Higher June 2024 Paper 1 Q18
18 Show that \(0.\dot{1}\dot{5} + 0.2\dot{2}\dot{7}\) can be written in the form \(\dfrac{m}{66}\) where \(m\) is an integer. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| Shown | M1 | for \(0.1515\ldots + 0.22727\ldots\ (= 0.37878\ldots\) or \(0.3\dot{7}\dot{8})\) |
| M1 | for finding two correct recurring decimals that when subtracted would result in a terminating decimal or integer, eg \((1000x - 10x =)\ 378.7878\ldots - 3.7878\ldots\ (= 375)\) or \(\dfrac{375}{990}\) or \((100x - x =)\ 37.8787\ldots - 0.37878\ldots\ (= 37.5)\) or \(\dfrac{37.5}{99}\) | |
| C1 | for correct working leading to \(\dfrac{25}{66}\) | |
| OR | ||
| M1 | for start of a method to convert \(0.1515\ldots\) or \(0.22727\ldots\) to a fraction, eg \(100x = 15.1515\ldots\) or \(\dfrac{15}{99}\) or \(\dfrac{5}{33}\) oe or \(10y = 2.2727\ldots\) or \(100y = 22.7272\ldots\) or \(1000y = 227.2727\ldots\) or \(\dfrac{225}{990}\) or \(\dfrac{22.5}{99}\) or \(\dfrac{5}{22}\) oe | |
| M1 | for a method to convert \(0.1515\ldots\) and \(0.22727\ldots\) to fractions, eg \((100x - x =)\ 15.1515\ldots - 0.1515\ldots\ (= 15)\) or \(\dfrac{15}{99}\) or \(\dfrac{5}{33}\) oe and \((1000y - 10y =)\ 227.2727\ldots - 2.2727\ldots\ (= 225)\) or \((100y - y =)\ 22.7272\ldots - 0.22727\ldots\ (= 22.5)\) or \(\dfrac{225}{990}\) or \(\dfrac{22.5}{99}\) or \(\dfrac{5}{22}\) oe | |
| C1 | for correct working leading to \(\dfrac{25}{66}\) |
Additional guidance
Recurring decimal notation acceptable for this mark
Recurring decimal notation acceptable for both M marks