Higher June 2019 Paper 2R Q24
24 A box contains marbles.
4 of the marbles are red.
The rest of the marbles are yellow.
Antonia takes at random a marble from the box and does not replace it.
Sergio then takes at random a marble from the box.
The probability that Antonia and Sergio both take a yellow marble is 0.7
Work out how many marbles were originally in the box.
Show your working clearly.
(5)
| Scheme | Marks |
|---|---|
| \(\dfrac{x - 4}{x} \times \dfrac{x - 5}{x - 1} = 0.7\) | M2 |
| \(3x^2 - 83x + 200\) (= 0) oe | A1 |
\(\dfrac{83 \pm \sqrt{83^2 - (4 \times 3 \times 200)}}{2 \times 3}\) or \((3x - 8)(x - 25)\) (=0) or \((x - 83/6)^2 + 200/3 - 83^2/36\) (=0) | M1 |
| Working required Answer: 25 | A1 |
| (5) | |
| (5 marks) |
Notes
M2: If not M2 then M1 for either \(\dfrac{x - 4}{x}\) or \(\dfrac{x - 5}{x - 1}\)
A1: Rearrangement of their quadratic to the form \(ax^2 + bx + c\) (= 0)
M1: 1st step in solving the correct 3 term quadratic
A1: Accept 25 only (dep on M3 if using algebra)
| Scheme | Marks |
|---|---|
\(y\) = yellow marbles \(\dfrac{y}{y + 4} \times \dfrac{y - 1}{y + 3} = 0.7\) | M2 |
| \(3y^2 - 59y - 84\) (= 0) oe | A1 |
\(\dfrac{59 \pm \sqrt{59^2 - (4 \times 3 \times -84)}}{2 \times 3}\) or \((3y + 4)(y - 21)\) or \((y - 59/6)^2 - 84/3 - 59^2/36\) (=0) \(y = 21\) | M1 |
| Working required 21+4 Answer: 25 | A1 |
Notes
M2: If not M2 then M1 for either \(\dfrac{y}{y + 4}\) or \(\dfrac{y - 1}{y + 3}\)
A1: Rearrangement of their quadratic to the form \(ay^2 + by + c\) (= 0)
M1: 1st step in solving the correct 3 term quadratic