Mixture & Alligation: The Complete Human-Friendly Guide

If you’ve ever solved a “milk and water” problem, you’ve already met mixtures. But mixture problems are not just about milk and water—they show up in profit/loss, averages, chemical fertilizers, and even multi-vessel transfers. The good news? Almost every mixture problem is built on a few simple ideas. Once you understand them, you can solve even the advanced ones without memorizing hundreds of questions.

This blog is a complete concept revision based on a full set of mixture & alligation notes—from basics to advanced. I’ve changed the example numbers so you can see the concepts fresh, but the logic remains exactly the same.

1. The Absolute Basics: What Is a Mixture?

A mixture is a homogeneous combination of two or more components. “Homogeneous” means the ratio is the same throughout. If you take a spoonful from anywhere in the mixture, the ratio of components is identical.

Key Rule

If you remove a part of a homogeneous mixture, the removed part has the same ratio as the original. And if you only remove (without adding anything), the remaining mixture keeps the same ratio.

Example:
A 120 L mixture has spirit and water in the ratio 7:5.
Total parts = 12. One part = 10 L.
Spirit = 70 L, Water = 50 L.

If you remove 30 L, the remaining 90 L still has spirit:water = 7:5.
Removed part also has 7:5.

2. Adding One Component

When you add only one component (say water), the other component (say milk) stays the same. That unchanged component becomes your anchor.

Example: Add Water

A 90 L mixture has milk and water in the ratio 5:1. How much water must be added to make the ratio 10:3?

Initial: Milk = 75 L, Water = 15 L.
Milk stays 75 L.
Let added water = x.

Example: Add Milk

A 60 L mixture has milk:water = 3:2. How much milk must be added to make it 7:3?

Initial: Milk = 36 L, Water = 24 L.
Water stays 24 L.
Let added milk = x.

3. Vaporization and Evaporation

When water evaporates, only water reduces. The solute (sugar, salt, milk solids) remains unchanged.

Example:
A mixture has sugar and water in the ratio 7:11. When 36 L of water is vaporized, the quantities become equal. Find the initial quantity.

Sugar = 7 parts, Water = 11 parts.
After vaporization, water becomes 7 parts.
Water reduced by 4 parts = 36 L → 1 part = 9 L.
Total initial = 18 parts = 162 L.

4. Alligation: The Magic Cross

Alligation is a shortcut to find the ratio in which two ingredients must be mixed to get a desired mean value. It’s just weighted average in disguise.

Formula

Let cheaper value = x, dearer value = y, mean = z.

Diagram

text

Cheaper (x)                  Dearer (y)

        \                    /

          \                /

            Mean (z)

          /                    \

    (y - z)                  (z - x)

Example 1: Milk Percentages

Mixture P has 30% milk, Mixture Q has 70% milk. What ratio should they be mixed to get 50% milk?

Example 2: Prices

Two types of sugar cost Rs.40/kg and Rs.60/kg. What ratio to mix to get a mixture worth Rs.48/kg?

Example 3: Profit and Adulteration

A milkman sells a milk-water mixture at Rs.6 per litre and earns a 50% profit. If pure milk costs Rs.8 per litre, find the milk:water ratio.

CP of mixture = SP / (1 + Profit%) = 6 / 1.5 = Rs.4.
Water costs Rs.0.
Alligation: Milk (8), Water (0), Mean (4).

Example 4: Averages

In a class, boys average 60 kg, girls average 40 kg, and the whole class averages 48 kg. If there are 50 students, how many boys?

Total parts = 5. 1 part = 10.
Boys = 2 × 10 = 20.

5. Multiple Replacement

This is where things get interesting. If you repeatedly remove a fixed fraction of a mixture and replace it with another component (usually water), the amount of the non-added component decreases exponentially.

Formula

where = number of operations.

Example 1: Basic Replacement

A vessel has 100 L pure milk. 20 L is removed and replaced with water. This is done twice. How much milk is left?

Fraction removed = 20/100 = 1/5.

Example 2: Finding the Removed Quantity

72 L pure milk. L is removed and replaced with water twice. Finally, water is 40 L. Find .

Final milk = 72 – 40 = 32 L.

