Foundation January 2022 Paper 2R Q16
16
(a) Show that \(\dfrac{3}{8} \div \dfrac{27}{32} = \dfrac{4}{9}\) (2)
(b) Show that \(\dfrac{5}{6} - \dfrac{3}{8} = \dfrac{11}{24}\) (2)
| Scheme | Marks |
|---|---|
| eg \(\dfrac{3}{8} \times \dfrac{32}{27}\) or \(\dfrac{12}{32} \div \dfrac{27}{32}\) | M1 |
eg \(\dfrac{3}{8} \times \dfrac{32}{27} = \dfrac{96}{216} = \dfrac{4}{9}\) or \(\dfrac{12}{32} \div \dfrac{27}{32} = \dfrac{12}{27} = \dfrac{4}{9}\) or eg \(\dfrac{\cancel{3}^{\,1}}{\cancel{8}_{\,1}} \times \dfrac{\cancel{32}^{\,4}}{\cancel{27}_{\,9}} = \dfrac{4}{9}\) Answer: Shown | A1 |
| (2) |
Notes
M1: Inverting \(\dfrac{27}{32}\) and changing to multiply
or writing both factions with the same denominator.
A1: Conclusion to \(\dfrac{4}{9}\) - either sight of the result of the multiplication eg \(\dfrac{96}{216}\) or \(\dfrac{48}{108}\) or \(\dfrac{24}{54}\) must be seen
or fully correct cancelling must be seen prior to multiplication
NB use of decimals scores no marks.
| Scheme | Marks |
|---|---|
| eg \(\dfrac{40}{48} - \dfrac{18}{48}\) or \(\dfrac{20}{24} - \dfrac{9}{24}\) | M1 |
eg \(\dfrac{40}{48} - \dfrac{18}{48} = \dfrac{22}{48} = \dfrac{11}{24}\) or \(\dfrac{20}{24} - \dfrac{9}{24} = \dfrac{11}{24}\) Answer: Shown | A1 |
| (2) | |
| (4 marks) |
Notes
M1: for correct fractions with a common denominator of 24 or a multiple of 24
A1: dep M1 for a correct answer from fully correct working.