Mixture and Alligation

We begin with the easier type. A mixture is given in ratio or percentage form along with one actual quantity, and you have to find another quantity. Nothing is being mixed with anything yet; you only have to read the composition correctly.

From a ratio. If milk : water = a : b and the total is M, then milk = aa + b × M and water = ba + b × M. If only one component is known, scale it: water = milk × ba.

From a percentage. If a 50-litre mixture has 30% oil, oil = 15 litres and the other liquid = 35 litres. If you are told "the oil is 15 litres and it is 30% of the mixture", the whole mixture is 150.30 = 50 litres.

A known part and the rest. Suppose 4 litres of syrup are mixed with x litres of water and the syrup is 16% of the mixture. The syrup is the only thing you know, so first find the whole mixture (4 ÷ 0.16 = 25 litres). Then water = 25 − 4 = 21 litres. This last subtraction is exactly where students lose the mark.

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Same solution, different vessels. If two vessels hold the same solution, the ratio of the ingredients is the same in both. Find the ratio from the vessel where both amounts are given and use it in the other.

Taking out a part. When some mixture is removed, the ratio of what remains is unchanged; only the total changes.

Take Radhika's 48-litre fruit punch, with juice and soda in the ratio 5 : 3. Juice = 58 × 48 = 30 litres and soda = 18 litres. She pours out 16 litres for her friends, so 32 litres remain, still in the ratio 5 : 3, and the juice left is 58 × 32 = 20 litres.

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Here the ingredients and their quantities are known and you have to find the value of the mixture: its price per kg, its percentage of fat or acid, its temperature, or its cost per litre.

The method never changes. Divide the total value by the total quantity:

Mean = q₁v₁ + q₂v₂ + q₃v₃ + …q₁ + q₂ + q₃ + …

Prices. 4 kg at ₹55 and 6 kg at ₹70 give mean = 4 × 55 + 6 × 7010 = 220 + 42010 = ₹64 per kg.

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Percentages. Mix 8 litres of 15% acid with 12 litres of 35% acid. The acid in the mixture = 1.2 + 4.2 = 5.4 litres out of 20, which is 27%. Directly, 8 × 15 + 12 × 3520 = 120 + 42020 = 27%. Percentages are averaged exactly like prices, with the quantities as weights.

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Ratios instead of quantities. If the varieties are mixed in the ratio 2 : 3, use 2 and 3 as the weights. The actual total (60 litres, 5 kg, anything) makes no difference to the mean price.

Price given for a bundle. "₹250 for 5 kg" means ₹50 per kg, and "3 candies for ₹15" means ₹5 per candy. Bring every price to one unit before you average.

Temperature follows the same rule. Radhika pours 1 litre of tea at 80 °C into 3 litres of water at 40 °C. The temperature after mixing = 1 × 80 + 3 × 404 = 2004 = 50 °C, a weighted average once again.

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Everything else in this chapter grows out of this type. Two things with values c (cheaper) and d (dearer) are to be mixed so that the mixture has the value m, and the question asks for the ratio.

Cheaper : Dearer = (d − m) : (m − c)

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Draw the cross exactly as in the figure. Put c and d on top and m in the middle, subtract diagonally, and read the ratio at the bottom. The ₹32 rice and the ₹47 rice give ₹38 rice when they are mixed in the ratio (47 − 38) : (38 − 32) = 9 : 6 = 3 : 2.

Percentages work the same way. To get a 20% sugar syrup from a 12% syrup and a 32% syrup, mix them in (32 − 20) : (20 − 12) = 12 : 8 = 3 : 2.

Water is free and 0%. To bring milk worth ₹45 per litre down to ₹40 per litre by adding water, take water at ₹0: water : milk = (45 − 40) : (40 − 0) = 5 : 40 = 1 : 8. In the same way, pure acid is 100% and pure water is 0%.

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Read the order of the ratio from the question. The cross gives cheaper : dearer. If the question asks for "first : second" and the first one is the dearer, turn the ratio round. Most wrong answers in this type are the right ratio written in reverse, and that reversed ratio is usually waiting among the options.

From ratio to quantity. Once the ratio is known, any one quantity gives the rest. If the ₹32 and ₹47 rice go in as 3 : 2 and you have 18 kg of the ₹32 rice, you need 12 kg of the ₹47 rice.

