Higher June 2024 Paper 3 Q10
10 There are only red counters and yellow counters in bag A.
\[\text{number of red counters} : \text{number of yellow counters} = 3 : 5\]There are only green counters and blue counters in bag B.
The number of counters in bag B is half the number of counters in bag A.
Given that there are \(x\) red counters in bag A,
use algebra to show that the total number of counters in bag A and bag B is \(4x\) (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| Result shown | M1 | for method to find the number of yellow counters in bag A, eg \(x \div 3 \times 5\ \left(= \dfrac{5x}{3}\right)\) or for method to find the total number of counters in bag A eg \(x \div 3 \times 8\ \left(= \dfrac{8x}{3}\right)\) or for starting to work with ratio using algebra eg \(3y\), \(5y\) |
| M1 | (dep) for method to find the total number of counters in bag B, eg \(\left(x + \text{``}\dfrac{5x}{3}\text{''}\right) \div 2\ \left(= \dfrac{4x}{3}\right)\) or \(\text{``}\dfrac{8x}{3}\text{''} \div 2\ \left(= \dfrac{4x}{3}\right)\) or \((3y + 5y) \div 2\ (= 4y)\) | |
| C1 | for complete method showing that total number of counters in bag A and bag B is \(4x\), eg \(\dfrac{8x}{3} + \dfrac{4x}{3} = 4x\) or \(3y + 5y + 4y = 12y\) and \(12y \div 3y \times x = 4x\) |
Additional guidance
Could use any letter other than \(y\) apart from \(x\)
For the method marks condone decimals that are rounded or truncated to 1dp
For the C mark only accept values that are shown to be recurring and allow \(3.\dot{9}x = 4x\)