Example 3: Wine Replacement

80 L pure wine. 25% is removed and replaced with water. Repeated 3 times. How much wine is left?

25% = 1/4.

Key tip: Always track the component that is never added.

6. Complex Multi-Vessel Problems

These problems involve several vessels, each with its own ratio, and sometimes transfer between vessels.

Example: Three Vessels

Vessels A, B, C have quantities in ratio 1:1:2.
A has milk:water = 2:1.
B has water:milk = 1:2 (so milk:water = 2:1).
C has milk:water = 3:1.
When all are mixed, milk exceeds water by 40 L. Find total quantity in A and B.

Let A = x, B = x, C = 2x.
A: M = 2x/3, W = x/3
B: M = 2x/3, W = x/3
C: M = 1.5x, W = 0.5x
Total M = 17x/6, Total W = 7x/6
Difference = 10x/6 = 5x/3 = 40 ⇒ x = 24.
A + B = 2x = 48 L.

Example: Transfer Between Jars

Jar A has 60 L milk:water = 5:1.
Jar B has milk:water = 3:2.
12 L from A is poured into B. Now the difference between milk and water in B is 18 L. Find initial milk in B.

12 L from A: Milk = 10 L, Water = 2 L.
Let B initially have milk = 3k, water = 2k.
After transfer: Milk = 3k + 10, Water = 2k + 2.
Difference = (3k+10) – (2k+2) = k + 8 = 18 ⇒ k = 10.
Initial milk in B = 3k = 30 L.

7. Miscellaneous Advanced Concepts

Replacement with Another Mixture

A 200 L mixture has acid:spirit = 3:5. 50 L is removed and replaced with a mixture having acid:spirit = 1:1. Find the final ratio.

Initial: Acid = 75, Spirit = 125.
Remove 50 L (3:5): Acid removed = 18.75, Spirit removed = 31.25.
Remaining: Acid = 56.25, Spirit = 93.75.
Add 50 L (1:1): Acid = 25, Spirit = 25.
Final: Acid = 81.25, Spirit = 118.75 = 13:19.

Chemical / Fertilizer Mixtures

AS fertilizer has N=20%, P=60%, K=20%.
AP fertilizer has only N and P.
A mixture of AS and AP has N=30%, P=65%, K=5%.
Find N:P in AP.

K comes only from AS.
0.20 × AS = 0.05 × Total ⇒ AS = Total/4, AP = 3/4 Total.
Let Total = 100. AS = 25, AP = 75.
N from AS = 5. Total N = 30 ⇒ N from AP = 25.
P from AS = 15. Total P = 65 ⇒ P from AP = 50.
N:P in AP = 25:50 = 1:2.

Multi-Level Mixing

Item K is made by mixing Chemical A and B in ratio 7:5.
Chemical A = X:Y = 2:5.
Chemical B = Y:Z = 3:2.
756 units of K are mixed with water so that Y concentration becomes 45%. How much water?

Y in A = 5/7. Y in B = 3/5.
K = A:B = 7:5.
Y in K = (7×5/7 + 5×3/5) / 12 = (5 + 3)/12 = 8/12 = 2/3.
Y in 756 units = 504.
Final concentration = 45% = 0.45.

Weighted Averages with Sections

Four sections A1, A2, A3, A4 have average marks 44%, 52%, 70%, 79%.
Overall average = 65%.
(A1+A2) average = 47%, (A2+A3) average = 65%.
Find A1:A4.

From (A1,A2)=47%:

From (A2,A3)=65%:

Combine: A1:A2:A3 = 25:15:39.
Let A4 = x. Overall average 65% gives:

Solving gives A1:A4 = 2:3.

8. Quick Recap & Formula Sheet

Final Words

Mixture and alligation problems look scary because they combine ratios, percentages, and equations. But they all boil down to a few ideas:

1. Identify what doesn’t change (milk when adding water, solute when evaporating, etc.).
2. Use alligation when two things are mixed to get a mean.
3. Use the replacement formula when the same operation repeats.
4. For complex problems, write equations for each component.

Practice with different numbers, and soon you’ll see the same patterns everywhere. Happy solving!