Hard math question -- I think

[From: ] [author: ] [Date: 12-05-01] [Hit: ]

t where they meet will also be the same :-(If Z ran the distance, d then L has to run the distance (1/4) + d t = d / 40 = [(1/4) + d ] / 505d = 4.[(1/4) + d] 5d = 1 + 4dd = 1At the point (1/4) + 1 = 5 / 4 miles = 1.25 miles from where they started, the L will overtake the Z.---------------------------------------…Hope this helps!......

t = d / 50 = [(1/4) - d ] / 40 ------------ multiply 200 both sides

4d = 5.[(1 / 4) - d]

4d = (5/4) - 5d ----------------------------- add 5d both sides

9d = 5 / 4 ----------------------------------- divide 9 both sides

d = 5 / 36

At the point 5 / 36 miles = 0.138 miles from where they started, the L will overtake the Z.
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If the L is on the chase the Z and they start running at the same, then the time, t where they meet will also be the same :-(
If Z ran the distance, d then L has to run the distance (1/4) + d
t = d / 40 = [(1/4) + d ] / 50
5d = 4.[(1/4) + d]
5d = 1 + 4d
d = 1
At the point (1/4) + 1 = 5 / 4 miles = 1.25 miles from where they started, the L will overtake the Z.
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Hope this helps!

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Think of it this way. When the lion overtakes the zebra they will be at the same position.
The lions position is always 50t were t=time in hours. The zebra's has a head start so it's position will be 1/4 +40t. Setting them equal to each other to get 50t=40t+1/4. Solve it and you should be able to derive the rest.

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when the lion ears the zebra in

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