Foundation June 2023 Paper 2 Q6
6. Philip has 250 coins in his money box.
- 22% are £1 coins.
- \(\dfrac{2}{5}\) are 20p coins.
- The rest are 10p coins.
(a) How many £1 coins are there in Philip’s money box? [2]
(b) How many 20p coins are there in Philip’s money box? [2]
(c) What is the total value of the coins in Philip’s money box? [4]
| Scheme | Marks | Comments |
|---|---|---|
| 0.22 × 250 | M1 | Or equivalent complete and correct method. |
| = 55 ISW (for calculating value) | A1 | Allow M1 A1 for an answer of £55 |
| Scheme | Marks | Comments |
|---|---|---|
| 250 ÷ 5 × 2 | M1 | Or equivalent complete and correct method. May be seen in stages. Note: M0 if method seen but then spoiled e.g. 250 ÷ 5 × 2 − 55 |
| = 100 ISW (for calculating value) | A1 | If no marks, award SC1 for unsupported £20 |
| Scheme | Marks | Comments |
|---|---|---|
| 250 – 55 – 100 or (1 – 0.22 – 0.4) × 250 oe | M1 | FT 250 – ‘their 55’ – ‘their 100’, provided ‘their 55’ + ‘their 100’ < 250 and are a whole number of coins |
| = 95 (10p coins) | A1 | FT. Sight of £9.50 implies M1 A1. May be seen in working. |
| 55 × (£)1 + 100 × (£) 0.20 + 95 × (£) 0.10 (= £55 + £20 + £9.50) or 55 × 100(p) + 100 × 20(p) + 95 × 10(p) (= 5500p + 2000p + 950p) | M1 | FT ‘their derived 95’, ‘their 55’ and ‘their 100’, provided ‘their 55’ + ‘their 100’ < 250 and are a whole number of coins Allow M1 A0 if inconsistent place value used unless corrected. |
| = (£) 84.5(0) or 8450(p) | A1 | FT If units are given, they must be correct. |
| (8) |