Mixtures & Alligation

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high importance~1 Q in Tier 120 formulas⚡ 15 shortcuts5 subtopics

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Milk–Water Ratio & Profit by Adulteration

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⏱ 4 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

The seller's game: mix in free water (or a cheap filler) and sell the blend at the good item's price.

Sold at cost price, the gain is the free part per unit of the real item:

$\text{gain \%} = \dfrac{\text{water}}{\text{milk}} \times 100$ .

For a target ratio, freeze the constant ingredient and solve for the addition. When two priced items are blended, the blend's cost price is the weighted mean, and profit acts on it.

The milkman's trick

Water is free. The milkman adds it to milk and sells the blend at the milk price. Every litre sold brings milk-price money, and part of each litre cost him nothing.

\text{gain\%} = \frac{\text{water}}{\text{milk}} \times 100

Water equal to one-fifth of the milk → gain $= \frac{1}{5} \times 100 = 20\%$ .

Rule: The gain is water per unit of MILK, not per unit of mixture.

Water as a per cent of the mixture

'Water is 25% of the mixture' reads differently: milk is 75%, so gain $= \frac{25}{75} = 33\frac{1}{3}\%$ . Convert a mixture-per-cent into water-per-milk before using the formula.

Watch: Options usually include both readings — 25% and $33\frac{1}{3}\%$ . Compute, do not guess.

Working to a target gain

For a wanted gain of g%, set water : milk = g : 100.

25% gain → water : milk = $25 : 100 = 1 : 4$ .
$16\frac{2}{3}\%$ gain → $1 : 6$ .

One litre of water per four litres of milk gives exactly 25%.

Blend cost price and profit

With two real, priced ingredients the blend's cost price is the weighted mean:

CP = \frac{q_1 c_1 + q_2 c_2}{q_1 + q_2}

Rice at ₹30 and ₹40 mixed 2 : 3 → CP = $\frac{60 + 120}{5} = ₹36$ . Sold at ₹45, the gain is $\frac{9}{36} = 25\%$ . For a wanted profit of 25%, fix $SP = 1.25 \times CP$ and work backwards.

Adulterating with a cheaper item

Ghee at ₹120 is cut with oil at ₹60 and sold at ₹120 for a 50% gain.

The blend's CP must be $\frac{120}{1.5} = ₹80$ . Alligate: ghee : oil $= (80 - 60) : (120 - 80) = 1 : 2$ . Verify: $\frac{120 \times 1 + 60 \times 2}{3} = ₹80$ .

Tip: The costly item takes the smaller share when the target CP sits below the midpoint. Always recompute the CP from your ratio.

Selling above cost price

The trick gets worse when the blend also sells at a markup. Chain the factors: each free litre multiplies the money, and the markup multiplies again.

Water equal to 20% of the milk AND a 10% markup: $1.20 \times 1.10 = 1.32$ → a 32% gain. Water alone would give 20%; the markup alone 10%; together they multiply, never add.

Rule: $\text{gain factor} = \left(1 + \frac{W}{M}\right)\left(1 + \frac{\text{markup}}{100}\right)$ . Subtract 1 for the gain per cent.

A discount can eat the gain

A discount works the same chain in reverse. Water at 25% of the milk, then a 10% discount: $1.25 \times 0.9 = 1.125$ — the milkman still pockets 12.5%. Options quoting 25% or 15% are the two traps.

Markup and discount on a mixture

A blended item marked up and then discounted still ends at one selling price. Chain the factors: $1.25 \times 0.9 = 1.125$ → a 12.5% gain on a CP of ₹40, selling at ₹45.

Profit chips act on the blend's CP; water additions act on the cost per sold litre. Keep the two bases apart.

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

Gain per cent from added water

How to spot it:

Water is added to milk (or a cheap filler to a good item) and the blend is sold at cost price — find the gain %.

\text{gain\%} = \frac{\text{water}}{\text{milk}} \times 100

Method

Express water per unit of milk (convert 'p% of mixture' into p/(100−p)).
Multiply by 100.
Sanity check: water equal to half the milk → 50% gain.

Why it works:

The milk's cost comes back in full; the water volume is pure margin.

Try this

A milkman mixes water equal to one-fifth of the milk and sells the mixture at cost price. Find his gain per cent.

Show solution

Water : milk = $1 : 5$ .
Gain = $\frac{1}{5} \times 100 = 20\%$ .

Answer

20%

Type 2very common2 practice Q

Water ratio for a target gain

How to spot it:

'What ratio of water to milk gives a gain of x% when sold at cost?'

W : M = g : 100

Method

Set water : milk = g : 100 and reduce.
State the ratio in the direction the options use.
Check with one litre: (1 + M) litres sold for the price of M.

Why it works:

The gain per cent IS the water-to-milk ratio in per cent form.

Try this

A milkman wants a 25% gain by adding water and selling at cost price. Find the ratio of water to milk in his mixture.

Show solution

$W : M = 25 : 100$ .
= $1 : 4$ .

Answer

1 : 4

Type 3common2 practice Q

Blend cost price and profit

How to spot it:

Two priced ingredients mixed in a given ratio; the selling price or profit % is asked.

