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A sum of money on compound interest amounts to Rs. 10648 in 3 years and Rs. 9680 in 2 years. The rate of interest per annum is = ?
Answer & Solution
Correct Answer:
Option
B
Let the sum be Rs. P and rate of interest be R% per annum. Then,
$$\eqalign{ & P{\left( {1 + \frac{R}{{100}}} \right)^2} = 9680\,.....\,\left( 1 \right) \cr & P{\left( {1 + \frac{R}{{100}}} \right)^3} = 10648\,.....\,\left( 2 \right) \cr} $$
On dividing equation (2) by (1) :
$$\eqalign{ & 1 + \frac{R}{{100}} = \frac{{10648}}{{9680}} \cr & \Rightarrow \frac{R}{{100}} = \frac{{10648}}{{9680}} - 1 \cr & \Rightarrow \frac{R}{{100}} = \frac{{10648 - 9680}}{{9680}} \cr & \Rightarrow \frac{R}{{100}} = \frac{{968}}{{9680}} \cr & \Rightarrow \frac{R}{{100}} = \frac{1}{{10}} \cr & \Rightarrow R = \frac{1}{{10}} \times 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 10\% \cr} $$
$$\eqalign{ & P{\left( {1 + \frac{R}{{100}}} \right)^2} = 9680\,.....\,\left( 1 \right) \cr & P{\left( {1 + \frac{R}{{100}}} \right)^3} = 10648\,.....\,\left( 2 \right) \cr} $$
On dividing equation (2) by (1) :
$$\eqalign{ & 1 + \frac{R}{{100}} = \frac{{10648}}{{9680}} \cr & \Rightarrow \frac{R}{{100}} = \frac{{10648}}{{9680}} - 1 \cr & \Rightarrow \frac{R}{{100}} = \frac{{10648 - 9680}}{{9680}} \cr & \Rightarrow \frac{R}{{100}} = \frac{{968}}{{9680}} \cr & \Rightarrow \frac{R}{{100}} = \frac{1}{{10}} \cr & \Rightarrow R = \frac{1}{{10}} \times 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 10\% \cr} $$
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Login968 in terms of 9680 because it increases percentage increase=Diff/Lower value=968/9680=10%