Numeracy Higher November 2020 Unit 1 Q2
2. Albert, Terri and Gareth are going camping.
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]
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 + 2 OCW]

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]
| 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) |
| 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 |
| Organisation and communication | OC1 | For OC1, candidates will be expected to: • present their response in a structured way • explain to the reader what they are doing at each step of their response • lay out their explanations and working in a way that is clear and logical • write a conclusion that draws together their results and explains what their answer means |
| Writing | W1 | For W1, candidates will be expected to: • show all their working • make few, if any, errors in spelling, punctuation and grammar • use correct mathematical form in their working • use appropriate terminology, units, etc. |
| 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 | A1 | |
| (Overall length =) 3.2 (m) | 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 |
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 | A1 | |
| (Overall length =) 3.2 (m) | A1 | FT ‘their 1.2’ + 2 provided at least M1 previously awarded |
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 |