CAT 2020 Set-2 | Question: 68

Question

$\text{A}$ and $\text{B}$ are two points on a straight line. Ram runs from $\text{A}$ to $\text{B}$ while Rahim runs from $\text{B}$ to $\text{A}.$ After crossing each other, Ram and Rahim reach their destinations in one minute and four minutes, respectively. If they start at the same time, then the ratio of Ram’s speed to Rahim’s speed is  

• A. $2$
• B. $\sqrt{2}$
• C. $2\sqrt{2}$
• D. $\frac{1}{2}$

Answer 1

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 $2.$

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$\textbf{Short Method:}$

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$: \text{A}$