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kristo · Mar 19, 2007, 08:51 AM

Here it is:
Two persons started moving towards the other town at the same time, when they met, one person had walked 6 kilometers more than other person. After meeting they continued walking towards the other town at the same speed, first person arrived at the town 4 hours after meeting, second person 9 hours after meeting.

Hope you understood it.
$\{ \frac{s}{v_1}=\frac{s+6}{v_2} \\ \frac{s}{v_2}+\frac{s+6}{v_1}=13$
That's what I got, I'm sure it's wrong, but I couldn't think of any other way to do it since there are 2 different speeds.
So yeah, help please

asterisk_man · Mar 19, 2007, 09:39 AM

I was going to assume that they both started at the same town but they can't if they meet after different walking distances. Is there another sentence that goes before what you wrote that helps define the problem?

Capuchin · Mar 19, 2007, 09:43 AM

Yeah I'm certain you haven't given enough info, I don't really understand it.

kristo · Mar 19, 2007, 11:33 AM

No, they are starting from different towns, person A from town A is going to town B and person B from town B going to town A, they meet on the way.
So person A arrived at town B 4 hours after meeting and person B arrived at town A 9 hours after meeting.
Sorry for my English lol, hope you understand it now.

kristo · Mar 20, 2007, 03:29 AM

I think I solved it..
$\{ \frac{s}{v_1}=\frac{s+6}{v_2} \\ \frac{s}{v_2}=4 \\ \frac{s+6}{v_1}=9$
Which solves for s=12, v1=2, v2=3

Capuchin · Mar 20, 2007, 03:40 AM

S being the distance between the two towns? So 12km? So they must meet at 3km along the line for the first case to be true. This doesn't make sense for the speeds you defined, as one has walked 9 km and the other 3km in the same time

Or am I missing something?

kristo · Mar 20, 2007, 04:38 AM

S being the distance person A walked until meeting.
The distance between town A and B is s + s + 6 which is 30km.

Capuchin · Mar 20, 2007, 04:44 AM

Okay that makes sense then :)