Radhika has juice concentrates of 15% and 40% and needs a 25% drink. 15% : 40% = (40 − 25) : (25 − 15) = 15 : 10 = 3 : 2. If she uses 600 mL of the 15% one, she needs 400 mL of the 40% one.

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Now the mean is known and the ratio is known, but one of the ingredients has an unknown price or strength. You are running the weighted-average formula backwards.

Basic method: total value. Total value of the mixture = (total quantity) × (mean). Subtract the values you know, and what is left belongs to the unknown ingredient.

Say 3 kg of tea at ₹140 is mixed with 2 kg of another tea and the mixture is worth ₹152 per kg. Total value = 5 × 152 = ₹760 and the known part = 3 × 140 = ₹420. So the 2 kg of the second tea are worth ₹340, which is ₹170 per kg.

Alligation method (two ingredients). Read the cross backwards. The ratio 3 : 2 must equal (d − 152) : (152 − 140) = (d − 152) : 12. So 2 parts = 12, 1 part = 6, and d − 152 = 3 × 6 = 18, giving d = ₹170. The answer is the same and the numbers are smaller.

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Three ingredients in the ratio 1 : 1 : 2. Rice at ₹a and ₹b mixed with a third variety in the ratio 1 : 1 : 2 gives a mean m. Then
a + b + 2x = 4m ⇒ x = 4m − a − b2
There is no need to learn this by heart. Write "total value = total quantity × mean" and solve.

Balance method (gain = loss). Each ingredient is either above the mean or below it. The total shortfall of the cheaper ones below the mean equals the total excess of the dearer ones above it. With 1 : 1 : 2, mean ₹100 and known prices ₹82 and ₹94, the shortfall is 18 + 6 = 24. The 2 parts of the third variety must together be 24 above the mean, that is 12 each, so the third variety costs ₹112.

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Radhika mixes equal amounts of a 24% salt solution and an unknown solution and gets 30%. With equal amounts the mean sits exactly in the middle, so the unknown solution is 30 + 6 = 36%.

A trader mixes things and sells the mixture. Profit and Loss comes in, but its basic rule does not change: profit or loss is always on the cost price. Here the cost price is the mean cost of the mixture.

Step 1: find the cost price of the mixture, the weighted average of the costs per unit.
Step 2: compare it with the selling price and use Profit% = SP − CPCP × 100, or SP = CP × 100 + profit%100.

Rice at ₹44 and ₹59 per kg mixed 2 : 1 costs 88 + 593 = ₹49 per kg. Sold at ₹56, the profit is 749 = 17 = 1427%.

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When the cost is hidden in a selling price. Suppose tea is "sold at ₹150 at a loss of 25%". Recover the cost first: CP = 1500.75 = ₹200. Whenever a question tells you the ingredients were sold at a profit or a loss, do this step before you average.

Prices for a group of items. "4 for ₹10" means ₹2.50 each. Convert every buying and selling price to the price of one item first.

The dishonest milkman. A milkman who adds water and sells the whole mixture at the cost price of pure milk earns money on the water. If he adds w litres of water to M litres of milk, he is paid for M + w litres but has spent money only on M:

So 1 litre of water for every 6 litres of milk gives a profit of 16 = 1623%. It works in reverse too: a profit of 25% = 14 means water : milk = 1 : 4.

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Radhika's uncle mixes 1 litre of water with every 9 litres of milk and sells at the cost price of milk. His profit is 19 = 1119%. Water is 1 litre out of 10, that is 10% of the mixture. The profit is counted on the milk he paid for, not on the mixture.

Adulterated milk sold at a different price. The rule above needs the mixture to be sold at the cost price of pure milk. If the selling price per litre is something else, go back to the basic step and find what one litre of the mixture really costs the seller. If milk costs ₹c per litre and a fraction f of the mixture is milk, then cost of 1 litre of mixture = f × c, because the water costs nothing. Compare this with the selling price per litre.

Radhika's uncle buys milk at ₹60 per litre and mixes 1 litre of water with every 4 litres of milk. A litre of this mixture costs him 45 × 60 = ₹48. Sold at ₹54 per litre, his profit is 648 = 18 = 12.5%. The idea also runs backwards: if he sells the mixture at ₹48 per litre and has neither profit nor loss, the milk in it must be 4860 = 45, so one-fifth of the mixture is water.

