CMAT 2015 Solved Question Paper

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