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Question
The interest on a sum of Rs 2000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is Rs 163.20.
Solution
Let the time period be n years.
Then, we have:
\[\text{ CI = P }\left( 1 + \frac{R}{100} \right)^n - P\]
\[163 . 20 = 2, 000 \left( 1 + \frac{4}{100} \right)^n - 2, 000\]
\[2, 163 . 20 = 2, 000 \left( 1 . 04 \right)^n \]
\[ \left( 1 . 04 \right)^n = \frac{2, 163 . 20}{2, 000}\]
\[ \left( 1 . 04 \right)^n = 1 . 0816\]
\[ \left( 1 . 04 \right)^n = \left( 1 . 04 \right)^2 \]
On comparing both the sides, we get:
n = 2
Thus, the required time is two years.
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