How to Solve Milk and Water Mixture Problems
Solve milk and water mixture problems using the repeated-replacement formula and alligation, with a worked example and practice questions.
Expected Interview Answer
Milk and water mixture problems track milk (or water) quantity as a fraction of total volume through repeated replacement, using the key result that after n replacements of x units from a total of V units with pure water, remaining milk = V × (1 − x/V)^n.
The replacement formula generalizes the two-liquid split: each time x units of the mixture are removed and replaced with pure water, the milk fraction remaining is multiplied by (1 − x/V), so after n such operations the milk quantity is V(1−x/V)^n. For a one-time addition of pure water (not replacement), simply add the new water to the total while milk stays unchanged, then recompute the milk percentage as milk/(new total). Concentration or purity problems phrase the same idea as a percentage: X% milk means X% of the volume is milk, and mixing two known-purity solutions is solved with alligation. Always distinguish “replaced with water” (uses the repeated-dilution formula) from “water simply added” (milk quantity unchanged, only the total grows).
- The (1−x/V)^n formula solves every repeated-replacement milk problem directly
- Distinguishing “replace” from “add” avoids the most common setup error
- Alligation handles mixing two different-purity solutions cleanly
AI Mentor Explanation
A squad of 20 players is 100% experienced initially; each season, 4 experienced players (1/5 of the squad) are replaced with rookies, so after 3 seasons the experienced fraction remaining is (1−4/20)^3 = (0.8)^3 = 0.512, meaning about 51.2% of the squad is still experienced — exactly the repeated-dilution formula milk-water problems use. If instead 4 new roster spots were simply added without removing anyone, the experienced count would stay fixed while only the total squad size grows, a completely different calculation.
Worked example (repeated replacement)
Fraction removed each time
- 8/40 = 1/5
After 2 replacements
- (1 − 1/5)^2 = (4/5)^2 = 16/25
Milk remaining
- 40 × 16/25 = 25.6 litres
Step-by-Step Explanation
Step 1
Identify replace vs add
"Replaced with water" uses the dilution formula; "water added" leaves milk quantity unchanged.
Step 2
Apply the replacement formula
Milk remaining = V × (1 − x/V)^n for n replacements of x units from total V.
Step 3
Handle a simple addition
New total = old total + water added; milk percentage = old milk ÷ new total.
Step 4
Use alligation for mixing two purities
Blend two known-percentage solutions using the alligation cross-difference ratio.
What Interviewer Expects
- Correct distinction between “replaced with water” and “water added”
- Accurate application of the (1−x/V)^n replacement formula
- Correct recomputation of milk percentage after dilution
- Using alligation when blending two known-purity milk solutions
Common Mistakes
- Applying the replacement formula when water was simply added, not replaced
- Using the wrong fraction x/V in the replacement formula
- Forgetting to raise the fraction to the power n for repeated operations
- Confusing milk percentage with milk quantity after the total volume changes
Best Answer (HR Friendly)
“The first thing I check is whether the water is being added on top or replacing part of the mixture — those are two completely different calculations. For repeated replacement, I use milk remaining equals the original volume times one minus the replaced fraction, raised to the power of the number of replacements. If water is just being added, the milk quantity itself never changes — only the total volume grows, so I recompute the percentage as the unchanged milk over the new total. When two different-purity solutions are blended, I reach for alligation.”
Follow-up Questions
- How do you find the number of replacements needed to reach a target purity?
- How does this problem change if a different volume is removed each time?
- How would you find the original milk quantity given the final diluted amount?
- How do you combine the replacement formula with alligation in a single problem?
MCQ Practice
1. A 50-litre container is full of milk. 10 litres are withdrawn and replaced with water once. The milk remaining is?
Milk remaining = 50 × (1 − 10/50) = 50 × 0.8 = 40 litres.
2. A container has 20 litres of milk. 10 litres of water are added (not replaced). What is the new milk percentage?
Milk stays at 20L; new total = 30L; percentage = 20/30 = 66.7%.
3. A 27-litre container of milk undergoes the replacement process (9 litres replaced with water) 3 times. Milk remaining is closest to?
Fraction removed each time = 9/27 = 1/3. Remaining = 27×(2/3)^3 = 27×8/27 = 8 litres.
Flash Cards
Repeated replacement formula? — Milk remaining = V × (1 − x/V)^n for n replacements of x units.
Difference between “replace” and “add”? — Replace dilutes milk via the formula; adding water leaves milk quantity unchanged, only total grows.
How to recompute percentage after adding water? — Unchanged milk quantity divided by the new total volume.
How to blend two different-purity milk solutions? — Use the alligation rule on their percentages to find the mixing ratio.