Mixtures and alligation is the chapter where you answer one fundamental question: when you combine two things of different "strengths" (price, concentration, speed — anything), what does the blend look like?
Here is the plain picture. You have a ₹20/kg rice and a ₹30/kg rice. You mix some amount of each and get a blend that costs ₹24/kg. The alligation rule tells you exactly how much of each you used — without any equation.
The word alligation comes from the Latin alligare, to bind together. In Indian classroom language, you will often hear this called the cross method (आड़ा तरीका), because you literally draw a cross on paper and subtract diagonally.
The mental model: think of a seesaw. The mixture price (₹24) sits at the fulcrum. The cheaper ingredient (₹20) sits on the left, the costlier one (₹30) on the right. The heavier side needs a smaller quantity to balance — so the side farther from the mixture gets the smaller quantity. Your ratio is determined by how far each ingredient is from the mixture value.
This same logic applies to:
The power of alligation is that it converts a messy percentage/fraction problem into a single subtraction on each arm of the cross. That is why SSC MTS setters love it — it looks like a calculation-heavy problem but it is a 20-second solve if you know the cross.
You have two ingredients A and B with values a and b respectively (a < b), mixed to give a mixture of value m, where a < m < b.
Draw the cross:
a b
\ /
m
/ \
(b - m) (a - m)*
(take the absolute value on the lower right: m - a)
(b - m) : (m - a)That is the entire rule. The diagonal differences give you the ratio.
Micro-example: Rice at ₹20 and ₹30 mixed to give ₹24/kg.
30 - 24 = 6 → quantity of A24 - 20 = 4 → quantity of BCheck: (3×20 + 2×30) / 5 = (60 + 60)/5 = 24. Correct.
The mixture value m is a weighted average:
m = (a × qA + b × qB) / (qA + qB)
Rearranging:
m(qA + qB) = a·qA + b·qB
m·qA - a·qA = b·qB - m·qB
qA(m - a) = qB(b - m)
qA / qB = (b - m) / (m - a)
There it is — the cross formula is just algebra in disguise. Understanding this means you can apply it even in non-standard forms where the question does not look like a mixture problem.
Type 1 — Price Mixing (Most Common)
Two goods at different prices are mixed to get a mixture of a given price. Use the alligation cross directly on prices.
Type 2 — Solution Concentration
A milk-water solution at 60% milk is mixed with pure milk (100%) to get 75% milk. Here, treat the "concentration percentage" as the value in the cross.
(100 - 75) : (75 - 60) = 25 : 15 = 5 : 3Type 3 — Removal and Replacement (Dilution)
This is the trickiest type. You start with a container, remove some mixture, and replace with one of the components (usually water).
The formula for final concentration after one round:
Final fraction = Initial fraction × (1 - removed/total)
If you remove r litres from a T-litre mixture and replace with water:
Remaining alcohol = Initial alcohol × (T - r) / T
For multiple rounds of removal and replacement:
Final alcohol = Initial alcohol × [(T - r)/T]^n
where n is the number of rounds.
On your rough sheet, physically draw the letter X. Write the two ingredient values at the top-left and top-right corners, the mixture value in the center. Subtract diagonally (always bigger minus smaller — both differences are positive). Read the bottom corners as the ratio. This prevents sign errors that happen when you do it mentally.
Micro-example: ₹15 and ₹25 mixed to give ₹21.
Standard equation method: ~45 seconds. X-cross: ~10 seconds.
In milk-water or alcohol-water problems, you never need to convert anything. Pure milk = 100, pure water = 0. If a solution is 60% milk, its "value" is 60. Apply the cross directly on these numbers.
Micro-example: Add pure milk (100) to a 60% solution to get 75%.
Students who set up two-variable equations here take 90 seconds. Cross method: 15 seconds.
In removal-replacement problems, track only the ingredient that is NOT being replaced (usually alcohol or milk). After removing r litres from T litres and replacing with water:
Remaining pure ingredient = Previous amount × (T - r) / T
You do one multiplication per round. No need to track water separately — total volume stays T, so water = T minus alcohol.
Micro-example: 50 L mixture, 30 L alcohol. Remove 10 L, replace with water.
Setting up two equations and solving: ~60 seconds. Single multiplication: ~10 seconds.
