SSC GD Constable 6 Feb 2025 Shift 2 Paper

Relative speed $($ same direction $)=\left|{\mathit{S}}_1-{\mathit{S}}_2\right|$
Distance $=$ Speed × Time
$1\mathit{k}\mathit{m}∕\mathit{h}\mathit{r}=\frac{5}{18}\mathit{m}∕\mathit{s}$
When two trains cross each other, the total distance covered is the sum of their lengths $\left({\mathit{L}}_1+{\mathit{L}}_2\right)$.
Calculation:
Relative speed $=75\mathit{k}\mathit{m}∕\mathit{h}\mathit{r}-30\mathit{k}\mathit{m}∕\mathit{h}\mathit{r}=45\mathit{k}\mathit{m}∕\mathit{h}\mathit{r}$
Convert relative speed to $\mathit{m}∕\mathit{s}$ :
$\Rightarrow 45\mathit{k}\mathit{m}∕\mathit{h}\mathit{r}=45\times \frac{5}{18}\mathit{m}∕\mathit{s}$

Let the length of the slow train be ${\mathit{L}}_2$.
Total distance $={\mathit{L}}_1+{\mathit{L}}_2=950+{\mathit{L}}_2$ meters
Using Distance $=$ Speed × Time:
$\Rightarrow 950+{\mathit{L}}_2=12.5\times 108$
$\Rightarrow 950+{\mathit{L}}_2=1350$
$\Rightarrow {\mathit{L}}_2=1350-950$
$\Rightarrow {\mathit{L}}_2=400\mathit{m}$