Higher November 2018 Paper 1 Q16
16 Prove algebraically that \(0.2\dot{5}\dot{6}\) can be written as \(\dfrac{127}{495}\) (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| Proof with \(\dfrac{127}{495}\) | M1 | 0.25656... or 0.2 + 0.05656.. or \((10 \times 0.2\dot{5}\dot{6} =)\ 2.\dot{5}\dot{6}\) or 2.5656… or \((100 \times 0.2\dot{5}\dot{6} =)\ 25.\dot{6}\dot{5}\) or 25.6565…or \((1000 \times 0.2\dot{5}\dot{6} =)\ 256.\dot{5}\dot{6}\) or 256.5656… |
| M1 | for finding two correct recurring decimals that when subtracted would result in a terminating decimal or integer, eg. 256.5656….. – 2.5656….. or 25.6565….. – 0.25656….. or \(256.\dot{5}\dot{6} - 2.\dot{5}\dot{6}\) or \(25.\dot{6}\dot{5} - 0.2\dot{5}\dot{6}\) or for \(\dfrac{254}{990}\) or \(\dfrac{25.4}{99}\) | |
| C1 | full proof seen with \(\dfrac{127}{495}\) |