CP = \frac{q_1 c_1 + q_2 c_2}{q_1 + q_2}, \quad SP = CP\left(1+\frac{g}{100}\right)

Method

Compute the weighted CP of the blend.
Apply the profit factor, or divide SP by CP to find the gain.
Keep fractions — ₹36.25 answers are legitimate.

Why it works:

Profit acts on the blend's average cost, never on one ingredient alone.

Try this

Rice at ₹30 per kg and ₹40 per kg are mixed in the ratio 2 : 3 and the mixture is sold at ₹45 per kg. Find the profit per cent.

Show solution

CP = $\frac{2 \times 30 + 3 \times 40}{5} = \frac{180}{5} = ₹36$ .
Gain per kg = $45 - 36 = ₹9$ .
Profit = $\frac{9}{36} \times 100 = 25\%$ .

Answer

25%

Type 4common2 practice Q

Adulteration for a target gain

How to spot it:

A costly item is cut with a cheaper one; the selling price and target gain fix the mixing ratio.

\text{blend CP} = \frac{SP}{1 + g/100}; \quad \text{alligate on } c_1, c_2

Method

Convert the gain into the required blend CP.
Alligate between the two cost prices to reach that CP.
Verify by recomputing the weighted CP.

Why it works:

The blend CP that yields the gain pins the ratio uniquely.

Try this

Ghee costing ₹120 per kg is mixed with oil costing ₹60 per kg and the mixture is sold at ₹120 per kg for a 50% gain. Find the mixing ratio of ghee to oil.

Show solution

Blend CP = $\frac{120}{1.5} = ₹80$ .
Ghee : oil = $(80 - 60) : (120 - 80) = 20 : 40 = 1 : 2$ .
Check: $\frac{120 + 120}{3} = ₹80$ .

Answer

1 : 2

Type 5occasional2 practice Q

Mixture sold through markup and discount

How to spot it:

The blended item is marked up and then discounted — the net profit on the blend's CP is asked.

\text{net} = \left(1+\frac{m}{100}\right)\left(1-\frac{d}{100}\right) - 1

Method

Find the blend's CP per unit.
Multiply the markup and discount factors for the net factor.
Profit % = (net factor − 1) × 100.

Why it works:

Markup and discount are chained factors on top of the mixture arithmetic.

Try this

A mixture costs ₹40 per kg. It is marked 25% above cost and sold at a 10% discount. Find the profit per cent.

Show solution

Markup factor = $1.25$ ; discount factor = $0.9$ .
Net = $1.25 \times 0.9 = 1.125$ .
Profit = $12.5\%$ (SP = ₹45).

Answer

12.5%

Formula sheet

Adulteration gain

\text{gain\%} = \frac{\text{water}}{\text{milk}} \times 100

Target ratio

W : M = g : 100

For a gain of g% when sold at cost price.

Blend cost price

CP = \frac{q_1 c_1 + q_2 c_2}{q_1 + q_2}, \quad SP = CP\left(1 + \frac{g}{100}\right)

Alloy rebuild

\text{metal} = \text{alloy weight} \times \frac{\text{share}}{\text{sum of shares}}

Shortcuts that save time

⚡ Water over milk is the gain

Free litres per honest litre — that ratio, in per cent, is the profit.

Example

A milkman mixes 1 litre of water with every 5 litres of milk and sells the mixture at the cost price of milk. Find his gain per cent.

Show solution

Water : milk = $1 : 5$ .
Gain = $\frac{1}{5} \times 100 = 20\%$ .

Answer

20%

⚡ Hit the target ratio by addition

Keep the bigger quantity constant and solve for the addition.

Example

36 litres of a mixture has milk and water in the ratio 5 : 1. How much water must be added to make the ratio 5 : 3?

Show solution

Milk = $36 \times \frac{5}{6} = 30$ L, fixed; water = 6 L.
At 5 : 3, water must be $30 \times \frac{3}{5} = 18$ L.
Add $18 - 6 = 12$ L.

Answer

12 litres

⚡ Rebuild the alloy, then change one metal

Ratio to grams, adjust one column, re-ratio.

Example

An alloy of 40 g contains zinc and copper in the ratio 5 : 3. How much copper must be added to make the ratio 5 : 7?

Show solution

Zinc = $25$ g, copper = $15$ g.
At 5 : 7, copper = $25 \times \frac{7}{5} = 35$ g.
Add $35 - 15 = 20$ g.

Answer

20 g

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Computing profit on the mixture volume.

Gain = water ÷ milk. The milk's cost is what is recovered.

Mistake 02

Reading 'water is p% of the mixture' as a p% gain.

Convert first: gain = p ÷ (100 − p) × 100.

Mistake 03

Adding water and stretching the milk quantity too.

Only the water column and the total grow.

Mistake 04

Adding to the alloy total without re-checking the other metal's leg.

Freeze the unchanged metal, then solve for the new amount.

Mistake 05

Quoting a ratio without recomputing the blend CP.

Verify: weighted CP from your ratio must hit the target.

Quick revision

Read this the night before the exam.

Gain % = water ÷ milk × 100, when sold at cost price.
Water p% of mixture → gain = p ÷ (100 − p) × 100.
Target gain g% → water : milk = g : 100.
Blend CP = weighted mean; SP = CP × (1 + g/100).
Adulteration: alligate to the CP that yields the gain.
Verify every ratio by recomputing the CP.

Practice: 12 questions

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

Topic test · 12 questions

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

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