O-LevelMathematicsAlgebra and graphs4024/22 · Oct/Nov 2021 · Q914 Marks

The diagram shows an open box in the shape of a cuboid. The height of the box is x cm. The width of the box is 5 cm more than its height. The length of the box is two times its width. [Figure showing an open cuboid with height x, width x+5, length 2(x+5)]

(a)Write down expressions, in terms of x, for the width and the length of the box.[2]
(b)The external surface area of the open box is 210 cm². Form an equation in x and show that it simplifies to 4x²+25x-80 = 0 .[4]
(c)Solve the equation 4x²+25x-80 = 0 . Show your working and give your answers correct to 2 decimal places.[3]
(d)Calculate the volume of the box.[2]
(e)The box is filled with chocolates. The mass of the chocolates is 250 g, correct to the nearest 10 grams. The total mass of the box and chocolates is 262 g, correct to the nearest gram. Calculate the lower bound of the mass of the box.[3]

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 2021). The mark or mark type is shown in brackets.

[2] B1 for [W = ] x + 5 or B1FT for [L = ] 2×their algebraic W
[M2] FT their algebraic expressions in x for length and width B1FT for two different areas seen e.g. two of (x + 5)×2(x+5), x(x + 5), x × 2(x + 5) or 2((x + 5)×2(x+5) + x(x + 5) + x × 2(x + 5)) [M1] Set equal to 210 and expansion of brackets. Must have three different areas from width and length of form ax + b, a and b ≠ 0 [A1] Correct simplification to 4x²+25x-80=0
[B2] B1 for √25²-4×4×-80 oe or (-25 ± [...])/2×4 oe or (-25/8 ± √[25/8]² - (-80/4)) oe or (x + 25/8)² = (25/8)² - (-80/4)) [B1] 2.33 and -8.58
[2] M1 for their 2.33 × (their 2.33 + 5) × 2(their 2.33 + 5)
[3] B1 for 255 and 261.5 seen M1 for their 261.5 – their 255

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.