Important Formulas – Mixtures and Alligations

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  1. Alligation

    It is the rule which enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a specified price.

  2. Mean Price

    Mean price is the cost price of a unit quantity of the mixture

 

  1. Suppose a container contains x of liquid from which y units are taken out and replaced by water.After n operations, the quantity of pure liquid = [x(1−y/x)n]

 

  1. Rule of Alligation

    If two ingredients are mixed, then

    (Quantity of cheaper/Quantity of dearer)=(P. of dearer – Mean Price)/(Mean price – C.P. of cheaper)

Cost Price(CP) of a unit quantity
of cheaper (c)
Cost Price(CP) of a unit quantity
of dearer (d)
  
Mean Price
(m)
(d – m)(m – c)
  1. => (Cheaper quantity) : (Dearer quantity) = (d – m) : (m – c)

 

Solved Examples

Level 1

1.A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
A. 26 litresB. 29.16 litres
C. 28 litresD. 28.2 litres

 

 

Answer : Option B

Explanation :

Assume that a container contains x of liquid from which y units are taken out and replaced

by water. After n operations, the quantity of pure liquid

=[x(1−y/x)n]

Hence milk now contained by the container = 40(1−4/40)3=40(1−1/10)3

40×9/10×9/10×9/10=(4×9×9×9)/100=29.16

2.A jar full of whiskey contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohols and now the Percentage of alcohol was found to be 26%. The quantity of whisky replaced is
A. 43B. 34
C. 32D. 23

 

 

Answer : Option D

Explanation :

Concentration of alcohol in 1st Jar = 40%

Concentration of alcohol in 2nd Jar = 19%

After the mixing, Concentration of alcohol in the mixture = 26%

By the rule of alligation,

Concentration of alcohol in 1st JarConcentration of alcohol in 2nd Jar
40%19%
Mean
26%
714

Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2

 

3.In what ratio should rice at Rs.9.30 per Kg be mixed with rice at Rs. 10.80 per Kg so that the mixture be worth Rs.10 per Kg ?
A. 7 : 8B. 8 : 7
C. 6 : 7D. 7 ; 6

 

 

Answer : Option B

 

Explanation :

By the rule of alligation, we have

 

Cost of 1 kg rice of 1st kindCost of 1 kg rice of 2nd kind
9.310.80
Mean Price
10
10.8-10 = .810 – 9.3 = .7

Required ratio = .8 : .7 = 8 : 7.

 

 

 

 4.In what ratio must water be mixed with milk costing Rs.12 per litre in order to get a mixture worth of Rs.8 per litre?
A. 1 : 3B. 2 : 2
C. 1 : 2D. 3 : 1

 

 

Answer : Option C

 

Explanation :

By the rule of alligation, we have

Cost Price of 1 litre of waterCost Price of 1 litre of milk
012
Mean Price
8
12-8=48-0=8

Required Ratio = 4 : 8 = 1 : 2

 

 

 

5.In 1 kg mixture of iron and manganese 20% of manganese. How much iron should be added so that the proportion of manganese becomes 10%
A. 1.5 KgB. 2 Kg
C. .5 KgD. 1 Kg

 

 

 

 

Answer : Option D

 

Explanation :

By the rule of alligation, we have

Percentage concentration of
manganese in the mixture : 20
Percentage concentration of
manganese in pure iron : 0
Percentage concentration of manganese in the final mixture
10
10 – 0 = 1020 – 10 = 10

=> Quantity of the mixture : Quantity of iron = 10 : 10 = 1 : 1

Given that Quantity of the mixture = 1 Kg

Hence Quantity of iron to be added = 1 Kg

 

6.A trader has 1600Kg of sugar. He sells a part at 8% profit and the rest at 12% profit. If he gains 11% on the whole , find the quantity sold at 12%.
A. 1200 KgB. 1400 Kg
C. 1600 KgD. 800 Kg

 

 

 

Answer : Option A

 

Explanation :

By the rule of alligation, we have

 

% Profit by selling part1% Profit by selling part2
812
Net % Profit
11
12 – 11 = 111 – 8 = 3

