4MA0 Higher June 2011 Paper 4 Q18
18 Show that the recurring decimal \(0.\dot{3}9\dot{6} = \dfrac{44}{111}\)
(2)
| Scheme | Marks |
|---|---|
| \(x = 0.396396...\) \(1000x = 396.396....\) \(999x = 396\) | M1 |
| \(\dfrac{44}{111}\) | A1 |
| (Total 2 marks) |
Notes
A1: must reach \(\frac{396}{999}\) or equivalent fraction (but not \(\frac{44}{111}\))