4MA0 Higher June 2014 Paper 4 Q20
20 Use algebra to show that the recurring decimal \(0.3\dot{8} = \dfrac{7}{18}\)
(2)
| Scheme | Marks |
|---|---|
\(x = 0.3888888\ldots\) \(10x = 3.88888\ldots\) \(9x = 3.5\) \(x = \dfrac{3.5}{9}\) | M1 |
| \(x = \dfrac{3.5}{9}\) | A1 |
| (2 marks) |
Notes
M1: for method as far as attempting to subtract
eg \(100x = 38.88888\ldots\), \(10x = 3.88888\ldots\), \(90x = 35\), \(x = \dfrac{35}{90}\)
eg \(1000x = 388.8888\ldots\), \(10x = 3.88888\ldots\), \(990x = 385\), \(x = \dfrac{385}{990}\)
A1: must reach \(\dfrac{3.5}{9}\) or equivalent fraction or \(18x = 7\) before reaching \(\dfrac{7}{18}\)