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2011-8-16 16:01 jackson1998

equation(����������)



1)   If the length of each side of a square is increase by 3 cm ,then the perimter  becomes 4 times the
original perimter .Find
(a)     the original perimter of the square ,
(b)    the new area of the square
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2)   There are some marbles in a blue bag and a red bag ,and the number of the marbles in the blue bag is 2 times that of the red bag. If 6 marbles are taken out from the blue bag and put into the red bag, the
number of marbles in the blue bag will be 3 more than half that of thered bag .Find the total number of
marbles in the two bags.

[[i] �����̫�� jackson1998 �� 2011-8-16 16:19 �s�� [/i]]

2011-8-16 16:10 �ְ��[�@1

�^�_ #1 jackson1998 �����l



F.1 Maths???

2011-8-16 16:11 jackson1998
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2011-8-16 16:14 keroroEX

�^�_ #1 jackson1998 �����l



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2011-8-16 16:18 Amuro
1
a)
Let x be the original perimeter of the square
x+3*4 = 4x
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b)
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2
Let x be the number of marbles in red bag
2x-6 = [(x+6)/2]+3

2011-8-16 16:23 kingking
1a) let the orignal length be xcm
4x=orignal perimeter
4(x+3)=new perimeter
4x X4=4x+12
12=4x X3
4=4x
1=x
the orignal perimeter is 4x1=4(cm)
1b)
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let x be the amount of marbles in the red bag
2x-6=(x+6)/2+3
2x-9=(x+6)/2
4x-18=x+6
4x-24=x
3x=24
x=8

[[i] �����̫�� kingking �� 2011-8-16 16:34 �s�� [/i]]

2011-8-16 16:25 kimwong3252000
1)   If the length of each side of a square is increase by 3 cm ,then the perimter  becomes 4 times the
original perimter .Find
(a)     the original perimter of the square ,
Let x be the original length.
4(x+3)=16x
4x+12=16x
        x=1
original perimter=1x4=4cm
(b)    the new area of the square
4x4=16cm^2

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2)   There are some marbles in a blue bag and a red bag ,and the number of the marbles in the blue bag is 2 times that of the red bag. If 6 marbles are taken out from the blue bag and put into the red bag, the number of marbles in the blue bag will be 3 more than half that of the red bag .Find the total number of
marbles in the two bags.

Let x be the no of marbles in blue bag.

2x-6=(x+6)/2+3
4x-12=x+12
     x=8

blue: 16
red: 8

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2011-8-16 16:27 Amuro

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2011-8-16 16:37 jackson1998
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2011-8-16 20:02 UniCorn
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2011-8-16 20:39 Amuro

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2011-8-16 21:32 ���@��
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2011-8-16 21:33 jackson1998
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2011-8-16 21:54 Freedom_X10a
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