Two guns were fired from the same place at an interval of 10

If the person were static he would hear ("meet") the sound in 10.5 minutes, but he heard ("met") the sound in 10 minutes, and that's because during these 10 minutes the person was moving towards the sound. Thus these 10 minutes of person moving "saved" the sound its 0.5 minutes of moving.

OR: let's say (imagine) the distance between the shooting point and person when the second shot was made was \(d\). We are told that this distance would be covered in 10.5 minutes if person were static: \(d=10.5s\) (\(s\) speed of sound).

But because person was was moving towards the sound they "met" in 10 minutes, which means that the distance \(d\) was covered in 10 minutes at their combined speed. Their combined speed was \(s+t\) (\(t\) the speed of person/train): \(d=10(s+t)\).

\(d=10.5s=10(s+t)\) --> \(0.5s=10t\).

Hope it helps.