SSC CGL 24 Aug 2021 Shift 2 Paper

We know that, $\mathit{A}=\mathit{P}{\left(1+\frac{\mathit{R}}{100}\right)}^{\mathit{t}}$
Principal for the second year
$=$ Amount of 1 st year $-4800$
Principal $=12000{\left(1+\frac{\mathit{R}}{100}\right)}^1-4800$
Now, put this principal value in
$\mathit{A}=\mathit{P}{\left(1+\frac{\mathit{R}}{100}\right)}^{\mathit{t}}$
where, $\mathit{P}$ is given above, and $\mathit{A}=\mathit{₹}9240$,
$\mathit{t}=1\mathit{y}\mathit{r}$
$\Rightarrow \mathit{A}=\mathit{P}{\left(1+\frac{\mathit{r}}{100}\right)}^{\mathit{t}}$
$\Rightarrow 9240=\left[12000{\left(1+\frac{\mathit{r}}{100}\right)}^1-4800\right]$
$\left(1+\frac{\mathit{r}}{100}\right)$
$\Rightarrow 9240=\left(12000+120\mathit{r}-4800\right)$
$\left(1+\frac{\mathit{r}}{100}\right)$
$\Rightarrow 9240=\left(120\mathit{r}+7200\right)\left(1+\frac{\mathit{r}}{100}\right)$
$\Rightarrow 9240=120\mathit{r}+1\cdot 2{\mathit{r}}^2+7200+72\mathit{r}$
$\Rightarrow 12{\mathit{r}}^2+192\mathit{r}-2040=0$
$\Rightarrow [$ Divide by 1.2]
$\Rightarrow {\mathit{r}}^2+170\mathit{r}-10\mathit{r}-1700=0$
$\Rightarrow \mathit{r}\left(\mathit{r}+170\right)-10\left(\mathit{r}+170\right)=0$
$\Rightarrow \left(\mathit{r}+170\right)\left(\mathit{r}-10\right)=0$
$\Rightarrow \mathit{r}=10\%$