Mixture & alligation: balance two ingredients

01 / THE STARTING POINT

Before you begin

What you need: Weighted averages, ratios and percentages.

Calculate a mixture's weighted mean
Derive the alligation ratio
Handle removal and replacement

02 / BUILD THE CONCEPT

Understand the idea

The mixture preserves totals

Combine quantities by adding the amount of the relevant ingredient or the total cost. Divide by the total quantity to obtain the new concentration or mean price. Units and concentration bases must match.

A target mean sits between its sources

Mixing positive quantities of two strengths a < b gives a mean strictly between them. A target outside that interval is impossible without adding another ingredient or changing the process.

Alligation is a balance equation

For x units at a and y units at b with mean m, ax + by = m(x + y). Rearranging gives (m − a)x = (b − m)y, so x:y = (b − m):(m − a).

03 / A DIFFERENT WAY TO SEE IT

Opposite gaps set the quantities

Low: ₹20/kg
Target: ₹26/kg
High: ₹30/kg
Low : high = 4 : 6

The target is closer to ₹30, so the mixture needs more of the higher-priced ingredient: 2:3.

04 / FROM IDEA TO ANSWER

A method you can reuse

KEEP THIS HANDY

Quantity at a : quantity at b = (b − m) : (m − a)

For a < m < b, matching units and a weighted-mean mixture model.

Write the cheaper/weaker value a, dearer/stronger value b and target m.
Check a < m < b and take opposite differences.
Simplify the ratio; if a total quantity is given, divide it into the resulting parts.

A useful insight

If a fraction f is removed and replaced with a substance containing none of the original ingredient, the original amount after n well-mixed rounds is initial × (1 − f)ⁿ.

05 / WATCH THE METHOD WORK

Worked examples

Read the question first. Try a step yourself, then compare your reasoning.

EXAMPLE 01

Find the mixing ratio

Mix rice at ₹20/kg and ₹30/kg to average ₹26/kg.

Low-side quantity ∝ 30 − 26 = 4.
High-side quantity ∝ 26 − 20 = 6.
4:6 simplifies to 2:3.

Answer2:3

EXAMPLE 02

Weighted concentration

Mix 2 L of 10% solution and 3 L of 20% solution. Find concentration, assuming volumes add.

Ingredient amounts: 2 × 0.10 = 0.2 L and 3 × 0.20 = 0.6 L.
Total ingredient = 0.8 L; total volume = 5 L.
0.8/5 × 100% = 16%.

Answer16%

EXAMPLE 03

Replace part of a mixture

A 10 L tank contains pure milk. Remove 2 L and replace with water; mix and repeat once. Milk left?

First removal leaves 10 × 0.8 = 8 L milk.
After mixing, the second removal takes 20% of the remaining milk.
Milk left = 8 × 0.8 = 6.4 L.

Answer6.4 L

06 / YOUR TURN

Check your understanding

Quick questions, with explanations. These are for self-study and do not affect your account score.

0 of 3 questions checked

07 / TAKE THE NEXT STEP

Put your understanding to work

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