Alligation or Mixture questions with answers and concepts

Learn alligation or mixture with concept notes and worked examples, then practise focused questions with explained answers. Build accuracy for placement and competitive aptitude tests before trying timed mode.

113 practice questionsPage updated 11 Nov 2026Review metadata pendingPractice questions

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Alligation is a rule that enables us to find the ratio in which two or more ingredients at given prices must be mixed to produce a mixture of a desired price. It is the reverse of weighted average and is highly useful in solving mixture problems, especially those involving different concentrations, costs, or qualities being blended together.

Key formulas and rules

6 rules

Alligation Rule: $\frac{Quantity of Cheaper}{Quantity of Dearer} = \frac{d - m}{m - c}$ (where $c < m < d$ )

Mean Price formula: $m = \frac{c _{1} q _{1} + c _{2} q _{2}}{q _{1} + q _{2}}$

Mixing Two Varieties: $\frac{Q _{A}}{Q _{B}} = \frac{Rate _{B} - Mean Rate}{Mean Rate - Rate _{A}}$

Repeated Replacement: $Liquid Left = Initial \times (1 - \frac{x}{V})^{n}$ (replacing $x$ litres from volume $V$ , $n$ times)

Ratio after $n$ replacements: $\frac{Pure Liquid Left}{Total Volume} = (1 - \frac{x}{V})^{n}$

Profit on water dilution: If pure milk costs $C P$ and water is free, $Profit % = \frac{Water}{Milk} \times 100$

Key concepts

Understanding the Alligation Cross

The alligation cross is a visual method where the cheaper price (c) is placed at the bottom left, dearer price (d) at the top left, and mean price (m) in the centre. The differences (d - m) and (m - c) are calculated diagonally to give the mixing ratio. This method is faster than algebraic equations for simple two-ingredient mixtures.

Types of Mixture Problems

1. Mixing two different priced items (grains, liquids)
2. Dilution problems (adding water to milk, acid solutions)
3. Successive replacement (removing and replacing liquid multiple times)
4. Mixing two different concentration solutions (acid-water mixtures)
5. Profit/loss in mixtures when selling at certain rates

The Replacement Formula

When a container has 'T' litres of pure liquid and 'x' litres is taken out and replaced with water, this process repeated 'n' times: Final pure liquid = T x (1 - x/T)^n. This exponential decay formula is crucial for repeated dilution scenarios. For example, if 10 litres is removed from 100 litres and replaced 3 times, pure liquid remaining = 100 x (0.9)^3 = 72.9 litres.

Weighted Average vs Alligation

Weighted average calculates the mean of mixed quantities: Mean = (q p + q p ) / (q + q ). Alligation reverses this - given the mean, it finds the ratio q :q . Alligation is essentially the weighted average formula rearranged to solve for ratios rather than the mean. Use alligation when you know the desired mean and individual prices; use weighted average when you know quantities and want the mean.

Milk and Water Problems

Common pattern: A vessel contains milk and water in ratio a:b. When x litres is removed and replaced with the other liquid, find the new ratio. Method: Calculate amount of milk/water removed, subtract from original, add replacement. For successive operations, apply the replacement formula. Example: In 60 litres with milk:water = 2:1 (40:20), if 10 litres milk added and 10 litres removed, new ratio calculation requires careful step-by-step tracking.

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Worked examples

Example 1

In what ratio must rice at 30 per kg be mixed with rice at 40 per kg so that the mixture is worth 35 per kg?

Solution

Step 1: Apply the rule of alligation.

\frac{Quantity of Cheaper}{Quantity of Dearer} = \frac{d - m}{m - c}

Step 2: Substitute $c = 30$ , $d = 40$ , and $m = 35$ .

\frac{Quantity of Cheaper}{Quantity of Dearer} = \frac{40 - 35}{35 - 30} = \frac{5}{5} = \frac{1}{1}

Answer: 1 : 1.

Example 2

A merchant has 100 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. Find the quantity sold at 18% profit.

Solution

Step 1: Apply alligation on profit percentages.

\frac{Q _{8%}}{Q _{18%}} = \frac{18 - 14}{14 - 8} = \frac{4}{6} = \frac{2}{3}

Step 2: Divide total 100 kg in the ratio $2 : 3$ .

Q_{18%} = \frac{3}{2 + 3} \times 100 kg = \frac{3}{5} \times 100 = 60 kg

Answer: 60 kg.

