Peter will reach B (70 km) then turn around and meet Paul (x km) who has traveled most of the way to B (70-x km). The total distance that the two of them cover 70 + 70 - x + x = 140.

So the two are covering 140 km of total distance, we need to know at what rate they do so. Since the two of them are moving towards each other, working together, we can combine their speeds. They are closing that 140 km gap at a rate of 52 + 39 = 91 km/hr.

140 km / 91 km/hr = 20/13 hrs. It takes the 20/13 of an hour to move 140 km total distance and meet in the middle.

In 20/13 hrs, at 39 kph, Paul will have traveled (20/13)hrs * 39 kph = 60 km.

Since Paul has traveled 60 km when he meets Peter, he is 70 - 60 = 10 km from point B.

thanks Ali for the different and timesaving approach.

I did it with the basic formula, time=dis/speed
since they are meeting after sometime say T. thefore i compared the timetaken by both peter and paul.
70+x is covered by peter and 70-x is covered by paul.

(70+x)/52=(70-x)/39
this gives x=10.

aa2kash,

that was a clever way...and the quickest way i think
thanks for that solution

Cheers,
Vivek

Thank you Eli and aakash for your nice approach to this problem. I completely got the approach now.

Thank you once again!

Hi aa2kash,

You have equated "T" ( time) for both the equations? can it be done ?

The time taken to travel cannot be equal for both peter and paul .. Correct me if I am wrong here?

Regards,
Viju

@Viju
Its mention in the question that they start simultaneously. It means they when they are meeting they have taken the same time to travel their respective distances.So, we can equate the time taken by both of them.
Hope it clears your query now.

Hi aa2kash,

Yep,it clarified. Thank you