A certain amount of water was poured into a 300-litre container, and the remaining portion of the container was filled with milk. Then an amount of this solution was taken out from the container, which was twice the volume of water that was earlier poured into it, and water was poured to refill the container again. If the resulting solution contains 72% milk, then the amount of water, in litres, that was initially poured into the container was:

Question

Accepted Answer

1. **Let the initial quantity of water in the container be $$ x $$:**  
   - Total capacity of the container = 300 litres.  
   - Initial quantity of milk = $$ 300 - x $$.

2. **Amount of mixture taken out:**  
   - Twice the volume of water initially poured = $$ 2x $$.  
   - Fraction of mixture taken out = $$ \frac{2x}{300} $$.

3. **Milk retained after removal of mixture:**  
   - Fraction of mixture retained = $$ 1 - \frac{2x}{300} $$.  
   - Milk retained = $$ (300 - x) 	imes \left(1 - \frac{2x}{300}ight) $$.

4. **Final condition after refilling with water:**  
   - After refilling with water, total milk retained = $$ 72\% $$ of 300 litres = 216 litres.

5. **Equation for milk retained:**  
   $$
   (300 - x) 	imes \left(1 - \frac{2x}{300}ight) = 216
   $$

6. **Simplify and solve for $$ x $$:**  
   Expanding and simplifying:  
   $$
   (300 - x) 	imes (1 - \frac{2x}{300}) = 216
   $$  
   $$
   90000 - 300x - 600x + 2x^2 = 216 	imes 300
   $$  
   Simplified quadratic equation:  
   $$
   2x^2 - 900x + 25200 = 0
   $$

7. **Solve quadratic equation:**  
   Factorize:  
   $$
   x^2 - 450x + 12600 = 0
   $$  
   Roots: $$ x = 420 $$ or $$ x = 30 $$.

8. **Reject inadmissible root ($$ x = 420 > 300$$):**  
   The total content cannot exceed the container's capacity. Hence, $$ x = 30 $$.

Thus, the initial quantity of water poured into the container is **30 litres**.