Higher June 2018 Paper 2R Q17
17
(a) Use algebra to show that \(0.4\dot{3}\dot{6} = \dfrac{24}{55}\) (2)
(b) Show that \(\dfrac{\sqrt{20} + \sqrt{80}}{\sqrt{3}}\) can be expressed in the form \(\sqrt{a}\) where \(a\) is an integer.
Show your working clearly. (3)
Show your working clearly. (3)
| Scheme | Marks |
|---|---|
| eg \(x = 0.4\dot{3}\dot{6}\) and \(100x = 43.6\dot{3}\dot{6}\) or \(10x = 4.\dot{3}\dot{6}\) and \(1000x = 436.\dot{3}\dot{6}\) | M1 |
\(99x = 43.2\), \(x = \dfrac{43.2}{99}\) or \(990x = 432\), \(x = \dfrac{432}{990}\) Answer: show | A1 |
| (2) |
Notes
M1: eg two decimals that when subtracted give a finite decimal
A1: ‘show that’ completed: arrives at the given answer from correct working.
| Scheme | Marks |
|---|---|
| M1 | |
| M1dep | |
| \(\sqrt{60}\) | A1 |
| (3) | |
| (5 marks) |
Notes
M1: for \(\sqrt{20} = 2\sqrt{5}\) and \(\sqrt{80} = 4\sqrt{5}\) or \(\dfrac{\sqrt{20} + \sqrt{80}}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}\) or \(\dfrac{\sqrt{20} + 2\sqrt{20}}{\sqrt{3}}\)
M1dep: for \(\dfrac{6\sqrt{15}}{3}\) or \(2\sqrt{15}\) or \(\dfrac{3\sqrt{60}}{3}\) oe
A1: dep on M2, accept \(a = 60\)