I found it easy, apart from the question 20 (the rectangle one), so I'm hoping to be in the 90s.

Could anyone explain that question though? Most of my year couldn't do it, apparently it is to do with simultanious equations though?

The length of the rectangle was x, there was a diagonal cutting it into two triangles which was 8 and the perimeter was 20. Show that x^2 - 10x + 18 = 0.

this question is a lot easier than it looks. If you draw out the rectangle with everything you know on it you'll notice you don't know one of the sides. Lets call this side 'y'

You know that the perimeter of the rectangle is 20 therefore X+X+Y+Y=20, so 2X+2Y=20. this can be rearanged to get Y=10-X.

You also know that the hypotenuse of each triangle is 8. As it is a right-angled triangle you can use pythagoras to form another equation. X^2+Y^2=64. then you just solve the simultaneous equations by substituting in Y=10-X.