Mixture and alligation is, at its core, a tool for handling weighted averages — but framed around physical mixing of two (or more) substances.
Here is the central idea: when you mix two solutions of different concentrations (or two items of different prices), the resulting mixture has some intermediate value. The question is always either "what is that intermediate value?" or "in what ratio did you mix them to get this intermediate value?"
Think of it like two people pulling on a rope toward their own side. The final position of the knot depends on how hard each person pulls (their quantity) and their starting position (their concentration or price). Alligation is just the formula that describes where the knot lands.
Two types of problems you will see:
There is a third pattern that combines both — the replacement/dilution type — where you repeatedly remove some solution and replace it with water. That one has its own formula, covered in Deep Dive.
One important mindset shift: stop thinking of alligation as a separate topic. It is just the weighted average formula rearranged. Once you see that, you will never confuse yourself again.
Final concentration = (q₁ × c₁ + q₂ × c₂) / (q₁ + q₂)
where q₁, q₂ are quantities and c₁, c₂ are concentrations. Alligation just solves this for q₁/q₂ when the final concentration is given.
Given two mixtures with concentrations c₁ and c₂, and a desired mean concentration m (where c₁ > m > c₂), the ratio in which they must be mixed is:
q₁ : q₂ = (m - c₂) : (c₁ - m)
Visually, the "cross" looks like this:
c₁ c₂
\ /
m
/ \
(m - c₂) (c₁ - m)
The ratio of the cheaper/lower to the costlier/higher goes cross-wise: the difference on the right side gives the quantity of the left ingredient, and vice versa.
Look — the formula says the ingredient with the higher value gets mixed in proportion to how far the mean is from the lower value. This is exactly the lever principle (moment balance). The mean divides the gap between c₁ and c₂ in inverse proportion to the quantities mixed.
Example: Mix 18% methanol solution with a b% solution in ratio 4:5 to get 16%. Then:
4/5 = (16 - b) / (18 - 16) = (16 - b) / 2
So 8 = 16 - b, giving b = 14.4%. (This is PYQ id 6a297fea9548c79b8100a95d — solved in one line with the cross.)
When ratios are given explicitly, don't use the alligation cross — just compute total quantity of the component directly.
Total fat = q₁ × c₁ + q₂ × c₂
Final % = Total fat / (q₁ + q₂) × 100
This is faster than setting up the cross when you already know the mixing quantities.
These are extremely common in SSC CHSL. The setup: a solution with x% of a substance exists in V litres; you add pure water to bring concentration to y%. How much water to add?
Let water added = w. The amount of the substance doesn't change (you're only adding water):
(x/100) × V = (y/100) × (V + w)
x × V = y × (V + w)
w = V × (x - y) / y
This is derivable in 10 seconds from first principles. You don't need to memorise it — derive it every time. The key insight: the substance quantity is fixed; only the total volume changes.
When k litres are removed from a V-litre container and replaced with water, and this is done n times:
Remaining milk / Initial milk = ((V - k) / V)ⁿ
This is the compound interest / depreciation formula in disguise. Each iteration multiplies the concentration by (V - k)/V.
Example: 70 L wine-water (5:2). Remove 14 L and replace with 14 L water. One iteration, so:
Remaining wine fraction = (70 - 14)/70 = 56/70 = 4/5
Initial wine = 50 L. Final wine = 50 × 4/5 = 40 L. Total = 70 L. Water = 30 L. Ratio = 40:30 = 4:3.
When three containers contribute to a combined mixture, compute the quantity of each component container-by-container, sum them up, then take the ratio. There is no shortcut here — just clean tabular arithmetic. Do not attempt to apply the alligation cross to three containers simultaneously; it leads to errors.
Step 1: Find actual volume in each container (capacity × fill percentage).
Step 2: Find quantity of milk in each container (volume × milk fraction from ratio).
Step 3: Sum all milk, sum all water. Take ratio.
When you know both concentrations (c₁, c₂) and the mean (m), write:
Ratio = (m − c₂) : (c₁ − m)
Micro-example: Mix 30% and 10% solutions to get 18%. Ratio = (18−10):(30−18) = 8:12 = 2:3. Done.
Standard method (set up equation, solve for ratio): ~40 seconds. Cross rule: ~8 seconds. You save 32 seconds per question.
When water is added to a solution, the substance (milk, alcohol, etc.) does not change. Anchor your equation on that constant.
If initial volume V has x% substance, and you add w litres water for y% concentration: w = V(x − y)/y
Micro-example: 70 L, 10% water initially. Add w litres to make 25% water. But here the variable part IS water — so anchor on milk (which is fixed). Milk = 90% of 70 = 63 L. After adding w: 63/(70+w) = 75/100 → 252 = 210 + 3w → w = 14.
Two-step approach via direct equation: ~50 seconds. Anchor method with milk constant: ~20 seconds.
After n replacements of k litres from V litres, the fraction of original substance remaining is ((V−k)/V)ⁿ.
Micro-example: 70 L, remove 14 L once. Remaining fraction = (56/70)¹ = 4/5. Multiply by initial quantity of substance. No simultaneous equations needed.
Expanded calculation (track removal each step): 4 sub-steps. Formula: 1 multiplication. Saves 3 sub-steps per replacement.
In profit/mixture problems where two items of different prices are mixed to a target average price, apply exactly the same cross. No separate formula.
Micro-example: Rs. 20/kg item mixed with Rs. 30/kg item to sell at Rs. 24/kg. Ratio = (24−20):(30−24) = 4:6 = 2:3 (cheaper : costlier).
Eliminates the need to remember a "price mixture" formula separately. One cross rule covers both concentration and price problems.
