Management Aptitude Test 2017 Paper

(b) ₹ 20,000
Let, the amount borrowed be 'P' .
Compound interest (Cl) = P$\left[{\left(1+\frac{\mathit{R}}{100}\right)}^{\mathit{n}}-1\right]$
⇒ (Cl) = P$\left[{\left(1+\frac{10}{100}\right)}^3-1\right]$
⇒ (Cl) = P$\left[{\left(\frac{110}{100}\right)}^3-1\right]$
⇒ (Cl) = P$\left(\frac{1331-1000}{1000}\right)$
⇒ CI = $\frac{331\mathit{P}}{1000}$
and simple interest (SI) = $\frac{\mathit{P}\times 10\times 3}{100}=\frac{30\mathit{P}}{100}$
According to the question, Cl - SI = 620
⇒ $\frac{331\mathit{P}}{1000}-\frac{30\mathit{P}}{100}$ = 620
⇒ $\frac{\left(331-300\right)\mathit{p}}{1000}$ = 620
⇒ 31P = 620000
⇒ P = ₹ 20000
∴ Principal amount is ₹ 20000