Foundation November 2020 Paper 1R Q17
17 Show that \(\quad 3\dfrac{3}{4} \times \dfrac{7}{9} = 2\dfrac{11}{12}\)
(3)
| Scheme | Marks |
|---|---|
| e.g. \(\dfrac{15}{4}\) | M1 |
| e.g. \(\dfrac{\cancel{15}^{\,5}}{4} \times \dfrac{7}{\cancel{9}_{\,3}}\) OR \(\dfrac{105}{36}\) oe | M1 |
e.g. \(\dfrac{\cancel{15}^{\,5}}{4} \times \dfrac{7}{\cancel{9}_{\,3}} = \dfrac{35}{12} = 2\dfrac{11}{12}\) or \(\dfrac{15}{4} \times \dfrac{7}{9} = \dfrac{105}{36} = \dfrac{35}{12} = 2\dfrac{11}{12}\) or \(\dfrac{15}{4} \times \dfrac{7}{9} = \dfrac{105}{36} = 2\dfrac{33}{36} = 2\dfrac{11}{12}\) Working required Answer: shown | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for \(3\dfrac{3}{4}\) expressed as an improper fraction
M1: correct cancelling or multiplication of numerators and denominators without cancelling
A1: dep on M2, for conclusion to \(2\dfrac{11}{12}\) from correct working – either sight of the result of the multiplication e.g. \(\dfrac{105}{36}\) oe must be seen
or correct cancelling prior to the multiplication to \(\dfrac{35}{12}\)
NB: use of decimals scores no marks