=>Quantity of part1 : Quantity of part2 = 1 : 3

Given that total quantity = 1600 Kg

Hence quantity of part2 (quantity sold at 12%) = 1600×3/4=1200

 

 

7.How many litres of water must be added to 16 liters of milk and water contains 10% water to make it 20% water in it
A. 4 litreB. 2 litre
C. 1 litreD. 3 litre

 

 

Answer : Option B

 

Explanation :

By the rule of alligation, we have

% Concentration of water
in pure water : 100
% Concentration of water
in the given mixture : 10
Mean % concentration
20
20 – 10 = 10100 – 20 = 80

=> Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8

Here Quantity of the mixture = 16 litres

=> Quantity of water : 16 = 1 : 8

Quantity of water = 16×1/8=2 litre

 

8.A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is
A. 300B. 400
C. 600D. 500

 

 

Answer : Option C

 

Explanation :

By the rule of alligation,we have

Profit% by selling 1st partProfit% by selling 2nd part
818
Net % profit
14
18-14=414-8=6

=> Quantity of part1 : Quantity of part2 = 4 : 6 = 2 : 3

Total quantity is given as 1000Kg

So Quantity of part2 (Quantity sold at 18% profit) = 1000×3/5=600Kg

 

Level 2

 

1.Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety of tea in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, what is the price of the third variety per kg ?
A. Rs.182.50B. Rs.170.5
C. Rs.175.50D. Rs.180

 

 

Answer : Option C

Explanation :

Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed in the ratio 1 : 1

So their Average price = (126+135)2=130.5

Hence let’s consider that the mixture is formed by mixing two varieties of tea.

one at Rs. 130.50 per kg and the other at Rs. x per kg in the ratio 2 : 2, i.e.,

1 : 1. Now let’s find out x.

By the rule of alligation, we can write as

Cost of 1 kg of 1st kind of teaCost of 1 kg of 2nd kind of tea
130.50x
Mean Price
153
(x – 153)22.50

(x – 153) : 22.5 = 1 : 1

=>x – 153 = 22.50

=> x = 153 + 22.5 = 175.5

 

2.A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
A. 5litres, 7 litresB. 7litres, 4 litres
C. 6litres, 6 litresD. 4litres, 8 litres

 

 

Answer : Option C

Explanation :

Let cost of 1 litre milk be Rs.1

Milk in 1 litre mix in 1st can = 3/4 litre

Cost Price(CP) of 1 litre mix in 1st can = Rs.3/4

Milk in 1 litre mix in 2nd can = 1/2 litre

Cost Price(CP) of 1 litre mix in 2nd can = Rs.1/2

Milk in 1 litre of the final mix=5/8

Cost Price(CP) of 1 litre final mix = Rs.5/8

=> Mean price = 5/8
By the rule of alligation, we can write as

CP of 1 litre mix in 2nd canCP of 1 litre mix in 1st can
1/23/4
Mean Price
5/8
3/4 – 5/8 = 1/85/8 – 1/2 = 1/8

=> mix in 2nd can :mix in 1st can = 1/8 : 1/8 = 1:1

ie, from each can, 12×12=6 litre should be taken

 

3.Two vessels A and B contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively. Find the ratio in which these mixture be mixed to obtain a new mixture in vessel C containing spirit and water in the ration 8 : 5 ?
A. 3: 4B. 4 : 3
C. 9 : 7D. 7 : 9

 

 

Answer : Option D

 

Explanation :

 

Let Cost Price(CP) of 1 litre spirit be Rs.1

Quantity of spirit in 1 litre mixture from vessel A = 5/7

Cost Price(CP) of 1 litre mixture from vessel A = Rs. 5/7

Quantity of spirit in 1 litre mixture from vessel B = 7/13

Cost Price(CP) of 1 litre mixture from vessel B = Rs. 7/13

Quantity of spirit to be obtained in 1 litre mixture from vessel C = 8/13

Cost Price(CP) of 1 litre mixture from vessel C = Rs. 8/13 = Mean Price

By the rule of alligation, we can write as

 

