Mixtures & Alligation

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Replacement & Repeated Operations

Remove some mixture and replace it with another liquid — each operation multiplies the original liquid by a shrink factor.

Equal-volume replacements (take r litres from a vessel of C litres, any number of times): $\text{original liquid left} = C\left(1 - \frac{r}{C}\right)^n$

Different replacement strengths: multiply the surviving fractions: $\left(1 - \frac{r_1}{C}\right)\left(1 - \frac{r_2}{C}\right)\cdots$

Partial removal of one ingredient (drawn mixture changes only the drawn part): remove the ingredient's proportional share: lost = removed $\times \frac{\text{ingredient fraction}}{1}$ .

The ratio of original : added liquid after n operations is $\left(1-\frac{r}{C}\right)^n : \left[1 - \left(1-\frac{r}{C}\right)^n\right]$ .

📖 Notes 🧩 5 question types ⚡ Formulas 🎯 15 practice Q

Detailed notes

The repeated-replacement formula

A vessel holds C litres of pure milk (or any component). x litres are drawn out and replaced with water; this is done n times. Because the mixture is uniform, each operation removes the same fraction $\frac{x}{C}$ of whatever is in the vessel. After n rounds the milk left is $\text{milk} = C\left(1 - \frac{x}{C}\right)^n$ 80 L, 8 L drawn and replaced, twice: $80 \times \left(\frac{9}{10}\right)^2 = 64.8$ L. The water is simply $C - \text{milk}$ (the total never changes when you replace what you remove).

Why the fraction form works

Each operation multiplies the remaining milk by $\left(1 - \frac{x}{C}\right)$ — the milk is drawn proportionally, the water topping-up adds no milk. Multiplying is why powers appear; the same one-liner handles two, three or four rounds without any table.

One operation only

A single draw-replace needs no formula: subtract x litres of the component directly, add x of the other. From 60 L of milk, remove 12 L, add water: milk 48, water 12 → 4:1. A second operation then multiplies by $\frac{48}{60}$ : milk $= 48 \times \frac{48}{60} = 38.4$ → ratio $16:9$ . Note $\frac{48}{60} = 1 - \frac{12}{60}$ — consistent with the formula.

Removing from a mixed vessel

If the vessel is already a mixture in ratio m:w, a draw of x litres takes each component in proportion — the ratio inside the vessel is unchanged; only the volume shrinks. Adding anything afterwards is what shifts the ratio. From milk:water 7:3 in 80 L (56, 24), remove 20 L (14, 6) — still 7:3 in 60 L; add 20 L water → 42:38.

Reverse problems

Given the final quantity, extract x or n:

$C\left(1-\frac{x}{C}\right)^2 = 32$ with C = 50 → $\left(1-\frac{x}{50}\right)^2 = \frac{16}{25}$ → $x = 10$ L.
Final ratio gives the fraction: wine : water 16:65 → wine fraction $\frac{16}{81} = \left(\frac{2}{3}\right)^4$ → four operations with $\frac{x}{C} = \frac{1}{3}$ . Take roots — square for two operations, fourth root for four — and match against perfect powers.

Common slips

Subtracting x litres of milk every round even after the vessel is diluted (that is the $\left(1-\frac{x}{C}\right)^n$ trap).
Forgetting the total stays C when you replace, so 'milk : water' pairs milk with $C -$ milk.
Using addition instead of multiplication across rounds.

Quick revision

$n$ rounds: milk $= C\left(1-\frac{x}{C}\right)^n$ ; water $= C -$ milk.
One round: plain subtraction; two rounds: multiply twice.
Drawing from a uniform mixture preserves its internal ratio.
Reverse: match the final fraction to a perfect power.
Total is constant when you replace what you remove.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Repeated replacement (formula)very common2 practice Q

How to spot it:

x litres drawn and replaced with water, n times, from C litres of pure milk/wine — find what remains.

