Iterations: WJEC GCSE Intermediate June 2022 Unit 2 Q14

Intermediate June 2022 Unit 2 Q14

WJECIntermediateCurrent spec6 marksIterations Volume & Surface Area

14. This cuboid has:

length = 5 cm
width = \(x\) cm
height = \((x^2 + 3)\) cm
volume = 132 cm³.

Cuboid with length 5, width x and height x² + 3

Diagram not drawn to scale

(a) Show that \(5x^3 + 15x = 132\). [1]

(b)

(i) A solution of the equation \[5x^3 + 15x = 132\] lies between 2 and 3.
Use the method of trial and improvement to find this solution correct to 1 decimal place.
You must show all your working. [4]

(ii) Hence, find the height of the cuboid. [1]

Height of the cuboid = ………………… cm

Mark scheme (a)

Scheme	Marks	Comments
Any intention of length × width × height = 132 e.g. \(5x\,(x^2 + 3) = 132\) \(5 \times x \times\ (x^2 + 3) = 132\) or \(5x \times (x^2 + 3) = 132\) or equivalent	B1	Must be = 132. May be seen in parts. Do not allow missing brackets e.g. \(5 \times x \times x^2 + 3 = 132\).

Mark scheme (b)

Scheme

Marks

Comments

(i)
One correct evaluation \(2 \leqslant x \leqslant 3\)

Correct evaluation regarded as enough to identify if <132 or >132. If evaluations not seen accept ‘too high’ or ‘too low’.
Look out for testing \(5x^3 + 15x - 132 = 0\) or \(x^3 + 3x = 26{\cdot}4\) or equivalent

2 correct evaluations \(2{\cdot}55 \leqslant x \leqslant 2{\cdot}75\),
(one value < 132, one value > 132)

2 correct evaluations \(2{\cdot}55 \leqslant x \leqslant 2{\cdot}65\),
(one value < 132, one value > 132)

\(x\)	\(5x^3 + 15x\)
2	70
2.1	77.805
2.2	86.24
2.3	95.335
2.4	105.12
2.5	115.625	2.55	121.1568…
2.6	126.88	2.65	132.798…..
2.7	138.915	2.75	145.234…...
2.8	151.76
2.9	165.445
3	180

(The rows 2.6, 2.7 and the entry 2.65 are shaded and bold in the original mark scheme.)

\(x = 2{\cdot}6\)

(ii)
An answer in the range 9·76 to 10·16 (cm)

Answer may be shown on the diagram.

FT ‘their 2·6’² + 3.
FT 132 ÷ (5 × ‘their \(x\)’).