Higher June 2021 Paper 1 Q16
16 Use algebra to show that the recurring decimal \(0.28\dot{1}\dot{3} = \dfrac{557}{1980}\)
(2)
| Scheme | Marks |
|---|---|
| eg \(10\,000x = 2813.13\ldots\) \(\phantom{10\,0}100x = 28.13\ldots\) or \(1000x = 281.313\ldots\) \(\phantom{100}10x = 2.813\ldots\) or \(100x = 28.1313\ldots\) \(\phantom{10}x = 0.2813\ldots\) oe | M1 |
eg \(10\,000x - 100x = 2813.13\ldots - 28.1313\ldots = 2785\) and \(\dfrac{2785}{9900} = \dfrac{557}{1980}\) or \(1000x - 10x = 281.313\ldots - 2.81313\ldots = 278.5\) and \(\dfrac{278.5}{990} = \dfrac{557}{1980}\) or \(100x - x = 28.1313\ldots - 0.281313\ldots = 27.85\) and \(\dfrac{27.85}{99} = \dfrac{557}{1980}\) or eg \(10\,000x - 100x = 13.1313\ldots - 0.1313\ldots = 13\) and \(0.28 + \dfrac{13}{9900} = \dfrac{28 \times 99 + 13}{9900} = \dfrac{2785}{9900} = \dfrac{557}{1980}\) oe Answer: shown | A1 |
| (2) | |
| (2 marks) |
Notes
M1: For 2 recurring decimals that when subtracted give a whole number or terminating decimal (27.85 or 278.5 or 2785 etc)
eg \(10\,000x = 2813.13\ldots\) and \(100x = 28.1313\ldots\)
or \(1000x = 281.313\ldots\) and \(10x = 2.81313\ldots\)
or \(100x = 28.1313\ldots\) and \(x = 0.281313\ldots\)
with intention to subtract.
(if recurring dots not shown then showing at least one of the numbers to at least 6sf)
or \(0.28 + 0.00\dot{1}\dot{3}\) and eg \(100x = 0.1313\ldots\), \(10\,000x = 13.1313\ldots\) with intention to subtract.
A1: for completion to \(\dfrac{557}{1980}\) 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)