Foundation November 2020 Paper 2R Q11
11 Show that \(\quad \dfrac{5}{12} + \dfrac{3}{8} = \dfrac{19}{24}\)
(2)
| Scheme | Marks |
|---|---|
\(\dfrac{10}{24} + \dfrac{9}{24}\) or \(\dfrac{10n}{24n} + \dfrac{9n}{24n}\) or eg \(\dfrac{40 + 36}{96} \left(= \dfrac{76}{96}\right)\) | M1 |
\(\dfrac{10}{24} + \dfrac{9}{24} = \dfrac{19}{24}\) or eg \(\dfrac{40 + 36}{96} = \dfrac{76}{96} = \dfrac{19}{24}\) Working required Answer: clearly shown | A1 |
| (2) | |
| (2 marks) |
Notes
M1: for writing a sum, and each fraction with a common denominator, eg \(\dfrac{10}{24} + \dfrac{9}{24}\)
A1: dep on M1
continued to clearly show given result