Higher June 2023 Paper 3 Q20
20 Prove algebraically that \(0.1\dot{2}\dot{3}\) can be written as \(\dfrac{61}{495}\) (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| Proof | M1 | for \((10x =)\ 1.2323\ldots\) or \((100x =)\ 12.323\ldots\) or \((1000x =)\ 123.23\ldots\) |
| M1 | (dep M1) for a method using two recurring decimals that leads to a terminating decimal difference, using correct multiples of \(x\) eg \((1000x - 10x =)\ 123.23\ldots - 1.23\ldots\ (= 122)\) or \(\dfrac{122}{990}\) or \((100x - x =)\ 12.323\ldots - 0.123\ldots\ (= 12.2)\) or \(\dfrac{12.2}{99}\) | |
| C1 | for completing algebra to \(\dfrac{61}{495}\) |
Additional guidance
Any recurring notation acceptable throughout.
Proofs with terminating decimals (at least 5 figures) score M1M1C0