Grouped Data: WJEC GCSE Numeracy Higher June 2024 Unit 2 Q3

Numeracy Higher June 2024 Unit 2 Q3

WJECHigherCurrent spec12 marksGrouped Data Prisms Pythagoras

(a)

50 people living by the sea were asked how often they went for a walk along the sea wall each week.

The results were as follows:

Number of walks each week	Frequency
0 to 2	8
3 to 5	12
6 to 8	20
9 to 13	4
14 to 18	6

Calculate an estimate of the mean number of walks per person each week. [4]

(b)

High tide in the morning is, on average, 35 minutes later each day.
The morning high tide on 3rd March was at 08:03.
At what time was the morning high tide on 1st March? [1]

(c)

A new concrete sea-defence wall is to be built.
The sea-defence wall will have a uniform cross-section.
A sketch of the new wall is shown below.

Sketch of a prism of length 50 m whose cross-section is a right-angled trapezium: parallel sides 12·6 m (top) and 18·8 m (bottom), a vertical side at right angles to the base, and a sloping side of 7·6 m; a dashed vertical line from the top of the sloping side meets the base at a right angle

Diagram not drawn to scale

Calculate the volume of concrete needed to make this new sea-defence wall.
You must show all your working. [7]

From the formula list: Area of trapezium \(= \dfrac{1}{2}(a+b)h\); Volume of prism = area of cross-section × length

Mark scheme (a)

Scheme	Mark	Comments
Midpoints 1, 4, 7, 11, 16	B1
1 × 8 + 4 × 12 + 7 × 20 + 11 × 4 + 16 × 6 (= 8 + 48 + 140 + 44 + 96) ( = 336)	M1	FT ‘their midpoints’ provided at least 4 lie within the appropriate group, including bounds throughout Use of lower bounds gives 276 Use of upper bounds gives 396
÷ 50	m1
6.72 or 6.7 (walks)	A1	Allow rounded to 7 (walks) from appropriate working Use of lower bounds gives 5.5(2 walks) or 6 (walks) Use of upper bounds gives 7.9(2 walks) or 8 (walks)

Scheme	Mark	Comments
06(:)53 or 6(:)53 a.m.	B1	Allow 06(:)53 a.m. or 6(:)53 Do not accept (0)6(:)53 p.m.

Mark scheme (c)

Scheme	Mark	Comments
(Height)\(^2\) = 7.6\(^2\) – (18.8 – 12.6)\(^2\) or (Height)\(^2\) = 7.6\(^2\) – 6.2\(^2\) or cos\(^{-1}\)(\(\dfrac{18.8-12.6}{7.6}\)) = 35(.3345…)° and sin 35(.33…)° = \(\dfrac{\text{Height}}{7.6}\) or tan 35(.33…)° = \(\dfrac{\text{Height}}{6.2}\)	M2	M1 for sight of 18.8 – 12.6 with any attempt to use Pythagoras’ Theorem (including summing rather than subtraction) cos\(^{-1}\)(\(\dfrac{18.8-12.6}{7.6}\)) = 35(.3345…)° and an attempt to use sine or tan
Height\(^2\) = 19.32 or (Height = ) √19.32 or (Height =) 7.6 × sin 35(.33…)° or (Height =) 6.2 × tan 35(.33…)°	A1
(Height =) 4.39(54... m) or 4.4(m)	A1	Do not allow 4.3(m) from premature rounding of 35.3345…° May be implied in further working Provided at least M2 previously awarded, FT from √’their 19.32’ provided < 7.6 (m)
(Volume of concrete) ½ × (12.6 + 18.8) × 4.4 × 50 or ½ × 31.4 × 4.4 × 50 or ½ × (18.8 – 12.6) × 4.4 × 50 + 12.6 × 4.4 × 50 or ½ × 6.2 × 4.4 × 50 + 12.6 × 4.4 × 50	M2	FT ‘their derived 4.4’ provided ‘their derived 4.4’ < 7.6 ‘their derived 4.4’ ≠ 6.2 ‘their derived 4.4’ ≠ ‘their 18.8 – 12.6’ May be seen in stages, e.g. with ‘× 50’ in further working M1 for any one of the following: (Area of cross-section) ½ × (12.6 + 18.8) × 4.4 (= 69.08 or 69.1m\(^2\)) ½ × (18.8 – 12.6) × 4.4 + 12.6 × 4.4 (= 13.64m\(^2\) + 55.44m\(^2\)) (Volume cuboid) 12.6 × 4.4 × 50 (= 55.44 × 50 = 2772 m\(^3\)) (Volume triangular prism) ½ × (18.8 – 12.6) × 4.4 × 50 (= 13.64 × 50 = 682 m\(^3\))
(Volume of concrete) Answer in the range 3450 (m\(^3\)) to 3455 (m\(^3\))	A1	FT from previous M2 only and ‘their derived 4.4’ from an attempt to use Pythagoras’ Theorem or cosine followed by sine On FT from M2, allow a similar range from rounding or truncation If previous M0 A0, award SC1 for an answer of 4867 (m\(^3\)) from ‘their 4.4’ = 6.2