Higher June 2022 Paper 1R Q24
24 Elliot has \(x\) counters.
Each counter has one red face and one green face.
Elliot spreads all the counters out on a table and sees that the number of counters showing a red face is 5
Elliot then picks at random one of the counters and turns the counter over.
He then picks at random a second counter and turns the counter over.
The probability that there are still 5 counters showing a red face is \(\dfrac{19}{32}\)
Work out the value of \(x\)
Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
| \(\dfrac{5}{x} \times \dfrac{(x - 4)}{x}\) oe or \(\dfrac{(x - 5)}{x} \times \dfrac{6}{x}\) oe | M1 |
| \(\dfrac{5}{x} \times \dfrac{(x - 4)}{x} + \dfrac{(x - 5)}{x} \times \dfrac{6}{x}\) oe | M1 |
| \(19x^2 - 352x + 1600\;(= 0)\) oe or \(19x^2 - 352x = -1600\) oe | M1 |
\((x - 8)(19x - 200)\;(= 0)\) or \((x =)\;\dfrac{--352 \pm \sqrt{(-352)^2 - (4 \times 19 \times 1600)}}{2 \times 19}\) or \(19\left[\left(x - \dfrac{176}{19}\right)^2 - \left(\dfrac{176}{19}\right)^2\right] + 1600\;(= 0)\) | M1 |
| Working required Answer: 8 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for a correct expression for P(R,G) or P(G,R)
M1: for a correct expression for P(R,G) + P(G,R)
M1: for a correct equation in the form \(ax^2 + bx + c\;(= 0)\) oe or \(ax^2 + bx = -c\) oe
M1: for solving their 3-term quadratic equation using any correct method - if factorising, allow brackets which expanded give 2 out of 3 terms correct (if using formula or completing the square allow one sign error and some simplification – allow as far as \(\dfrac{352 \pm \sqrt{123904 - 121600}}{38}\) oe or \(19\left(x - \dfrac{176}{19}\right)^2 - \dfrac{576}{19}\;(= 0)\) oe)
A1: cao, dep on M2. Do not award if non-integer solution also given.
8 must come from correct working.