maths 問題，超急!!!!!20POINTS!!!!!!

Let $x,$y,$z be the cost of the three prizes respectively.

x+y+z=960-------------1
y=(4/5)x------------------2
z=(3/4)y------------------3

From 1: x=960-y-z
Sub x into 2: y=4/5(960-y-z)

y=4/5(960-y-z)-----------4
z=(3/4)y--------------------3

Sub z into 4: y=4/5{960-y-[(3/4)y)]}
y=768-4/5y-3/5y
2又2/5y=768
y=320

Sub y into 3: z=(3/4)*320
z=240

Sub z into 1: x+240+320=960
x=400



The price of three prize are $400,$320 and $240 respectively.

2010-10-25 22:33:13 補充：
早講

Let x be the number of apples and y be the number of children.

3y+80=x----------------1
4y-30=x-----------------2

Sub x into 2: 4y-30=3y+80
y=110
Sub y into 1: 330+80=x
x=410

The number of apples and chilren are 410 and 110 respectively.

1.x + 4/5x + (4/5x)3/4 = 960
2+2/5X = 960
12/5x = 960
x = 400


2.3y+80=x----------------1
4y-30=x-----------------2
320 x 3/4)=240

Let a =first prize ; b =second prize ; c = third prize

a+b+c=960
b=4/5 a
c=3/4 b

a+b+c=960 => a + (4/5 a )+ (3/4 b ) = 960
=> a + 4/5 a + 3/4(4/5 a) = 960
=> a + 4/5 a +3/5 a = 960
=> ( 5 + 4 + 3 )/5 a = 960
=> 12/5 a = 960
=> a = ( 960 x 5 )/12
=> a = 400

b=4/5 a => b =4/5 (400)
=> b = 320

c= 3/4 b => c = 3/4 (320)
=> c = 240

^ ^再見!

Let x be the value of the first prize

x + 4/5x + (4/5x)3/4 = 960
2+2/5X = 960
12/5x = 960
x = 400

.
. .The first prize is $400,the second prize is $(400 x 4/5)=320 and the third prize is $(320 x 3/4)=240

2010-10-25 22:23:58 補充：
呢個方法應該易D :)