Higher November 2020 Paper 2 Q20
20 A bag contains \(X\) counters.
There are only red counters and blue counters in the bag.
There are 4 more blue counters than red counters in the bag.
Finty takes at random 2 counters from the bag.
The probability that Finty takes 2 blue counters from the bag is \(\dfrac{3}{8}\)
Work out the value of \(X\).
Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
\(\dfrac{\left(\frac{X + 4}{2}\right)}{X}\left(= \dfrac{X + 4}{2X}\right)\) or \(\dfrac{\left(\frac{X + 4}{2}\right) - 1}{X - 1}\left(= \dfrac{X + 2}{2X - 2}\right)\) or eg, where \(b\) = number of blue counters \(\dfrac{b}{2b - 4}\) or \(\dfrac{b - 1}{2b - 5}\) or eg, where \(r\) = number of red counters \(\dfrac{r + 4}{2r + 4}\) or \(\dfrac{r + 3}{2r + 3}\) | M1 |
eg \(\dfrac{X + 4}{2X} \times \dfrac{X + 2}{2X - 2}\) or eg \(\dfrac{b}{2b - 4} \times \dfrac{b - 1}{2b - 5}\) or eg \(\dfrac{r + 4}{2r + 4} \times \dfrac{r + 3}{2r + 3}\) | M1 |
| eg \(8(X^2 + 6X + 8) = 3(4X^2 - 4X)\) or eg \(8b(b - 1) = 3(2b - 4)(2b - 5)\) or eg \(8(r + 4)(r + 3) = 3(2r + 4)(2r + 3)\) | M1 |
| Eg \(4X^2 - 60X - 64\;(= 0)\) or \(X^2 - 15X - 16\;(= 0)\) oe or eg \(4b^2 - 46b + 60\;(= 0)\) or \(2b^2 - 23b + 30\;(= 0)\) oe or eg \(4r^2 - 14r - 60\;(= 0)\) or \(2r^2 - 7r - 30\;(= 0)\) oe | M1 |
| Working required Answer: 16 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for making a correct start by finding the probability of the first counter being blue for their method
M1: oe correct calculation for 2 blue (using one variable)
M1: dep for a correct equation with no algebraic fractions eg could have \(X^2 + 6X + 8 = \dfrac{3}{8}(4X^2 - 4X)\)
M1: for rearranging their equation to a correct 3 term quadratic
A1: cao dep on M4