Aptitude 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

**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