This is the harder kind of mixture question. Two varieties are mixed, the mixture is sold at some price with some profit, and you must find the ratio or a quantity.

Everything rests on one rule: alligation works only on cost prices. The selling price of the mixture is not its mean cost, so the work goes in three steps.

Step 1: convert the selling price of the mixture into its cost price.
CP = SP × 100100 + profit%
Step 2: apply the alligation cross with the two costs and this mean cost.
Step 3: use the ratio to get the quantity, if a quantity is asked.

Rice costs ₹36 and ₹54 per kg, and the mixture is sold at ₹55 with a 25% profit. Mean cost = 55 × 100125 = ₹44. The cross gives ₹36 : ₹54 = (54 − 44) : (44 − 36) = 10 : 8 = 5 : 4. If 20 kg of the ₹36 rice is used, 16 kg of the ₹54 rice is needed.

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Look at the selling price ₹55. It is more than both costs. Put ₹55 in the cross and it does not even lie between 36 and 54. That is your warning to remove the profit first.

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Fractions save time. A profit of 10% means CP = SP × 1011; 20% means × 56; 25% means × 45; 30% means × 1013; 40% means × 57; 15% means × 2023.

Radhika's shop mixes lentils at ₹48 and ₹63 and sells the mixture at ₹66 with a 20% profit. Mean cost = 66 × 56 = ₹55, so the ratio = (63 − 55) : (55 − 48) = 8 : 7.

A solution of known strength is given. Pure water, pure acid, pure milk or some other single component is added to it, or some water evaporates. You have to find how much was added, or the new strength.

Everything here is Habit 1 from the foundation: find the quantity that does not change.

Water added to an acid solution: the acid stays the same.
80 litres of 25% acid has 20 litres of acid. To make it 16%, these 20 litres of acid must now be 16% of the new total. New total = 200.16 = 125 litres, so water added = 125 − 80 = 45 litres.

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Pure acid added: now the water stays the same.
80 litres of 25% acid has 60 litres of water. To make it 40% acid, the water must be 60% of the new total. New total = 600.60 = 100 litres, so acid added = 20 litres.

Evaporation: water goes out, the salt (or sugar, or pulp) stays.
50 litres of 12% salt water has 6 litres of salt. For a strength of 20%, the new total is 60.20 = 30 litres, so 20 litres of water must evaporate.

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Fresh and dry fruit follow the same rule. Drying removes only water, so the pulp (the part that is not water) stays the same.

The alligation view. Adding water is mixing with a 0% solution, and adding pure acid is mixing with a 100% solution. In the first case, 25% solution : water = (16 − 0) : (25 − 16) = 16 : 9. Since 80 litres ↔ 16 parts, water = 9 parts = 45 litres.

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Radhika boils 20 litres of sugarcane juice with 9% sugar until the sugar is 15%. The sugar stays at 1.8 litres, so the juice left = 1.80.15 = 12 litres. She boiled away 8 litres of water, and the sugar never moved.

The idea is the same as in the last type, but the strength is written as a ratio. A typical question reads: "Milk : water = 7 : 3. How much water should be added to make it 7 : 4?"

Method: make the unchanged part equal.
Water is added, so the milk does not change. Write both ratios with the same number for milk.
Old 7 : 3, new 7 : 4. Milk is already 7 in both, so water went from 3 parts to 4 parts, which means 1 part of water is added.
Now bring in the litres. If the mixture was 50 litres, 10 parts = 50 and 1 part = 5 litres, so 5 litres of water are added.

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If the unchanged part is not equal in the two ratios, make it equal with the LCM.
Juice : water = 5 : 3 becomes 2 : 1 after juice is added. Water is unchanged, so make it 3 in both: 5 : 3 and 6 : 3. Juice went from 5 parts to 6 parts, and 1 part was added.

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When the added quantity is given and the original is asked, the added parts must match the added litres. Take acid : water = 4 : 3; after 9 litres of water are added, the ratio becomes 4 : 6. Water rose by 3 parts = 9 litres, so 1 part = 3 litres and the original 7 parts = 21 litres.

