Numeracy Higher November 2021 Unit 2 Q4
4. In this question, you will be assessed on the quality of your organisation, communication and accuracy in writing.
Elwyn makes 10 full jugs of fresh lemonade.
Each jug is cylindrical with internal measurements as follows:
- height 28 cm,
- base radius 5 cm.
Elwyn’s drinking glasses each hold a serving of 170 cm\(^3\) of lemonade.
He serves 80 people lemonade.
How many full jugs of lemonade does Elwyn have left over?
You must show all your working. [5 + 2 OCW]
| Scheme | Marks | Comments |
|---|---|---|
| (Volume of a jug) π × 5\(^2\) × 28 Answer in the range 2198 (cm\(^3\)) to 2200 (cm\(^3\)) or 700π (cm\(^3\)) | M1 A1 | May be implied in further working |
| (Number of jugs needed) 170 × 80 ÷ 2199.(...) | M1 | FT ‘their derived volume of a jug’ provided > ‘their 170 × 80’ ÷10 |
| 6.1(84… jugs) or 6.2 (jugs) or 6 (jugs) 3 (full jugs left over) | A1 A1 | FT 10 – ‘their 6.18…’ (depends on previous M1) Note: For final M and A marks, allow if found from listing the capacity of a number of jugs |
| 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. |
Alternative
4. Alternative method 1:
| Scheme | Marks | Comments |
|---|---|---|
| (Volume of 10 jugs) 10 × π × 5\(^2\) × 28 Answer in the range 21980 (cm\(^3\)) to 22000 (cm\(^3\)) or 7000π (cm\(^3\)) | M1 A1 | May be implied in further working |
| (Volume left over = volume 10 jugs – 80 servings) = 10 × π × 5\(^2\) × 28 – 80 × 170 | M1 | (= 21980 – 13600) FT ‘their derived volume of 10 jugs’ provided > ‘their 170 × 80’ (Note: Correct answer is the range 8380 to 8394 cm\(^3\)) |
| (Number of jugs left over) 8380 ÷ 2199.(...) | m1 | (= 3.81...) FT ‘their derived volume of 10 jugs’ provided > ‘their 170 × 80’ |
| 3 (full jugs left over) | A1 | Note: For final M and A marks, allow if found from listing the capacity of a number of jugs |
Alternative
4. Alternative method 2:
| Scheme | Marks | Comments |
|---|---|---|
| (Volume of a jug) π × 5\(^2\) × 28 Answer in the range 2198 (cm\(^3\)) to 2200 (cm\(^3\)) or 700π (cm\(^3\)) | M1 A1 | May be implied in further working |
| (Number of jugs left over) 10 – 80 ÷ (2199.(..) ÷ 170) (= 10 – 80 ÷ 12.935...) | M2 | FT ‘their derived volume of a jug’ provided > ‘their 170 × 80’ ÷10 M1 for sight of 80 ÷ (2199.(..) ÷ 170) (=6.1(84...)) |
| 3 (full jugs left over) | A1 | Do not allow A1 from truncation of 12.9(...) to 12 |
Alternative
4. Alternative method 3:
| Scheme | Marks | Comments |
|---|---|---|
| (Volume of 10 jugs) 10 × π × 5\(^2\) × 28 Answer in the range 21980 (cm\(^3\)) to 22000 (cm\(^3\)) or 7000π (cm\(^3\)) | M1 A1 | May be implied in further working |
| (Number of glasses not needed) 10 × π × 5\(^2\) × 28 ÷ 170 – 80 | M1 | (= 21980 ÷ 170 – 80 = 49.29...) FT ‘their derived volume of 10 jugs’ > ‘their 170 × 80’ |
| (Number of jugs left over) 49.29... ÷ (2199.(...)÷170) or 49.29.... ÷ 12.9... | m1 | (= 3.81...) FT ‘their 10 × π × 5\(^2\) × 28 ÷ 170 – 80’ |
| 3 (full jugs left over) | A1 | Do not allow A1 from truncation of 12.9(...) to 12 |