Answer : Option D

Explanation :

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Solution 1
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Let the sum be Rs.x
$MF#%\begin{align}&\text{Amount after 2 years on Rs.x at 4% per annum when interest is compounded annually }\\\\&=\text{x}\left(1 + \dfrac{4}{100}\right)^2 = \text{x}\left(\dfrac{104}{100}\right)^2\\\\&\text{Compound Interest = }\text{x}\left(\dfrac{104}{100}\right)^2 - x \\\\&= x\left[\left(\dfrac{104}{100}\right)^2 - 1\right] = x\left[\left(\dfrac{26}{25}\right)^2 - 1\right] = x\left[\dfrac{676}{625} - 1\right] = x\left[\dfrac{51}{625} \right] = \dfrac{51x}{625}\\\\\\\\ &\text{Simple Interest = }\dfrac{\text{PRT}}{100} = \dfrac{x \times 4 \times 2}{100} = \dfrac{2x}{25}\end{align}$MF#%
Given that difference between compound interest and simple interest is Rs.1
$MF#%\begin{align}&\Rightarrow \dfrac{51x}{625} - \dfrac{2x}{25} = 1\\\\ &\Rightarrow \dfrac{51x - 50x}{625} = 1\\\\ &\Rightarrow x = 625\end{align}$MF#%
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Solution 2
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The difference between compound interest and simple interest on Rs. P for 2 years at R% per annum
$MF#%= \text{P}\left(\dfrac{\text{R}}{100}\right)^2$MF#%