Foundation June 2023 Paper 2 Q21
21 Nancy has some coins with a total value of 85 pence.
She has only 2 pence coins and 5 pence coins.
The ratio
number of 2 pence coins : number of 5 pence coins = 1 : 3
Nancy has more 5 pence coins than 2 pence coins.
How many more?
(4)
| Scheme | Marks |
|---|---|
2 and 15 seen or \(1 \times 2\) (+) \(3 \times 5\) (= 17) or \(2x + 15x\) (= 85) or \(\dfrac{2}{3}y + 5y\) (= 85) or \(0.25t \times 2 + 0.75t \times 5\) (= 85) | M1 |
\(85 \div (2 + 15)\) (= 5) or at least two pairs of multiples of the values of 2 and 15 (eg 4, 30; 6, 45…..) or 10(p) (and) 75(p) or 10 : 75 or \(5 \times 2\) and \(15 \times 5\) \(2 \times 5 + 5 \times 3 \times 5\) or 20 coins or \(17x = 85\) (\(x = 5\)) or \(\dfrac{17}{3}y = 85\) (\(y = 15\)) or \(4.25t = 85\) (\(t = 20\)) | M1 |
| 5 (2p coins) and 15 (5p coins) or 5 : 15 (if clearly identified (or used) as the key ratio eg not just part of a list) or \((3 - 1) \times 5\) or eg \(15 - 5\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 10 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: For 2 and 15 oe seen or 17 or a correct equation in one unknown for number of 2p coins (\(x\)) or number of 5p coins (\(y\)) or total number of coins (\(t\))
M1: assumes previous M1 for number of 2p coins or number of 5p coins or total number of coins or value of 2p coins and value of 5p coins
may be clearly listed eg
2 555
2 555
2 555
2 555
2 555
with no ambiguity
M1: Correct number of 2p coins and 5p coins or a sum to find the difference in number of coins
A1: SCB1 if no other marks awarded for 21.25 in working or on answer line