Foundation June 2023 Paper 1 Q15
15 Finley has 72 sweets.
Finley gives
- 25% of the sweets to Alex
- \(\dfrac{1}{6}\) of the sweets to Umi.
Show that Finley has \(\dfrac{7}{12}\) of the sweets left. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(0.25 \times 72\) oe may be implied by 18 | 1 | Do not award 4 marks unless fully correct | |
| \(72 \div 6\) oe may be implied by 12 | 1 | Note \(\dfrac{5}{12}\) with no working scores 0 as it may come from \(1 - \dfrac{7}{12}\) | |
| 72 – their (18 + 12) or 18 + 12 + 42 = 72 | 1 | If start with \(\dfrac{7}{12}\) = 42 must show 18 + 12 = 30 | |
| \(\dfrac{42}{72} = \dfrac{7}{12}\) or \(\dfrac{7}{12} \times 72 = 42\) oe Accept 42 is \(\dfrac{7}{12}\) of 72 | 1 | Accept equivalent alternative methods e.g. 1 for \(\dfrac{(18+12)}{72} = \dfrac{5}{12}\) and 1 for \(1 - \dfrac{5}{12} = \dfrac{7}{12}\) | |
| Alternative method | Alternative method | ||
| [25% =] \(\dfrac{1}{4}\) oe | 1 | \(\dfrac{1}{6} = 0.1666\) to 0.167 | |
| \(\dfrac{1}{4} + \dfrac{1}{6}\) | 1 | 0.25 + 0.1666 to 0.167 | Accept equivalent percentages to the same accuracy |
| \(1 -\) their \(\left(\dfrac{1}{4} + \dfrac{1}{6}\right)\) oe | 1 | \(1 -\) their (0.25 + 0.1666 to 0.17) | Penalise 0.17 on the second but not the third mark |
| \(\dfrac{7}{12}\) from use of common denominator | 1 | 0.583[3] = \(\dfrac{7}{12}\) | |