SSC CGL 20 Aug 2021 Shift 1 Paper

Given, amount $=\mathit{₹}81840$,
Time $=3\mathit{y}$ r and $\mathit{R}=\mathit{x}\%$
We know that, $\mathit{A}=\mathit{P}+\frac{\mathit{P}\mathit{R}\mathit{T}}{100}$
$\Rightarrow 81840=\mathit{P}+\frac{\mathit{P}\times \mathit{R}\times 3}{100}$
$\Rightarrow 81840=\mathit{P}+\frac{3\mathit{P}\mathit{R}}{100}$ . . . (i)
Given, $\mathit{A}=\mathit{₹}92400$, Time $=5\mathit{y}\mathit{r}$ and $\mathit{R}=\mathit{x}\%$
Again, $\mathit{A}=\mathit{P}+\frac{\mathit{P}\times \mathit{R}\times \mathit{T}}{100}$
$\Rightarrow 92400=\mathit{P}+\frac{\mathit{P}\times \mathit{R}\times 5}{100}...$ (ii)
Now, subtracting Eqs. (i) and (ii)
$\Rightarrow 92400-81840$
$=\mathit{P}+\frac{\mathit{P}\times \mathit{R}\times 5}{100}-\mathit{P}-\frac{3\mathit{P}\mathit{R}}{100}$
$\Rightarrow 10560=\frac{2\mathit{P}\mathit{R}}{100}$

Put value of $\mathit{P}\mathit{R}$ from Eq. (iii) in Eq. (i)
$81840=\mathit{P}+\frac{5280\times 100\times 3}{100}$
$\Rightarrow \mathit{P}=81840-15840$
$\Rightarrow \mathit{P}=66000$
Put in Eq. (iii) $\mathit{P}\mathit{R}=5280\times 100$
$\mathit{R}=\frac{5280\times 100}{66000}$
$\Rightarrow \mathit{R}=8\%$
Now, if rate increase by $\left(\mathit{x}+2\right)\%$
$=8+2=10\%$
According to the question,
$\mathit{A}=\mathit{P}+\mathit{S}\mathit{l}$
$\Rightarrow 2\mathit{P}-\mathit{P}=\mathit{S}\mathit{I}$

$\Rightarrow \mathit{P}=\frac{\mathit{P}\times 10\times \mathit{T}}{100}$