Question 1203699: A student leaves Oakville Mall and runs down Trafalgar Road towards Sheridan College (they are late for class!) as another student leaves Sheridan College and walks towards Oakville Mall. The distance from Oakville Mall to Sheridan College is 1080m. The people pass each other 6 minutes later. The jogger runs 1m/s faster than the person walking. At what distance away from Oakville Mall do the students pass each other?
Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39625)   (Show Source):
Answer by ikleyn(52855)  (Show Source):
You can put this solution on YOUR website! .
A student leaves Oakville Mall and runs down Trafalgar Road towards Sheridan College
(they are late for class!) as another student leaves Sheridan College and walks towards Oakville Mall.
The distance from Oakville Mall to Sheridan College is 1080m.
The people pass each other 6 minutes later. The jogger runs 1m/s faster than the person walking.
At what distance away from Oakville Mall do the students pass each other?
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What is written in the post by @josgarithmetic is gibberish.
Ignore it as if you never have seen it.
I came to bring a correct solution.
Let x be the rate of the runner in meters per second;
then the rate of the walking student is (x-1) m/s, according to the problem.
The travel time is 6 minutes, or 6*60 = 360 seconds.
Write the total distance equation
360x + 360*(x-1) = 1080 meters.
Simplify and find x
360x + 360x - 360 = 1080
720x = 1080 + 360 = 1440
x = 1440/720 = 2.
Thus the rate of the runner is 2 meters per second.
The problem asks for the distance traveled by the runner in 6 minutes.
It is 6*60*2 = 720 meters.
ANSWER. The distance from Oakville Mall to the point where they met each other is 720 meters.
Solved.
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After seeing my solution, @josgarithmetic re-wrote his solution,
but still did not answer the problem's question.
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
The distance is 1080m.
The time required for them to pass each other is 6 minutes, or 360 seconds.
So their combined rate is 1080/360 = 3m/s.
Formal algebra should not be needed to see that, if their combined rate is 3m/s and the jogger's rate is 1m/s faster than the walker's, then the walker's rate is 1m/s and the jogger's rate is 2m/s.
Then with the jogger's rate equal to twice the walker's rate, it should be easy to determine that the jogger covers 2/3 of the 1080m, or 720m.
ANSWER: they meet 720m from Oakville Mall.
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