Question

Points $A$ and $B$ on a national highway are at a distance of $120km$ from each other. One car starts from $A$ and another from $B$ at the same time. The car which starts from $A$ moves with a constant speed along the direction $AB$ and the second moves with a constant speed in the same direction from $B$. The first car overtakes the second car in $4$ hours. However, if the second car moves from $B$ towards $A$ then the two meet each other after $2$ hours. Find the speed of the car which starts from $A$.

• A. $45km/hr$
• B. $65km/hr$
• C. $85km/hr$
• D. None of these

Answer 1

The correct option is A $45km/hr$

Let the speed of the car starting from point $A$ be $x$ and speed of the car starting from point $B$ be $y$ km/hr and distance $AB=120$ km.

If both cars move in same direction $AB$, first car overtakes second car in $4$ hrs. So distance covered by first car in $4$ hrs is $120$ km more to the distance covered by second car. Distance = speed $\times$ time.

$4x=4y+120\Rightarrow 4x-4y=120\Rightarrow 4(x-y)=120\Rightarrow x-y=\frac{120}{4}\Rightarrow x-y=30....\left(1\right)$

So, if they move towards each other, they meet in $2$ hrs.

Therefore, distance covered by first and second car in $2$ hrs is $120$ km

$2x+2y=120\Rightarrow 2(x+y)=120\Rightarrow x+y=\frac{120}{2}\Rightarrow x+y=60....\left(2\right)$

Adding equations 1 and 2 we get:

$(x-y)+(x+y)=30+60\Rightarrow 2x=90\Rightarrow x=\frac{90}{2}\Rightarrow x=45$

Hence, the speed of the car which starts from $A$ is $45$ km/hr.