Mixtures & Alligation

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Replacement & Repeated Operations

⏱ 5 min read🧩 5 question types🎯 15 practice Q

The idea in one minute

Take r litres from a vessel of C litres and replace with water, n times. Each round the original liquid keeps the same fraction:

$\text{left} = C\left(1 - \dfrac{r}{C}\right)^n$ and water = C − left.

One round is plain subtraction. Drawing from a mixture takes both ingredients in proportion, so the inside ratio does not move. The total volume stays C whenever you replace what you removed.

One round, in plain numbers

A vessel holds 60 L of pure milk. 12 L are drawn out and replaced with water. Now: milk 48 L, water 12 L → ratio $4 : 1$ .

No formula is needed for a single round — subtract the drawn milk, add the drawn volume as water.

Rule: 'Drawn and replaced' keeps the TOTAL at C. Only the split between the liquids changes.

The repeated-replacement formula

Do the same draw again and the vessel is no longer pure. The second draw takes out milk AND water in proportion, so the milk keeps the same fraction each round:

\text{milk left} = C\left(1 - \frac{x}{C}\right)^n, \qquad \text{water} = C - \text{milk}

80 L vessel, 8 L drawn and replaced, twice: milk = $80 \times \left(\frac{9}{10}\right)^2 = 64.8$ L. The second round on the 60 L example: $48 \times \frac{48}{60} = 38.4$ L milk → ratio $38.4 : 21.6 = 16 : 9$ .

Tip: One power per round. Never subtract x litres of milk twice — after the first round the draw is part water.

Drawing from an already-mixed vessel

Draw from a uniform mixture and you take each ingredient in the vessel's own ratio. The ratio inside does not move; only the volume shrinks. The addition afterwards is what shifts the ratio.

80 L of milk : water = 7 : 3 holds 56 L milk, 24 L water. Draw 20 L: it takes 14 milk and 6 water, leaving 42 : 18 — still 7 : 3 in 60 L. Now add 20 L of water: $42 : 38 = 21 : 19$ .

Watch: Replacing with the SAME ingredient works the other way. From 50 L of 4 : 1 (40 milk, 10 water), draw 10 L (8 milk, 2 water) and replace with pure milk: $42 : 8 = 21 : 4$ .

Working backwards

Given the final ratio, take roots.

Wine : water ends at $16 : 65$ after four equal draws. Wine fraction = $\frac{16}{81} = \left(\frac{2}{3}\right)^4$ , so each round kept $\frac{2}{3}$ : $1 - \frac{x}{C} = \frac{2}{3}$ , and with $x = 8$ , $C = 24$ L.
Two rounds from a 50 L vessel leave 32 L of milk: $\left(1 - \frac{x}{50}\right)^2 = \frac{16}{25}$ → $x = 10$ L.

Water by the complement

Water never needs its own formula. Water = C − milk, so with kept fraction $k = 1 - \frac{x}{C}$ , after n rounds:

\text{milk} : \text{water} = k^n : (1 - k^n)

On the 80 L vessel ( $k = \frac{9}{10}$ , two rounds): milk $\frac{81}{100}$ , water $\frac{19}{100}$ → $81 : 19$ — 64.8 L against 15.2 L, the same numbers as before.

Shift the ratio by one replacement

A vessel is at 7 : 5. Nine litres of mixture are replaced with water and it becomes 7 : 9. Find the starting milk.

Total T: milk goes from $\frac{7T}{12}$ to $\frac{7T}{16}$ . The draw removes $9 \times \frac{7}{12}$ of milk:

\frac{7T}{12} - \frac{63}{12} = \frac{7T}{16} \Rightarrow T = 36, \text{ so milk} = 21 \text{ L}

Tip: Match the final fraction to a perfect power ( $\frac{4}{9}, \frac{16}{25}, \frac{16}{81}$ ). Exam numbers are built to land on them.

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

Repeated replacement (formula)

How to spot it:

x litres drawn and replaced with water, n times, from C litres of pure liquid — what remains is asked.

\text{left} = C\left(1 - \frac{x}{C}\right)^n

Method

Compute the kept fraction $1 - \frac{x}{C}$ .
Raise it to the number of operations.
Multiply by C; water = C − milk if asked.

Why it works:

Each round removes the same fraction of whatever original liquid is still present.

Try this

A vessel contains 80 litres of pure milk. 8 litres are drawn and replaced with water, and the operation is repeated once more. Find the milk left.

Show solution

Kept fraction = $1 - \frac{8}{80} = \frac{9}{10}$ .
Two rounds: $\left(\frac{9}{10}\right)^2 = \frac{81}{100}$ .
Milk = $80 \times \frac{81}{100} = 64.8$ L.

Answer

64.8 litres

Type 2very common2 practice Q

One or two draws — ratio form

How to spot it:

A single draw-replace, or a second round from the diluted vessel — the final ratio is asked.

\text{after round 2: } \text{milk}:\text{water} = \left(1-\tfrac{x}{C}\right)^2 : \left[1-\left(1-\tfrac{x}{C}\right)^2\right]

Method

Round 1: subtract x litres of the pure liquid, add x of water.
Round 2: multiply the survivor by the kept fraction again.
Pair the milk with (C − milk) for the ratio.

Why it works:

Only the original liquid keeps shrinking; water absorbs the difference.

Try this

From 60 litres of pure milk, 12 litres are removed and replaced with water. Find the ratio of milk to water now.

Show solution

Milk = $60 - 12 = 48$ L.
Water = $12$ L.
Ratio = $48 : 12 = 4 : 1$ .

Answer

4 : 1

Type 3common2 practice Q

Reverse — find x or the capacity

How to spot it:

The final quantity or ratio is given; the drawn amount, the capacity, or the number of rounds is asked.

