Foundation January 2021 Paper 1 Q16
16 Show that \(\;\dfrac{5}{6} - \dfrac{3}{8} = \dfrac{11}{24}\)
(2)
| Scheme | Marks |
|---|---|
| e.g. \(\dfrac{20}{24}\) and \(\dfrac{9}{24}\) or \(\dfrac{40}{48}\) and \(\dfrac{18}{48}\) or \(\dfrac{20n}{24n}\) and \(\dfrac{9n}{24n}\) | M1 |
\(\dfrac{20}{24} - \dfrac{9}{24} = \dfrac{11}{24}\) \(\dfrac{40}{48} - \dfrac{18}{48} = \dfrac{22}{48} = \dfrac{11}{24}\) \(\dfrac{20n}{24n} - \dfrac{9n}{24n} = \dfrac{11n}{24n} = \dfrac{11}{24}\) Answer: Shown | A1 |
| (2) | |
| (2 marks) |
Notes
M1: for finding a common denominator with at least one fraction correct
A1: dep on M1, for a complete correct method leading to \(\dfrac{11}{24}\)