Percentage targets are ratios too. "50% water" means water : milk = 1 : 1, and "20% water" means 1 : 4.

Equation method (for checking). For alcohol : water = 7 : 3 in 50 litres, alcohol = 35 and water = 15. Adding x litres of water to get 7 : 4 means 3515 + x = 74 ⇒ 15 + x = 20 ⇒ x = 5.

Radhika's sherbet has syrup and water in the ratio 3 : 8. She adds 6 litres of syrup and the ratio becomes 5 : 8. Water stays at 8 parts, so the 2 extra parts of syrup are the 6 litres she added. 1 part = 3 litres, and she had 11 parts = 33 litres of sherbet at first.

Now each vessel already holds a mixture (milk and water, acid and water, juice and water), and the vessels are poured together.

Case 1: the quantities in the vessels are known and the new ratio is asked.
Find the milk and the water in each vessel separately and add them. 15 litres of a 2 : 1 milk–water mixture has 10 L milk and 5 L water; 12 litres of a 3 : 1 mixture has 9 L milk and 3 L water. Together that is 19 L milk and 8 L water, a ratio of 19 : 8.

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Equal capacities. If the vessels are equal but no size is given, take each vessel as the LCM of the ratio totals. For 2 : 1 and 3 : 1, take each as 12 litres: milk (8 + 9) and water (4 + 3) give 17 : 7. You can also add the fractions: milk = 23 + 34 = 1712, water = 13 + 14 = 712.

Case 2: the ratio in which two mixtures must be mixed. This is alligation again, but the value in the cross is the fraction of one component (say milk) in each mixture. Never put the ratios themselves (like 2 : 1) in the cross; change them into fractions of the whole first.

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In the figure the milk fractions are 23 and 12, and we need 35. Over the denominator 30 they become 20, 15 and 18. The cross gives (18 − 15) : (20 − 18) = 3 : 2, so the 2 : 1 mixture and the 1 : 1 mixture are taken in the ratio 3 : 2.

Case 3: a quantity from a third vessel. When some litres from two vessels are fixed and the third vessel has to bring the mixture to a target, first find the total milk and water already there. Then see how much extra of one component is needed and how much of it each litre of the third vessel supplies.

Radhika pours 6 litres of a 1 : 2 juice–water drink and 8 litres of a 5 : 3 drink into one jug. Juice = 2 + 5 = 7 litres and water = 4 + 3 = 7 litres, so the jug has juice and water in the ratio 1 : 1.

Pouring back and forth. In some questions a part of vessel A is poured into vessel B, and then a part of the new mixture in B is poured back into A. There is no formula to learn here. Keep a small table of milk and water for each vessel and update it after every pouring. One fact does all the work: whatever you pour out of a vessel has the same ratio as the vessel has at that moment.

Radhika has 30 litres of milk in jug A and 40 litres of water in jug B. She pours 10 litres from A into B, so B now holds 10 L milk and 40 L water: 50 litres in the ratio 1 : 4. She pours 10 litres of this back into A. Those 10 litres carry 15 × 10 = 2 L milk and 8 L water. So A ends with 20 + 2 = 22 L milk and 8 L water (ratio 11 : 4), and B keeps 8 L milk and 32 L water (ratio 1 : 4).

Look at the figure. When A starts with pure milk, B with pure water, and both end with their original volumes, the milk that has gone into B is equal to the water that has come into A. Here both are 8 litres. This single check catches most calculation slips.

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These questions chain two or three operations. Some mixture is taken out or sold, then one or two components are added, sometimes in a tank holding three liquids. No new formula is needed, only careful bookkeeping. Keep a small table with one column for each component and update it after every step.

Rule 1: when mixture is taken out (or sold), each component goes out in the original ratio.
Rule 2: when a pure component is added, only that column changes.
Rule 3: when equal amounts of two components are added, both columns grow by the same number, so the difference between them stays the same.

Here is rule 3 at work. A drink has 30 L syrup and 18 L water. Equal amounts of syrup and water are added and the ratio becomes 3 : 2. The difference 30 − 18 = 12 does not change, and in 3 : 2 the difference is 1 part, so 1 part = 12. Syrup becomes 36, which means 6 litres of each were added.

