Drawing & Using Graphs: WJEC GCSE Numeracy Higher November 2018 Unit 2 Q10

Numeracy Higher November 2018 Unit 2 Q10

WJECHigherCurrent spec9 marksDrawing & Using Graphs Forming Equations Volume & Surface Area

10. Gardeners can apply weedkiller to large areas of land by using a spray gun.
Weedkiller is stored in a large bottle that gardeners carry on their backs, and this feeds the spray gun.

A gardening company has designed the bottle shown below.
It consists of a hollow cylinder and cone that are joined.

Bottle made of a vertical cylinder with a cone joined underneath, point downwards

Diagram not drawn to scale

The base radius of the cone and the radius of the cylinder are both 9 cm.
The height of the cylinder is four times the vertical height of the cone.

The bottle has been designed so that it has a capacity of 10 litres.

From the formula list: Volume of cone \(= \dfrac{1}{3}\pi r^2 h\)

(a) Calculate the total height of the bottle. [7]

Total height of the bottle = …………… cm

(b) Weedkiller is poured into the bottle at a constant rate.
The cone is full of weedkiller after 2 seconds.
The bottle as a whole is full after 26 seconds.
Using the axes below, sketch a graph of the height of weedkiller in the bottle during the 26 seconds it takes to fill. [2]

Blank axes: height of weedkiller (cm) against time (seconds), with 2 and 26 marked on the time axis, four equally spaced notches on the height axis and a dashed line labelled Bottle full

Mark scheme (a)

Scheme	Marks	Comments
		Accept any notation or word(s) for ‘height’
(Volume of cone =) \(\dfrac{1}{3}\) × π × 9² × h	B1	Or 27πh, or 84…h
(Volume of cylinder =) π × 9² × 4h	B1	Or 324πh, or 1017…h
10 000 = \(\dfrac{1}{3}\)×π×9²×h + π×9²×4h	M2	(Or 10 000 = 351πh) M1 for either: the sum of 2 terms equated to 10 000 with 1 term correct, OR the sum of 2 correct terms equated to 10ⁿ where n is an integer and n > 0
h = \(\dfrac{10\,000 \times 3}{13 \times \pi \times 9^2}\) or \(\dfrac{10\,000}{1102\ldots}\) or \(\dfrac{10\,000}{351\pi}\)	m1	FT from M1
h = 9∙0 (cm) to 9∙1 (cm)	A1	No FT from M1m1
(Total height =) 45∙3 (cm) to 45∙5 (cm) or 45 (cm)	B1	FT from M1 or M2 for 5 × ‘their 9∙0(68…)’

Alternative

Alternative methods:

Scheme	Marks	Comments
\(\dfrac{1}{3}\) × π × 9² × \(\dfrac{h}{4}\)	B1	Or 6.75πh, or 21…h
π × 9² × h	B1	Or 81πh, or 254…h
10 000 = \(\dfrac{1}{3}\)×π×9²×\(\dfrac{h}{4}\) + π×9²×h	M2	(Or 10 000 = 351πh/4) M1 for either: the sum of 2 terms equated to 10 000 with 1 term correct, OR the sum of 2 correct terms equated to 10ⁿ where n is an integer and n > 0
h = \(\dfrac{10\,000 \times 12}{13 \times \pi \times 9^2}\) or \(\dfrac{10\,000}{275\ldots}\) or \(\dfrac{40\,000}{351\pi}\)	m1	FT from M1
h = 36∙2 (cm) to 36∙3 (cm)	A1	No FT from M1m1
45∙3 (cm) to 45∙5 (cm) or 45 (cm)	B1	FT from M1 or M2 for \(\dfrac{5}{4}\) × ‘their 36∙2(746…)’

Alternative

Alternative methods:

Scheme	Marks	Comments
Vol of cone: vol of cylinder = 1:12 10(000) ÷ 13	B1 M1	This M1 implies previous B1
(vol of cone =) 0∙76(9… litres) or 0∙77 or 769(∙…cm³) or 770 (cm³)	A1
(vol of cone =) \(\dfrac{1}{3}\) × π × 9² × h	B1	Or 27πh, or 84…h
\(\dfrac{1}{3}\) × π × 9² × h = 769(∙…)	m1
(h =) 9∙0 (cm) to 9∙1 (cm)	A1
(Total height =) 45∙3 (cm) to 45∙5 (cm) or 45(cm)	B1	FT 5 × ‘their 9∙0(68…)’ provided previous M1 awarded

Mark scheme (b)

Scheme	Marks	Comments
A curve of decreasing gradient for the first 2 seconds with positive gradient at time 2 seconds A connecting straight line of constant gradient for the last 24 seconds up to the ‘container full’ line The intention for the 2 sections of their graph to meet at (2, 1^st notch)	B2	Allow the intention of the straight line B2 for meeting all 3 conditions B1 for meeting any 2 conditions, with their graph starting at (0,0) and ending at (26, full)