Higher June 2025 Paper 1R Q16
16 Use algebra to show that the recurring decimal \(0.6\dot{1}\dot{2} = \dfrac{101}{165}\)
(2)
| Scheme | Marks |
|---|---|
| (10000\(x\) =) 6121.21... eg (100\(x\)=) 61.21... (1000\(x\) =) 612.12... or (10\(x\)=) 6.12... (100\(x\) =) 61.212... or (\(x\)=) 0.612... | M1 |
eg 10000\(x\) − 100\(x\) = 6121.21… − 61.21…=6060 (9900\(x\) = 6060) and \(\dfrac{6060}{9900} = \dfrac{101}{165}\) or 1000\(x\) – 10\(x\) = 612.12... – 6.12... = 606 (990\(x\) = 606) and \(\dfrac{606}{990} = \dfrac{101}{165}\) or 100\(x\) – \(x\) = 61.212... – 0.612... = 60.6 (99\(x\) = 60.6) and \(\dfrac{60.6}{99} = \dfrac{101}{165}\) oe OR 0.6 +… and (1000\(x\) –100\(x\) = 990\(x\) = 12) and \(0.6 + \dfrac{12}{990} = \dfrac{0.6 \times 990 + 12}{990} = \dfrac{101}{165}\) oe 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 60.6 or 606 or 6060 etc)
eg
(1000\(x\) =) 612.12... and (10\(x\) =) 6.12....
or
(100 000\(x\) =) 61 212.12.… and (1000\(x\) =) 612.…
or
(100\(x\) =) 61.212.… and (\(x\) =) 0.612.…
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 5sf)
or \(\dfrac{6}{10} + 1000x(12.12) - 10x(0.12)\)
A1: for completion to \(\dfrac{101}{165}\) dep on M1 and must use algebra for this final mark to be awarded
[allow for instance 99\(x\) = 60.6 and then \(\dfrac{606}{990} = \dfrac{101}{165}\)]
No algebra used gets a maximum of 1 mark