Mixtures & Alligation

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Rule of Alligation

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⏱ 5 min read🧩 5 question types🎯 14 practice Q

The idea in one minute

Alligation mixes two ingredients of known 'strength' — price, percentage, average — to hit a target mean.

Subtract diagonally: cheaper : dearer = (dearer − mean) : (mean − cheaper).

Each ingredient's quantity pairs with the OTHER ingredient's distance from the mean. The mean must sit between the two strengths. It works on prices, concentrations, marks, ages, rates and profit per cent — anywhere a weighted average applies.

The rule of alligation

Two ingredients: a cheaper one at strength C and a dearer one at strength D. Mix them to get a mean M, with C < M < D.

\text{cheaper} : \text{dearer} = (D - M) : (M - C)

Picture it: write C and D on top, M in the middle. Subtract diagonally. Each quantity takes the arm that points at the OTHER ingredient.

Example: Tea at ₹320 and ₹250 must blend to ₹300. Cheaper : dearer = $(320-300):(300-250) = 20:50 = 2:5$ .

Rule: Check C < M < D first. A mean outside the two strengths is impossible — the question or the setup is misread.

Why it works

Let the quantities be x (cheaper) and y (dearer). The blend's value is $\frac{Cx + Dy}{x+y} = M$ . Cross-multiply:

x(D - M) = y(M - C)

That is exactly the two arms. Alligation is one linear equation drawn as a picture.

Water is free; pure is 100%

Dilution questions are alligation with one strength at 0. Strengthening questions use 100% for the pure ingredient.

Tip: Water to be added to 60 L of milk at ₹24/L to sell the blend at ₹20/L: milk : water = $(24-20):(20-0) = 4:20 = 1:5$ . So water = $60 \div 5 = 12$ L.

Alligation on averages

Any two group averages and their overall average alligate the same way.

Boys average 70 marks, girls 75, class 72 → boys : girls = $(75-72):(72-70) = 3:2$ .
A class of 50 averages 42 kg; boys 45 kg, girls 40 kg → boys : girls = $2 : 3$ , so girls = 30.

Watch: For speeds over EQUAL DISTANCES, the average is not the plain mean — alligate carefully or use the harmonic mean.

Scaling to a known quantity

The arms give a ratio. When one real quantity is known, price one part and scale.

25 kg of sugar at ₹24 is mixed with sugar at ₹42 to average ₹30. Arms: cheaper : dearer = $(42-30):(30-24) = 2:1$ . The cheaper side is double, so dearer sugar = $25 \div 2 = 12.5$ kg.

Tip: Always verify: rebuild the mean from your quantities before choosing the option.

Three ingredients, one mean

With three prices, hunt a pair whose alligation hits the mean (or a price already AT the mean).

Rice at ₹30, ₹36 and ₹42 must average ₹36. The ₹30 and ₹42 kinds mix 1 : 1 to $\frac{30+42}{2} = ₹36$ — and any amount of the ₹36 kind can join without moving the mean. So the answer is 1 : any : 1 in that order.

Tip: Equal quantities of two prices average to their midpoint. Spot the pair first, then let the mean-priced item ride along.

Equal quantities average to the midpoint

40 kg at ₹6 and 40 kg at ₹7 average ₹6.50 exactly — the arms are equal, so the ratio is 1 : 1. When a question quotes equal amounts, the mean falls in the middle with no work at all.

Rates and profit per cents

Percentages alligate as numbers. ₹12,000 is split at 5% and 8% simple interest to earn ₹750 a year.

Mean rate = $\dfrac{750}{12000} \times 100 = 6.25\%$ . Arms: $(8 - 6.25):(6.25 - 5) = 1.75 : 1.25 = 7 : 5$ . So ₹7,000 at 5% and ₹5,000 at 8%. Check: $350 + 400 = 750$ .

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Two-ingredient alligation for the ratio

How to spot it:

Two unit prices are given with the target mean price; the mixing ratio is asked.

D : C = (M - C) : (D - M)

Method

Identify cheaper, dearer and mean; check C < M < D.
Subtract diagonally: (D − M) and (M − C).
State the ratio, scaling by any known quantity.

Why it works:

The weighted-mean equation rearranges into exactly these arms.

Try this

In what ratio should tea at ₹320 per kg be mixed with tea at ₹250 per kg to get a mixture worth ₹300 per kg?

Show solution

Arms: $320 - 300 = 20$ and $300 - 250 = 50$ .
Cheaper : dearer = $20 : 50 = 2 : 5$ .

Answer

2 : 5

Type 2very common2 practice Q

Mixing solutions by strength

How to spot it:

Two solutions of given concentration are mixed to reach a target strength.

\frac{\text{dilute}}{\text{strong}} = \frac{S - M}{M - W}

Method

Alligate on the strength percentages exactly as on prices.
The arm (S − M) belongs to the dilute solution.
Split any given final volume by the ratio.

Why it works:

Strength is the 'price' per litre of solution.

Try this

Solutions of 20% and 50% acid are mixed to make a 30% solution. In what ratio are they mixed?

Show solution

Arms: $50 - 30 = 20$ and $30 - 20 = 10$ .
Dilute : strong = $20 : 10 = 2 : 1$ .

