Management Aptitude Test Dec 2013 Paper

(c) Let the sum be P, rate r % per annum and time t yr.
According to the question,
$\left[\mathit{P}{\left(1+\frac{\mathit{r}}{100}\right)}^2-\mathit{P}\right]$ - $\frac{\mathit{P}\times \mathit{r}\times 2}{100}$ = 20
⇒ P + $\frac{2}{100}\mathit{P}\mathit{r}+\frac{\mathit{P}{\mathit{r}}^2}{{\left(100\right)}^2}$ - $\mathit{P}-\frac{2}{100}\mathit{P}\mathit{r}$ = 20
∴ $\frac{\mathit{P}{\mathit{r}}^2}{{\left(100\right)}^2}$ = 20 .....(i)
and
$\left[\mathit{P}{\left(1+\frac{\mathit{r}}{100}\right)}^3-\mathit{P}\right]$- $\frac{\mathit{P}\times \mathit{r}\times 3}{100}$ = 61
⇒ P + $\frac{3}{100}\mathit{P}\mathit{r}+\frac{\mathit{P}{\mathit{r}}^3}{{\left(100\right)}^3}+\frac{3\mathit{P}{\mathit{r}}^2}{{\left(100\right)}^2}$ - $\mathit{P}-\frac{3}{100}\mathit{P}\mathit{r}$ = 61
∴ $\frac{\mathit{P}{\mathit{r}}^3}{{\left(100\right)}^3}+\frac{3\mathit{P}{\mathit{r}}^2}{{\left(100\right)}^2}$ = 61 .....(ii)
Using Eq. (i) in Eq. (ii), we get
⇒ $\frac{\mathit{r}}{100}\times \frac{\mathit{P}{\mathit{r}}^2}{{\left(100\right)}^2}+3\times \frac{\mathit{P}{\mathit{r}}^2}{{\left(100\right)}^2}$ = 61
⇒ $\frac{\mathit{r}}{100}$ x20 + 3 x 20 = 61    [from Eq. (i)]
⇒ $\frac{\mathit{r}}{5}$ = 1 ⇒ r = 5
Now,
$\frac{\mathit{P}{\mathit{r}}^2}{{\left(100\right)}^2}$ = 20 ⇒ $\frac{\mathit{P}\times 5^2}{{\left(100\right)}^2}$ = 20
⇒ $\frac{\mathit{P}\times 25}{100\times 100}$ = 20
⇒ P = 400 x 20
= ₹ 8000