Higher November 2024 Paper 2 Q15
15 Use algebra to show that \(0.7\dot{6}\dot{3} = \dfrac{42}{55}\)
(2)
| Scheme | Marks |
|---|---|
| eg (10 000\(x\) =) 7636.36… (100\(x\) =) 76.36… or (1000\(x\) =) 763.63… (10\(x\) =) 7.63… or (100\(x\) =) 76.363… (\(x\) =) 0.763… oe | M1 |
eg \(10\,000x - 100x = 7636.36\ldots - 76.36\ldots = 7560\) and \(\dfrac{7560}{9900} = \dfrac{42}{55}\) or \(1000x - 10x = 763.63\ldots - 7.63\ldots = 756\) \((990x = 756)\) and \(\dfrac{756}{990} = \dfrac{42}{55}\) or \(100x - x = 76.363\ldots - 0.763\ldots = 75.6\) and \(\dfrac{75.6}{99} = \dfrac{42}{55}\) \((99x = 75.6)\) or \(\dfrac{7}{10} + \dfrac{63}{990} = \dfrac{7 \times 99 + 63}{990} = \dfrac{42}{55}\) Working required Answer: shown | A1 |
| (2) | |
| (2 marks) |
Notes
M1: For 2 recurring decimals that when subtracted give a whole number or terminating decimal with intention to subtract.
(ie give 75.6 or 756 or 7560 etc)
eg (10 000\(x\) =) 7636.36... and (100\(x\) =) 76.36....
or (1000\(x\) =) 763.63… and (10\(x\) =) 7.63…
or (100\(x\) =) 76.363… and (\(x\) =) 0.763…
with intention to subtract.
(if recurring not shown then showing at least one of the numbers to at least 5sf)
or \(\dfrac{7}{10} + 1000x(63.63) - 10x(0.63)\)
A1: for completion to \(\dfrac{42}{55}\) dep on M1 and must use algebra for this final mark to be awarded
[allow for instance \(99x = 75.6\) and then \(\dfrac{756}{990} = \dfrac{42}{55}\)]
No algebra used gets a maximum of 1