SSC GD Constable 4 Feb 2025 Shift 3 Paper

Amount $\left(\mathit{A}\right)=\mathit{P}{\left(1+\mathit{r}∕100\right)}^{\mathit{t}}$, where:
$\mathit{A}=$ Amount after time t
P = Principal
$\mathit{r}=$ Rate of interest
t = Time period
Calculation:
Let the amount given to Ajay be ₹x, and the amount given to Prakash be ₹ ( $1224-\mathit{x}$ ).
According to the problem, Ajay and Prakash received the same amount after 18 and 19 years, so we can write the equation for their amounts:
$\mathit{x}{\left(1+4∕100\right)}^{18}=\left(1224-\mathit{x}\right){\left(1+4∕100\right)}^{19}$
$\Rightarrow \mathit{x}{\left(1.04\right)}^{18}=\left(1224-\mathit{x}\right){\left(1.04\right)}^{19}$
Now, divide both sides by ${\left(1.04\right)}^{18}$ :
$\Rightarrow \mathit{x}=\left(1224-\mathit{x}\right)\left(1.04\right)$
$\Rightarrow \mathit{x}=1224\times 1.04-\mathit{x}\times 1.04$
$\Rightarrow \mathit{x}+\mathit{x}\times 1.04=1224\times 1.04$
$\Rightarrow \mathit{x}\left(1+1.04\right)=1224\times 1.04$
$\Rightarrow \mathit{x}\times 2.04=1224\times 1.04$
$\Rightarrow \mathit{x}=\left(1224\times 1.04\right)∕2.04$
$\Rightarrow \mathit{x}=1272.96∕2.04$
$\Rightarrow \mathit{x}\approx \mathit{₹}624$
Amount given to Prakash $=1224-624=600$