Foundation June 2025 Paper 1R Q18
18 Show that \(2\dfrac{1}{4} \times 1\dfrac{5}{7} = 3\dfrac{6}{7}\)
(3)
| Scheme | Marks |
|---|---|
| eg \(\dfrac{9}{4}\) and \(\dfrac{12}{7}\) | M1 |
eg \(\dfrac{9}{\cancel{4}_1} \times \dfrac{\cancel{12}^3}{7}\) OR \(\dfrac{9}{4} \times \dfrac{12}{7} = \dfrac{108}{28}\) oe eg \(\dfrac{63}{28} \times \dfrac{48}{28} = \dfrac{3024}{784}\) | M1 |
eg \(\dfrac{9}{\cancel{4}_1} \times \dfrac{\cancel{12}^3}{7} = \dfrac{27}{7} = 3\dfrac{6}{7}\) or \(\dfrac{9}{4} \times \dfrac{12}{7} = \dfrac{108}{28} = \dfrac{27}{7} = 3\dfrac{6}{7}\) or \(\dfrac{9}{4} \times \dfrac{12}{7} = \dfrac{108}{28} = 3\dfrac{24}{28} = 3\dfrac{6}{7}\) or \(\dfrac{9}{4} \times \dfrac{12}{7} = \dfrac{63}{28} \times \dfrac{48}{28} = \dfrac{3024}{784} = \dfrac{27}{7} = 3\dfrac{6}{7}\) or \(\dfrac{9}{4} \times \dfrac{12}{7} = \dfrac{63}{28} \times \dfrac{48}{28} = \dfrac{3024}{784} = 3\dfrac{672}{784} = 3\dfrac{6}{7}\) or correct working to \(\dfrac{27}{7}\) and writing \(3\dfrac{6}{7} = \dfrac{27}{7}\) Working requiredAnswer: shown | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for \(2\dfrac{1}{4}\) and \(1\dfrac{5}{7}\) expressed as improper fractions
M1: correct cancelling or multiplication of numerators and denominators without cancelling
A1: dep on M2, for conclusion to \(3\dfrac{6}{7}\) from correct working – either sight of the result of the multiplication e.g. \(\dfrac{108}{28}\) oe must be seen
or correct cancelling prior to the multiplication to \(\dfrac{27}{7}\)
NB: use of decimals scores no marks unless as a check