Circles & Sectors: WJEC GCSE Numeracy Higher November 2024 Unit 2 Q8

Numeracy Higher November 2024 Unit 2 Q8

WJECHigherCurrent spec17 marksCircles & Sectors Manipulating Formulae/ Changing the Subject Prisms

(a)

Signs-R-Us is a company that makes national speed limit signs for roads.
The national speed limit sign is white with a black strip across it.

Signs-R-Us needs to calculate the area of a sign that is white and the area that is black.

The diagram below shows some of the dimensions of the national speed limit sign.
The sign is a circle of radius 30 cm, centre O.
The centre of the black region is also at O.
The black region is not a rectangle.

Circle with centre O and a shaded diagonal strip across it; two dashed radii of 30 cm from O to the ends of the strip edges on the circle, with an angle of 160° between them at O

Diagram not drawn to scale

Calculate the area of the sign that is white.
Hence, calculate the area of the sign that is black. [6]

Area that is white = …………… cm\(^2\)
Area that is black = …………… cm\(^2\)

(b)

Signs-R-Us makes temporary road signs for roadworks.
These signs need to be moved easily.
For each sign, Signs-R-Us makes a base with a vertical hole drilled through it, so that a pole can be placed inside.

A road sign on a pole standing in a base; the base is labelled

Diagram not drawn to scale

Signs-R-Us makes these bases for different-sized poles.
The radius of a hole is related to the mass of a base.
This is shown by the formula,

\[r = \sqrt[3]{\dfrac{M}{125(46 - 3\pi)}}\]

where:
\(r\) is the radius in centimetres, and
\(M\) is the mass in kilograms.

Rearrange this formula to make \(M\) the subject. [2]

(c)

As well as the national speed limit sign, Signs-R-Us makes 3 other road signs.
The table below shows the number of road signs Signs-R-Us hopes to make every month.

Type of sign	Stop	Give way	One way	National speed limit
Number to be made each month	148	185	115	87

Each month, Signs-R-Us plans to take a stratified sample of 60 signs to check the quality of their production.

Calculate the number of each type of road sign that Signs-R-Us should include in their sample.
You must show all your working. [3]

Type of sign	Stop	Give way	One way	National speed limit
Number in the sample	……………	……………	……………	……………

(d)

Signs-R-Us also makes a speed bump, as shown below.
It has a uniform cross-sectional area, consisting of a quarter circle, a rectangle and a right-angled triangle.

Speed bump: a prism 2 m wide whose cross-section is a quarter circle of radius h, then a rectangle of height h (Length of the rectangle marked along the top), then a right-angled triangle sloping down to the ground; the total length of the speed bump is marked along the bottom

Diagram not drawn to scale

The height of the speed bump is shown as \(h\) in the diagram above.
The length of the rectangle is 4 times the height.
The total length of the speed bump is 8 times the height.

The width of the speed bump is 2 m, and its volume is 0·1 m\(^3\).

Calculate the height of the speed bump. [6]

From the formula list: Volume of prism = area of cross-section × length; In any triangle \(ABC\): Area of triangle \(= \dfrac{1}{2}ab\sin C\)

Mark scheme (a)

Scheme	Marks	Comments
(Area that is white =) \(2 \times \left(\dfrac{160}{360} \times \pi \times 30^2 - \dfrac{1}{2} \times 30^2 \times \sin 160\right)\) or equivalent [ 2×((1256 to 1257 (or 400\(\pi\))) – (153.9(09…) to 154))]	M3	Note: 2 × \(\dfrac{(\sin 80 \times 30) \times (\cos 80 \times 30)}{2}\) is equivalent to \(\dfrac{1}{2} \times 30^2 \times \sin 160\) M2 for any one of the following: \(2 \times \dfrac{160}{360} \times \pi \times 30^2 - \dfrac{1}{2} \times 30^2 \times \sin 160\) (= 2358 to 2360.1) \(\dfrac{160}{360} \times \pi \times 30^2 - 2 \times \dfrac{1}{2} \times 30^2 \times \sin 160\) (= 948 to 949.2) \(\dfrac{160}{360} \times \pi \times 30^2 - \dfrac{1}{2} \times 30^2 \times \sin 160\) (= 1102 to 1103.1) M1 for any one of the following: sight of \(\dfrac{160}{360} \times \pi \times 30^2\) (= 1256 to 1257 (or 400\(\pi\))) sight of \(\dfrac{1}{2} \times 30^2 \times \sin 160\) (= 153.9(09…) to 154)
= 2204 to 2206.2 (cm\(^2\))	A1	CAO
(Area that is black =) \(\pi \times 30^2\) − (2204 to 2206.2) [(2826 to 2827.8) − (2204 to 2206.2)]	M1	FT their white area provided at least M1 previously awarded OR a correct full alternative method (see below)
= 621.6 to 622 (cm\(^2\))	A1

Alternative

Alternative method:

Scheme	Marks	Comments
(Area that is black =) \(2 \times \left(\dfrac{180-160}{360} \times \pi \times 30^2 + \dfrac{1}{2} \times 30^2 \times \sin 160(^\circ)\right)\) [ 2×((157 to 157.1 (or 50\(\pi\))) + (153.9(09…) to 154))]	M3	Note: 2 × \(\dfrac{(\sin 80 \times 30) \times (\cos 80 \times 30)}{2}\) is equivalent to \(\dfrac{1}{2} \times 30^2 \times \sin 160\) M2 for any one of the following: \(2 \times \dfrac{180-160}{360} \times \pi \times 30^2 + \dfrac{1}{2} \times 30^2 \times \sin 160\) (= 467.9 to 468.2) \(\dfrac{180-160}{360} \times \pi \times 30^2 + 2 \times \dfrac{1}{2} \times 30^2 \times \sin 160\) (= 464.8 to 465) \(\dfrac{180-160}{360} \times \pi \times 30^2 + \dfrac{1}{2} \times 30^2 \times \sin 160\) (= 310.9 to 311.1) M1 for any one of the following: sight of \(\dfrac{180-160}{360} \times \pi \times 30^2\)(= 157 to 157.1(or 50\(\pi\))) sight of \(\dfrac{1}{2} \times 30^2 \times \sin 160\) (= 153.9(09…) to 154)
= 621.8 to 622.2 (cm\(^2\))	A1	CAO
(Area that is white =) \(\pi \times 30^2\) − (621.8 to 622.2) [(2826 to 2827.8) − (621.8 to 622.2)]	M1	FT their black area provided at least M1 previously awarded OR a correct full alternative method (see above)
= 2204.2 to 2205.6(cm\(^2\))	A1