CP of 1 litre mixture from vessel ACP of 1 litre mixture from vessel B
5/77/13
Mean Price
8/13
8/13 – 7/13 = 1/135/7 – 8/13 = 9/91

=> Mixture from Vessel A : Mixture from Vessel B = 1/13 : 9/91 = 7 : 9 = Required Ratio

 

4.How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per Kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per Kg ?
A. 60 KgB. 63 kg
C. 58 KgD. 56 Kg

 

 

Answer : Option B

Explanation :

Selling Price(SP) of 1 Kg mixture= Rs. 9.24

Profit = 10%

Cost Price(CP) of 1 Kg mixture= 100×SP /(100+Profit%) =100×9.24/(100+10)

=100×9.24/110=92.4/11=Rs.8.4

By the rule of alligation, we have

CP of 1 kg sugar of 1st kindCP of 1 kg sugar of 2nd kind
Rs. 9Rs. 7
Mean Price
Rs.8.4
8.4 – 7 = 1.49 – 8.4 = .6

ie, to get a cost price of 8.4, the sugars of kind1 and kind2 should be mixed in the

ratio 1.4 : .6 = 14 : 6 = 7 : 3

Let x Kg of kind1 sugar is mixed with 27 kg of kind2 sugar

then x : 27 = 7 : 3

⇒x/27=7/3

⇒x=(27×7)/3=63

5.A container contains a mixture of two liquids P and Q in the ratio 7 : 5. When 9 litres of mixture are drawn off and the container is filled with Q, the ratio of P and Q becomes 7 : 9. How many litres of liquid P was contained in the container initially?
A. 23B. 21
C. 19D. 17

 

 

Answer : Option B

Explanation :

Let’s initial quantity of P in the container be 7x

and initial quantity of Q in the container be 5x

Now 9 litres of mixture are drawn off from the container

Quantity of P in 9 litres of the mixtures drawn off = 9×7/12=63/12=21/ 4

Quantity of Q in 9 litres of the mixtures drawn off = 9×5/12=45/12=1/54

HenceQuantity of P remains in the mixtures after 9 litres is drawn off =7x−21/4

Quantity of Q remains in the mixtures after 9 litres is drawn off =5x−15/4

Since the container is filled with Q after 9 litres of mixture is drawn off,

Quantity of Q in the mixtures=5x−15/4+9=5x+21/4

Given that the ratio of P and Q becomes 7 : 9

⇒(7x−21/4):(5x+21/4)=7:9

⇒(7x−21/4)(5x+21/4)=7/9

63x−(9×21/4)=35x+(7×2/14)

28x=(16×21/4)

x=(16×21)/(4×28)

litres of P contained in the container initially = 7x=(7×16×21)/(4×28)=(16×21)/(4×4)=21

 

6.A dishonest milkman sells his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture?
A. 25%B. 20%
C. 22%D. 24%

 

 

Answer : Option B

Explanation :

Let CP of 1 litre milk = Rs.1

Given that SP of 1 litre mixture = CP of 1 Litre milk = Rs.1

Given than Gain = 25%

Hence CP of 1 litre mixture = 100/(100+Gain%)×SP

=100(100+25)×1

=100/125=4/5

By the rule of alligation, we have

CP of 1 litre milkCP of 1 litre water
10
CP of 1 litre mixture
4/5
4/5 – 0 = 4/51- 4/5 = 1/5

=> Quantity of milk : Quantity of water = 4/5 : 1/5 = 4 : 1

Hence percentage of water in the mixture = 1/5×100=20%,

Alligation

Alligation is a method of finding the proportions of two or more ingredients that must be mixed to obtain a desired mixture. It is based on the principle that the cost of a mixture is equal to the sum of the costs of the ingredients in it.

To use alligation, first list the ingredients and their costs per unit of weight or volume. Then, draw a line and label one end with the cost of the cheapest ingredient and the other end with the cost of the most expensive ingredient. For each ingredient, draw a line segment from its cost to the line. The point where the line segments intersect is the alligation point.

