F4 math problem~!!

Let x km/h and y km/h be the speed of Eric and Frank respectively.

(a)Then,
90/y - 90/x = 1/2.................(1)
90/x = 30/y + 60/(y+10)......................(2)

(b)Sub (2) into (1),
90/y - (30/y + 60/(y+10)) = 1/2
90/y - 30/y - 60/(y+10) = 1/2
60/y - 60/(y+10) = 1/2..............(3)

(c)From (3),
60/y - 60/(y+10) = 1/2
(60/y - 60/(y+10))(2y(y+10)) = (1/2)(2y(y+10))
120(y+10) - 120y = y(y+10)
120y+1200 - 120y = y^2 + 10y
1200 = y^2 +10y
y^2 + 10y -1200 = 0
(y - 30)(y + 40) = 0
y = 30 or y = -40 (rejected)
therefore y = 30

Frank takes (90/30) hours = 3 hours to cycle from Kowloon to Tung Chung.
As Eric would arrive 30mins earlier than Frank, Eric should take 2.5hours to cycle from Kowloon to Tung Chung.

(d)As y = 30, Frank need 1hour to cycle 30km.

Eric travels:
(90/2.5) * 1
=39(km)

39 - 30
=9
That is the distance between two boys when Frank starts to increase his speed is 9km.

Let the speed of Eric and Frank be x km/hr and y km/hr respectively

Part (a)
30 minutes = 0.5 hr
This implies 90/y - 90/x = 1/2 ---- (i)
And 90/x = 30/y + (90-30)/(y+10)
i.e. 90/x = 30/y + 60/(y+10) ---- (ii)

Part (b)
Eliminate 90/x from (i) and (ii), we have
90/y - 1/2 = 30/y + 60/(y+10)
i.e. 60/y - 60/y+10 = 1/2

Part (c)
From (b), 60/y - 60/y+10 = 1/2
i.e. y^2 + 10y - 1200 = 0
y = 30 or -40(rejected)
Put y = 30 into (i)
90/30 - 90/x = 1/2, implies x = 36
So the time taken for Eric to cycle from Kowloon to Tung Chung = 90/36 = 2.5hrs

Part (d)
Time taken for Frank to cycle 30km = 30/30 = 1hr
Distance travelled by Eric in 1hr = 1 * 36 = 36km
So distance between the two boys when Frank starts to increase his speed = 36 - 30 = 6km

Hope the above information can help you =)

2006-12-20 18:48:43 補充：
Your answer in part (d) is actually wrong...=.=
And you comment that you are a maths teacher...
I doubt the reliability