Higher January 2022 Paper 2 Q2
2 Show that \(6\dfrac{3}{4} \div 2\dfrac{4}{7} = 2\dfrac{5}{8}\)
(3)
| Scheme | Marks |
|---|---|
| eg \(\dfrac{27}{4}\) and \(\dfrac{18}{7}\) | M1 |
\(\dfrac{27}{4} \times \dfrac{7}{18}\) oe or eg \(\dfrac{189}{28} \div \dfrac{72}{28}\) | M1 |
eg \(\dfrac{27}{4} \times \dfrac{7}{18} = \dfrac{189}{72} = \dfrac{21}{8} = 2\dfrac{5}{8}\) or \(\dfrac{27}{4} \times \dfrac{7}{18} = \dfrac{189}{72} = 2\dfrac{45}{72} = 2\dfrac{5}{8}\) or \(\dfrac{\cancel{27}^{\,3}}{4} \times \dfrac{7}{\cancel{18}_{\,2}} = \dfrac{21}{8} = 2\dfrac{5}{8}\) or \(\dfrac{189}{28} \div \dfrac{72}{28} = \dfrac{189}{72} = 2\dfrac{45}{72} = 2\dfrac{5}{8}\) oe if the student clearly shows \(2\dfrac{5}{8} = \dfrac{21}{8}\) then they only need to complete the LHS to \(\dfrac{21}{8}\) (often done in 1st line of working) Answer: shown | A1 |
| (3) | |
| (3 marks) |
Notes
M1: Both fractions expressed as improper fractions.
M1: Invert 2nd fraction or for both fractions expressed as equivalent fractions with denominators that are a common multiple of 4 and 7
(seeing this stage gains M2)
A1: dep M2 conclusion to \(2\dfrac{5}{8}\) from correct working – either sight of the result of the multiplication e.g. \(\dfrac{189}{72}\) must be seen then cancelled or correct cancelling prior to the multiplication with \(\dfrac{21}{8}\) seen.
NB entire solution using decimals scores no marks.