The proportion of the cheapest ingredient to the proportion of the most expensive ingredient is given by the ratio of the lengths of the line segments from the alligation point to the two ends of the line. For example, if the alligation point is 2 units from the left end of the line and 4 units from the right end of the line, then the proportion of the cheapest ingredient to the proportion of the most expensive ingredient is 2:4.

Concentration

The concentration of a solution is a measure of how much solute is dissolved in a given amount of solvent. There are many different ways to express concentration, but the most common are percent concentration, molarity, and molality.

Percent concentration is the amount of solute in a solution expressed as a percentage of the total volume of the solution. For example, a solution that is 10% sugar has 10 grams of sugar dissolved in 100 grams of solution.

Molarity is the number of moles of solute per liter of solution. For example, a solution that is 1 molar (M) sugar has 1 mole of sugar dissolved in 1 liter of solution.

Molality is the number of moles of solute per kilogram of solvent. For example, a solution that is 1 molal (m) sugar has 1 mole of sugar dissolved in 1 kilogram of water.

Dilution

Dilution is the process of adding solvent to a solution to reduce the concentration of the solute. The amount of solvent added is equal to the volume of the solution times the desired concentration divided by the initial concentration.

For example, to dilute a 1 M solution of sugar to a 0.5 M solution, you would add 1 liter of water to 1 liter of the original solution.

Percent Composition

The percent composition of a mixture is the percentage of each component in the mixture. To calculate the percent composition, first find the mass of each component in the mixture. Then, divide the mass of each component by the total mass of the mixture and multiply by 100%.

For example, if a mixture contains 10 grams of sugar and 20 grams of water, the percent composition of sugar is 50% and the percent composition of water is 50%.

Solute

The solute is the substance that is dissolved in a solution. The solvent is the substance that dissolves the solute.

For example, in a solution of sugar water, the sugar is the solute and the water is the solvent.

Solution

A solution is a homogeneous mixture of two or more substances. The substances in a solution are called solutes and the substance that dissolves the solutes is called the solvent.

solutions can be classified into two types: unsaturated and saturated. An unsaturated solution is one that can dissolve more solute. A saturated solution is one that cannot dissolve any more solute.

Weight by Weight (W/W)

Weight by weight (W/W) is a method of expressing the concentration of a solution. In W/W, the concentration is expressed as the weight of the solute per weight of the solution.

For example, a solution that is 10% sugar by weight has 10 grams of sugar per 100 grams of solution.

Weight by Volume (W/V)

Weight by volume (W/V) is a method of expressing the concentration of a solution. In W/V, the concentration is expressed as the weight of the solute per volume of the solution.

For example, a solution that is 10% sugar by volume has 10 grams of sugar per 100 milliliters of solution.

Volume by Volume (V/V)

Volume by volume (V/V) is a method of expressing the concentration of a solution. In V/V, the concentration is expressed as the volume of the solute per volume of the solution.

For example, a solution that is 10% sugar by volume has 10 milliliters of sugar per 100 milliliters of solution.

Mixtures

A mixture is a material made up of two or more substances that are mixed together but are not chemically combined. The substances in a mixture can be solids, liquids, or gases.

There are two main types of mixtures: homogeneous mixtures and heterogeneous mixtures.

A homogeneous mixture is a mixture in which the components are evenly distributed throughout the mixture. For example, salt water is a homogeneous mixture because the salt is evenly distributed throughout the water.

A heterogeneous mixture is a mixture in which the components are not evenly distributed throughout the mixture. For example, a mixture of sand and water is a heterogeneous mixture because the sand particles are not evenly distributed throughout the water.

Alligation

Alligation is a method of finding the proportions of two or more ingredients that must be mixed to obtain a desired mixture.

The alligation rule states that the difference between the percentage of the more expensive ingredient in the first mixture and the percentage of the more expensive ingredient in the second mixture is equal to the difference between the percentage of the less expensive ingredient in the first mixture and the percentage of the less expensive ingredient in the second mixture.

For example, if you want to make a mixture that is 50% A and 50% B, and you have two mixtures, one that is 70% A and 30% B and one that is 30% A and 70% B, you would use the alligation rule to find the proportions of the two mixtures that you need to mix.

