Higher June 2025 Paper 2 Q18
18 Use algebra to show that \(0.\dot{7}0\dot{2} = \dfrac{26}{37}\)
(2)
| Scheme | Marks |
|---|---|
| (1000\(x\) =) 702.702… eg (\(x\)=) 0.702… (10000\(x\) =) 7027.02… or (10\(x\)=) 7.02… (100000\(x\) =) 70270.20… or (100\(x\)=) 70.20… | M1 |
eg 1000\(x\) − \(x\) = 702.70… − 0.702…= 702 (999\(x\) = 702) and \(\dfrac{702}{999} = \dfrac{26}{37}\) or 10000\(x\) – 10\(x\) = 7027.02… – 7.02… = 7020 (9990\(x\) = 7020) and \(\dfrac{7020}{9990} = \dfrac{26}{37}\) or 100000\(x\) –100\(x\) = 70270.20… – 70.20 = 70200 (99900\(x\) = 70200) and \(\dfrac{70200}{99900} = \dfrac{26}{37}\) Working required Answer: shown | A1 |
| (2) | |
| (2 marks) |
Notes
M1: for 2 recurring decimals that when subtracted give a whole number or terminating decimal with intention to subtract. (ie give 702 or 7020 or 70200 etc)
eg
(1000\(x\) =) 702.702… and (\(x\) =) 0.702….
or
(10000\(x\)) = 7027.02.… and (10\(x\)) = 7.02.…
or
(100000\(x\) =) 70270.20.… and (100\(x\) =) 70.20.…
with intention to subtract
\(x\) is not required to award this mark
(if recurring dots not shown in both numbers then showing at least one of the numbers to at least 6sf)
NB Accept bar notation for dot notation to indicate recurring decimals
A1: for completion to \(\dfrac{26}{37}\) dep on M1 and must use algebra for this final mark to be awarded
No algebra used gets a maximum of 1 mark