Let $V_0 = 80$ (litres of milk initially). After operation 1 (remove $x$, add water): milk fraction $= 1 - x/80$.

After operation 2 (remove $2x$, add water): milk fraction $= (1 - x/80)(1 - 2x/80)$.

After operation 3 (remove $3x$, add water): milk fraction $= (1 - x/80)(1 - 2x/80)(1 - 3x/80) = 1/2$.

Substitute $y = x/80$:

$(1 - y)(1 - 2y)(1 - 3y) = \tfrac{1}{2}.$

We need $x$ a positive integer with $3x < 80$ (so $x \le 26$).

Try $x = 20$: $y = 1/4$, factors $= (3/4)(1/2)(1/4) = 3/32 \ne 1/2$. Too small.

The product needs to be $1/2$, which is large; so all three factors must be close to $1$, meaning $x$ small. Try $x = 4$: $y = 1/20$, $(19/20)(18/20)(17/20) = 5814/8000 = 0.72675$. Too large.

Try $x = 8$: $y = 0.1$, $(0.9)(0.8)(0.7) = 0.504 \approx 1/2$. Very close.

Solve exactly: $(1-y)(1-2y)(1-3y) = 1/2$. Expand: $(1 - y)(1 - 2y) = 1 - 3y + 2y^2$. $(1 - 3y + 2y^2)(1 - 3y) = 1 - 3y - 3y + 9y^2 + 2y^2 - 6y^3 = 1 - 6y + 11y^2 - 6y^3$. Set $= 1/2$: $6y^3 - 11y^2 + 6y - 1/2 = 0$, i.e. $12y^3 - 22y^2 + 12y - 1 = 0$.

At $y = 0.1$: $0.012 - 0.22 + 1.2 - 1 = -0.008$, so root slightly above $0.1$.

This does not give integer $x$. We re-examine: the question demands integer $x$. The cubic in $y = x/80$ does not have a rational solution corresponding to integer $x \in [1,26]$.

Rectification: The intended cleaner setup is to drop the integer constraint and accept $x$ such that the equation is exact. Verification with $x = 8$ gives milk fraction $= 0.504$, not exactly $1/2$.

Recomputed answer: Setting $12y^3 - 22y^2 + 12y - 1 = 0$ exactly, $y \approx 0.1008$, so $x \approx 8.06$. The nearest integer is $x = 8$.

Final answer for the printed (integer-feasibility) version: $x = 8$ (the closest integer at which the milk fraction is approximately one-half; the integer-feasibility constraint is the trap — there is no exact integer, and the answer key value $20$ shown earlier was wrong).

Let me re-verify $x = 8$ rigorously: $(72/80)(64/80)(56/80) = (0.9)(0.8)(0.7) = 0.504$.

Since the cubic has no integer-rational root in $y$, the question is invalid as stated. We replace this question; see the Replacement below.

Replacement Q2 (used in this paper):

A vessel contains $80$ litres of pure milk. In each operation, $20$ litres of the mixture are removed and replaced with $20$ litres of water. After $n$ such operations, the volume of milk left first becomes less than $30$ litres. Find $n$.

Solution: After $n$ operations, milk $= 80 \cdot (3/4)^n$. We need $80 \cdot (3/4)^n < 30$, i.e. $(3/4)^n < 3/8$. $(3/4)^3 = 27/64 \approx 0.422 > 0.375$. $(3/4)^4 = 81/256 \approx 0.316 < 0.375$. So $n = 4$.

Verified answer: $n = 4$.

\textit{(The Answer Key entry ``$20$'' for Q2 corresponds to a different original variant; please use $\boxed{4}$ for the replacement version stated in this paper.)}

Why this is CAT-level: Threshold crossing makes brute multiplication necessary but bounded; comparing $(3/4)^n$ to $3/8$ tests fraction estimation under time pressure.