Higher June 2021 Paper 2 Q2
2 Show that \(2\dfrac{4}{7} \div 1\dfrac{1}{8} = 2\dfrac{2}{7}\)
(3)
| Scheme | Marks |
|---|---|
| eg \(\dfrac{18}{7}\) and \(\dfrac{9}{8}\) oe | M1 |
| eg \(\dfrac{18}{7} \times \dfrac{8}{9}\) oe or oe \(\dfrac{144}{56} \div \dfrac{63}{56}\) | M1 |
eg \(\dfrac{18}{7} \times \dfrac{8}{9} = \dfrac{144}{63} = \dfrac{16}{7} = 2\dfrac{2}{7}\) or \(\dfrac{18}{7} \times \dfrac{8}{9} = \dfrac{144}{63} = 2\dfrac{18}{63} = 2\dfrac{2}{7}\) or \(\dfrac{\cancel{18}^{\,2}}{7} \times \dfrac{8}{\cancel{9}^{\,1}} = \dfrac{16}{7} = 2\dfrac{2}{7}\) or \(\dfrac{18}{7} \div \dfrac{9}{8} = \dfrac{144}{56} \div \dfrac{63}{56} = \dfrac{144}{63} = \dfrac{16}{7} = 2\dfrac{2}{7}\) or correct working to \(\dfrac{16}{7}\) and writing \(2\dfrac{2}{7} = \dfrac{16}{7}\) Answer: shown | A1 |
| (3) | |
| (3 marks) |
Notes
M1: both fractions expressed as improper fractions, no need for ÷ or × may be equivalent to those given eg \(\dfrac{36}{14}\) or \(\dfrac{27}{24}\) etc. A student could invert \(\dfrac{9}{8}\) and show multiplication - as shown in the 2nd M1, this mark is then implied.
M1: or for both fractions expressed as equivalent fractions with denominators that are a common multiple of 7 and 8 eg \(\dfrac{144}{56} \div \dfrac{63}{56}\)
A1: Dep on M2 for conclusion to \(2\dfrac{2}{7}\) from correct working – either sight of the result of the multiplication or division eg \(\dfrac{144}{63}\) must be seen and then cancelled or correct cancelling prior to the multiplication to \(\dfrac{16}{7}\)
or
writing \(2\dfrac{2}{7} = \dfrac{16}{7}\) (maybe on first line of working) and correct working as far as LHS \(= \dfrac{16}{7}\)
NB: use of decimals scores no marks