4400 Higher November 2006 Paper 3 Q21
21 \(\dfrac{1}{3}\) of the people in a club are men.
The number of men in the club is \(n\).
(a) Write down an expression, in terms of \(n\), for the number of people in the club. (1)
Two of the people in the club are chosen at random.
The probability that both these people are men is \(\dfrac{1}{10}\)
(b) Calculate the number of people in the club. (5)
| Scheme | Marks |
|---|---|
| \(3n\) oe | B1 |
| (1) |
Notes
B1: Accept eg \(n + 2n\)
| Scheme | Marks |
|---|---|
| \(n - 1\), \(3n - 1\) seen | B2 |
| \(\dfrac{1}{3} \times \dfrac{n - 1}{3n - 1} = \dfrac{1}{10}\) oe inc \(\dfrac{n}{3n} \times \dfrac{n - 1}{3n - 1} = \dfrac{1}{10}\) | M1 |
| \(10(n - 1) = 3(3n - 1)\) oe inc \(10n(n - 1) = 3n(3n - 1)\) | M1 |
| (\(n = 7\)) Answer: 21 | A1 |
| (5) | |
| (6 marks) |
Notes
B2: B1 for each
M1: for correct equation
M1: for correctly removing fractions
A1: cao