Foundation June 2023 Paper 2 Q14
14 A student is buying some gifts for their friends.
The gifts are shown below with the prices.
The student has £50 to spend.
They first buy 6 key rings and 2 wallets.
They then buy badges with the remainder of the money.
(a) Work out the maximum number of badges that the student can buy.
You must show your working. [5]
You must show your working. [5]
(b) Work out the amount of money they have left over. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 4 [badges] with correct working | 5 | M3 for 50 – \((6 \times 3.50 + 2 \times 7.50)\) oe or M2 for \(6 \times 3.50 + 2 \times 7.50\) oe or M1 for \(6 \times 3.50\) or \(2 \times 7.50\) oe AND M1 for their (50 – \((6 \times 3.50 + 2 \times 7.50)) \div 2.99\) oe If 0 or 1 scored, instead award SC2 for 4 [badges] with no or insufficient working or If 0 scored SC1 for 14 with no or insufficient working | ‘Correct working’ requires full evidence of at least M2M1 M3 implied by 14, 15, 21 seen or 14, 36 seen M2 implied by 36 M1 implied by 21 or 15 Accept their (50 – \(6 \times 3.50 + 2 \times 7.50) \div 3\) Implied by list 2.99, 5.98, 8.97, 11.96, [14.95, …….] up to one less than their 14 Condone one arithmetic slip. or 3, 6, 9, 12, [15, ……] up to one less than their 14 or Embedded e.g. 4 x 2.99 = 11.96 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 2.04 | 2 | M1 for their 14 – their ((a) \(\times\) 2.99) oe | their ((a) \(\times\) 2.99) could be seen in part (a) and must be \(\lt\) their 14 M1 can be implied by a correct FT answer to their 14 – their ((a) \(\times\) 2.99) |