Foundation November 2022 Paper 3 Q18
18 11 identical full tins of red paint hold a total of 3630 ml
All the paint from 4 of these tins is poured into an empty bucket.
The bucket can hold 2500 ml
Tins of white paint each hold 140 ml
Can all the white paint from 9 tins be added to the bucket?
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – capacity of 9 tins of white paint and 4 tins of red paint compared with the 2500 ml bucket capacity | ||
| \(3630 \div 11\) or 330 or \(9 \times 140\) or 1260 | M1 | oe |
| their \(330 \times 4\) or 1320 or \(2500 -\) their 1260 or 1240 or \(2500 -\) their \(330 \times 4\) or 1180 | M1dep | oe \(3630 \times \dfrac{4}{11}\) is M2 their 330 and their 1260 must be from correct methods |
| their \(1260 +\) their 1320 or 2580 or \(2500 -\) their 1320 and their 1260 or their 1180 and their 1260 or \(2500 -\) their 1260 and their 1320 or their 1240 and their 1320 | M1dep | oe eg \(2500 - 1320\) or 1180 and \(1180 - 140 - 140 - 140 - 140 - 140 - 140 - 140 - 140 - 140\) or −80 their 1180, their 1240, their 1260 and their 1320 must be from correct methods |
| 2580 and No or 1180 and 1260 and No or 1240 and 1320 and No or (−)80 and No | A1 | oe eg1 No, there is 80 too much eg2 No, only 60 ml of the last tin will fit into the bucket |
| Alternative method 2 – The number of tins of white or red paint that can be added to 4 tins of red or 9 tins of white paint to fill the 2500 ml bucket | ||
| \(3630 \div 11\) or 330 or \(9 \times 140\) or 1260 | M1 | oe |
| their \(330 \times 4\) or 1320 or \(2500 -\) their 1260 or 1240 or \(2500 -\) their \(330 \times 4\) or 1180 | M1dep | oe \(3630 \times \dfrac{4}{11}\) is M2 their 330 and their 1260 must be from correct methods |
| \(\dfrac{2500 - \text{their } 1320}{140}\) or \(\dfrac{\text{their } 1180}{140}\) or [8.4, 8.43] or \(\dfrac{2500 - \text{their } 1320}{9}\) or \(\dfrac{\text{their } 1180}{9}\) or 131(.1…) or \(\dfrac{2500 - \text{their } 1260}{\text{their } 330}\) or \(\dfrac{\text{their } 1240}{\text{their } 330}\) or [3.75, 3.8] or \(\dfrac{2500 - \text{their } 1260}{4}\) or \(\dfrac{\text{their } 1240}{4}\) or 310 | M1dep | oe their 330, their 1180, their 1240, their 1260 and their 1320 must be from correct methods |
| [8.4, 8.43] and No or [3.75, 3.8] and No or 131(.1…) and No or 310 and No | A1 | oe |
| Alternative method 3 – 4 tins of red paint as a proportion of 2500 ml added to 9 tins of white as a proportion of 2500 ml | ||
| \(3630 \div 11\) or 330 or \(9 \times 140\) or 1260 | M1 | oe |
| \(\dfrac{\text{their } 330 \times 4}{2500}\) or 0.52(8) or 0.53 or \(\dfrac{\text{their } 1260}{2500}\) or 0.504 or 0.5(0) | M1dep | oe their 330 and their 1260 must be from correct methods |
| \(\dfrac{\text{their } 330 \times 4}{2500}\) or 0.52(8) or 0.53 and \(\dfrac{\text{their } 1260}{2500}\) or 0.504 or 0.5(0) | M1dep | oe |
| \(0.528 + 0.504 = 1.032\) and No | A1 | oe eg1 \(0.53 + 0.5 = 1.03\) and No eg2 \(52(\%) + 50(\%) \gt 100(\%)\) and No |
| Alternative method 4 – 4 tins of red paint as proportion of 2500 ml compared with the volume of the bucket remaining after 9 tins of white added as a proportion of 2500 ml | ||
| \(3630 \div 11\) or 330 or \(9 \times 140\) or 1260 | M1 | oe |
| \(\dfrac{\text{their } 330 \times 4}{2500}\) or 0.52(8) or 0.53 or \(\dfrac{2500 - \text{their } 1260}{2500}\) or 0.49(6) or 0.5(0) | M1dep | oe their 330 and their 1260 must be from correct methods |
| \(\dfrac{\text{their } 330 \times 4}{2500}\) or 0.52(8) or 0.53 and \(\dfrac{2500 - \text{their } 1260}{2500}\) or 0.49(6) or 0.5(0) | M1dep | oe their 330 and their 1260 must be from correct methods |
| \(0.528 \gt 0.496\) and No | A1 | oe eg1 \(0.53 \gt 0.5\) and No eg2 \(52(\%) \gt 50(\%)\) and No |
Additional guidance
| Up to M3 may be awarded for correct work, with no or incorrect answer, even if this is seen amongst multiple attempts | |
| Allow working in other units eg litres but units must be consistent for the 3rd mark | |
| No may be implied eg1 2580 and there is 80 (ml) too much paint eg2 8.4 tins so 9 tins is too much | |
| 2580 and No | M1M1M1A1 |
| 1180 and 1260 and No | M1M1M1A1 |
| 1240 and 1320 and No | M1M1M1A1 |
| 80 and No | M1M1M1A1 |
| Condone \(1180 - 1260 = 80\) and No | M1M1M1A1 |
| Condone an incorrect statement after the correct answer seen eg 1180 and 1260 and −80 and No, there is 60ml left in the 9th tin | M1M1M1A1 |