Question

The distance between two stations $A$ and $B$ is $60$ km. Two cars start together from $A$ and $B$. If both the cars go in the same direction, they meet after $3$ hours and if they go in the opposite directions, they meet after $30$ minutes. Assuming that car $A$ has greater speed and both the cars have uniform speeds, find their speeds.

• A. $A=40$ km/hr, $B=70$ km/hr
• B. $A=70$ km/hr, $B=50$ km/hr
• C. $A=10$ km/hr, $B=50$ km/hr
• D. $A=60$ km/hr, $B=70$ km/hr

Answer 1

The correct option is B $A=70$ km/hr, $B=50$ km/hr

When they travel in same direction, as speed of $A$ is more than $B$, suppose they meet when $B$ travels for $x$ m, then $A$ will have travelled $(60+x)$ m in the same time.
Since, speed $=\frac{Distance}{Time}$

Speed of car at $A$ $=\frac{x+60}{3}$

And Speed of car at $B$ $=\frac{x}{3}$
And when they travel in the opposite direction, suppose they meet when $A$ travels for $y$ m, then $B$ will have travelled $(60-y)$ m in the same time.

Now, $30$ minutes $=\frac{1}{2}$ hour

Speed of car at $A$ $=\frac{y}{\frac{1}{2}}=2y$

And Speed of car at $B$ $=\frac{60-y}{\frac{1}{2}}=2(60-y)$

Equating the speeds of the cars in both the cases,
Speed of car at $A$, $\frac{x+60}{3}=2y$
$x+60=6y$
$=>x-6y=-60$ --- (1)
Speed of car at $B$, $\frac{x}{3}=2(60-y)$
$=>x+6y=360$ --- (2)
Adding equations $1$ and $2$, we get $2x=300=>x=150$
So, Speed of car at $A$ $=\frac{x+60}{3}=\frac{210}{3}=70km/hr$

Speed of car at $B$ $=\frac{x}{3}=\frac{150}{3}=50km/hr$