Calculate the average accurate titre volume of aqueous silver nitrate (AgNO₃).

The accurate titre readings are $30.0 \text{ cm}^3$ and $29.8 \text{ cm}^3$.

$$\text{Average titre} = \frac{30.0 \text{ cm}^3 + 29.8 \text{ cm}^3}{2} = \frac{59.8 \text{ cm}^3}{2} = 29.9 \text{ cm}^3$$

The average titre is $29.9 \text{ cm}^3$.

Therefore, $\boxed{29.9}$ $\text{cm}^3$ titre (t) of aqueous silver nitrate reacts with $25.0 \text{ cm}^3$ of aqueous hydrochloric acid.

The reaction is a precipitation reaction where silver ions react with chloride ions.

$$\text{AgNO}_3(\text{aq}) + \text{HCl}(\text{aq}) \longrightarrow \text{AgCl}(\text{s}) + \text{HNO}_3(\text{aq})$$

From the equation, the mole ratio of $\text{AgNO}_3$ to HCl is $1:1$.

Given concentration of $\text{AgNO}_3 = 0.1 \text{ M}$.
Volume of $\text{AgNO}_3$ used (average titre) $= 29.9 \text{ cm}^3 = 29.9 \times 10^{-3} \text{ dm}^3$.

$$\text{Moles of AgNO}_3 = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of AgNO}_3 = 0.1 \text{ mol/dm}^3 \times 29.9 \times 10^{-3} \text{ dm}^3$$ $$\text{Moles of AgNO}_3 = 2.99 \times 10^{-3} \text{ mol}$$

Since the mole ratio of $\text{AgNO}_3 : \text{HCl}$ is $1:1$:

$$\text{Moles of HCl} = \text{Moles of AgNO}_3 = 2.99 \times 10^{-3} \text{ mol}$$

Volume of HCl used $= 25.0 \text{ cm}^3 = 25.0 \times 10^{-3} \text{ dm}^3$.

$$\text{Concentration of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl}}$$ $$\text{Concentration of HCl} = \frac{2.99 \times 10^{-3} \text{ mol}}{25.0 \times 10^{-3} \text{ dm}^3}$$ $$\text{Concentration of HCl} = \frac{2.99}{25.0} \text{ mol/dm}^3$$ $$\text{Concentration of HCl} = 0.1196 \text{ mol/dm}^3$$

The concentration of hydrochloric acid is $\boxed{\text{0.1196 M}}$.