Answer : Option A

Explanation :

Let the sum be P and Rate of Interest be R% per annum
Simple Interest on Rs.P for 2 years = 80
$MF#%\dfrac{\text{PRT}}{100} = 80\\\\ \dfrac{\text{PR}\times 2}{100} = 80\\\\ \dfrac{\text{PR}}{50} = 80\\\\ \text{PR} = 4000 \quad \color{#F00}{\text{--- (equation 1)}}$MF#%
$MF#%\text{Compound Interest = }\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} - \text{P} \\\\= \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^2 - \text{P} \\\\ = \text{P}\left[\left(1 + \dfrac{\text{R}}{100}\right)^2 - 1\right]
= \text{P}\left[\left(1 + \dfrac{\text{2R}}{100} + \dfrac{\text{R}^2}{10000}\right) - 1\right]
= \text{P}\left( \dfrac{\text{2R}}{100} + \dfrac{\text{R}^2}{10000}\right) \\\\ = \dfrac{\text{2PR}}{100} + \dfrac{\text{PR}^2}{10000}\\\\ = \dfrac{\text{2PR}}{100} + \dfrac{\text{PR} \times \text{R}}{10000}\\\\ = \dfrac{2 \times 4000}{100} + \dfrac{4000 \times \text{R}}{10000} \quad \color{#F00}{\text{ (∵ substituted the value of PR from equation 1)}}\\\\ = 80 + 0.4\text{R}$MF#%
Given that compound interest = Rs.80.80
=> 80 + 0.4R = 80.80
=> 0.4R = 0.80
$MF#%=> \text{R} = \dfrac{0.80}{0.4} = 2\%$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#%