S.1 Rate 急!急!急!

Let the width of the inner track be a m;
Let the length of the inner track be b m;
Let the width of the path be x m;
Thus, the width of the outer track = a + 2x m ;
Thus, the length of the outer track = b + 2x m;
Accordingly, the perimeter of the inner track = 2a + 2b m;
Accordingly, the perimeter of the outer track = [2 (a + 2x) + 2(b + 2x)] m;
Thus, Perimeter difference between outer and inner tracks =
= [2 (a + 2x) + 2(b + 2x)] m - (2a + 2b) m
= 2a + 4x + 2b + 4x - 2a - 2b m
= 8x m
That is, 8x / 18 = (5 * 1000) / (60 * 60)
x = (5 * 1000 * 18) / [(60 * 60) * (8)]
x = 3.125 m
Ans.: The width of the path is 3.125m.
I hope I can help.

2007-05-28 09:48:53 補充：
(path) is used instead in order not to confuse with the inner and upper (tracks).Hope you can understand the equation 8x/18 = (5000)/(3600);It is always (distance / time) (left hand side = right hand side).

Let y be the distance from the centre of the park to the outer track and
x be the width of the track.
Hence the distance from the centre of the park to the inner track is (y-x).

His running speed = 5km/hr = 25m/18s

The total length of outer track
=(2y)(4)
=8y

The total length of the inner track
=[ 2 (y - x) ] (4)
=8y-8x

Since it takes him 18s more to finish running a round along the outer track than the inner one.
Therefore , we have
(8y) / (25m/18s) -(8y-8x) / (25m/18s) = 18s
8y-8y+8x = (18s )(25m/18s)
8x = 25m
x=3.125m

i.e. The width of the track is 3.125m.