Mark scheme (b)

Scheme	Marks	Comments
\(M = 125r^3(46 - 3\pi)\) or \(M = 5750r^3 - 375\pi r^3\) or \(M = 4571.75r^3\) to \(4572.5r^3\)	B2	Mark final answer B1 for \(r^3 = \dfrac{M}{125(46 - 3\pi)}\) or \(r^3 = \dfrac{M}{5750 - 375\pi}\) or \(r^3 = \dfrac{M}{4571.75 \text{ to } 4572.5}\) or B1 for \(125r^3(46 - 3\pi)\) or \(5750r^3 - 375\pi r^3\) or \(4571.75r^3\) to \(4572.5r^3\) (omitting M =)

Mark scheme (c)

Scheme	Marks	Comments
60 × (number made) ÷ (148+185+115+87) or 60 × (number made) ÷ 535	M1	Sight of this calculation for any type of sign Allow (number of signs to be made) ÷ 8·9(166…)
(List of unrounded answers) 16.5(98…) or 16.6, 20.7(47…), 12.8(97…) or 12.9, 9.7(57…) or 9.8	A1	OR A1 for final answers of 17, 21, 13, 10
(Number in the sample =) 16, 21, 13, 10	A1	Award M1A1A1 for 16, 21, 13, 10 provided no incorrect work seen If M1A0 awarded, A1 on FT from their unrounded answers for equivalent difficulty i.e. provided: their correctly rounded answers sum to 59 or 61 AND the correct numbers in the sample are given from their unrounded answers, including any decisions regarding rounding up or down AND the numbers in the sample add to 60

Mark scheme (d)

Scheme	Marks	Comments
Sight of base of triangle = 3h	B1	Any letter or word may be used for the height Needs to be convincing. May be seen on diagram
\(\left(\dfrac{\pi \times h^2}{4} + 4h^2 + \dfrac{3h^2}{2}\right) \times 2 = 0.1\) or \(\dfrac{2\pi \times h^2}{4} + 8h^2 + \dfrac{6h^2}{2} = 0.1\) or (12.57 to 12.571)h\(^2\) = 0.1 or equivalent	M2	Ignore a place value error from an incorrect attempt to convert m\(^3\) into cm\(^3\) and/or m into cm for M and m marks only but A0 Allow omission of ×2 for M2 or M1 and possibly m1 M1 for: the sum of appropriate terms equated to 0.1, with no more than 1 error in the terms \(\left(\dfrac{\pi \times h^2}{4} + 4h^2 + \dfrac{3h^2}{2}\right) \times 2\) or equivalent
h\(^2\) = \(\dfrac{0.1}{2(\frac{\pi}{4} + 5.5)}\) or equivalent	m1	FT if possible from M1 provided h\(^2\) in every term Note: \(\pi\)/4 + 5.5 = 6.285 to 6.2855 \(\pi\)/2 + 11 = 12.57 to 12.571
h = 0.089 to 0.0892 (m) or equivalent	A2	CAO Ignore an incorrect attempt to convert to cm or mm Accept 0.09 (m) from correct working A1 for h = \(\sqrt{\dfrac{0.1}{2(\frac{\pi}{4} + 5.5)}}\) or \(\sqrt{\dfrac{1}{5\pi + 110}}\) or h = \(\sqrt{0.00795\ldots}\)

Alternative

Alternative method:

Scheme	Marks	Comments
Base of trapezium = 7h	B1	Any letter or word may be used for the height Needs to be convincing. May be seen on diagram
\(\left(\dfrac{\pi \times h^2}{4} + \dfrac{(4h + 7h) \times h}{2}\right) \times 2 = 0.1\) or \(\dfrac{2\pi \times h^2}{4} + 11h^2 = 0.1\) or (12.57 to 12.571)h\(^2\) = 0.1 or equivalent	M2	Ignore a place value error from an incorrect attempt to convert m\(^3\) into cm\(^3\) and/or m into cm for M and m marks only but A0 Allow omission of ×2 for M2 or M1 and possibly m1 M1 for: the sum of appropriate terms equated to 0.1, with no more than 1 error in the terms \(\left(\dfrac{\pi \times h^2}{4} + \dfrac{(4h + 7h) \times h}{2}\right) \times 2\) or equivalent
h\(^2\) = \(\dfrac{0.1}{(\frac{\pi}{2} + 11)}\) or equivalent	m1	FT if possible from M1 provided h\(^2\) in every term Note: \(\pi\)/2 + 11 = 12.57 to 12.571
h = 0.089 to 0.0892 (m) or equivalent	A2	CAO Ignore an incorrect attempt to convert to cm or mm Accept 0.09 (m) from correct working A1 for h = \(\sqrt{\dfrac{0.1}{2(\frac{\pi}{4} + 5.5)}}\) or \(\sqrt{\dfrac{1}{5\pi + 110}}\) or h = \(\sqrt{0.00795\ldots}\)