Higher January 2023 Paper 2 Q1
1 Show that \(\;3\dfrac{5}{7} \div 1\dfrac{5}{8} = 2\dfrac{2}{7}\)
(3)
| Scheme | Marks |
|---|---|
| \(\dfrac{26}{7}, \dfrac{13}{8}\) oe | M1 |
| \(\dfrac{26}{7} \times \dfrac{8}{13}\) oe or eg \(\dfrac{208}{56} \div \dfrac{91}{56}\) | M1 |
eg \(\dfrac{26}{7} \times \dfrac{8}{13} = \dfrac{208}{91} = \dfrac{16}{7} = 2\dfrac{2}{7}\) or \(\dfrac{26}{7} \times \dfrac{8}{13} = \dfrac{208}{91} = 2\dfrac{26}{91} = 2\dfrac{2}{7}\) or \(\dfrac{\cancel{26}^{\,2}}{7} \times \dfrac{8}{\cancel{13}^{\,1}} = \dfrac{16}{7} = 2\dfrac{2}{7}\) or \(\dfrac{208}{56} \div \dfrac{91}{56} = \dfrac{208}{91} = \dfrac{16}{7} = 2\dfrac{2}{7}\) or correct working to \(\dfrac{16}{7}\) and writing \(2\dfrac{2}{7} = \dfrac{16}{7}\) (usually on the first line of working) working required Answer: shown | A1 |
| (3) | |
| (3 marks) |
Notes
M1: both fractions expressed as improper fractions, no need for ÷ or × may be equivalent to those given eg \(\dfrac{52}{14}, \dfrac{26}{16}\) etc. A student could invert \(\dfrac{13}{8}\) and show multiplication - as shown in the 2nd M1, this mark is then implied.
M1: or for both fractions expressed as equivalent fractions with denominators that are a common multiple of 7 and 8 eg \(\dfrac{208}{56} \div \dfrac{91}{56}\)
A1: dep on M2
NB: use of decimals scores no marks (unless used as a check)