O-LevelMathematicsAlgebra and graphs4024/11 · Oct/Nov 2025 · Q2410 Marks

The diagram shows a rectangle and a triangle. [Figure 24.1] The length of the rectangle is x cm and the area of the rectangle is 30 cm². The triangle has sides of length (x-1) cm, (x+2) cm and (2x-3) cm.

(a)Write down an expression, in terms of x, for the width of the rectangle.[1]
(b)The perimeter of the rectangle is equal to the perimeter of the triangle. Form an equation in x and show that it simplifies to x²-x-30 = 0.[4]
(c)Solve the equation x²-x-30 = 0.[3]
(d)Work out the perimeter of the rectangle.[2]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

[1] (a) 30/x
[M1] (b) 2(their 30/x + x) = x-1+x+2+2x-3 or 60+2x² = 4x²-2x oe
[FT] their 30/x
[M2] Correct removal of their algebraic fraction and their algebraic brackets
[M1 dep] for correctly reducing their equation to 4 terms e.g. 2x + 60/x = 4x-2 oe or for correct removal of their algebraic fraction 60+2x² = x(4x-2) oe dependent on equation with fraction with algebraic denominator
[A1] After full working shown and no errors or omissions
[3] (c) 6, -5
[M2] for (x-6)(x+5) or x = (1±√121)/2 or M1 for x(x+5)-6(x+5) or x(x-6)+5(x-6) or (x+a)(x+b) with ab=-30 or a+b=-1 or x = (1±√1-4*1*(-30))/2*1
[2] (d) 22
[M1] for 2(their 30/x + their 6) oe or 4x/their 6 - 2 oe

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