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Question
At what rate percent will a sum of Rs 1000 amount to Rs 1102.50 in 2 years at compound interest?
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Solution
\[A = P \left( 1 + \frac{R}{100} \right)^n \]
\[1102 . 50 = 1000 \left( 1 + \frac{R}{100} \right)^2 \]
\[\frac{1102 . 50}{1000} = \left( 1 + 0 . 01R \right)^2 \]
\[ \left( 1 + 0 . 01R \right)^2 = 1 . 1025\]
\[ \left( 1 + 0 . 01R \right)^2 = \left( 1 . 05 \right)^2 \]
On comparing both the sides, we get:
1 + 0.01R = 1.05
0.01R = 0.05
R = 5
Thus, the required rate percent is 5.
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