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Unknown initial quantity. Take the initial amounts as multiples of k (like 7k, 8k, 5k), write each component in terms of k after every step, and use the one condition the question gives ("B is 25 litres more than A", "the new ratio is 13 : 15").

Tanks from which mixture is taken out. If the mixture is a : b : c and some of it is taken out, the remaining mixture R is still a : b : c. So write everything in terms of R instead of the original total, and at the end add back what was taken out.

Radhika has 40 L of a drink with juice and water in the ratio 5 : 3. She sells 8 L, so 32 L remain with juice 20 and water 12. She then adds 4 L of juice, and the ratio becomes 24 : 12 = 2 : 1.

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Here a part of the mixture is taken out and replaced by the same amount of something else: water, a pure liquid, or another mixture. The total volume comes back to what it was.

Do it in two steps.
Step 1: taking out. If V litres are drawn from M litres, a fraction VM of the mixture goes out, and every component loses the same fraction VM. What remains of each component is (1 − VM) of it.
Step 2: putting in. Add the replacing liquid to its own column.

Take 64 L of milk and water in the ratio 5 : 3 (milk 40, water 24), and replace 16 L by water. 1664 = 14 goes out and 34 remains: milk 30, water 18. Adding 16 L of water gives milk 30 and water 34, so the ratio is 15 : 17.

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Replaced by water. In the end only the component that is not water goes down, to (1 − VM) of what it was. So the strength of milk (or acid) gets multiplied by (1 − VM). If 15 of a 75% acid is replaced by water, the strength becomes 75 × 45 = 60%.

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Replaced by the pure liquid. Now the other component (water) is multiplied by (1 − VM), and the pure liquid fills the rest.

Replaced by another mixture. After step 1, add the amount of each component that the new mixture brings in.

Two different replacements one after another. Repeat the two steps and keep the table up to date.

Radhika has 30 L of lemonade with 20% sugar syrup. She takes out 6 L and pours in 6 L of plain water. 630 = 15 goes out, so the syrup is 45 of 6 L = 4.8 L, and the strength is 20 × 45 = 16%.

Now the replacement has been done and the final ratio is given. The question asks how much was replaced, or how much was there at the start.

Method 1: the untouched component. Choose the component that is not being poured in. It only gets multiplied by (1 − f), where f is the fraction replaced, so compare its old and new share of the total.

Milk : water = 5 : 3, so milk = 58 of the vessel. Some mixture is replaced by water and the ratio becomes 1 : 1, so milk = 12. Milk only shrinks: 58(1 − f) = 12 ⇒ 1 − f = 45 ⇒ f = 15. So 20% of the mixture was replaced.

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Method 2: alligation, because replacement is mixing. Taking out a fraction f and pouring in a new liquid is the same as mixing (1 − f) of the old mixture with f of the new liquid. Alligate on the strength of the component: old mixture : new liquid = (target − new) : (old − target), with the differences taken positive. Here milk is 58 in the old mixture and 0 in water, with target 12. Old : water = 12 : 18 = 4 : 1, so 15 was replaced.

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Method 3: unknown starting total. If the ratio is a : b and V litres are removed and refilled with B, take the total as (a + b)k. After removal, A = ak(1 − V(a + b)k) and B gains V. It is usually quicker to note that A loses aa + b × V litres and nothing else happens to A.

Radhika's jug has 48 litres of juice and water in the ratio 5 : 1. She replaces some of it with water and the ratio becomes 5 : 3. The juice share goes from 56 to 58, so 1 − f = 58 ÷ 56 = 34 and f = 14. She replaced 12 litres.

A vessel holds x litres of pure milk (or wine, spirit, acid). Each time, y litres are drawn out and the vessel is filled up with water. This is done n times. How much milk is left?

Look at one operation. Drawing y litres out of x takes away the fraction yx of whatever milk is there, so the milk left is (1 − yx) times what it was, and the water brings no milk back. The next operation does the same to the new amount. After every operation the milk is multiplied by the same factor (1 − yx), just as money grows under compound interest, only here it shrinks.

Milk left after n operations = x(1 − yx)ⁿ

Since the total stays x, the concentration of milk after n operations is (1 − yx)ⁿ, and water = total − milk.

