How to Solve Mixture Replacement Problems
Solve mixture replacement aptitude problems with the geometric dilution formula, a worked example, and practice questions with answers.
Expected Interview Answer
When a fraction of a mixture is repeatedly removed and replaced with a pure substance, the amount of the original ingredient left after n operations is Initial × (1 − x/V)^n, where x is the volume removed each time and V is the total volume.
Each replacement removes the same fraction x/V of whatever is currently in the container, including any of the original substance already diluted by prior replacements. Because the fraction removed is constant, the surviving quantity of the original substance shrinks geometrically rather than linearly, which is exactly why the formula uses a power of n instead of a simple subtraction. The container’s total volume V never changes across operations, only its composition does. Reading a replacement problem correctly means identifying V, x, and n before touching the formula.
- One formula covers any number of replacement rounds
- Avoids the common trap of subtracting a fixed amount each round
- Extends directly to milk-water, alloy, and solution-strength problems
AI Mentor Explanation
A squad of 15 replaces 3 players with new recruits before every series, and those departures are drawn proportionally from whoever is currently in the squad, including recruits added in earlier rounds. After several series, the fraction of the squad still holding an original founding member is (12/15) raised to the number of series, not a straight subtraction of 3 each time. This geometric shrinkage is the exact mechanism behind mixture replacement: a constant fraction removed repeatedly compounds down, it does not decline in a straight line.
Worked example
Setup
- V = 40, x = 8, n = 2
Formula
- Milk = V × (1 − x/V)^n
Result
- 40 × 0.64 = 25.6 liters
Step-by-Step Explanation
Step 1
Identify V, x, n
Total volume V, amount replaced each round x, number of rounds n.
Step 2
Apply the formula
Remaining original = V × (1 − x/V)^n.
Step 3
Compute the fraction removed
x/V is the constant proportion lost every single round.
Step 4
Raise to the power n
Multiply the surviving fraction by itself once per round, never subtract linearly.
What Interviewer Expects
- Correct identification of V, x, and n from the problem statement
- Use of the geometric formula rather than linear subtraction
- Recognizing the fraction removed stays constant across rounds
- Ability to extend the formula to unequal replacement amounts per round
Common Mistakes
- Subtracting x liters of original substance from V each round instead of a proportional fraction
- Forgetting that replaced liquid also gets partially removed in later rounds
- Using n as a multiplier instead of an exponent
- Mixing up which substance the question asks for (original vs replacement)
Best Answer (HR Friendly)
“The trick is realizing that every replacement removes the same fraction of whatever is currently in the container, so the original substance does not decline by a fixed amount each round, it declines geometrically. I find the constant fraction removed, x over V, then raise (1 minus that fraction) to the power of the number of rounds, and multiply by the initial quantity to get what remains.”
Follow-up Questions
- How would the formula change if a different amount were replaced each round?
- How do you find the number of replacements needed to reduce the original substance below a given percentage?
- How does this formula relate to exponential decay in other contexts?
- How would you solve this if two different substances were both being diluted simultaneously?
MCQ Practice
1. A container has 81 liters of pure liquid. 27 liters are removed and replaced with water three times. How much pure liquid remains?
Remaining = 81 × (1 − 27/81)^3 = 81 × (2/3)^3 = 81 × 8/27 = 24 liters.
2. A tank has 50 liters of milk. 10 liters are withdrawn and replaced with water once. What fraction of the tank is now milk?
Fraction remaining = (1 − 10/50) = 40/50 = 4/5.
3. In the replacement formula V × (1 − x/V)^n, what does n represent?
n is the count of times the withdraw-and-replace operation is performed.
Flash Cards
Mixture replacement formula? — Remaining original = V × (1 − x/V)^n.
Why geometric, not linear? — Each round removes the same fraction of the current mix, including previously added replacement.
What stays constant across rounds? — The total volume V and the fraction removed each round, x/V.
What if x differs each round? — Multiply (1 − x1/V), (1 − x2/V), ... individually instead of raising one term to a power.