Higher January 2023 Paper 1R Q13
13 Use algebra to show that \(\;0.3\dot{8}\dot{1} = \dfrac{21}{55}\)
(2)
| Scheme | Marks |
|---|---|
| eg \(10\,000x = 3818.18\ldots\) \(\phantom{eg 10\,0}\underline{100x = \phantom{38}38.18\ldots}\) or \(1000x = 381.818\ldots\) \(\phantom{or 10}\underline{10x = \phantom{38}3.818\ldots}\) or \(100x = 38.1818\ldots\) \(\phantom{or 10}\underline{x = \phantom{3}0.3818\ldots}\) oe | M1 |
eg \(10\,000x - 100x = 3818.18\ldots - 38.1818\ldots = 3780\) \((9900x = 3780)\) and \(\dfrac{3780}{9900} = \dfrac{21}{55}\) or eg \(1000x - 10x = 381.818\ldots - 3.81818\ldots = 378\) \((990x = 378)\) and \(\dfrac{378}{990} = \dfrac{21}{55}\) or eg \(100x - x = 38.1818\ldots - 0.381818\ldots = 37.8\) \((99x = 37.8)\) and \(\dfrac{37.8}{99} = \dfrac{21}{55}\) or eg \(10\,000x - 100x = 18.1818\ldots - 0.181818\ldots = 18\) and \(0.38 + \dfrac{18}{9900} = \dfrac{38 \times 99 + 18}{9900} = \dfrac{3780}{9900} = \dfrac{21}{55}\) oe Answer: shown | A1 |
| (2) | |
| (2 marks) |
Notes
M1: For selecting 2 correct recurring decimals that when subtracted give a whole number or terminating decimal (37.8 or 378 or 3780 etc)
eg \(10\,000x = 3818.18\ldots\) and \(100x = 38.1818\ldots\)
or \(1000x = 381.818\ldots\) and \(10x = 3.81818\ldots\)
or \(100x = 38.1818\ldots\) and \(x = 0.381818\ldots\)
with intention to subtract.
(if recurring dots not shown then showing at least one of the numbers to at least 5 sf)
or \(0.38 + 0.00\dot{1}\dot{8}\) and eg \(100x = 0.1818\ldots\), \(10\,000x = 18.1818\ldots\) with intention to subtract.
A1: for completion to \(\dfrac{21}{55}\) dep on M1
(NB: this is a “use algebra to show that…” question, so we need to see algebra as well as seeing all the stages of working to award full marks)