Higher June 2019 Paper 2 Q23
23 Boris has a bag that only contains red sweets and green sweets.
Boris takes at random 2 sweets from the bag.
The probability that Boris takes exactly 1 red sweet from the bag is \(\dfrac{12}{35}\)
Originally there were 3 red sweets in the bag.
Work out how many green sweets there were originally in the bag.
Show your working clearly.
(5)
| Scheme | Marks |
|---|---|
RG and GR method \(\dfrac{3}{t} \times \dfrac{t - 3}{t - 1}\) or \(\dfrac{t - 3}{t} \times \dfrac{3}{t - 1}\) or RR and GG method \(\dfrac{3}{t} \times \dfrac{2}{t - 1}\) or \(\dfrac{t - 3}{t} \times \dfrac{t - 4}{t - 1}\) | M1 |
\(\dfrac{3}{t} \times \dfrac{t - 3}{t - 1} + \dfrac{t - 3}{t} \times \dfrac{3}{t - 1} = \dfrac{12}{35}\) or \(2 \times \dfrac{3}{t} \times \dfrac{t - 3}{t - 1} = \dfrac{12}{35}\) oe or \(\dfrac{3}{t} \times \dfrac{2}{t - 1} + \dfrac{t - 3}{t} \times \dfrac{t - 4}{t - 1} = \dfrac{23}{35}\) | M1 |
| e.g. \(2t^2 - 37t + 105\) (= 0) or allow \(2t^2 - 37t = -105\) | A1 |
e.g. \((2t - 7)(t - 15) = 0\) e.g. \(t = \dfrac{-(-37) \pm \sqrt{(-37)^2 - 4 \times 2 \times 105}}{2 \times 2}\) e.g. \(2\left(\left(t - \dfrac{37}{4}\right)^2 - \left(\dfrac{37}{4}\right)^2\right) = -105\) | M1 |
| Working required Answer: 12 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for one correct product
M1: dep on M1 for a correct equation
A1: (dep on M2) writing the correct quadratic expression in form \(ax^2 + bx + c\) (= 0)
allow \(ax^2 + bx = c\)
M1: (dep on A1) for a complete method to solve the 3-term quadratic equation (allow one sign error and some simplification – allow as far as \(\dfrac{37 \pm \sqrt{1369 - 840}}{4}\)) or
Can be implied by answers of 15 (and \(\dfrac{7}{2}\))
A1: (dep on A1) cao
| Scheme | Marks |
|---|---|
RG and GR method \(\dfrac{3}{x + 3} \times \dfrac{x}{x + 2}\) or \(\dfrac{x}{x + 3} \times \dfrac{3}{x + 2}\) or RR and GG method \(\dfrac{3}{x + 3} \times \dfrac{2}{x + 2}\) or \(\dfrac{x}{x + 3} \times \dfrac{x - 1}{x + 2}\) | M1 |
\(\dfrac{3}{x + 3} \times \dfrac{x}{x + 2} + \dfrac{x}{x + 3} \times \dfrac{3}{x + 2} = \dfrac{12}{35}\) or \(2 \times \dfrac{3}{x + 3} \times \dfrac{x}{x + 2} = \dfrac{12}{35}\) oe or \(\dfrac{3}{x + 3} \times \dfrac{2}{x + 2} + \dfrac{x}{x + 3} \times \dfrac{x - 1}{x + 2} = \dfrac{23}{35}\) | M1 |
| e.g. \(2x^2 - 25x + 12\) (= 0) or allow \(2x^2 - 25x = -12\) | A1 |
e.g. \((2x - 1)(x - 12) = 0\) e.g. \(x = \dfrac{-(-25) \pm \sqrt{(-25)^2 - 4 \times 2 \times 12}}{2 \times 2}\) e.g. \(2\left(\left(x - \dfrac{25}{4}\right)^2 - \left(\dfrac{25}{4}\right)^2\right) = -12\) | M1 |
| Working required Answer: 12 | A1 |
Notes
M1: for one correct product
M1: dep on M1 for a correct equation
A1: (dep on M2) writing the correct quadratic expression in form \(ax^2 + bx + c\) (= 0)
allow \(ax^2 + bx = c\)
M1: (dep on A1) for a complete method to solve the 3-term quadratic equation (allow one sign error and some simplification – allow as far as \(\dfrac{25 \pm \sqrt{625 - 96}}{4}\)) or
can be implied by answers of 12 (and \(\dfrac{1}{2}\))
A1: (dep on A1) cao