4400 Higher November 2010 Paper 3 Q5
5
(a) Show that \(\dfrac{6}{7} \div 4 = \dfrac{3}{14}\) (2)
(b) Show that \(3\dfrac{2}{5} - 1\dfrac{2}{3} = 1\dfrac{11}{15}\) (3)
| Scheme | Marks |
|---|---|
| \(\frac{6}{7} \times \frac{1}{4}\) | M1 |
| \(\frac{6}{28}\) or \(\frac{3}{7} \times \frac{1}{2}\) | A1 |
| (2) |
Notes
M1: or \(\frac{6}{7} \div \frac{28}{7}\)
A1: answer \(\equiv \frac{3}{14}\) (but not \(= \frac{3}{14}\)) or cancelling
| Scheme | Marks |
|---|---|
| \(\frac{51}{15}\) and \(\frac{25}{15}\) any multiple of 15 valid | M1 |
| \(\frac{51}{15} - \frac{25}{15}\) correct fractions subtracted | M1 |
| \(\frac{26}{15}\) | A1 |
| (3) | |
| (5 marks) |
Notes
M1: \(\frac{6}{15}\) and \(\frac{10}{15}\)
M1: dep \(-\frac{4}{15}\) or \(\frac{6}{15} - \frac{10}{15}\)
A1: (dep on M2) \(2 - \frac{4}{15}\) oe (but not \(1\frac{11}{15}\))