We can draw the diagram for better understanding.
Let the distance between $\text{A} \& \text{B}$ be $\text{‘D’} \; \text{meter}.$
Let they meet after $\text{‘t’}$ seconds at point $\text{C}.$ And distance between $\text{A} \;\&\; \text{C}$ be $\text{‘K’} \; \text{meter}.$
Let the speed of Ram and Rahim be $\text{S}_{1}$ and $\text{S}_{2}$ respectively.
We know that, $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$
Now,
- $\text{S}_{1} = \frac{\text{K}}{\text{t}} \quad \longrightarrow (1) $
- $\text{S}_{2} = \frac{\text{D-K}}{\text{t}} \quad \longrightarrow (2) $
Also,
- $\text{S}_{1} = \frac{\text{D}}{\text{t+60}} \quad \longrightarrow (3) $
- $\text{S}_{2} = \frac{\text{D}}{\text{t+240}} \quad \longrightarrow (4) $
And,
- $\text{S}_{1} = \frac{\text{D-K}}{\text{60}} \quad \longrightarrow (5) $
- $\text{S}_{2} = \frac{\text{K}}{\text{240}} \quad \longrightarrow (6) $
Divide equation $(1)$ by $(2),$ we get.
$\frac{\text{S}_{1}}{\text{S}_{2}} = \dfrac{\left(\frac{\text{K}}{\text{t}} \right)}{ \left(\frac{\text{D-K}}{\text{t}} \right)}$
$\Rightarrow \frac{\text{S}_{1}}{\text{S}_{2}} = \left( \frac{\text{K}}{\text{t}} \right) \times \left( \frac{\text{t}}{\text{D-K}} \right) $
$ \Rightarrow \frac{\text{S}_{1}}{\text{S}_{2}} = \left( \frac{\text{K}}{\text{D-K}} \right) \quad \longrightarrow (7) $
Divide equation $(5)$ by $(6),$ we get
$ \frac{\text{S}_{1}}{\text{S}_{2}} = \dfrac{ \left( \frac{\text{D-K}}{60} \right)} { \left( \frac{\text{K}}{240} \right)} $
$ \Rightarrow \frac{\text{S}_{1}}{\text{S}_{2}} = \left( \frac{ \text{D-K}}{60} \right) \times \left( \frac{240}{\text{K}} \right) $
$ \Rightarrow \frac{\text{S}_{1}}{\text{S}_{2}} = \frac{ 4 \left( \text{D-K} \right)} {\text{K}} \quad \longrightarrow (8) $
Now, multiply equation $(7)$ and equation $(8),$ we get
$ \left( \frac{\text{S}_{1}}{\text{S}_{2}} \right) \left( \frac{\text{S}_{1}}{\text{S}_{2}} \right) = \left( \frac{\text{K}}{\text{D-K}} \right) \left[ \frac{ 4 \left( \text{D-K} \right)} {\text{K}} \right] $
$ \Rightarrow \left( \frac{\text{S}_{1}}{\text{S}_{2}} \right)^{2} = 4 $
$ \Rightarrow \frac{\text{S}_{1}}{\text{S}_{2}} = \frac{2}{1} $
$ \Rightarrow \boxed{\text{S}_{1} : \text{S}_{2} = 2 : 1} $
$\therefore$ The ratio of Ram’s speed to Rahim’s speed is $2.$
$\textbf{Short Method:}$
Assume two objects $\text{A}$ and $\text{B}$ start at the same time in opposite directions from $\text{P}$ and $\text{Q}$ respectively. After passing each other, $\text{A}$ reaches $\text{Q}$ in $\text{‘a’} \; \text{seconds}$ and $\text{B}$ reaches $\text{P}$ in $\text{‘b’} \; \text{seconds}.$ Then, $ \boxed{\text{Speed of A} : \text{Speed of B} = \sqrt{\text{b}} : \sqrt{\text{a}}}$
Now, the required ratio $ = \sqrt{4} : \sqrt{1} = 2 : 1.$
Correct Answer$: \text{A}$