SOLUTION: Steven left Town A and walked towards Town B at a speed of 100m/min. At the same time, Jason and Melvin started from Town B and walked towards Town A at a speed of 80m/min and 75m/

Question 389197: Steven left Town A and walked towards Town B at a speed of 100m/min. At the same time, Jason and Melvin started from Town B and walked towards Town A at a speed of 80m/min and 75m/min respectively. If Steven met Melvin six minutes after passing Jason, find the distance between Town A and Town B.
Found 2 solutions by mananth, josmiceli:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
Steven left Town A and walked towards Town B at a speed of 100m/min. At the same time, Jason and Melvin started from Town B and walked towards Town A at a speed of 80m/min and 75m/min respectively. If Steven met Melvin six minutes after passing Jason, find the distance between Town A and Town B.
...
Steven ----- 100m/min
...
Jason--------80m/min
...
Melvin-------75m/min
......
Let distance between towns be x m
Steven ----- 100m/min
Jason--------80m/min
combined speed = 180 m.min
Time taken = x/180 minutes.
...
Steven ----- 100m/min
Melvin-------75m/min
combine speed = 175 m/min
..
Time = x/175
...
Difference in time of meeting = 6 minutes.
x/175 - x/180 = 6
(180x-175x)/175*180=6
5x=175*180*6
5x=189,000
/5
x= 37,800 meters
OR 37.8 km
...
m.ananth@hotmail.ca

Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Stephen meets Jason in min
Stephen went m and
Jason went m
I can write 2 equations:
m
m
Now add the equations:
m
Let , the distance between the towns

In 8 more min, stephen goes m further
m , so he went
total distance to meet Melvin
Now I want to know how far Melvin went in min
He went m
m
Note that the distance between the towns

and, from before, I said , so

and, if

m
The distance between towns A and B is 50.4 km
check answer:

m
and

m
m
OK
m

OK

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