Aptitude Overflow

+1 vote

Use the following data: A and B are running along a circular course of radius 7 km in opposite directions such that when they meet they reverse their directions and when they meet, A will run at the speed of B and vice-versa, Initially, the speed of A is thrice the speed of B. Assume that they start from $M_{0}$ and they first meet at $M_{1}$, then at $M_{2}$, next at $M_{3}$, and finally at $M_{4}$.

What is the shortest distance between $M_{1}$ and $M_{2}$?

- $11$ km
- $7 \sqrt{2}$ km
- $7$ km
- $14$ km

0 votes

Assuming, Initially $A's$ speed is $3x\hspace{0.1cm}km/hr$ & $B's$ speed is $x\hspace{0.1cm}km/hr$.

So, in a certain amount of time, $A$ is travelling $\dfrac{3}{4}^{th} $ distance of circle, whereas in that time $B$ is travelling only $\dfrac{1}{4}^{th} $ distance of circle.

let's see the pictorial representation of the events.

I'm going to divide the circle into $4$ equal halves,as if in a certain period of time $A$ is covering $3$ units of distance & in the same amount of time $B$ is covering $1$ unit of distance. then the circle should be divided into $(3+1)=4$ equal halves.

$A$ & $B$ is starting from $M_0$ & Both are running opposite direction of each other & as $A's$ speed is $3x\hspace{0.1cm} km/hr$ & $B's$ speed is $x km/hr$, $A$ will cover $3$ units, whereas $B$ will cover $1$ units & both will meet at $M_1$ point .

Now $A$ & $B's$ speed will be reversed as per the criterion. $A's$ speed will be $x\hspace{0.1cm}km/hr$ & $B's$ speed will be $3x\hspace{0.1cm}km/hr$

After meeting this point , $A's$ speed will be $3x\hspace{0.1cm}km/hr$ & $B's$ speed will be $x\hspace{0.1cm}km/hr$

After meeting this point , $A's$ speed will be $x\hspace{0.1cm}km/hr$ & $B's$ speed will be $3x\hspace{0.1cm}km/hr$

$\text{And, Finally, they'll meet at}$ $M_4$.

Now, the question ask us to find the shortest distance between $M_1$ & $M_2$

Applying Pythagoras Theorem,

$(OM_{1})^{2}+(OM_{2})^{2}= (M_{1}M_{2})^2$

Or, $7^2+7^2 =(M_{1}M_{2})^2$

Or, $98 =(M_{1}M_{2})^2$

Or, $\sqrt{98} =(M_{1}M_{2})$

Or, $(M_{1}M_{2})= 7\sqrt{2}$

$\color{green}{\text{∴ The shortest distance between}}$ $\color{red}{M_{1}}$ $\color{green}{and}$ $\color{red}{M_{2}}$ $\color{green}{is}$ $\color{orange}{7\sqrt{2}\hspace{0.1cm}km}$

2,831 questions

1,261 answers

496 comments

40,660 users