4MA0 Higher January 2012 Paper 3 Q17
17 Show that the recurring decimal \(0.1\dot{7} = \dfrac{8}{45}\)
(2)
| Scheme | Marks |
|---|---|
| \(x = 0.1777....\) and \(10x = 1.777..\) \(9x = 1.6\) | M1 |
| 16/90 oe | A1 |
| (Total 2 marks) |
Notes
M1: See at least 3 sevens or recurring symbol. Condone omission of \(x\).
Accept \(10x = 1.777..\) and \(100x = 17.77..\)
A1: Must be integers in numerator and denominator but not 8 & 45
N.B for 0.1777 = 1/10 + 0.0777..
(0.777 needs to be shown to be 7/90 to gain first M1)