Alligation is the method that tells you in what ratio to mix two things so the final blend hits a target price, concentration, or quality. It is the inverse of the weighted average — instead of finding the average from known ratios, you find the ratio from the known average.
Think of it this way: you run a tea stall. You have cheap tea at ₹180/kg and premium tea at ₹280/kg. A customer wants a blend that costs ₹267/kg (roughly). How many kilos of each do you take? You cannot guess — alligation gives you the exact answer in under 20 seconds.
The same logic applies to mixing liquids (milk and water, glycerin and water, alcohol solutions), blending grains at different prices, or even combining groups with different averages. The "ingredient" being mixed does not matter — the mathematics is identical.
Here is the underlying intuition. The cheaper ingredient pulls the average down; the costlier one pulls it up. The ratio in which they are mixed is inversely proportional to how far each is from the target average. Close to the target means more of it; far from the target means less of it. This is the cross-difference principle.
A good analogy: imagine a see-saw. The target average is the fulcrum. The cheap price sits on the left, the expensive price sits on the right. To balance the see-saw, the left side takes a weight equal to the right side's distance from the fulcrum, and vice versa. That is alligation in one image.
There is also a second class of problems — replacement problems — where you repeatedly draw some mixture out and refill with one ingredient. These use a slightly different formula but the same underlying logic. Both types appear in UP Police Constable papers, so you need to handle both fluently.
You have two ingredients with values c (cheap/lower) and d (dear/higher), and you want a mixture with mean value m, where c < m < d.
Draw the cross:
c d
\ /
m
/ \
(d−m) (m−c)
(d − m) : (m − c)That is it. The ratio of the cheaper to the dearer is (higher value − mean) : (mean − lower value).
Example. Rice at ₹11/kg and ₹21/kg, target ₹17/kg.
This took four arithmetic operations. No equations needed.
Look — sometimes the mixture problem gives you a selling price plus a profit percentage, and you must first find the cost price of the mixture before applying alligation.
Cost price of mixture = Selling price ÷ (1 + profit%)
In the tea problem: SP = ₹320 at 20% profit → CP = 320/1.2 = ₹266.67. Now apply alligation between ₹180 and ₹280 with mean ₹266.67.
When the question gives you an existing ratio and says "we add X litres of one ingredient, ratio changes to Y", set up one variable for the multiplier.
Structure: Let initial quantities be a·k and b·k for a ratio a:b. After adding, equate the new ratio. You get one linear equation in k. Solve.
Example. Milk:Water = 3:4. Add 10L milk and 10L water. New ratio = 4:5.
When you draw out a fraction f of a mixture and replace with pure water (or any one ingredient), the concentration of the original liquid after n replacements is:
Where:
C_0 = initial concentration of the ingredient you are trackingf = fraction of mixture removed each timen = number of replacementsFor a single replacement (n = 1):
Example. Glycerin:Water = 3:1. Initial glycerin concentration = 3/4. We want it to become 1/2. Remove fraction x and replace with water.
So you draw off 1/3 of the mixture.
This is a third pattern: the existing mixture has ratio a:b, you add only one ingredient (say water), and the ratio changes to a:c (note: the amount of milk/glycerin does not change).
Since the first ingredient is unchanged, just track it.
This is usually easier than setting up alligation here.
Draw a literal X on your rough sheet. Write the cheaper value top-left, dearer value top-right, mean in the centre. Subtract diagonally downward — top-right minus centre goes bottom-left (ratio of cheaper), centre minus top-left goes bottom-right (ratio of dearer). You will never mix up which difference goes with which ingredient. Standard method (algebraic setup): ~50s. Cross method: ~15s.
Whenever you see "sold at ₹X with Y% profit", your first pen-stroke must be CP = SP ÷ (1 + Y/100). Do not try to alligation the selling price directly — it will give a wrong ratio. Eliminate that mistake by making CP calculation a reflex. Once you have CP, the cross takes 10 seconds. Total: ~25s vs ~60s for someone who re-reads the question twice after getting confused.
When a fraction f is drawn off once and replaced by pure water, the remaining concentration of original liquid = initial concentration × (1 − f). Substitute directly: no simultaneous equations needed. For the glycerin problem — initial 3/4, target 1/2 — write (3/4)(1−f) = 1/2 and solve in 2 steps. Standard equation-based approach: 4–5 steps. Substitution: 2 steps.
When both ingredients are added equally (10L milk AND 10L water), set the initial parts as a·k and b·k. The equal addition cancels out elegantly — you always end up with one linear equation in k. Cross-multiply the new ratio, expand, and subtract. This takes exactly 3 arithmetic lines. Recognise the pattern and you will never need more than 30 seconds on these questions.