Take 50 litres of pure milk with 10 litres replaced by water three times. Factor = 1 − 1050 = 45. Milk left = 50 × (45)³ = 50 × 64125 = 25.6 litres, and water = 24.4 litres.

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If the vessel starts with a mixture, apply the factor to the component that is not being added. A 70% acid solution with 110 replaced by water twice ends at 70 × (0.9)² = 56.7% acid.

Reverse questions. If the final ratio of milk to water is given, find the milk fraction and take the n-th root. Milk : water = 27 : 37 after 3 operations means milktotal = 2764 = (34)³. So 1 − yx = 34, and the quantity drawn each time is 14 of the vessel.

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Different amounts each time. If y₁, y₂, y₃ litres are drawn in the three operations, the factors multiply: milk = x(1 − y₁x)(1 − y₂x)(1 − y₃x).

Radhika has a 20-litre can of pure juice. She takes out 4 litres and fills the can with water, and does this twice. Factor = 45, so the juice left = 20 × 1625 = 12.8 litres.

An alloy is nothing but a mixture of metals, so every method of this chapter works here. Only the words change: "melted together" means mixed, and "weight" means quantity.

1. A metal in a given weight of alloy. 63 kg of an alloy with copper : zinc = 4 : 3 has 47 × 63 = 36 kg of copper.

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2. Two alloys melted together: the new ratio. Find each metal in each alloy and add, exactly as with milk and water in vessels. With equal weights, take each alloy as the LCM of the ratio totals, or add the fractions.

3. Mixing ratio for a target composition. Alligate on the fraction of one metal. With gold at 34 and 25 and a target of 12, the values over 20 are 15, 8 and 10, so first : second = (10 − 8) : (15 − 10) = 2 : 5.

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4. Adding a pure metal. The other metals do not change, which is Habit 1 once more. 60 kg of brass with 20% zinc has 48 kg of copper. To make the zinc 25%, copper must be 75% of the new weight: 480.75 = 64 kg, so 4 kg of zinc is added.

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5. Alloys with three or more metals. Some metals appear in only one alloy. When equal weights of two alloys are melted together, a metal present only in the second alloy gets half of its share there. If lead is 25 of the second alloy and it is melted with an equal weight of a lead-free alloy, lead becomes 15 of the new alloy.

6. Weight versus value. If the question gives the prices of the metals, the cost of each metal = weight × price. Ratio of costs = ratio of weights × ratio of prices.

Radhika's silver chain weighs 30 g with silver : copper = 9 : 1, so silver = 27 g and copper = 3 g. She melts it with another 30 g chain in the ratio 4 : 1. Silver = 27 + 24 = 51 g and copper = 3 + 6 = 9 g, so the new ratio is 51 : 9 = 17 : 3.

Alligation is not really about milk and rice. It works whenever a combined value is the weighted average of two group values. The weights are whatever the values are measured "per".

Averages. In a class the boys average 62 marks, the girls average 74 and the whole class averages 70. Boys : girls = (74 − 70) : (70 − 62) = 4 : 8 = 1 : 2. If there are 36 students, 12 are boys.

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Prices of items. Pens at ₹12 and ₹20 average ₹15 over 40 pens. Cheaper : dearer = 5 : 3, so there are 25 pens at ₹12 and 15 at ₹20.

Speeds. Average speed is per unit of time, so alligation gives the ratio of the times, not of the distances. If 40 km/h and 60 km/h average out to 48 km/h, time at 40 : time at 60 = 12 : 8 = 3 : 2.

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Percentage rates on different bases. Simple interest at 6% and 10% giving 7.5% overall means the principals are in the ratio 2.5 : 1.5 = 5 : 3. Profit percentages on two articles giving one overall profit put the cost prices in the alligation ratio. Growth rates of men and women giving the growth of the whole population put the initial numbers of men and women in that ratio.

Temperature. When the same liquid at different temperatures is mixed and no heat is lost, the final temperature is the weighted average with the quantities as weights.

Radhika averages 72 in 6 tests and 84 in the next few tests, and her overall average becomes 76. First tests : next tests = (84 − 76) : (76 − 72) = 8 : 4 = 2 : 1. There were 6 first tests, so there were 3 more. When people or objects are being counted, make sure the ratio gives whole numbers.