Question:
A container has a mixture of milk and water in the ratio 5 : 2. If 14 liters of this mixture is removed and replaced with water, the ratio becomes 5 : 4. What is the initial quantity of the mixture?
A container has a mixture of milk and water in the ratio 5 : 2. If 14 liters of this mixture is removed and replaced with water, the ratio becomes 5 : 4. What is the initial quantity of the mixture?
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In mixture replacement problems, focus on the substance whose quantity is not directly being replenished. The final amount of that substance is equal to its initial amount in the reduced volume. For this problem: Final Milk = (Initial Milk Proportion) \(\times\) (Initial Volume - Volume Removed). This can often lead to a faster solution.
Updated On: Jul 4, 2026
- 63 L
- 35 L
- 49 L
- 56 L
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The Correct Option is A
Approach Solution - 1
Approach: The key insight is that milk is only ever removed, never added. So track milk through the operation and let the changed ratio pin down the total.
Step 1: Name the quantities.
Let initial milk \(= 5x\) L and water \(= 2x\) L, so total \(= 7x\) L.
Step 2: Remove 14 L of mixture.
The 14 L leaves in the 5:2 ratio, so milk removed \(= 14 \times \frac{5}{7} = 10\) L and water removed \(= 14 \times \frac{2}{7} = 4\) L.
Now milk \(= 5x - 10\), water \(= 2x - 4\).
Step 3: Add 14 L of water.
Milk is untouched: \(5x - 10\). Water becomes \((2x - 4) + 14 = 2x + 10\).
Step 4: Apply the new ratio 5 : 4.
\[ \frac{5x - 10}{2x + 10} = \frac{5}{4}. \]
Cross-multiply:
\[ 4(5x - 10) = 5(2x + 10) \implies 20x - 40 = 10x + 50 \implies 10x = 90 \implies x = 9. \]
Step 5: Total mixture.
\[ 7x = 7 \times 9 = 63 \text{ L}. \]
Check: milk \(= 45 - 10 = 35\), water \(= 18 + 10 = 28\), and \(35 : 28 = 5 : 4\). Correct.
Final Answer: \(\boxed{63\text{ L}}\). (Option 1)
Step 1: Name the quantities.
Let initial milk \(= 5x\) L and water \(= 2x\) L, so total \(= 7x\) L.
Step 2: Remove 14 L of mixture.
The 14 L leaves in the 5:2 ratio, so milk removed \(= 14 \times \frac{5}{7} = 10\) L and water removed \(= 14 \times \frac{2}{7} = 4\) L.
Now milk \(= 5x - 10\), water \(= 2x - 4\).
Step 3: Add 14 L of water.
Milk is untouched: \(5x - 10\). Water becomes \((2x - 4) + 14 = 2x + 10\).
Step 4: Apply the new ratio 5 : 4.
\[ \frac{5x - 10}{2x + 10} = \frac{5}{4}. \]
Cross-multiply:
\[ 4(5x - 10) = 5(2x + 10) \implies 20x - 40 = 10x + 50 \implies 10x = 90 \implies x = 9. \]
Step 5: Total mixture.
\[ 7x = 7 \times 9 = 63 \text{ L}. \]
Check: milk \(= 45 - 10 = 35\), water \(= 18 + 10 = 28\), and \(35 : 28 = 5 : 4\). Correct.
Final Answer: \(\boxed{63\text{ L}}\). (Option 1)
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Approach Solution -2
Approach (componendo-dividendo): Once the ratio equation is set up, a classical ratio trick shortens the algebra instead of cross-multiplying and expanding.
Step 1: With milk \( =5x-10 \) and water \( =2x+10 \) after the replacement, the new ratio gives:
\[ \frac{5x-10}{2x+10} = \frac{5}{4}. \]
Step 2: Apply componendo-dividendo, add and subtract numerator and denominator on both sides:
\[ \frac{(5x-10)+(2x+10)}{(5x-10)-(2x+10)} = \frac{5+4}{5-4} \implies \frac{7x}{3x-20} = 9. \]
Step 3: Solve the simplified linear equation:
\[ 7x = 9(3x-20) = 27x-180 \implies 20x = 180 \implies x=9. \]
Step 4: Total mixture \( =7x = 63 \) L.
Final Answer: \( \boxed{63\text{ L}} \). (Option 1)
Step 1: With milk \( =5x-10 \) and water \( =2x+10 \) after the replacement, the new ratio gives:
\[ \frac{5x-10}{2x+10} = \frac{5}{4}. \]
Step 2: Apply componendo-dividendo, add and subtract numerator and denominator on both sides:
\[ \frac{(5x-10)+(2x+10)}{(5x-10)-(2x+10)} = \frac{5+4}{5-4} \implies \frac{7x}{3x-20} = 9. \]
Step 3: Solve the simplified linear equation:
\[ 7x = 9(3x-20) = 27x-180 \implies 20x = 180 \implies x=9. \]
Step 4: Total mixture \( =7x = 63 \) L.
Final Answer: \( \boxed{63\text{ L}} \). (Option 1)
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