Mixtures & Alligation

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Milk–Water Ratio & Profit by Adulteration

The seller's game: mix in free (or cheap) water and sell at the milk price.

Profit % when water is mixed and the mixture is sold at cost: $\text{profit\%} = \frac{\text{water}}{\text{milk}} \times 100 = \frac{w}{m} \times 100$

To reach a wanted ratio, fix the constant ingredient and solve for the added one — one linear equation, never a system.

To alter an alloy, compute each metal from the current ratio first, then add/subtract the named metal and rebuild.

📖 Notes 🧩 5 question types ⚡ Formulas 🎯 12 practice Q

Detailed notes

Water sells for nothing — that is the whole trick

A milkman adds water to milk and sells the blend at the cost price of milk: every litre sold brings milk-price rupees, but part of the litre cost him nothing. His gain comes purely from the free water. $\text{gain\%} = \frac{\text{water}}{\text{milk}} \times 100$ Water : milk 1:5 → gain 20%. Water is 25% of the mixture → milk is 75%, so gain $= \frac{25}{75} = 33\frac{1}{3}\%$ . The two readings differ; the formula uses water per unit milk.

Working backwards

For a target gain g%, set $\frac{W}{M} = \frac{g}{100}$ : to gain 25%, water : milk = 1:4 (one litre of water per four of milk). To gain $16\frac{2}{3}\%$ , $W : M = 1:6$ . Options often include the inverted ratio and the mixture-fraction misread — compute, don't guess.

Blends that must earn a profit

When two real ingredients are mixed, the blend's cost price per kg is the weighted mean: $CP = \frac{q_1 c_1 + q_2 c_2}{q_1 + q_2}$ Selling price follows: for a 25% gain, $SP = 1.25 \times CP$ . Rice ₹30 and ₹40 mixed 2:3 → CP ₹36; sell at ₹45 → 25% profit. Reverse: a required profit fixes the SP, and the alligation arms fix the mixing ratio that makes it possible.

Adulteration with a cheaper item

Ghee at ₹120 cut with oil at ₹60, sold at ₹120 for a 50% gain: the blend's CP must be $\frac{120}{1.5} = ₹80$ , so alligate: ghee : oil $= (80-60):(120-80) = 1:2$ . Verify: $\frac{120 \times 1 + 60 \times 2}{3} = ₹80$ ✓ — the costly item takes the smaller share when the target CP sits below the two-price midpoint. The lesson: always verify the blend's CP from your ratio before quoting the answer.

Marked price chains on a mixture

A marked-up-then-discounted sale still ends at one SP: chips multiply ( $1.25 \times 0.9 = 1.125$ → 12.5% net gain). Compute the mixture's CP first, then run the markup chain on it.

A worked case end to end

80 kg of rice at ₹24 is mixed with 40 kg at ₹36, and the trader wants 25%. Blend CP $= \frac{1920 + 1440}{120} = ₹28$ ; SP $= 28 \times 1.25 = ₹35$ . Every adulteration question is this skeleton with one slot hidden — the ratio, the SP, or the gain. Fix any two and the third follows by the same three lines. Note the asymmetry: profit chips act on the blend's CP, while water additions act on the cost per sold litre — keep the two bases apart.

Quick revision

Water : milk = g : 100 for gain g% when sold at cost.
'Water is p% of the mixture' → gain $= \frac{p}{100-p} \times 100$ .
Blend CP = weighted mean; SP from the target profit.
Adulteration: alligate to hit the CP that yields the gain.
Verify every ratio by recomputing the CP.

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: Gain % from added watervery common2 practice Q

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

Express water per unit milk (convert 'p% of mixture' → $\frac{p}{100-p}$ ).
Multiply by 100.
Sanity: water equal to half the milk → 50% gain.

Why: the milk's cost is fully recovered; the water volume is pure margin.

Example: A milkman mixes water equal to one-fifth of the milk and sells the mixture at cost price. His gain per cent is:

$\frac{1}{5} \times 100 = 20\%$ .

Type 2: Find the water ratio for a target gainvery common2 practice Q

How to spot it:

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

W : M = g : 100

Set $\frac{W}{M} = \frac{g}{100}$ and reduce.
State the ratio in the units the options use (water:milk vs milk:water).
Cross-check with one litre: (1+M) litres sold for M litres' cost.

Why: the gain per cent is the water-to-milk ratio in per cent form.

Example: A milkman wants a 25% gain by adding water and selling at cost price. The ratio of water to milk in his mixture is:

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

Type 3: Blend CP and profit on the mixturecommon2 practice Q

How to spot it:

Two priced ingredients mixed in a given ratio; the mixture's 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)

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

Why: profit acts on the blend's average cost, not on either ingredient.

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

CP $= \frac{2 \times 30 + 3 \times 40}{5} = ₹36$ → gain $= \frac{9}{36} \times 100 = 25\%$ .

Type 4: Adulteration for a target gaincommon2 practice Q

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 must equal } \frac{SP}{1 + g/100};\quad \text{alligate on } C_1, C_2

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

Why: the blend CP that yields the gain pins the ratio uniquely.

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

Blend CP must be $₹80$ → ghee : oil $= (80-60):(120-80) = 1:2$ . Check: $\frac{120 \times 1 + 60 \times 2}{3} = ₹80$ ✓.

Type 5: Mixture sold through markup and discountoccasional2 practice Q

How to spot it:

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

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

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

Why: markup-discount is a chip chain on top of the mixture arithmetic.

Example: A mixture costing ₹40 per kg is marked 25% above cost and sold at a 10% discount. The profit per cent is:

$1.25 \times 0.9 = 1.125$ → 12.5% profit (SP ₹45).

Formulas

Adulteration profit

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

Ratio target

\frac{m}{w + x} = \text{wanted ratio} \Rightarrow x

Alloy rebuild

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

SP of mixture

\text{SP} = (\text{milk}) \times \text{milk price} + (\text{water}) \times 0

Shortcut tricks

⚡ Water-over-milk is the profit

Adulteration profit = free litres per honest litre.

Example: A milkman mixes 1 litre of water with every 5 litres of milk and sells the mixture at the cost price of milk. His profit per cent is:

$\frac{1}{5} \times 100 = 20\%$ .

⚡ Hit the target ratio by addition

Keep the bigger quantity constant; 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?

Milk 30 L stays; water must reach $30 \times \frac{3}{5} = 18$ ⇒ add $18 - 6 = 12$ litres.

⚡ Rebuild the alloy, then change one metal

Convert ratio to grams, adjust, 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?

Zinc 25 g, copper 15 g; want zinc : copper $= 5 : 7$ ⇒ copper $= 35$ ⇒ add $20$ g.

Where students lose marks

Computing profit on the mixture instead of the milk cost (profit = water/milk, sold at cost).
Adding water and expecting the milk quantity to stretch — only water grows.
In alloy problems, adding to the total and to one metal without updating the other ratio leg.
Misreading 'ratio 5 : 7' as 'add 7 parts' instead of solving for the new copper amount.

Practice sets — 12 questions

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

Topic test · 12 questions

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