\left(1 - \frac{x}{C}\right)^n = \text{final fraction of the original liquid}

Method

Divide the final amount of the original liquid by C.
Take the n-th root; match it to a perfect power.
Solve $1 - \frac{x}{C} = \text{root}$ for the unknown.

Why it works:

The formula runs backwards once you spot the power.

Try this

A cask is full of wine. 8 litres are drawn and replaced with water, and the operation is performed three more times. The final ratio of wine to water is 16 : 65. Find the capacity of the cask.

Show solution

Wine fraction = $\frac{16}{16+65} = \frac{16}{81}$ .
$\frac{16}{81} = \left(\frac{2}{3}\right)^4$ , so each round kept $\frac{2}{3}$ .
$1 - \frac{8}{C} = \frac{2}{3} \Rightarrow C = 24$ L.

Answer

24 litres

Type 4common2 practice Q

Drawing from an already-mixed vessel

How to spot it:

The vessel starts as a mixture, not a pure liquid; a draw is followed by an addition.

\text{the draw splits in the vessel's own ratio}

Method

Convert the starting ratio into litres of each ingredient.
Split the drawn volume in that same ratio — the ratio does not move.
Add the new ingredient to its column and rebuild the ratio.

Why it works:

A uniform mixture leaves in its own proportions.

Try this

80 litres of a milk-water mixture in the ratio 7 : 3 has 20 litres drawn off, and then 20 litres of water are added. Find the new ratio.

Show solution

Start: 56 milk, 24 water. Draw takes 14 and 6.
After draw: 42 milk, 18 water (still 7 : 3).
Add 20 water: $42 : 38 = 21 : 19$ .

Answer

21 : 19

Type 5occasional2 practice Q

Replace with the same ingredient (enriching)

How to spot it:

The vessel is topped up with the pure version of one ingredient, so the ratio tightens instead of diluting.

\text{remove in the ratio, then add the pure component}

Method

Split the removed volume by the current ratio.
Subtract those amounts from their columns.
Add the removed volume of the pure ingredient to its column.

Why it works:

The topping-up choice decides which column swells.

Try this

From 50 litres of a milk-water mixture in the ratio 4 : 1, 10 litres are removed and replaced with pure milk. Find the new ratio.

Show solution

Start: 40 milk, 10 water. Draw takes 8 and 2.
After draw: 32 milk, 8 water.
Add 10 milk: $42 : 8 = 21 : 4$ .

Answer

21 : 4

Formula sheet

Repeated equal replacement

\text{left} = C\left(1 - \frac{x}{C}\right)^n

Milk : water after n rounds

\left(1-\frac{x}{C}\right)^n : \left[1-\left(1-\frac{x}{C}\right)^n\right]

Proportional removal

\text{lost} = \text{drawn volume} \times \frac{\text{ingredient share}}{\text{total}}

Unequal draws

C \prod_k \left(1 - \frac{x_k}{C}\right)

Multiply one factor per round when the draws differ.

Shortcuts that save time

⚡ One fraction per round

Each round multiplies the original liquid by (1 − x/C). Multiply the fractions; never subtract x twice.

Example

From a vessel full of milk, 8 litres are drawn and replaced with water. The vessel holds 80 litres. Find the ratio of milk to water after this single operation.

Show solution

Milk left = $80 - 8 = 72$ L.
Water = $8$ L.
Ratio = $72 : 8 = 9 : 1$ .

Answer

9 : 1

⚡ Two rounds: square the kept fraction

The complement of the milk is the water — no separate calculation.

Example

8 litres are drawn from a cask full of wine and replaced with water; this is done once more. The ratio of wine to water is then 16 : 9. Find the capacity of the cask.

Show solution

Wine fraction = $\frac{16}{16+9} = \frac{16}{25}$ .
$\left(1 - \frac{8}{C}\right)^2 = \frac{16}{25} \Rightarrow 1 - \frac{8}{C} = \frac{4}{5}$ .
$\frac{8}{C} = \frac{1}{5} \Rightarrow C = 40$ L.

Answer

40 litres

⚡ Draws keep the inside ratio

A uniform draw removes both liquids in proportion, so the ratio survives the withdrawal. Only the refill changes it.

Example

A vessel has milk and water in the ratio 7 : 5. Nine litres of the mixture are removed and replaced with water, making the ratio 7 : 9. How much milk was in the vessel initially?

Show solution

Draw removes $\frac{7}{12}$ of 9 L = milk.
$\frac{7T}{12} - \frac{63}{12} = \frac{7T}{16} \Rightarrow T = 36$ L.
Milk = $\frac{7}{12} \times 36 = 21$ L.

Answer

21 litres

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Subtracting x litres of milk in every round.

After round one the draw is part water. Multiply by (1 − x/C) each round instead.

Mistake 02

Shrinking the water as well.

Only the original liquid shrinks. Water = C − milk; the total stays C.

Mistake 03

Using (1 − x/C)ⁿ when the draws have different sizes.

Multiply one separate factor per round: (1 − x₁/C)(1 − x₂/C)…

Mistake 04

Changing the total volume after a replacement.

Draw r and pour back r: the vessel holds C litres throughout.

Mistake 05

Adding the margins in reverse questions.

Take the n-th root of the final fraction and match it to a perfect power.

Quick revision

Read this the night before the exam.

n equal rounds: milk = C(1 − x/C)ⁿ; water = C − milk.
One round: plain subtraction.
A uniform draw leaves the inside ratio unchanged.
Total stays C whenever you replace what you remove.
Reverse: final fraction → perfect power → root → solve.
Replace with the same ingredient to enrich, not dilute.

Practice: 15 questions

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

Topic test · 10 questions

Suggested time 7 min · wrong answers go to your mistake notebook automatically.

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