Numeracy Higher November 2021 Unit 2 Q6
6. Hannah is opening her own gym.
She has designed a logo that will be on her own brand of sports clothing.
She will sell the sports clothing in the gym.
The design of the symmetrical logo is shown below.
It consists of:
- two sectors of circles, both of radius 1·5 cm,
- a straight line of length 4 cm connecting the centres of the two sectors,
- a parallel straight line of length 4 cm connecting the lower vertices of the sectors.
Diagram not drawn to scale
The logo will be stitched onto all the sports clothing.
Calculate the perimeter of the logo. [4]
An enlarged version of the logo will be painted on the front wall of the gym.
All the lengths on this logo will be 200 times as long as the lengths in her original design.
Use your answer to (i) to calculate the area of the logo painted on the front wall of the gym.
Give your answer in m\(^2\). [3]
For the opening week, Hannah reduces the prices of her clothing range by 20%.
On the last day of the opening week, she reduces prices by a further 10%.
| Opening week offer: 20% off all clothing Last day promotion: all prices reduced by a further 10% |
On the last day of the opening week, Hannah has a sports top priced at £9.72.
Calculate the original price of the sports top before the two reductions in price. [3]
| Scheme | Marks | Comments |
|---|---|---|
| (Arc length =) \(\dfrac{145}{360}\) × \(\boldsymbol{\pi}\) × 3 × 2 or equivalent (= 29\(\boldsymbol{\pi}\)/12 or 7.58 to 7.6) | M2 | M1 for \(\dfrac{145}{360}\) × \(\boldsymbol{\pi}\) × 3 or equivalent (= 29\(\boldsymbol{\pi}\)/24 or 3.79 to 3.8) |
| + 1.5 + 1.5 + 4 + 4 | m1 | FT from M1 |
| = 18.58 to 18.6 (cm) | A1 | CAO. Accept an answer of 29\(\boldsymbol{\pi}\)/12 + 11 Allow a final answer of 19 (cm) provided 18.58 to 18.6 seen |
(i)
| Scheme | Marks | Comments |
|---|---|---|
| (Area of sector =) \(\dfrac{145}{360}\) × \(\boldsymbol{\pi}\) × 1.5\(^2\) (×2) (=29\(\boldsymbol{\pi}\)/16 or 5.68 to 5.7) | M1 | \(\dfrac{145}{360}\) × \(\boldsymbol{\pi}\) × 1.5\(^2\) (= 29\(\boldsymbol{\pi}\)/32 or 2.84(5…) to 2.85) |
| + 4 × 1.5 | m1 | |
| = 11.6(9…) (cm\(^2\)) | A1 | CAO. Accept values in the range 11.68 to 11.7 (cm\(^2\)) Accept an answer of 29\(\boldsymbol{\pi}\)/16 + 6 Allow a final answer of 11.6 Allow a final answer of 12 (cm\(^2\)) provided 11.6 or values from 11.68 to 11.7 seen |
(ii)
| Scheme | Marks | Comments |
|---|---|---|
| 11.6(9…) × 200\(^2\) ÷ 100\(^2\) or equivalent = 46.7(765…) (m\(^2\)) | M1 m1 A1 | FT ‘their 11.6(9…)’ from (b)(i) Accept values in the range 46.7 to 46.8 Accept an answer of 29\(\boldsymbol{\pi}\)/4 + 24 Allow a final answer of 46.4 (from use of 11.6) Allow 47 (m\(^2\)) from correct working If no marks awarded, then ‘their 11.6(9…)’ × 4 with a place value error implies M1m0A0 |
Alternative
6(b)(ii) Alternative method 1:
| Scheme | Marks | Comments |
|---|---|---|
| (Area of sector =) \(\dfrac{145}{360}\) × \(\boldsymbol{\pi}\) × 300\(^2\) (×2) + 800 × 300 | M1 | \(\dfrac{145}{360}\) × \(\boldsymbol{\pi}\) × 300\(^2\)×2 = 227650 to 227795 |
| ÷ 100\(^2\) or equivalent = 46.7(765…) (m\(^2\)) | m1 A1 | CAO. Accept values in the range 46.7 to 46.8 Accept an answer of 29\(\boldsymbol{\pi}\)/4 + 24 Allow 47 (m\(^2\)) from correct working |
Alternative
6(b)(ii) Alternative method 2:
| Scheme | Marks | Comments |
|---|---|---|
| (Area of sector =) \(\dfrac{145}{360}\) × \(\boldsymbol{\pi}\) × 3\(^2\) (×2) | M1 | \(\dfrac{145}{360}\) × \(\boldsymbol{\pi}\) × 3\(^2\)×2 = 22.765 to 22.8 |
| + 8 × 3 = 46.7(765…) (m\(^2\)) | m1 A1 | CAO. Accept values in the range 46.7 to 46.8 Accept an answer of 29\(\boldsymbol{\pi}\)/4 + 24 Allow 47 (m\(^2\)) from correct working |
| Scheme | Marks | Comments |
|---|---|---|
| (£)9.72 ÷ 0.9 OR ÷ 90 × 100 (=(£)10.8(0)) ÷ 0.8 OR ÷ 80 × 100 | M1 M1 | These M1 marks may be awarded in any order The intermediate answer if done in reverse order is (£)12.15 OR M2 for 9.72 ÷ 0.72 |
| = (£) 13.5(0) | A1 | CAO If no marks awarded, SC1 for an answer of (£)13.88 or (£)13.89 (from 9.72 ÷ 0.7) |