Higher January 2019 Paper 1 Q20
20 A bowl contains \(n\) pieces of fruit.
Of these, 4 are oranges and the rest are apples.
Two pieces of fruit are going to be taken at random from the bowl.
The probability that the bowl will then contain \((n - 6)\) apples is \(\dfrac{1}{3}\)
Work out the value of \(n\)
Show your working clearly.
(6)
| Scheme | Marks |
|---|---|
| \(\dfrac{n - 4}{n}\) or \(\dfrac{n - 5}{n - 1}\) | M1 |
| \(\dfrac{n - 4}{n} \times \dfrac{n - 5}{n - 1} = \dfrac{1}{3}\) | M1 |
| Eg \(3(n^2 - 9n + 20) = n(n - 1)\) or \(3n^2 - 27n + 60 = n^2 - n\) | M1 |
| Eg \(2n^2 - 26n + 60 = 0\) or \(n^2 - 13n + 30 = 0\) | M1 |
| Eg \((n - 10)(n - 3) = 0\) or \(\dfrac{--13 \pm \sqrt{(-13)^2 - 4 \times 1 \times 30}}{2 \times 1}\) | M1 |
| Working required Answer: 10 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: \(\dfrac{n - 4}{n}\) or \(\dfrac{n - 5}{n - 1}\)
M1: for the correct equation
M1: for a correct quadratic equation with fractions removed
M1: for a correct quadratic equation equal to 0
M1: dep on M2 ft for method to solve 3 term quadratic
A1: for correct answer from correct working
NB. Award M5A1 for an answer of 10 with justification e.g. \(\dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{1}{3}\)
Award M0A0 for an answer of 10 with no working and no justification