Example 3

A vessel contains 80 litres of milk. 8 litres of milk is drawn out and replaced with water. This replacement operation is performed three times in total. Find the quantity of pure milk remaining in the vessel.

Solution

Step 1: Apply the repeated dilution formula.

Remaining Pure Milk = x (1 - \frac{y}{x})^{n}

Step 2: Substitute initial volume $x = 80$ , replacement $y = 8$ , and operations $n = 3$ .

Remaining = 80 (1 - \frac{8}{80})^{3} = 80 (1 - \frac{1}{10})^{3} = 80 \times (0.9)^{3} = 80 \times 0.729 = 58.32 litres

Answer: 58.32 litres.

Example 4

How many kilograms of tea costing 200 per kg must be mixed with 40 kg of tea costing 280 per kg to earn a 25% profit by selling the mixture at 300 per kg?

Solution

Step 1: Find the mean cost price $m$ of the mixture.

Selling Price = m \times 1.25 = 300 ⟹ m = \frac{300}{1.25} = 240/ kg

Step 2: Apply alligation ratio.

\frac{Q _{cheaper}}{Q _{dearer}} = \frac{280 - 240}{240 - 200} = \frac{40}{40} = \frac{1}{1}

Step 3: Solve for $Q_{cheaper}$ given $Q_{dearer} = 40 kg$ .

Q_{cheaper} = 1 \times 40 = 40 kg

Answer: 40 kg.

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

A mixture contains alcohol and water in the ratio 4:3. If 5 litres of water is added, the ratio becomes 4:5. Find the quantity of alcohol in the mixture.

8 litres
10 litres
12 litres
15 litres

Answer: B. 10 litres

Explanation

Let alcohol = 4x, water = 3x
After adding 5 litres water: water = 3x + 5
New ratio: $4 x / (3 x + 5) = 4/5$
$20 x = 4 (3 x + 5) = 12 x + 20$
$8 x = 20, x = 2.5$
Alcohol = 4 x 2.5 = 10 litres

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 2

How many kg of rice at Rs. 50/kg must be mixed with 30 kg of rice at Rs. 70/kg to get a mixture worth Rs. 58/kg?

20 kg
25 kg
30 kg
35 kg

Answer: A. 20 kg

Explanation

Using alligation rule:
Cheaper (Rs.50/kg) vs Dearer (Rs.70/kg), Mean price = Rs.58/kg
Ratio (cheaper:dearer) = (70-58): $(58 - 50) = 12$ : $8 = 3$ :2
Dearer quantity = 30 kg, so cheaper = (2/3) x $30 = 20$ kg
Answer: 20 kg

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 3

A vessel contains 60 litres of milk. 6 litres of milk is taken out and replaced by water. This process is repeated once more. How much milk remains in the vessel?

48.6 litres
50 litres
52.4 litres
54 litres

Answer: A. 48.6 litres

Explanation

Using replacement formula:
Final milk = Initial x (1 - replaced/total)^n
= 60 x ( $1 - 6/60)^{2} = 60$ x (0.9)^2 = 60 x 0.81 = 48.6 litres

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Always identify which value is the 'mean' (target) price and which are the component prices before applying alligation.
For dilution problems with repeated removal and replacement, memorise and apply the formula: Final = Initial x (1 - x/T)^n to save time.
When dealing with percentages in alligation (like profit%, alcohol%, milk%), treat them as the 'prices' in the formula.
For three or more ingredients, solve using simultaneous equations rather than alligation - the cross method only works for two components.
In milk-water replacement problems, track only one component (usually milk) and calculate water as the remainder.
Check your answer by verifying: (Quantity x Price + Quantity x Price ) / (Quantity + Quantity ) = Mean Price

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Key concepts

Understanding the Alligation Cross

Types of Mixture Problems

The Replacement Formula

Weighted Average vs Alligation

Milk and Water Problems

Check the idea while it is fresh.Open a focused exercise without losing this article.Practise this concept

Worked examples

Example 1

In what ratio must rice at 30 per kg be mixed with rice at 40 per kg so that the mixture is worth 35 per kg?

Solution

Step 1: Apply the rule of alligation.

\frac{Quantity of Cheaper}{Quantity of Dearer} = \frac{d - m}{m - c}

Step 2: Substitute $c = 30$ , $d = 40$ , and $m = 35$ .

