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

towards Town A at a speed of 80 m/min & 75 m/min respectively. If Steven met Melvin 6 minutes after passing Jason, find the distance between Town A & Town B.

Expert Answers

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Steven walks from town A towards town B at 100m/min. Jason walks from town B towards town A at 80m/min and Melvin walks from town B towards town A at 75 m/min.

Let the distance between A and B be L. Let Steven meet Jason after time T.

In time T Steven has travelled a distance 100*T and Jason has walked a distance 80*T. As the two are coming from opposite directions, they have covered the distance between the two towns

=> 100T + 80T = L ...(1)

After 6 minutes Steven meets Melvin. Here Steven has walked (T + 6)*100 and Melvin was walked (T + 6)*75.

Again (T + 6)*100 + (T + 6)*75 = L

=> 100T + 600 + 75T + 450 = L

=> 175T + 1050 = L ...(2)

(1) and (2) give

180T = 175T + 1050

=> 5T = 1050

=> T = 1050/5

=> T = 210

Substitute in (2)

175T + 1050 = L

=> 175* 210 + 1050 = L

=> 37800 m

The distance between the two towns is 37.8 km.

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