Foundation November 2022 Paper 2 Q9
9 Hamish has saved 295 coins.
Each one is a 20p coin.
He gives an equal number of 20p coins to each of his 8 grandchildren.
He gives them as many coins as possible.
How much, in £, does he have left? [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Using number of coins left | ||
| \(295 \div 8\) or 36(.875) or 36.88 or 36.9 | M1 | oe implied by \((295 \div 20) \div 8\) or \(14.75 \div 8\) or 1.84… |
| their \(36 \times 8\) or 288 or their \(36.875 -\) their 36 or 0.8(75) or 0.88 | M1dep | oe their 36 must be an integer |
| \(295 -\) their 288 or their \(0.875 \times 8\) or 7 (coins left) | M1dep | oe implied by \(0.875 \times 20 \times 8\) or \(0.875 \times 160\) or 140 or 1.4 |
| 1.40 | A1 | |
| Alternative method 2 Using total value of coins given | ||
| \(295 \div 8\) or 36(.875) or 36.88 or 36.9 | M1 | oe implied by \((295 \div 20) \div 8\) or \(14.75 \div 8\) or 1.84… |
| their \(36 \times 20 \times 8\) or their \(36 \times 160\) or 5760 | M1dep | oe their 36 must be an integer |
| \(295 \times 20\) or 5900 | M1 | oe |
| 1.40 | A1 | |
| Alternative method 3 Using value of coins given to each child | ||
| \(295 \div 8\) or 36(.875) or 36.88 or 36.9 | M1 | oe implied by \((295 \div 20) \div 8\) or \(14.75 \div 8\) or 1.84… |
| their \(36 \times 20\) or 720 | M1dep | oe their 36 must be an integer |
| \(295 \div 8 \times 20\) or \(5900 \div 8\) or 737(.5) or 738 | M1dep | oe dep on 1st M1 only |
| 1.40 | A1 | |
Additional guidance
| Up to M3 may be awarded for correct work with no answer, or incorrect answer, even if this is seen amongst multiple attempts | |
| Use the scheme that awards most marks | |
| Methods are shown in pence but equivalent working may be in pounds | |
| NB 7 coins per child or (£)7, possibly from truncating £7.37 or £7.20 or from \(56 \div 8\), does not imply M3 in Alt 1. The 7 must be coins left | |
| Alt 3 740 or 7.4(0) with no method does not imply 737.5 or 7.375 | |
| In Alt 2 the 3rd mark is not dependent | |
| Note that the third mark in Alt 3 implies the first mark ie 737(.5) or 738 | M1M0M1 |