Let $t$ be TNYWR.

A water tank initially contains $x$ litres of water. Pablo adds $\frac{t}{2}$ litres of water to the tank, making the tank $20 \%$ full. After Pablo adds water to the tank, Chloe then adds $\frac{t^{2}}{4}$ litres of water to the tank, making the tank $50 \%$ full. What is the value of $x$ ?

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Let $V$ be the volume of the tank.

Using the second given piece of information, we know that when $\frac{t^{2}}{4}$ litres are added, the amount of water goes from $\frac{2 \mathrm{~V}}{10}$ to $\frac{5 \mathrm{~V}}{10}$, which means $\frac{t^{2}}{4}$ is the difference between these two amounts of water.

Therefore, $\frac{t^{2}}{4}=\frac{5 V-2 V}{10}=\frac{3 V}{10}$.

Solving for $V$ gives $V=\frac{10}{3} \times \frac{t^{2}}{4}=\frac{10 t^{2}}{12}=\frac{5 t^{2}}{6}$.

Since adding $\frac{t}{2}$ litres to the initial $x$ litres leads to the tank having $\frac{2 V}{10}$ litres in it, $x=\frac{2 V}{10}-\frac{t}{2}$.

Substituting $V=\frac{5 t^{2}}{6}$ gives $x=\frac{2}{10} \times \frac{5 t^{2}}{6}-\frac{t}{2}=\frac{t^{2}}{6}-\frac{t}{2}$.

Substituting $t=6$, we get $x=\frac{6^{2}}{6}-3=3$.