} \cr
& {\text{30000}}{\left( {1 + \frac{7}{{100}}} \right)^n} = 34347 \cr
& \Leftrightarrow {\left( {\frac{{107}}{{100}}} \right)^n} = \frac{{34347}}{{30000}} \cr
& \Leftrightarrow {\left( {\frac{{107}}{{100}}} \right)^n} = \frac{{11449}}{{10000}} = {\left( {\frac{{107}}{{100}}} \right)^2} \cr
& \therefore n = {\text{ 2 years}} \cr} $$
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