Milk Water Ratio Ee8Fc2

1. **State the problem:** We start with a solution that is $80\%$ milk and $20\%$ water. After adding some water, the milk to water ratio becomes $8:5$. Then, $39$ L of this new solution is replaced with pure milk, changing the ratio to $9:4$. We need to find the volume of water added initially. 2. **Define variables:** Let the initial volume of the solution be $V$ liters. Let the volume of water added initially be $x$ liters. 3. **Initial amounts:** Milk in initial solution = $0.8V$ Water in initial solution = $0.2V$ 4. **After adding water:** Total volume = $V + x$ Milk remains $0.8V$ (no milk added or removed yet) Water = $0.2V + x$ Given milk to water ratio is $8:5$, so: $$\frac{0.8V}{0.2V + x} = \frac{8}{5}$$ 5. **Solve for $x$ in terms of $V$:** Cross multiply: $$5 \times 0.8V = 8 \times (0.2V + x)$$ $$4V = 1.6V + 8x$$ $$4V - 1.6V = 8x$$ $$2.4V = 8x$$ $$x = \frac{2.4V}{8} = 0.3V$$ So, the volume of water added is $x = 0.3V$ liters. 6. **After replacing $39$ L of solution with pure milk:** The solution volume remains $V + x = V + 0.3V = 1.3V$ liters. Amount of milk and water in $39$ L removed: Milk removed = $39 \times \frac{8}{13}$ (since milk:water = 8:5, total parts = 13) $$= 39 \times \frac{8}{13} = 24$$ liters Water removed = $39 \times \frac{5}{13} = 15$ liters 7. **After removal:** Milk left = $0.8V - 24$ Water left = $(0.2V + 0.3V) - 15 = 0.5V - 15$ 8. **Add $39$ L pure milk:** Milk now = $0.8V - 24 + 39 = 0.8V + 15$ Water remains $0.5V - 15$ 9. **New ratio of milk to water is $9:4$:** $$\frac{0.8V + 15}{0.5V - 15} = \frac{9}{4}$$ 10. **Solve for $V$:** Cross multiply: $$4(0.8V + 15) = 9(0.5V - 15)$$ $$3.2V + 60 = 4.5V - 135$$ $$60 + 135 = 4.5V - 3.2V$$ $$195 = 1.3V$$ $$V = \frac{195}{1.3} = 150$$ liters 11. **Find $x$ (water added initially):** $$x = 0.3V = 0.3 \times 150 = 45$$ liters **Final answer:** The volume of water added initially is **45 liters**.