4MA0 Foundation January 2014 Paper 2R Q18
18
(a) Show that \(\dfrac{4}{5} + \dfrac{2}{3} = 1\dfrac{7}{15}\) (2)
(b) Show that \(2\dfrac{1}{4} \div 3\dfrac{1}{2} = \dfrac{9}{14}\) (3)
| Scheme | Marks |
|---|---|
| \(\dfrac{3 \times 4}{15} + \dfrac{5 \times 2}{15}\) or \(\dfrac{12}{15} + \dfrac{10}{15}\) | M1 |
| \(\dfrac{22}{15}\) | A1 |
| (2) |
Notes
M1: Any pair of correct fractions with a denominator a multiple of 15
A1: Dependent on first M1
In this question a correct answer does not on its own imply a correct method.
| Scheme | Marks |
|---|---|
| \(\dfrac{9}{4} \div \dfrac{7}{2}\) | M1 |
| \(\dfrac{9}{4} \times \dfrac{2}{7}\) oe | M1 |
| \(\dfrac{18}{28}\) | A1 |
| (3) | |
| (Total 5 marks) |
Notes
M1: May be implied by second M1
A1: Award A1 for \(\dfrac{9}{14}\) if cancelling seen to have taken place.
In this question a correct answer does not on its own imply a correct method.
Alt. method
| Scheme | Marks |
|---|---|
| \(\dfrac{9}{4} \div \dfrac{7}{2}\) | M1 |
| \(\dfrac{9}{4} \div \dfrac{14}{4}\) | M1 |
| \(\dfrac{9}{14}\) | A1 |
Notes
M1: May be implied by second M1
M1: Denominators must be the same.