Mixture and Alligation Problems: Always Use Solute ÷ Total, Never Parts Alone

"A 40-litre mixture of milk and water contains milk and water in the ratio 5 : 3. How much water must be added so that the new ratio becomes 5 : 4?"

"A 60-litre mixture is 20\% acid. How much water must be added to bring the concentration down to 12\%?"

"A vessel contains milk-water in 3 : 1. Twenty litres are drawn out and replaced with water. Now the ratio is 3 : 2. Find the original volume."

These are mixture problems, and they trap more students than almost any other ratio topic. The trap is structural: if you track the two parts separately — "milk and water" as if they were two independent quantities — you will keep losing track of which one changes when you pour something in or take something out.

The reflex that fixes every single mixture problem in one line is this.

The one rule

Track concentration as a single number: solute divided by total.

The solute is the thing you care about — usually the "active" ingredient. Milk in a milk-water mixture. Acid in an acid-water mixture. Salt in a salt-water solution. In profit-loss mixture problems, it's the substance whose proportion changes.
The total is the whole volume — solute plus solvent, or sum of all parts.
The concentration is \dfrac{\text{solute}}{\text{total}}, expressed as a fraction, decimal, or percentage.

When water is added, the solute stays the same; only the total grows. When the mixture is drawn out and replaced with water, the solute decreases by a known proportion while the total stays the same. Every operation on the mixture is an operation on this one ratio — not on the two parts separately.

Add water, and the solute stays fixed while the total grows. The concentration $\tfrac{\text{solute}}{\text{total}}$ drops from $\tfrac{5}{8}$ to $\tfrac{5}{9}$. Once you track this one ratio, every mixture problem reduces to finding how much water to add to move the concentration from one value to another.

Why "parts separately" fails: when you add pure water, both the water amount and the total change, but in such a way that the ratio water : milk is not the ratio you want. Working with three moving numbers (water, milk, total) creates space for errors. Working with one number (solute ÷ total) means there is one thing to track, and one unknown to solve for.

The canonical setup

Every single mixture-problem statement can be translated into this framework.

Step 1. Identify the solute (the tracked ingredient) and the total.

Step 2. Write down initial solute, initial total, initial concentration.

Step 3. Apply the operation (add water / add solute / draw and replace) and update solute and total separately — but keep your eyes on the concentration.

Step 4. Set up the final-concentration equation, solve.

Three worked examples

Example 1 — adding water. 40 L milk-water is in ratio 5 : 3. Add how much water so the ratio becomes 5 : 4?

Solute = milk. Initial milk = \tfrac{5}{8} \times 40 = 25 L. Initial total = 40 L. Initial concentration = \tfrac{25}{40} = \tfrac{5}{8}.
Final ratio 5 : 4 means milk : total = 5 : 9, so final concentration = \tfrac{5}{9}.
Let x = litres of water added. Milk stays fixed at 25. New total = 40 + x.
Equation: \dfrac{25}{40 + x} = \dfrac{5}{9}. Cross-multiply: 225 = 5(40 + x) = 200 + 5x, so x = 5 L.

Check: new total = 45 L, milk = 25 L, water = 20 L, ratio 25 : 20 = 5 : 4. Correct.

Example 2 — diluting acid. 60 L is 20\% acid. Add how much water to bring it to 12\%?

Acid = 0.20 \times 60 = 12 L. Initial total = 60.
Let x = water added. Acid stays 12. New total = 60 + x.
Equation: \dfrac{12}{60 + x} = 0.12. So 60 + x = \dfrac{12}{0.12} = 100, giving x = 40 L.

Check: 12 L in 100 L is 12\%. Correct.

Example 3 — draw and replace. A vessel of V litres is pure milk. Draw 20 L and replace with water. Do this twice. After the second replacement, the milk is \tfrac{9}{16} of the total. Find V.

Each operation multiplies milk by the factor \dfrac{V - 20}{V} (you remove a fraction of the mixture, so milk drops by that fraction).
After two operations: milk-fraction = \left(\dfrac{V - 20}{V}\right)^2 = \dfrac{9}{16}.
Taking square root: \dfrac{V - 20}{V} = \dfrac{3}{4} (taking the positive root because V > 20).
Cross-multiply: 4(V - 20) = 3V, so 4V - 80 = 3V, giving V = 80 L.

Check: after first draw, milk = \tfrac{3}{4} \times 80 = 60 L; after second, \tfrac{3}{4} \times 60 = 45 L; concentration 45/80 = 9/16. Correct.

The "draw and replace" formula — after n replacements, the remaining solute is V \cdot \left(\tfrac{V - a}{V}\right)^n — is worth memorising for JEE-level problems. It is a direct consequence of the one-rule framework.

The "alligation" shortcut

When you mix two solutions of different concentrations, there is a quick graphical method called alligation. If concentrations c_1 and c_2 are mixed in volumes V_1 and V_2 to give a mixture of concentration c, then

\frac{V_1}{V_2} = \frac{c_2 - c}{c - c_1}.

This is nothing more than the weighted-average identity written as a ratio: the mixture concentration c is a weighted average of c_1 and c_2, with weights V_1 and V_2. The "alligation" name is cosmetic — the reflex is still solute-over-total.

Alligation worked out

A merchant mixes 30\% coffee with 60\% coffee to make a 40\% blend. In what ratio?

c_1 = 0.30, c_2 = 0.60, c = 0.40.
\dfrac{V_1}{V_2} = \dfrac{0.60 - 0.40}{0.40 - 0.30} = \dfrac{0.20}{0.10} = \dfrac{2}{1}.

So the merchant must mix the 30\% coffee and 60\% coffee in the ratio 2 : 1. Quick sanity check: more of the weaker one is needed, because 0.40 is closer to 0.30 than to 0.60. The weight is inversely proportional to how close you are to the target.

The rule to carry

In every mixture problem, identify the one solute being tracked.
Write the concentration = solute ÷ total, not the parts separately.
Apply the operation: adding water changes total only; adding solute changes both; draw-and-replace scales the solute down by \tfrac{V - a}{V}.
Set up the equation in one unknown (volume to add, or initial volume), and solve.

Once this framework clicks, every mixture, alligation, dilution, or concentration problem collapses to the same two-line calculation. The moment you see "ratio of milk to water" or "percentage acid," your hand should reach for solute over total before reading the rest of the sentence.