\frac{Quantity of Cheaper}{Quantity of Dearer} = \frac{40 - 35}{35 - 30} = \frac{5}{5} = \frac{1}{1}

Answer: 1 : 1.

Example 2

A merchant has 100 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. Find the quantity sold at 18% profit.

Solution

Step 1: Apply alligation on profit percentages.

\frac{Q _{8%}}{Q _{18%}} = \frac{18 - 14}{14 - 8} = \frac{4}{6} = \frac{2}{3}

Step 2: Divide total 100 kg in the ratio $2 : 3$ .

Q_{18%} = \frac{3}{2 + 3} \times 100 kg = \frac{3}{5} \times 100 = 60 kg

Answer: 60 kg.

Example 3

Solution

Step 1: Apply the repeated dilution formula.

Remaining Pure Milk = x (1 - \frac{y}{x})^{n}

Step 2: Substitute initial volume $x = 80$ , replacement $y = 8$ , and operations $n = 3$ .

Remaining = 80 (1 - \frac{8}{80})^{3} = 80 (1 - \frac{1}{10})^{3} = 80 \times (0.9)^{3} = 80 \times 0.729 = 58.32 litres

Answer: 58.32 litres.

Example 4

How many kilograms of tea costing 200 per kg must be mixed with 40 kg of tea costing 280 per kg to earn a 25% profit by selling the mixture at 300 per kg?

Solution

Step 1: Find the mean cost price $m$ of the mixture.

Selling Price = m \times 1.25 = 300 ⟹ m = \frac{300}{1.25} = 240/ kg

Step 2: Apply alligation ratio.

\frac{Q _{cheaper}}{Q _{dearer}} = \frac{280 - 240}{240 - 200} = \frac{40}{40} = \frac{1}{1}

Step 3: Solve for $Q_{cheaper}$ given $Q_{dearer} = 40 kg$ .

Q_{cheaper} = 1 \times 40 = 40 kg

Answer: 40 kg.

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

A mixture contains alcohol and water in the ratio 4:3. If 5 litres of water is added, the ratio becomes 4:5. Find the quantity of alcohol in the mixture.

8 litres
10 litres
12 litres
15 litres

Answer: B. 10 litres

Explanation

Let alcohol = 4x, water = 3x
After adding 5 litres water: water = 3x + 5
New ratio: $4 x / (3 x + 5) = 4/5$
$20 x = 4 (3 x + 5) = 12 x + 20$
$8 x = 20, x = 2.5$
Alcohol = 4 x 2.5 = 10 litres

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 2

How many kg of rice at Rs. 50/kg must be mixed with 30 kg of rice at Rs. 70/kg to get a mixture worth Rs. 58/kg?

20 kg
25 kg
30 kg
35 kg

Answer: A. 20 kg

Explanation

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 3

A vessel contains 60 litres of milk. 6 litres of milk is taken out and replaced by water. This process is repeated once more. How much milk remains in the vessel?

48.6 litres
50 litres
52.4 litres
54 litres

Answer: A. 48.6 litres

Explanation

Using replacement formula:
Final milk = Initial x (1 - replaced/total)^n
= 60 x ( $1 - 6/60)^{2} = 60$ x (0.9)^2 = 60 x 0.81 = 48.6 litres

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Always identify which value is the 'mean' (target) price and which are the component prices before applying alligation.
For dilution problems with repeated removal and replacement, memorise and apply the formula: Final = Initial x (1 - x/T)^n to save time.
When dealing with percentages in alligation (like profit%, alcohol%, milk%), treat them as the 'prices' in the formula.
For three or more ingredients, solve using simultaneous equations rather than alligation - the cross method only works for two components.
In milk-water replacement problems, track only one component (usually milk) and calculate water as the remainder.
Check your answer by verifying: (Quantity x Price + Quantity x Price ) / (Quantity + Quantity ) = Mean Price

Key formulas and rules

Key concepts

Understanding the Alligation Cross

Types of Mixture Problems

The Replacement Formula

Weighted Average vs Alligation

Milk and Water Problems

Worked examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Example 4

Solution

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Ready to test your understanding?

Keep building the same skill family.

Key formulas and rules

Key concepts

Understanding the Alligation Cross

Types of Mixture Problems

The Replacement Formula

Weighted Average vs Alligation

Milk and Water Problems

Worked examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Example 4

Solution

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Ready to test your understanding?

Keep building the same skill family.