CMAT 2015 Solved Question Paper

(d) Let the principal be ₹ P and rate of interest be r.
Given, t = 2 yr
Amount = P$(1+{\frac{\mathit{r}}{100})}^{\mathit{t}}$ = P$(1+{\frac{\mathit{r}}{100})}^2$
Cl = Amount - Principal
= P$(1+{\frac{\mathit{r}}{100})}^2$ - P
According to the question,
P$(1+{\frac{\mathit{r}}{100})}^2$ - P = 5520$........\left(\mathit{i}\right)$
Again, principal = ₹ P, rate = r and time = 3 yr
SI = $\frac{\mathit{P}\times \mathit{t}\times \mathit{r}}{100}$ = $\frac{\mathit{P}\times 3\times \mathit{r}}{100}$
According to the question,
$\frac{\mathit{P}\times 3\times \mathit{r}}{100}$ = 7200 ⇒ $\frac{\mathit{r}}{100}=\frac{7200}{\mathit{P}\times 3}$
⇒ r = $\frac{2400\times 100}{\mathit{P}}$
∴ r = $\frac{240000}{\mathit{P}}$$.........\left(\mathit{i}\mathit{i}\right)$
On putting the value of r in Eq. (i). we have
P$(1+{\frac{240000}{\mathit{p}\times 100})}^2$ - P = 5520
⇒ P$(1+{\frac{2400}{\mathit{p}})}^2$ - P = 5520
⇒ P$[1+\left({\frac{2400}{\mathit{p}})}^2+\frac{4800}{\mathit{P}}\right]$ - P = 5520
⇒ P + $\frac{{2400}^2}{\mathit{P}}$ + 4800 - P = 5520
⇒ $\frac{{2400}^2}{\mathit{P}}$ = 55520 - 4800
⇒ $\frac{{2400}^2}{720}$ = P
∴ P = 8000
On putting the value of P in Eq (ii), we get
r = $\frac{240000}{8000}$ = 30
Therefore, rate of interest is 30% per annum.