Higher June 2019 Paper 1 Q1
1 Show that \(4\dfrac{2}{3} \div 1\dfrac{1}{9} = 4\dfrac{1}{5}\)
(3)
| Scheme | Marks |
|---|---|
| e.g. \(\dfrac{14}{3}\) and \(\dfrac{10}{9}\) | M1 |
| e.g. \(\dfrac{14}{3} \times \dfrac{9}{10}\) | M1 |
Working required e.g. \(\dfrac{14}{3} \times \dfrac{9}{10} = \dfrac{126}{30} = \dfrac{21}{5} = 4\dfrac{1}{5}\) or \(\dfrac{14}{3} \times \dfrac{9}{10} = \dfrac{126}{30} = 4\dfrac{6}{30} = 4\dfrac{1}{5}\) or \(\dfrac{\cancel{14}^{\,7}}{\cancel{3}_{\,1}} \times \dfrac{\cancel{9}^{\,3}}{\cancel{10}_{\,5}} = \dfrac{21}{5} = 4\dfrac{1}{5}\) or \(\dfrac{126}{27}, \dfrac{30}{27} = \dfrac{126}{30} = \dfrac{21}{5} = 4\dfrac{1}{5}\) Answer: Shown | A1 |
| (3) | |
| (3 marks) |
Notes
M1: Both fractions expressed as improper fractions
M1: or for both fractions expressed as equivalent fractions with denominators that are a common multiple of 3 and 9 eg. \(\dfrac{42}{9} \div \dfrac{10}{9}\) or \(\dfrac{126}{27}, \dfrac{30}{27}\)
A1: Dep on M2 for conclusion to \(4\dfrac{1}{5}\) from correct working – either sight of the result of the multiplication e.g. \(\dfrac{126}{30}\) must be seen or correct cancelling prior to the multiplication to \(\dfrac{21}{5}\)
NB: use of decimals scores no marks