Answer : Option C

Explanation :

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Solution 1
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Let the sum be P and rate of interest be R% per annum.
Amount after 2 year = 8240
$MF#%\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 8240\\\\ \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^2 = 8240 \quad \color{#F00}{\text{--- ( 1)}}$MF#%
Amount after 3 year = 9888
$MF#%\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 9888\\\\ \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^3 = 9888 \quad \color{#F00}{\text{--- (2)}}$MF#%
$MF#%\color{#F00}{\text{(2)÷(1) =>}} \dfrac{\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^3 }{\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^2} = \dfrac{9888}{8240}\\\\ 1 + \dfrac{\text{R}}{100} = \dfrac{9888}{8240}\\\\ \dfrac{\text{R}}{100} = \left(\dfrac{9888}{8240} - 1\right) = \dfrac{1648}{8240} = \dfrac{1}{5}\\\\ \text{R} = \dfrac{100}{5} = 20\%$MF#%
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Solution 2
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If a certain sum of money at compound interest amounts to Rs.x in t₁ years and Rs.y in t₂ years, then the rate of interest per annum can be given by
$MF#%\text{R} = \left[\left(\dfrac{y}{x}\right)^{\text{1}/(\text{t}_2-\text{t}_1)}- 1\right] \times 100 \%$MF#%