Higher June 2023 Paper 1R Q16
16 Use algebra to show that \(\;0.1\dot{7}\dot{6} = \dfrac{35}{198}\)
(2)
| Scheme | Marks |
|---|---|
| eg \(10\,000x = 1767.67\ldots\) \(\phantom{10\,0}100x = 17.67\ldots\) or \(1000x = 176.76\ldots\) \(\phantom{100}10x = 1.76\ldots\) or \(100x = 17.676\ldots\) \(\phantom{10}x = 0.176\ldots\) oe | M1 |
eg \(10\,000x - 100x = 1767.67\ldots - 17.67\ldots = 1750\) and \(\dfrac{1750}{9900} = \dfrac{35}{198}\) or \(1000x - 10x = 176.76\ldots - 1.76\ldots = 175\) and \(\dfrac{175}{990} = \dfrac{35}{198}\) or \(100x - x = 17.676\ldots - 0.176\ldots = 17.5\) and \(\dfrac{17.5}{99} = \dfrac{35}{198}\) or eg \(10x - x = 7.6767\ldots - 0.07676\ldots = 7.6\) and \(0.1 + \dfrac{7.6}{99} = \dfrac{0.1 \times 99 + 7.6}{99} = \dfrac{17.5}{99} = \dfrac{35}{198}\) oe Working required Answer: shown | A1 |
| (2) | |
| (2 marks) |
Notes
M1: for 2 recurring decimals that when subtracted give a whole number or terminating decimal (17.5 or 175 or 1750 etc)
eg \(10\,000x = 1767.67\ldots\) and \(100x = 17.676\ldots\)
or \(1000x = 176.76\ldots\) and \(10x = 1.7676\ldots\)
or \(100x = 17.676\ldots\) and \(x = 0.17676\ldots\)
with intention to subtract.
(if recurring dots not shown in both numbers then showing at least one of the numbers to at least 5sf)
or \(0.1 + 0.0\dot{7}\dot{6}\) and eg \(100x = 7.6767\ldots\), \(x = 0.07676\ldots\) with intention to subtract.
A1: for completion to \(\dfrac{35}{198}\) dep on M1