Time, Speed & Distance

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Two places $A$ and $B$ are separated by a distance of 200 m. Ajay and Jay have to start simultaneously from $A$, go to $B$ and return to $A$. In 10 s they meet at a place 10m from $B$. If Ajay is faster than Jay, in how much time, after they start, will Ajay return to $A$?


$19$ sec


\(\dfrac{200}{21}\) sec


\(\dfrac{400}{21}\) sec


\(\dfrac{190}{21}\) sec

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Option(C) is correct

Since Ajay is faster than Jay and they start together, to meet at 10 m from $B$, Ajay would have covered a distance from $A$ to $B$ and would meet Jay on his way back to $A$.

Jay would be on his way from $A$ to $B$.

So, Ajay covers $200+10=210$ m in 10 sec

Hence, Ajay's speed = 21 m/sec

So he will take \(\dfrac{190}{21}\)sec to cover the remaining 190 m.

The time required for Ajay to reach A will be:


(1) Comment(s)


total distance = 400m (200m A to B + 200m B to A)

from above we get speed of Ajay 21m/s

so simply T= distance/speed

400/21 s