2D Area & Perimeter: WJEC GCSE Numeracy Intermediate November 2020 Unit 1 Q8

Numeracy Intermediate November 2020 Unit 1 Q8

WJECIntermediateCurrent spec11 marks2D Area & Perimeter Percentages Ratio & Basic Proportion

8. Albert, Terri and Gareth are going camping.

(a) Albert, Terri and Gareth paid for a tent between them.
The amount they each paid for the tent was in the ratio 1 : 4 : 6 respectively.
Gareth paid £66.36 towards the tent.

Calculate the cost of the tent. [3]

(b) The charge to stay at a campsite has increased by 5% each year for the last two years.
Two years ago, the charge was £24 per night for a large tent and three people.

Calculate the current charge per night for a large tent and three people.
You must show all your working. [4]

Groundsheet: a rectangle 2·2 m wide with sides 2 m, topped by a trapezium with parallel top side 1·8 m; an arrow shows the overall length

Diagram not drawn to scale

From the formula list: Area of trapezium \(= \dfrac{1}{2}(a+b)h\)

The area of the groundsheet is 6·8 m².
The width of the groundsheet is 2·2 m.
Calculate the overall length of the groundsheet. [4]

Mark scheme (a)

Scheme	Marks	Comments
66.36 ÷ 6 × 11 or 66.36 ÷ 6 × (1 + 4 + 6) or 66.36 ÷ 6 + 4 × 66.36 ÷ 6 + 66.36 (= 11.06 + 44.24 + 66.36) or equivalent	M2	M1 for sight of 66.36 ÷ 6 or 11.06, or for sight of ‘11.6(0)’ (Note if ×10 seen, check if there is indication if this was derived from 1 + 4 + 6, if so accept for possible M2, if no evidence M0)
(£) 121.66	A1	CAO If no marks, award SC1 for an answer of (£)182.49 (from 11 × 66.36 ÷ 4)

Mark scheme (b)

Scheme	Marks	Comments
(First year increased charge) 24 × 0.05 + 24 or 24 + 24 ÷ 10 ÷ 2 or equivalent	M1
(£) 25.2(0)	A1	Accept 2520(p). Ignore units given
(Second year increased charge) 25.2(0) × 0.05 + 25.2(0) or 25.2(0) + 25.2(0) ÷ 10 ÷ 2 or equivalent	M1	FT ‘their 25.2(0)’
(Increased charge after 2 years is) (£) 26.46	A1	Accept 2646(p). If units are given they must be correct An answer of (£)26.4(0) (from 24 + 2 × 1.20) implies M1, A1, m0, A0 Sight of 24 × 1.05² implies M2, also award A1 for 24 × 1.1025

Mark scheme (c)

Scheme	Marks	Comments
½ ×(2.2 + 1.8) × height trapezium + 2.2 × 2 = 6.8	M2	M1 only if brackets omitted for sum of parallel sides in the overall calculation unless dealt with correctly in further working, OR M1 for ½ ×(2.2 + 1.8) × height trapezium (brackets must be given or any ‘missing brackets’ implied by correct interpretation)
2 × height trapezium = 2.4 or height trapezium = 1.2 (Overall length =) 3.2 (m)	A1 A1	FT ‘their 1.2’ + 2 provided at least M1 previously awarded If no marks, award SC1 for area of the trapezium as 2.4 (m²) provided not from incorrect working, e.g. 6.8 – (2.2 + 1.8 + (0).2 + (0).2) = 2.4 is SC0 6.8 – 2.2 × 2 = 2.4 is SC1

8(c) Alternative method 1:

Scheme	Marks	Comments
(2+ht trap)×2.2 – 2×½×[(2.2 – 1.8) ÷ 2] ×ht trap = 6.8	M2	M1 for ½×[(2.2 – 1.8) ÷ 2] ×height trapezium or 2×½×[(2.2 – 1.8) ÷ 2] × height trapezium
Height of trapezium = 1.2 (Overall length =) 3.2 (m)	A1 A1	FT ‘their 1.2’ + 2 provided at least M1 previously awarded

8(c) Alternative method 2:

Scheme	Marks	Comments
2×½×(2+overall length)× [(2.2 – 1.8) ÷ 2] + overall length×1.8 =6.8	M2	M1 for ½×(2+overall length)× [(2.2 – 1.8) ÷ 2] or 2×½×(2+overall length)× [(2.2 – 1.8) ÷ 2]
(Overall length =) 3.2 (m)	A2	A1 for 2 × Overall length = 6.4 or correct simplified equation in terms of overall length

The alternative method is printed in italics in the original.