After you get the ratio from alligation, verify in 5 seconds: multiply each value by its ratio weight, add, divide by total weight, and check it equals the mixture value. If it does not, you made a subtraction error. Catching this in the exam hall before you mark the answer saves you from a wrong attempt.
Micro-example from above: ratio 2:3 for ₹15 and ₹25. (2×15 + 3×25)/5 = (30 + 75)/5 = 105/5 = 21. Confirmed.
When you see a mixtures problem, run through this decision tree in about 5 seconds:
Step 1 — Identify the "value" being mixed.
Is it price per kg? Concentration percentage? Speed? That value goes into the cross.
Step 2 — Is there removal and replacement?
Yes → use the multiplication formula: remaining = initial × (T - r)/T. No → go to Step 3.
Step 3 — Are you mixing two containers to get a blend?
Yes → draw the X-cross. Subtract diagonally. Read ratio at the bottom.
Step 4 — Does the question give ratio and ask for price/concentration, or give price/concentration and ask for ratio?
Both directions work with the same cross — you just read it differently. If ratio is given, verify the mixture value. If mixture value is given, compute the ratio.
Step 5 — Sanity check.
The mixture value must lie strictly between the two ingredient values. If your answer puts it outside, you flipped the ratio.
Total time target: 45–60 seconds per question at SSC MTS level.
Why this question: The classic "price difference from mixture" phrasing trips up students who try to set absolute prices. Alligation does not need absolute prices — only differences.
Solving path:
Let mixture price = m. Superior variety costs m + 0.50. Inferior costs m - 0.75.
Draw the cross with superior at top-left, inferior at top-right, mixture m in center:
(m - 0.75) to m difference = 0.75(m + 0.50) to m difference = 0.50Ratio of superior : inferior = 0.75 : 0.50 = 3 : 2.
You never needed the actual value of m. The differences 0.75 and 0.50 are all the cross needs. Answer: 3:2.
Why this question: "How much pure liquid to add" is a standard SSC MTS template. Students who do not know the 100% substitution trick waste time on algebra.
Solving path: Original solution = 40 L at 60% milk. Pure milk = 100%. Target = 75%.
Cross:
(100 - 75) : (75 - 60) = 25 : 15 = 5 : 3So for every 5 L of original solution, you add 3 L of pure milk.
Original solution is 40 L → that is 40/5 = 8 units.
Pure milk to add = 8 × 3 = 24 L.
Answer: 24 L.
Why this question: Removal-and-replacement is the hardest sub-type. The direct multiplication formula cuts through in one line.
Solving path: Total = 50 L. Alcohol : Water = 3 : 2, so alcohol = 30 L, water = 20 L.
Remove 10 L of mixture (which is 3:2, so 6 L alcohol + 4 L water removed). Replace with 10 L water.
After removal and replacement:
Ratio = 24 : 26 = 12 : 13.
Alternatively with the formula: Remaining alcohol = 30 × (50 - 10)/50 = 30 × 4/5 = 24 L. Same result, one multiplication.
Putting the mixture value outside the two ingredient values. The cross only works when a < m < b. If your mixture price is higher than both ingredients, the problem is set up wrong — re-read it.
Flipping the ratio. The bottom-left number (from the left arm of the cross) is the ratio of the right ingredient, and vice versa. Students frequently assign the difference to the wrong side. The rule: each difference belongs to the opposite ingredient.
Treating "add pure liquid" as a mixture-of-two-solutions problem incorrectly. Pure milk is 100% milk, not some unknown. Always substitute 100 for pure milk and 0 for pure water — do not leave them as variables.
In removal-replacement, tracking both liquids instead of one. You only need to track the ingredient not being added (alcohol/milk). Total volume is constant, so the other component is always Total − tracked ingredient. Tracking both separately doubles your work and doubles your error risk.
Dividing by the wrong base in the concentration formula. After adding pure milk to 40 L to get 24 L more, the new total is 64 L, not 40 L. If you verify by checking (24 + 0.6×40) / 64 = (24 + 24)/64 = 48/64 = 75% — that confirms the answer. Students who forget to update the total volume get the wrong verify step.
Using alligation when the problem is "find final concentration after mixing unequal portions" without giving a target. In that case you already have the ratio and need to compute the weighted average directly — you do not need the cross at all.