For three-container problems, draw a 3-row table: (Container | Volume | Milk fraction | Milk | Water). Fill row by row. Sum the last two columns. This eliminates calculation errors from tracking multiple fractions in your head simultaneously.
Micro-example structure: Container A gives 21 L milk + 14 L water; B gives 20+4; C gives 14+4. Sum: 55 milk, 22 water. Ratio 55:22 = 5:2.
Without table, error rate on this type is high (mixing up fractions). Table reduces re-checking from ~2 times to 0.
Read the question. Identify which type it is — this takes 5 seconds and determines your entire approach.
Is a ratio being asked from two known concentrations and a known mean? Use the alligation cross: ratio = (m − c₂):(c₁ − m). Done in 8 seconds.
Is a concentration/value being asked from known quantities and concentrations?
Use direct weighted average: (q₁c₁ + q₂c₂)/(q₁+q₂). Done in 15 seconds.
Is water/pure substance being added to change a concentration? Anchor on the substance that does NOT change. Write one equation. Solve for the unknown volume.
Is there a repeated removal and replacement?
Apply ((V−k)/V)ⁿ × initial quantity. If n = 1, this simplifies to one multiplication.
Are three or more containers involved? Go row-by-row. Sum all milk, sum all water. Final ratio. Do not try to cross-alligation across three sources.
If you cannot immediately classify the problem, look for: what is fixed (a substance that doesn't change) and what is variable (a volume or concentration). Build your equation around the fixed quantity.
Why this question: Direct water-addition with the "fixed substance" anchor pattern. Most common type in SSC CHSL.
Solving path: Water is the variable part here — but milk is fixed at 90% of 70 = 63 L. After adding w litres of water: milk fraction must be 75% of new volume. So 63/(70+w) = 3/4 → 252 = 210 + 3w → w = 14 L. Alternatively, anchor on water directly: initial water = 7 L, final water fraction = 25%, so (7+w)/(70+w) = 1/4 → 28 + 4w = 70 + w → 3w = 42 → w = 14 L. Both anchors give the same answer — pick whichever is faster for you.
Why this question: Three-container combined ratio. Tests whether you can stay organised with multiple fractions simultaneously. A table is mandatory here.
Solving path: Container 1: 40 × 0.875 = 35 L; milk fraction 3/5; milk = 21 L, water = 14 L. Container 2: 30 × 0.8 = 24 L; milk fraction 5/6; milk = 20 L, water = 4 L. Container 3: 20 × 0.9 = 18 L; milk fraction 7/9; milk = 14 L, water = 4 L. Total milk = 55 L, water = 22 L. Ratio = 55:22 = 5:2. Note: the official key gives 35:9 — verify with the actual exam paper if sitting CHSL 2018.
Why this question: Single-replacement (n = 1) wine-water problem. Tests the replacement formula and careful initial-concentration reading.
Solving path: Initial wine = 5/7 × 70 = 50 L, water = 20 L. After removing 14 L of the 5:2 mixture: wine removed = 14 × 5/7 = 10 L, water removed = 4 L. Remaining: wine = 40 L, water = 16 L. Add 14 L pure water: water = 30 L. New ratio = 40:30 = 4:3. Alternatively, use replacement formula: remaining wine = 50 × (56/70) = 50 × 4/5 = 40 L. Same result, one fewer sub-step.
Why this question: Classical alligation — given mixing ratio and final concentration, find one unknown concentration. One-line solution with the cross.
Solving path: Weighted average equation: (4 × 18 + 5 × b)/9 = 16 → 72 + 5b = 144 → 5b = 72 → b = 14.4%. If you prefer the alligation cross: ratio 4:5 means (16 − b)/(18 − 16) = 4/5 → (16 − b)/2 = 4/5 → 16 − b = 8/5 = 1.6 → b = 14.4%. Either route, same answer. The weighted average equation is slightly faster here because the ratio 4:5 is already given.
Why this question: Direct weighted average — no alligation cross needed because both quantities and both concentrations are explicitly given. Tests whether you recognise you can skip the cross entirely.
Solving path: Total fat = 4 × 0.25 + 6 × 0.40 = 1 + 2.4 = 3.4 L in 10 L total. Fat% = 3.4/10 × 100 = 34%. No alligation cross needed. Do not set up a cross here — it adds a step you don't need when both concentrations and quantities are already given.
Applying alligation cross when quantities are given. If the problem tells you "4 L of 25% is mixed with 6 L of 40%", just compute the weighted average directly. Setting up a cross to find the ratio (which you already know) wastes time and introduces errors.
Forgetting to anchor on the fixed substance in dilution problems. When pure water is added, the milk (or alcohol, or salt) quantity stays constant. Students often write an equation where both numerator and denominator change in a way that doesn't account for this, leading to a wrong denominator.
Misreading "proportion of water increases to 25%" as "25% water is added". The target 25% refers to the final composition, not the amount being added. This mistake gives a completely different and wrong equation.
Not converting fill percentage before applying milk ratio. In multi-container problems (like the 3-container PYQ), the container is not full — it's filled to 87.5%, 80%, etc. Students often apply the milk:water ratio to the full capacity instead of the actual volume filled. Always compute actual volume first.
Wrong direction on the alligation cross. The higher concentration goes on top-left, lower on top-right, mean in the middle. The ratio is bottom-right : bottom-left (cross-difference). Swapping this gives the inverted ratio, which reverses your answer.
Treating n-step replacement as n separate independent mixtures. In repeated replacement, each step starts from the result of the previous step. The compound formula ((V−k)/V)ⁿ handles this automatically. Calculating step-by-step without the exponent is not wrong but is significantly slower and error-prone for n ≥ 2.