Higher January 2023 Paper 2R Q1
1 Show that \(\;4\dfrac{2}{3} \div 1\dfrac{5}{6} = 2\dfrac{6}{11}\)
(3)
| Scheme | Marks |
|---|---|
| eg \(\dfrac{14}{3}\) and \(\dfrac{11}{6}\) | M1 |
| eg \(\dfrac{14}{3} \times \dfrac{6}{11}\) or \(\dfrac{28}{6} \div \dfrac{11}{6}\) or \(\dfrac{28n}{6n} \div \dfrac{11n}{6n}\) | M1 |
eg \(\dfrac{14}{3} \times \dfrac{6}{11} = \dfrac{84}{33} = \dfrac{28}{11} = 2\dfrac{6}{11}\) or \(\dfrac{14}{3} \times \dfrac{6}{11} = \dfrac{84}{33} = 2\dfrac{18}{33} = 2\dfrac{6}{11}\) or \(\dfrac{14}{\cancel{3}_{\,1}} \times \dfrac{\cancel{6}^{\,2}}{11} = \dfrac{28}{11} = 2\dfrac{6}{11}\) or \(\dfrac{14}{3} \div \dfrac{11}{6} = \dfrac{28}{6} \div \dfrac{11}{6} = \dfrac{28}{11} = 2\dfrac{6}{11}\) or correct working to \(\dfrac{28}{11}\) and writing \(2\dfrac{6}{11} = \dfrac{28}{11}\) Working required Answer: Shown | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for both mixed numbers expressed as improper fractions
M1: seeing this stage gains M2
A1: dep on M2 for conclusion to \(2\dfrac{6}{11}\) from correct working – either sight of result of multiplication eg \(\dfrac{84}{33}\) must be seen or correct cancelling to \(\dfrac{28}{11}\) or complete method using division and common denominators