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

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

Why: each round removes the same fraction of whatever milk is present.

Example: 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.

$80 \times \left(\frac{9}{10}\right)^2 = 64.8$ litres.

Type 2: One or two draws — ratio formvery common2 practice Q

How to spot it:

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

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

Round 1: subtract x litres of milk, add x water — read the ratio.
Round 2: multiply the milk by the kept fraction again.
Pair the final milk with (C − milk).

Why: only milk keeps shrinking; water absorbs the difference.

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

Milk 48, water 12 → 4:1.

Type 3: Reverse — find x or the capacitycommon2 practice Q

How to spot it:

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

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

Divide the final milk by C to get the kept fraction.
Take the n-th root and match to a perfect power.
Solve $1 - \frac{x}{C} =$ root for the unknown.

Why: the formula is invertible once you spot the power.

Example: A cask full of wine: 8 litres drawn and replaced with water, the operation performed three more times. Wine : water is finally 16 : 65. Find the capacity.

Wine fraction $= \frac{16}{81} = \left(\frac{2}{3}\right)^4$ → $1 - \frac{8}{C} = \frac{2}{3}$ → C = 24 litres.

Type 4: Drawing from an already-mixed vesselcommon2 practice Q

How to spot it:

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

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

Convert the starting ratio to litres of each component.
Remove x litres split in that ratio — the ratio is unchanged.
Add the new ingredient to the required column and re-ratio.

Why: a uniform mixture yields its components proportionally.

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

Start 56, 24; draw removes 14, 6 → 42, 18; add 20 water → 42:38 = 21:19.

Type 5: Replace with the same component (enriching)occasional2 practice Q

How to spot it:

Mixture loses some of one component and is topped up with the other one — ratios tighten instead of dilute.

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

Split the removed volume by the current ratio.
Subtract from the matching columns; the total dips by x.
Add x litres of the topping component to its column.

Why: the topping-up choice decides which column swells.

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

Draw removes 8 milk, 2 water → 32, 8; add 10 milk → 42:8 = 21:4.

Formulas

Repeated equal replacement

C\left(1 - \frac{r}{C}\right)^n

Milk : water after n ops

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

Proportional removal

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

Chain of replacements

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

Shortcut tricks

⚡ The (1 − r/C)ⁿ shrink

One fraction per operation; multiply.

Example: From a vessel full of milk, 8 litres are drawn and replaced with water; the operation is repeated once more. The ratio of milk to water in the vessel (capacity 80 L) is:

Milk left $= 80 \times \frac{9}{10} = 72$ after one draw ⇒ milk : water $= 72 : 8 = 9 : 1$ .

⚡ Two operations, wine : water

Square the shrink factor; the complement is the water.

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. The capacity of the cask is:

$\left(1 - \frac{8}{C}\right)^2 = \frac{16}{25}$ ⇒ $1 - \frac{8}{C} = \frac{4}{5}$ ⇒ $C = 40$ litres.

⚡ Draw-and-refill with a ratio shift

Removing uniform mixture deletes each ingredient proportionally; only the drawn fraction matters.

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. The quantity of milk initially is:

Total T: milk goes from $\frac{7}{12}T$ to $\frac{7}{16}T$ ; drawn milk $= 9 \times \frac{7}{12}$ , so $\frac{7T}{12} - \frac{63}{12} = \frac{7T}{16}$ ⇒ $T = 36$ ⇒ milk $= 21$ litres.

Where students lose marks

Using $C\left(1 - \frac{r}{C}\right)^n$ when different volumes are drawn each time — multiply the varying factors instead.
Forgetting the vessel's total stays constant when you replace (drain + refill keeps capacity).
Computing water left by shrinking water too — added water enters whole, only the original liquid shrinks.
In ratio-shift problems, subtracting the drawn amount from the total as well as from the ingredient (total is unchanged after refill).

Practice sets — 15 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 10 questions

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