Let the specific heats of liquids $x, y, z$ be $c_x, c_y, c_z$ respectively.

Since equal masses are mixed, we can take each mass as $m$.

When $x$ at $10^\circ\text{C}$ is mixed with $y$ at $20^\circ\text{C}$, final temperature is $16^\circ\text{C}$.

Heat gained by $x$ = Heat lost by $y$: $m c_x (16-10) = m c_y (20-16)$ $6c_x = 4c_y$ $\frac{c_x}{c_y} = \frac{2}{3}$

When $y$ at $20^\circ\text{C}$ is mixed with $z$ at $30^\circ\text{C}$, final temperature is $26^\circ\text{C}$.

Heat gained by $y$ = Heat lost by $z$: $m c_y (26-20) = m c_z (30-26)$ $6c_y = 4c_z$ $\frac{c_z}{c_y} = \frac{3}{2}$

Express all specific heats in terms of $c_y$: $c_x = \frac{2}{3}c_y, \qquad c_z = \frac{3}{2}c_y$

Now mix $x$ at $10^\circ\text{C}$ and $z$ at $30^\circ\text{C}$. Let final temperature be $T$.

Heat gained by $x$ = Heat lost by $z$: $m c_x (T-10) = m c_z (30-T)$

Substitute $c_x = \frac{2}{3}c_y$ and $c_z = \frac{3}{2}c_y$: $\frac{2}{3}c_y(T-10) = \frac{3}{2}c_y(30-T)$

Cancel $c_y$: $\frac{2}{3}(T-10) = \frac{3}{2}(30-T)$

Solve: $4(T-10) = 9(30-T)$ $4T - 40 = 270 - 9T$ $13T = 310$ $T = \frac{310}{13} = 23.84^\circ\text{C}$

Compare with options:

• A: $28.32^\circ\text{C}$
• B: $25.62^\circ\text{C}$
• C: $23.84^\circ\text{C}$
• D: $20.28^\circ\text{C}$

Hence, the correct option is C.