Circles & Sectors: WJEC GCSE Numeracy Higher June 2018 Unit 1 Q14

Numeracy Higher June 2018 Unit 1 Q14

WJECHigherCurrent spec8 marksCircles & Sectors Direct & Inverse Proportion

14. The diagram shows the simplified model of part of an engine.
It shows a belt which runs around three circular cogs.
The engine rotates Cog 1.
Cog 1 rotates the belt, which then makes Cogs 2 and 3 rotate.

Belt around three cogs: straight sections AB = 12 cm, CD = 34 cm, EF = 29 cm; right angle at the centre of Cog 1 between radii to A and F; angle 120° at the centre of Cog 2 between radii to B and C; angle 150° at the centre of Cog 3 between radii to E and D

Diagram not drawn to scale

AB, CD and EF are straight sections of the belt.
AB = 12 cm, CD = 34 cm and EF = 29 cm.
The belt bends around the outer edges of the circular cogs, represented by the arcs BC, DE and AF.
The dimensions of the cogs are:

radius of Cog 1 = 6 cm
radius of Cog 2 = 4·5 cm
radius of Cog 3 = 3 cm

(a) What is the length of arc AF in terms of π?
Circle your answer. [1]

\(2\pi\) \(3\pi\) \(6\pi\) \(4\pi\) \(\dfrac{3\pi}{2}\)

(b) Calculate the total length of the belt.
Give your answer in terms of π in its simplest form. [4]

Total length of the belt = …………………… cm

(c)

Elen notices that when Cog 3 makes two revolutions, Cog 1 makes only one revolution, because the radius of Cog 3 is half the radius of Cog 1.

In one minute, Cog 3 makes 2400 revolutions.
Calculate the number of revolutions Cog 2 will make in one minute. [3]

Scheme	Marks	Comments
3π	B1

Mark scheme (b)

Scheme	Marks	Comments
Sight of \(\dfrac{120}{360}\)×2×π×4·5 OR \(\dfrac{150}{360}\)×2×π×3	B1	Or equivalents
\(\dfrac{120}{360}\)×2×π×4·5 + \(\dfrac{150}{360}\)×2×π×3	M1	Or equivalent
= 3π + \(^{5}/_{2}\)π OR 3π + 2·5π OR \(\dfrac{11\pi}{2}\) or 5·5π	A1	May be implied in further working or final answer
(Length of belt =) 75 + \(\dfrac{17\pi}{2}\) or 75 + 8·5π or \(\dfrac{150 + 17\pi}{2}\)	B1	FT their answer to (a) and ‘their 11π/2’ provided M1 awarded Needs to be in its simplest form If no marks awarded, SC2 if 3π/2 given in (a), leading to an answer of 75 + 17π/4 or 75 + 4·25π SC1 if 3π/2 given in (a), leading to an unsimplified version of 75 + 17π/4

Mark scheme (c)

Scheme	Marks	Comments
Use of \(\dfrac{3}{4{\cdot}5}\) or \(\dfrac{4{\cdot}5}{3}\) or equivalents with 2400	B1	e.g. ratio of 3 : 4·5 or 4·5 : 3 or equivalents B0 if 1·5 (4.5/3) comes from 3 + 1·5 = 4·5
2400 × \(\dfrac{3}{4{\cdot}5}\) or equivalent	M1
= 1600 (revolutions)	A1	Alternative method: B1 for use of 6/4.5 or 4.5/6 with 1200 M1 for 1200 × \(\dfrac{6}{4{\cdot}5}\) or equivalent A1 for 1600 (revolutions)