# Easy Time, Speed & Distance Solved QuestionAptitude Discussion

 Q. 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 $A$?
 ✖ A. $19$ sec ✖ B. $\dfrac{200}{21}$ sec ✔ C. $\dfrac{400}{21}$ sec ✖ D. $\dfrac{190}{21}$ sec

Solution:
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:

$=10+\dfrac{190}{21}=\dfrac{400}{21}$sec

## (1) Comment(s)

Shubham
()

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