A recurring decimal repeats one or more digits forever. A dot over a single digit marks it as the repeating one, e.g. $0.\dot{3} = 0.3333\dots$. For a repeating block of two or more digits, dots go over the first and last digit of the block only — $0.\dot{5}\dot{4} = 0.545454\dots$ and $0.7\dot{6}\dot{3} = 0.7636363\dots$ (the $7$ doesn't repeat; the block $63$ does).

Let $x$ equal the decimal, then multiply by $10^n$ where $n$ is the number of repeating digits, so the decimal part lines up exactly. Subtracting cancels the infinite tail.

Convert $0.\dot{2}\dot{7}$: Let $x = 0.272727\dots$. The block has 2 digits, so multiply by $100$: $100x = 27.272727\dots$. Subtract: $100x - x = 27 \Rightarrow 99x = 27 \Rightarrow x = \frac{27}{99} = \frac{3}{11}$.

This is where most marks are lost — picking the wrong pair of multipliers. The reliable rule: multiply once to put the decimal point just before the repeating block, and again to put it just after the first full block. Subtracting those two equations always cancels perfectly.

Convert $0.4\dot{7}$: Let $x = 0.4777\dots$. Multiply to sit before the block: $10x = 4.777\dots$. Multiply to sit after one full block: $100x = 47.777\dots$. Subtract: $100x - 10x = 43 \Rightarrow 90x = 43 \Rightarrow x = \frac{43}{90}$.

Nothing changes — you'll just end up with an improper fraction. $7.3\dot{2} = 7.3222\dots$: $10x=73.222\dots$, $100x=732.222\dots$, so $90x = 659$, giving $x = \frac{659}{90} = 7\frac{29}{90}$.

You can't do arithmetic directly on the dots — convert each one to a fraction first, then operate. For $0.\dot{3}\dot{9} \div 0.\dot{6}\dot{3}$: these convert to $\frac{13}{33}$ and $\frac{7}{11}$, so the division is $\frac{13}{33} \times \frac{11}{7} = \frac{143}{231} = \frac{13}{21}$.

$x$ is a digit, $1 \le x \le 9$. Prove algebraically that $0.\dot{0}\dot{x} = \dfrac{x}{99}$.

1. Let $y = 0.\dot{0}\dot{x} = 0.0x0x0x\dots$ — the repeating block is "$0x$" (2 digits).
2. Multiply by $10^2=100$ to shift one full block: $100y = 0x.0x0x\dots = x.\dot{0}\dot{x}$ (since "$0x$" as a 2-digit lead is just the value $x$).
3. Subtract the original: $100y - y = x \Rightarrow 99y = x$.
4. So $y = \dfrac{x}{99}$, as required.

This question type — an algebraic unknown inside the recurring decimal — is a classic Grade 9 differentiator: you treat $x$ exactly like a specific digit and follow the same subtraction method.