If only water is added (milk unchanged), fix the milk quantity as a constant. Initial milk = (ratio part of milk / total ratio parts) × total volume = constant. Write the same milk as a fraction of the new total using the new ratio. Equate. This avoids alligation entirely and gives you the answer in one equation. Saves you from drawing a cross unnecessarily — about 20s faster than applying alligation incorrectly to a one-ingredient-addition problem.
Read the question stem and classify in 10 seconds:
Is a fraction of mixture removed and replaced? → Use the replacement formula: C_new = C_old × (1 − f). Set equal to target concentration. Solve for f.
Are two separate ingredients mixed to hit a target price/concentration? → Draw the X cross. Compute (higher − mean) and (mean − lower). Write ratio. Done.
Is SP given with profit%? → Convert to CP first, then apply the cross.
Is a fixed amount added to an existing mixture, ratio changes? → Set initial parts as a·k and b·k. Write the new ratio as an equation. Solve for k. Find what the question asks.
Is only one ingredient added, and the other is unchanged? → Fix the unchanged ingredient's quantity. Equate using new ratio. Solve directly.
Never reach for simultaneous equations until you have confirmed none of the above patterns apply. In most UP Police Constable Quant questions, one of these five patterns covers it perfectly.
Why this question: Tests whether you can identify the alligation cross and use it directly, without setting up lengthy algebraic equations.
Solving path: Target mean = ₹17. Cross: (21 − 17) : (17 − 11) = 4 : 6 = 2 : 3. Cheaper (₹11) : Dearer (₹21) = 2 : 3. Answer: (B).
Why this question: Tests the replacement formula — the most common "fraction" type in mixture problems.
Solving path: Initial glycerin fraction = 3/4. Target = 1/2. Apply: (3/4)(1 − x) = 1/2. So 1 − x = (1/2) ÷ (3/4) = 2/3. Therefore x = 1/3. Answer: (B).
Why this question: Classic "add both ingredients equally" pattern. Tests the ratio-multiplier technique.
Solving path: Let milk = 3k, water = 4k. After adding 10L each: (3k + 10)/(4k + 10) = 4/5. Cross-multiply: 15k + 50 = 16k + 40. So k = 10. Total initial = 7k = 70. Answer: (B).
Why this question: Tests the "add only water" pattern — milk quantity stays fixed throughout.
Solving path: Let total mixture = 10x. Milk = 3x, water = 7x. Add 10L water: 3x / (7x + 10) = 1/3. So 9x = 7x + 10 → x = 5. Milk = 3 × 5 = 15 litres. Answer: (A).
Why this question: Tests alligation with a profit-percentage conversion step before applying the cross.
Solving path: CP of mixture = 320 ÷ 1.2 = ₹266.67. Cross: (280 − 266.67) : (266.67 − 180) = 13.33 : 86.67 = 1 : 6.5 = 2 : 13. Answer: (A).
Why this question: Tests adding only one ingredient (water) to change ratio from 12:13 to 12:15. Milk is unchanged.
Solving path: Milk in 2025L = (12/25) × 2025 = 972L. In new ratio 12:15, water needed = (15/12) × 972 = 1215L. Water already present = 2025 − 972 = 1053L. Water to add = 1215 − 1053 = 162 litres. Answer: (D).
Swapping the cross differences. The ratio of the cheaper ingredient = (higher value − mean), NOT (mean − lower value). Many students write it backwards and get an inverted ratio. Draw the X cross every single time until it is muscle memory.
Applying alligation to the selling price instead of the cost price. When a profit percentage is mentioned, you must convert SP to CP before using the cross. Using SP directly gives a wrong mean and therefore a wrong ratio.
Mixing up "which ingredient is unchanged" in replacement problems. In a replacement problem, the ingredient you are adding (say, water) changes in quantity. The one you are NOT adding is the one that decreases. Track the ingredient being drawn out, not the one being poured in.
Using the replacement formula when the question involves adding a fixed volume (not a fraction). The formula C × (1 − f) is for repeated fractional removal. If the question says "8 litres is drawn off from a 22-litre container", compute the fraction first: f = 8/22, then apply if needed. Do not directly plug in 8.
Forgetting to reduce the ratio. After computing the cross differences, always check for common factors. 4:6 is 2:3. Leaving it as 4:6 is not wrong mathematically, but it may not match any option, causing you to second-guess a correct answer.
Setting up ratio equations with the wrong total. When the question says "add 10L milk and 10L water", the new total is (old total + 20), not (old total + 10). A surprisingly common error under time pressure.