Answer

2 : 1

Type 3common2 practice Q

Alligation on averages (marks, ages, weights)

How to spot it:

Two groups with different averages combine to a known overall average — the ratio of sizes is asked.

\frac{n_1}{n_2} = \frac{A_2 - A}{A - A_1}

Method

Treat the group averages as the two prices and the overall as M.
Subtract to the arms; the arms give the sizes' ratio.
Scale to the total when a count is asked.

Why it works:

The overall average is the size-weighted mean of the group averages.

Try this

The average marks of boys in a class is 70 and of girls is 75. If the class average is 72, find the ratio of boys to girls.

Show solution

Arms: $75 - 72 = 3$ and $72 - 70 = 2$ .
Boys : girls = $3 : 2$ .

Answer

3 : 2

Type 4common2 practice Q

How much to add (alligation form)

How to spot it:

'How much of X must be added to a given amount of mixture to reach a strength or price?'

\frac{q_{\text{added}}}{q_{\text{existing}}} = \frac{M_{\text{old}} - M}{M - M_{\text{added}}}

Method

Give the added ingredient its strength: 0 for water, 100 for pure acid, its price for a priced item.
Alligate between the old mixture and the addition.
Solve for the unknown quantity.

Why it works:

Adding a zero-strength or full-strength ingredient is still just a two-ingredient mix.

Try this

How much pure acid must be added to 100 litres of a 30% acid solution to make it 50% acid?

Show solution

Old mixture 30, added acid 100, target 50.
Acid : solution = $(50 - 30):(100 - 50) = 20 : 50 = 2 : 5$ .
Acid = $100 \times \frac{2}{5} = 40$ L.

Answer

40 litres

Type 5common2 practice Q

Alligation on rates and profits

How to spot it:

Two investments at different interest rates, or items sold at different profit per cents, with a known combined result.

\frac{\text{amount}_1}{\text{amount}_2} = \frac{R_2 - R}{R - R_1}

Method

Find the overall percentage if it is not given (total gain ÷ total amount).
Alligate on the percentages to split the total.
Scale the ratio to the actual amounts.

Why it works:

The combined percentage is the amount-weighted mean of the two rates.

Try this

₹12,000 is invested partly at 5% and the rest at 8% simple interest. The total yearly interest is ₹750. Find the sum at 5%.

Show solution

Mean rate = $\frac{750}{12000} \times 100 = 6.25\%$ .
Arms: $(8 - 6.25):(6.25 - 5) = 1.75 : 1.25 = 7 : 5$ .
At 5%: $12000 \times \frac{7}{12} = ₹7000$ .

Answer

₹7,000

Formula sheet

Alligation ratio

\frac{\text{cheaper}}{\text{dearer}} = \frac{D - M}{M - C}

Weighted average

\bar{v} = \frac{n_1 v_1 + n_2 v_2}{n_1 + n_2}

Water as the free ingredient

\text{milk} : \text{water} = (m - 0) : (c - m),\ c > m

Shortcuts that save time

⚡ Cross the arms

Dearer-minus-mean and mean-minus-cheaper swap sides. Each quantity takes the other side's arm.

Example

In what ratio must rice at ₹36 per kg be mixed with rice at ₹24 per kg so that the mixture costs ₹30 per kg?

Show solution

Cheaper : dearer = $(36 - 30):(30 - 24)$ .
= $6 : 6 = 1 : 1$ .

Answer

1 : 1

⚡ Water has price zero

Dilution is alligation with one strength at 0.

Example

How much water must be added to 60 litres of milk worth ₹24 per litre so that the mixture is worth ₹20 per litre?

Show solution

Milk : water = $(24 - 20):(20 - 0) = 4 : 20 = 1 : 5$ .
Water = $60 \div 5 = 12$ L.

Answer

12 litres

⚡ Averages alligate too

Salary, age, marks, weight — any average splits into an alligation.

Example

A class of 50 has an average weight of 42 kg. The boys average 45 kg and the girls 40 kg. Find the number of girls.

Show solution

Boys : girls = $(42 - 40):(45 - 42) = 2 : 3$ .
Girls = $50 \times \frac{3}{5} = 30$ .

Answer

30 girls

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Reading the arms the same way round (cheaper with its own arm).

Each quantity pairs with the far arm: cheaper : dearer = (D − M) : (M − C).

Mistake 02

Alligating a mean outside the two strengths.

Check C < M < D first. If it fails, re-read the question.

Mistake 03

Giving water or a pure ingredient 'no value' and skipping it.

Water has strength 0; a pure ingredient has 100%. Both alligate normally.

Mistake 04

Alligating speeds over equal distances.

Equal distances average by the harmonic mean. Alligation fits equal times or linear quantities.

Mistake 05

Stopping at the ratio when a quantity was asked.

Price one part with the known quantity, then scale to the whole amount.

Quick revision

Read this the night before the exam.

cheaper : dearer = (D − M) : (M − C).
The mean must sit between the two strengths.
Water = strength 0; pure acid = 100%.
Works on prices, concentrations, marks, ages, rates, profit %.
Known quantity → price one part of the ratio, then scale.
Verify by recomputing the mean from your mix.

Practice: 14 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

Suggested time 5 min · wrong answers go to your mistake notebook automatically.

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