Higher January 2021 Paper 1R Q1
1 Show that \(3\dfrac{1}{5} \times 1\dfrac{5}{6} = 5\dfrac{13}{15}\)
(3)
| Scheme | Marks |
|---|---|
| e.g. \(\dfrac{16}{5}\) and \(\dfrac{11}{6}\) or \(\dfrac{96}{30}\) and \(\dfrac{55}{30}\) | M1 |
| e.g. \(\dfrac{\cancel{16}^8}{5} \times \dfrac{11}{\cancel{6}^3}\) or \(\dfrac{176}{30}\) or \(\dfrac{5280}{900}\) oe | M1 |
e.g. \(\dfrac{16}{5} \times \dfrac{11}{6} = \dfrac{176}{30} = \dfrac{88}{15} = 5\dfrac{13}{15}\) or \(\dfrac{16}{5} \times \dfrac{11}{6} = \dfrac{176}{30} = 5\dfrac{26}{30} = 5\dfrac{13}{15}\) or \(\dfrac{\cancel{16}^8}{5} \times \dfrac{11}{\cancel{6}^3} = \dfrac{88}{15} = 5\dfrac{13}{15}\) or \(\dfrac{96}{30} \times \dfrac{55}{30} = \dfrac{5280}{900} = \dfrac{88}{15} = 5\dfrac{13}{15}\) NB: a student can show initially that \(5\dfrac{13}{15} = \dfrac{88}{15}\) and they need to show that LHS = \(\dfrac{88}{15}\) Working required Answer: shown | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for two correct improper fractions
M1: correct cancelling or multiplication of numerators and denominators without cancelling
A1: Dep on M2 for conclusion to \(5\dfrac{13}{15}\) from correct working – either sight of the result of the multiplication e.g. \(\dfrac{176}{30}\) must be seen and equated to \(\dfrac{88}{15}\) or \(5\dfrac{26}{30}\)
or
correct cancelling prior to the multiplication to \(\dfrac{88}{15}\)
NB: use of decimals scores no marks