Higher November 2024 Paper 2 Q22
22 There are only red counters and yellow counters in a box.
\(\dfrac{3}{5}\) of the counters are red.
Sophie takes at random two counters from the box.
The probability that the two counters are the same colour is \(\dfrac{41}{80}\)
Work out the number of yellow counters in the box.
You must show all your working. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 26 | P1 | for a correct 2nd probability, eg \(\dfrac{2x - 1}{5x - 1}\) or \(\dfrac{3x - 1}{5x - 1}\) or \(\dfrac{2x}{5x - 1}\) or \(\dfrac{3x}{5x - 1}\) or \(\dfrac{\frac{3}{5}n - 1}{n - 1}\) or \(\dfrac{\frac{2}{5}n - 1}{n - 1}\) |
| P1 | for a correct product, eg \(\dfrac{3x}{5x} \times \dfrac{3x - 1}{5x - 1}\) or \(\dfrac{2x}{5x} \times \dfrac{2x - 1}{5x - 1}\) or \(\dfrac{3}{5} \times \dfrac{2x}{5x - 1}\) or \(\dfrac{2}{5} \times \dfrac{3x}{5x - 1}\) or \(\dfrac{3}{5} \times \dfrac{\frac{3}{5}n - 1}{n - 1}\) or \(\dfrac{2}{5} \times \dfrac{\frac{2}{5}n - 1}{n - 1}\) oe | |
| P1 | for process to form equation, eg \(\dfrac{3x}{5x} \times \dfrac{3x - 1}{5x - 1} + \dfrac{2x}{5x} \times \dfrac{2x - 1}{5x - 1} = \dfrac{41}{80}\) or \(2 \times \dfrac{3}{5} \times \dfrac{2x}{5x - 1} = \dfrac{39}{80}\) or \(\dfrac{3}{5} \times \dfrac{\frac{3}{5}n - 1}{n - 1} + \dfrac{2}{5} \times \dfrac{\frac{2}{5}n - 1}{n - 1}\) oe | |
| P1 | for process to eliminate fractions and reduce equation to linear or quadratic form, eg \(1040x - 400 = 1025x - 205\) or \(960x = 975x - 195\) or \(1040x^2 - 400x = 1025x^2 - 205x\) or \(960x^2 = 975x^2 - 195x\) or \(\dfrac{208}{5}n - 80 = 41n - 41\) or \(x = 13\) or \(n = 65\) | |
| A1 | cao |
Additional guidance
Award of this mark implies P2
Award of this mark implies P3