Percentages: WJEC GCSE Numeracy Intermediate June 2025 Unit 2 Q6

Numeracy Intermediate June 2025 Unit 2 Q6

WJECIntermediateCurrent spec7 marksPercentages Volume & Surface Area

6. Nina sells flowers.

(a)

At the start of last week, Nina spent £600 buying flowers to sell.

By the end of the week, she had taken £1140 from the sale of the flowers.

Calculate Nina’s percentage profit from the sale of these flowers. [2]

(b)

Nina wants to display her flowers.
She buys two glass vases from a website.
The two vases are different shapes.

Photographs of two glass vases: a cuboid and a cylinder

Diagrams not drawn to scale

Each vase can hold 1648 cm\(^3\) of water when it is full.

(i) The vase in the shape of a cuboid has a base measuring 10 cm by 8 cm.
Calculate the height of this vase. [2]

(ii)

The radius of the base of the cylindrical vase is 6 cm.

The website states that both of these vases have the same height.
Is the website correct?
You must show all your working. [3]

The website is correct ☐ The website is incorrect ☐

Mark scheme (a)

Scheme	Mark	Comments
(Percentage profit =) \(\dfrac{1140 - 600}{600}\) (× 100) or \(\dfrac{1140}{600}\) (× 100) – 1 (× 100) or 0.9 (× 100)	M1
90(%)	A1	If no marks, award SC1 for an answer of 190(%) (from \(\dfrac{1140}{600}\) × 100)

Mark scheme (b)

Scheme	Mark	Comments
(i) 10 × 8 × height = 1648 or (height =) \(\dfrac{1648}{10 \times 8}\)	M1	Allow for intention, e.g. 1648 ÷ 10 × 8, 10 × 8 × …… = 1648
20.6 (cm)	A1	Accept 21(cm) and allow 20(cm) provided not from incorrect working Allow an embedded answer.
(ii) Appropriate method, e.g. (Volume cylinder) \(\pi \times 6^2 \times \dfrac{1648}{10 \times 8}\) or \(\pi \times 6^2 \times 20.6\) (compared with 1648 cm\(^3\)) (Height cylinder) \(\pi \times 6^2 \times\) height = 1648 or (height =) \(\dfrac{1648}{\pi \times 6^2}\) (Area X-sections) \(\pi \times 6^2\) with 10 × 8 (Radius cylinder) \(\sqrt{\dfrac{10 \times 8}{\pi}}\) or \(\sqrt{\dfrac{1648}{\pi \times 20.6}}\) (compared with 6 cm)	M1	FT from (b)(i) for ‘their height of cuboid’ (‘their 20.6’) Accept 10 × 8 = 80 seen in (b)(i), not for embedded 10 × 8 alone
Correct evaluation for the method used, e.g. (Volume cylinder)(range) 2328(cm\(^3\)) to 2331(cm\(^3\)) (Heights) (range) 14.5(cm) to 15(cm) with (\(\dfrac{1648}{10 \times 8}\) =) 20.6 (cm) (Area X-sections) approximately 113 (cm\(^2\)) with (10 × 8 =) 80 (cm\(^2\)) (Radius cylinder) approximately 5 (cm)	A1	Allow height 14(cm), provided not from incorrect working. Answer for height may be from a trial
‘Incorrect’ stated or indicated	A1	Depends on M1 A1 previously awarded If no marks, due to error in not using \(\pi r^2\) as the base area (in area or volume calculation), however ‘their \(\pi r^2\)’ does include a product including \(\pi\) and 6 (or embedded as 12), but otherwise method, accuracy and conclusion correct for ‘their \(\pi r^2\)’, award SC1