O-LevelMathematicsAlgebra and graphs4024/22 · Oct/Nov 2019 · Q910 Marks

ABCD is a rectangle with AB = 10cm and AD = 12cm. PQ is parallel to AB and RS is parallel to AD. Y is the point of intersection of PQ and RS. BR = DP = x cm. A smaller rectangle and a square are shaded.

(a)(i)Write down an expression, in terms of x, for AR,[1]
(a)(ii)AP.[1]
(b)Area of shaded rectangle ARYP / Area of rectangle ABCD = 3/4. Show that x²-22x+30 = 0.[3]
(c)Solve the equation x²-22x+30 = 0. Show your working and give your answers correct to 2 decimal places.[3]
(d)Find the total unshaded area in rectangle ABCD.[2]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

[1] 10-x final answer
[1] 12-x final answer
[3] (10-x)(12-x)/(10×12) = 3/4
[3] M1 FT their algebraic AR and AP
[3] 120-12x-10x+x^2 seen
[3] M1 FT Correct expansion of their brackets
[3] Correct expansion and rearrangement to x^2-22x+30 = 0
[3] A1 Dep on correct AR and AP
[3] x = (-(-22)±√((-22)^2-4×1×30))/(2×1)
[3] B2
[3] B1 for (-22)^2-4×1×30 or for (-(-22)±√their 364)/(2×1)
[3] 1.46, 20.54
[3] B1
[2] 27.9 or 27.85 to 27.87...
[2] M1 for use of their 1.46 to find an area e.g. 1.46(10–1.46), 1.46(12–1.46), 1.46^2, etc.

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.