Higher June 2024 Paper 2R Q18
18 Use algebra to show that \(\;0.\dot{3}0\dot{6} = \dfrac{34}{111}\)
(2)
| Scheme | Marks |
|---|---|
| eg \(1000x = 306.306\ldots\) \(\underline{\phantom{100}x = \phantom{30}0.306\ldots}\) OR eg \(1\,000\,000x = 306\,306.(\ldots)\) \(\underline{\phantom{1\,00}1000x = \phantom{306\,}306.306\ldots}\) | M1 |
eg \(1000x - x = 306.306.... - 0.306306.... = 306\) and \(\dfrac{306}{999} = \dfrac{34}{111}\) or \(999x = 306\) and \(\dfrac{306}{999} = \dfrac{34}{111}\) OR eg \(1\,000\,000x - 1000x = 306306.(\ldots) - 306.306 = 306\,000\) and \(\dfrac{306\,000}{999\,000} = \dfrac{34}{111}\) or \(999\,000x = 306\,000\) and \(\dfrac{306\,000}{999\,000} = \dfrac{34}{111}\) Working required Answer: shown | A1 |
| (2) | |
| (2 marks) |
Notes
M1: M1 for two correct algebraic equations involving recurring decimals that when subtracted give a whole number or terminating decimal (306 or 306 000 etc) with intention to subtract.
eg
\(1000x = 306.306...\) and \(x = 0.306....\)
or
\(1\,000\,000x = 306306.(\ldots)\) and \(1000x = 306.306\)
(if recurring dots not shown in both numbers then showing at least one of the numbers to at least 6sf)
A1: for completion to \(\dfrac{34}{111}\) dep on M1