Higher November 2022 Paper 1 Q21
21 Convert \(\quad 0.6\dot{1} \quad\) to a fraction. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(10x = 6.11\ldots\) and \(x = 0.61\ldots\) or \(100x = 61.11\ldots\) and \(10x = 6.11\ldots\) | M1 | oe two powers of 10 |
| \(10x - x = 6.11\ldots - 0.61\ldots\) or \(9x = 5.5\) | M1dep | oe subtraction of powers of 10 eg \(100x - 10x = 61.1\ldots - 6.1\ldots\) |
| \(\dfrac{11}{18}\) or \(\dfrac{55}{90}\) or \(\dfrac{605}{990}\) | A1 | oe fraction |
| Alternative method 2 | ||
| \((0.6\dot{1} =)\ 0.6 + 0.0\dot{1}\) and \(10x = 0.11\ldots\) and \(x = 0.01\ldots\) or \(100x = 1.11\ldots\) and \(10x = 0.11\ldots\) | M1 | oe two powers of 10 |
| \(10x - x = 0.11\ldots - 0.01\ldots\) or \(9x = 0.1\) and \(\dfrac{6}{10} +\) their \(\dfrac{1}{90}\) | M1dep | oe subtraction of powers of 10, with \(x\) evaluated as a fraction and added to \(\dfrac{6}{10}\) eg \(1000x - 10x = 11.11\ldots - 0.11\ldots\) or \(990x = 11\) and \(\dfrac{3}{5} + \dfrac{11}{990}\) sum of correct fractions implies M1M1 |
| \(\dfrac{11}{18}\) or \(\dfrac{55}{90}\) or \(\dfrac{605}{990}\) | A1 | oe fraction |
Additional guidance
| Ignore incorrect simplification of a correct fraction eg \(\dfrac{605}{990}\) and \(\dfrac{121}{190}\) | M1M1A1 |
| Otherwise correct fraction with fraction(s) or decimal(s) as the numerator and/or denominator, eg \(\dfrac{5.5}{9}\) | M1M1A0 |