Numeracy Intermediate November 2021 Unit 2 Q9
9. 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 cm3 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]
Number of full jugs of lemonade left over is ……………
| Scheme | Marks | Comments |
|---|---|---|
| (Volume of a jug) π × 52 × 28 | M1 | |
| Answer in the range 2198 (cm3) to 2200 (cm3) or 700π (cm3) | 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) | A1 | |
| 3 (full jugs left over) | 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 |
Alternative
9. Alternative method 1:
| Scheme | Marks | Comments |
|---|---|---|
| (Volume of 10 jugs) 10 × π × 52 × 28 | M1 | |
| Answer in the range 21980 (cm3) to 22000 (cm3) or 7000π (cm3) | A1 | May be implied in further working |
| (Volume left over = volume 10 jugs – 80 servings) = 10 × π × 52 × 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 cm3) |
| (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
9. Alternative method 2:
| Scheme | Marks | Comments |
|---|---|---|
| (Volume of a jug) π × 52 × 28 | M1 | |
| Answer in the range 2198 (cm3) to 2200 (cm3) or 700π (cm3) | 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
9. Alternative method 3:
| Scheme | Marks | Comments |
|---|---|---|
| (Volume of 10 jugs) 10 × π × 52 × 28 | M1 | |
| Answer in the range 21980 (cm3) to 22000 (cm3) or 7000π (cm3) | A1 | May be implied in further working |
| (Number of glasses not needed) 10 × π × 52 × 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 × 52 × 28 ÷ 170 – 80’ |
| 3 (full jugs left over) | A1 | Do not allow A1 from truncation of 12.9(...) to 12 |