Higher June 2025 Paper 2 Q6
6 Here are two lists of numbers.
| List A | 276 | 400 | 157 | 139 | ||
|---|---|---|---|---|---|---|
| List B | 530 | 500 | 270 | \(x\) | 440 | 320 |
mean of list A : mean of list B = 3 : 5
Work out the value of \(x\). (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 370 | P1 | for finding the mean of list A, eg \((276 + 400 + 157 + 139) \div 4\ (= 243)\) OR an expression for the mean of list B, eg \((530 + 500 + 270 + x + 440 + 320) \div 6\ \left(= \dfrac{2060 + x}{6}\right)\) oe |
| P1 | for beginning to work with ratio, eg \(\text{``}243\text{''} \div 3\ (= 81)\) or \([\mathrm{A}] \div 3\) or \(\text{``}243\text{''} \times 5\ (= 1215)\) or \([\mathrm{A}] \times 5\) OR \(\text{``}\left(\dfrac{2060 + x}{6}\right)\text{''} \times 3\) or \([\mathrm{B}] \times 3\) or \(\text{``}\left(\dfrac{2060 + x}{6}\right)\text{''} \div 5\) or \([\mathrm{B}] \div 5\) | |
| P1 | for completing the work with ratio, eg \(\text{``}81\text{''} \times 5\ (= 405)\) or \([\mathrm{A}] \div 3 \times 5\) or \([\mathrm{B}] \times 3 \div 5\) or \(\text{``}\left(\dfrac{2060 + x}{6}\right)\text{''} \times \dfrac{3}{5}\) OR forms a suitable equation, eg \(\text{``}243\text{''} \times 5 = 3 \times \text{``}\left(\dfrac{2060 + x}{6}\right)\text{''}\) or \([\mathrm{A}] \times 5 = 3 \times \text{``}\left(\dfrac{2060 + x}{6}\right)\text{''}\) | |
| P1 | for working with mean of list B, eg \(\text{``}405\text{''} \times 6\ (= 2430)\) or \([\mathrm{A}] \div 3 \times 5 \times 6\) OR for process to remove brackets and denominator, eg \(\text{``}243\text{''} \times 5 \times 2 = \text{``}2060 + x\text{''}\) or \([\mathrm{A}] \times 5 \times 2 = \text{``}2060 + x\text{''}\) or \(2060 + x = \text{``}405\text{''} \times 6\) | |
| A1 | cao |
Additional guidance
[A] is what they believe to be the mean of A
[B] must be clearly their mean of B and be an expression including \(x\)