Symbiosis National Aptitude Test SNAP 2015 Paper

Given CI = ₹ 832, SI = ₹ 800 and T = 2 years
From formula,
CI = $\mathit{P}\left\{\left(1+\frac{\mathit{R}}{100}\right)\mathit{T}-1\right\}$
∴ 832 = $\mathit{P}\left\{{\left(1+\frac{\mathit{R}}{100}\right)}^2-1\right\}$
⇒ 832 = $\mathit{P}\left\{1+\frac{{\mathit{R}}^2}{10000}+\frac{2\mathit{R}}{100}-1\right\}$
⇒ 932 = $\mathit{P}\left\{\frac{{\mathit{R}}^2}{10000}+\frac{2\mathit{R}}{100}\right\}$ ... (i)
Using SI formula,
SI = $\frac{\mathit{P}\times \mathit{R}\times \mathit{T}}{100}$
⇒ 800 = $\frac{\mathit{P}\times \mathit{R}\times 2}{100}$
⇒ P = $\frac{40000}{\mathit{R}}$ ... (ii)
Now putting the value of P in Eq. (i) then,
932 = $\frac{40000}{\mathit{R}}\left(\frac{{\mathit{R}}^2}{10000}+\frac{2\mathit{R}}{100}\right)$
⇒ 83 = 4R + 800
⇒ 4R = 32
∴ R = $\frac{32}{4}$ = 8%