1. **State the problem:** Two acid solutions with concentrations 25% and 55% are mixed in ratio $m:n$ to form mixture $M$. 2. **Let the quantities be:** Amount of first solution = $m$ units Amount of second solution = $n$ units 3. **Initial acid concentration in mixture $M$: ** $$\frac{25m + 55n}{m+n}$$ 4. **After removing 20% of mixture $M$ and replacing it with pure water (0% acid):** The total volume remains $m+n$, but acid amount reduces because 20% is removed and replaced with no acid. Amount of acid after replacement: $$0.8 \times (25m + 55n)$$ New concentration is 33%, so: $$\frac{0.8(25m + 55n)}{m+n} = 33$$ 5. **Solve for $m:n$:** Multiply both sides by $m+n$: $$0.8(25m + 55n) = 33(m+n)$$ Simplify left side: $$0.8 \times 25m + 0.8 \times 55n = 33m + 33n$$ $$20m + 44n = 33m + 33n$$ Bring all terms to one side: $$20m + 44n - 33m - 33n = 0$$ $$-13m + 11n = 0$$ Therefore, $$13m = 11n$$ So, $$\frac{m}{n} = \frac{11}{13}$$ **Final answer:** $m:n = 11:13$