Higher June 2023 Paper 2 Q1
1 Show that \(\;4\dfrac{2}{3} \div 1\dfrac{1}{5} = 3\dfrac{8}{9}\)
(3)
| Scheme | Marks |
|---|---|
| eg \(\dfrac{14}{3}\) and \(\dfrac{6}{5}\) | M1 |
| \(\dfrac{14}{3} \times \dfrac{5}{6}\) oe or \(\dfrac{70}{15} \div \dfrac{18}{15}\) | M1 |
eg \(\dfrac{14}{3} \times \dfrac{5}{6} = \dfrac{70}{18} = \dfrac{35}{9} = 3\dfrac{8}{9}\) or \(\dfrac{14}{3} \times \dfrac{5}{6} = \dfrac{70}{18} = 3\dfrac{16}{18} = 3\dfrac{8}{9}\) or \(\dfrac{\cancel{14}^{\,7}}{3} \times \dfrac{5}{\cancel{6}_{\,3}} = \dfrac{35}{9} = 3\dfrac{8}{9}\) or \(\dfrac{14}{3} \div \dfrac{6}{5} = \dfrac{70}{15} \div \dfrac{18}{15} = \dfrac{70}{18} = \dfrac{35}{9} = 3\dfrac{8}{9}\) or correct working to \(\dfrac{35}{9}\) and writing \(3\dfrac{8}{9} = \dfrac{35}{9}\) (may be earlier in working) working required Answer: Shown | A1 |
| (3) | |
| (3 marks) |
Notes
M1: both fractions expressed as correct improper fractions, no need for ÷ or × may be equivalent to those given eg \(\dfrac{70}{15}\) or \(\dfrac{18}{15}\) etc. A student could invert \(\dfrac{6}{5}\) and go straight to the 2nd M1, this mark is then implied.
M1: For inverting 2nd fraction and showing intention to multiply or for both fractions expressed as correct equivalent fractions with the same denominator with intention to divide eg \(\dfrac{70}{15} \div \dfrac{18}{15}\)
A1: Dep on M2 for conclusion to \(3\dfrac{8}{9}\) from correct working – either sight of the result of the multiplication or division e.g. \(\dfrac{70}{18}\) must be seen or correct cancelling prior to the multiplication to \(\dfrac{35}{9}\) OR writing \(3\dfrac{8}{9} = \dfrac{35}{9}\) (maybe on first line of working) and correct working as far as LHS \(= \dfrac{35}{9}\)
NB: marks are awarded for use of fractions not decimals (but allow a decimal check of answer)