Maths. Quadratic equations (F4) >>10marks<

Q.2
Let number of children be x. So each will get 200/x chocolate.
For (x-2) children, each will get 200/(x-2) which is 5 more. That is
200/(x-2) - 200/x = 5
40/(x-2) - 40/x = 1
40x - 40(x-2) = x(x-2)
40x - 40x +80 = x^2 - 2x
x^2 - 2x - 80 = 0
(x+8)(x-10)= 0
x= -8 (rej.) and 10. So number of children = 10.
Q.3
Let sides of the box be x, y and z.
Therefore, x+2 = y+3 = z+5 since sides of a cube are equal.
So y = x-1 and z= x-3............(1)
Since volume increases by 720, therefore,
(x +2)(y+3)(z+5) - xyz = 720
(xy + 2y +3x + 6)(z+5) - xyz = 720
xyz +2zy + 3xz + 6z + 5xy + 10y + 15x + 30 - xyz = 720
2zy + 3xz + 6z + 5xy + 10y + 15x = 690...............(2)
Substitute (1) into (2), we get
2(x-1)(x-3) + 3x(x-3) + 6(x-3) + 5x(x-1) + 10(x-1) + 15x = 690
2x^2 + 6 - 8x + 3x^2 - 9x + 6x - 18 + 5x^2 - 5x + 10x - 10 + 15x = 690
10x^2 +9x -22 = 690
10x^2 + 9x - 712 = 0
(10x+89)(x-8)= 0
x= -89/10(rej) and 8.
Therefore, x=8, y= 7 and z = 5.
(a) Capacity of box is 8 x 7 x 5 = 280cm^3.
(b) Surface area = 2 x 8 x 7 + 2 x 8 x 5 + 2 x 5 x 7 = 112 + 80 + 70 = 262cm^2.
Q.1
Let length = x and width = y.
So xy= 864..............(1)
New dimension is (x+4)(y-3), since area is unchanged, that means
(x+4)(y-3) - xy = 0
xy + 4y - 3x - 12 -xy = 0
4y -3x -12 = 0
y= (3x + 12)/4. Substitute into (1)
x(3x+12)/4 = 864
3x^2 + 12x - 3456= 0
x^2 + 4x - 1152 = 0
(x-32)(x + 36) = 0
x= 32 and -36(rej.)
So y = 864/32 = 27. Dimension of field = 32m x 72m.