mathsssssssss

[anam]

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 the speed of 80m/min and 75 m/min respectively. If Steven met Melvin 6 mins after passing Jason, find the distance between town A and town B.

give explanatory ans. plz

 

[abcde]

Hi, I think this is the answer:
Let the time taken for Steven and Jason to meet be t min. So Steven and Melvin will meet at (t+6) min.
Distance traveled by Steven and Melvin in t min = Distance traveled by Steven and Melvin in (t+6) min.
=> 100t + 80t = 100(t+6) + 75(t+6)
=> 180t= 100t + 600 + 75t + 450.
=> 180t - 175t = 1050.
=> 5t = 1050.
=> t = 210 minutes.
So the distance between A and B is : 100(210) + 80(210) = 37 800 m or 37.8 km.
Hope this clears it

 

[AnGeL.Of.DaRkNeSs]

yeah i also think like abcde she solved it correct according to me.....correct me if i m wrong