The alligation rule can be used to find the proportions of any two or more ingredients that must be mixed to obtain a desired mixture.

Important Formulas

The following are some important formulas for mixtures and alligation:

  • The percentage of an ingredient in a mixture is equal to the mass of the ingredient divided by the total mass of the mixture, multiplied by 100%.
  • The mass of an ingredient in a mixture is equal to the percentage of the ingredient in the mixture, multiplied by the total mass of the mixture.
  • The volume of an ingredient in a mixture is equal to the percentage of the ingredient in the mixture, multiplied by the total volume of the mixture.
  • The density of a mixture is equal to the mass of the mixture divided by the volume of the mixture.
  • The specific gravity of a mixture is equal to the density of the mixture divided by the density of water.
  • The alligation rule states that the difference between the percentage of the more expensive ingredient in the first mixture and the percentage of the more expensive ingredient in the second mixture is equal to the difference between the percentage of the less expensive ingredient in the first mixture and the percentage of the less expensive ingredient in the second mixture.

Frequently Asked Questions

  1. What is a mixture?

A mixture is a material made up of two or more substances that are mixed together but are not chemically combined. The substances in a mixture can be solids, liquids, or gases.

  1. What are the two main types of mixtures?

The two main types of mixtures are homogeneous mixtures and heterogeneous mixtures.

A homogeneous mixture is a mixture in which the components are evenly distributed throughout the mixture. For example, salt water is a homogeneous mixture because the salt is evenly distributed throughout the water.

A heterogeneous mixture is a mixture in which the components are not evenly distributed throughout the mixture. For example, a mixture of sand and water is a heterogeneous mixture because the sand particles are not evenly distributed throughout the water.

  1. What is the alligation rule?

The alligation rule states that the difference between the percentage of the more expensive ingredient in the first mixture and the percentage of the more expensive ingredient in the second mixture is equal to the difference between the percentage of the less expensive ingredient in the first mixture and the percentage of the less expensive ingredient in the second mixture.

  1. What are some important formulas for mixtures and alligation?

The following are some important formulas for mixtures and alligation:

  • The percentage of an ingredient in a mixture is equal to the mass of the ingredient divided by the total mass of the mixture, multiplied by 100%.
  • The mass of an ingredient in a mixture is equal to the percentage of the ingredient in the mixture, multiplied by the total mass of the mixture.
  • The volume of an ingredient in a mixture is equal to the percentage of the ingredient in the mixture, multiplied by the total volume of the mixture.
  • The density of a mixture is equal to the mass of the mixture divided by the volume of the mixture.
  • The specific gravity of a mixture is equal to the density of the mixture divided by the density of water.
  • The alligation rule states that the difference between the percentage of the more expensive ingredient in
  1. A mixture is a material made up of two or more substances that are mixed together but are not chemically combined.
  2. A solution is a homogeneous mixture of two or more substances.
  3. A suspension is a heterogeneous mixture of a solid and a liquid.
  4. A colloid is a heterogeneous mixture of a solid and a liquid in which the solid particles are too small to be seen with the naked eye.
  5. An Alloy is a homogeneous mixture of two or more metals.
  6. A solution is a homogeneous mixture of two or more substances. The solute is the substance that is dissolved in the solvent. The solvent is the substance that does the dissolving.
  7. The concentration of a solution is the amount of solute that is dissolved in a given amount of solvent. The concentration can be expressed in molarity, molality, or percent by volume.
  8. Molarity is the number of moles of solute per liter of solution.
  9. Molality is the number of moles of solute per kilogram of solvent.
  10. Percent by volume is the volume of solute per 100 volumes of solution.
  11. Dilution is the process of adding more solvent to a solution to decrease the concentration of the solute.
  12. A saturated solution is a solution that contains the maximum amount of solute that can be dissolved at a given temperature.
  13. An unsaturated solution is a solution that contains less solute than the maximum amount that can be dissolved at a given temperature.
  14. A supersaturated solution is a solution that contains more solute than the maximum amount that can be dissolved at a given temperature.
  15. A precipitate is a solid that forms when two solutions are mixed.
  16. A double displacement reaction is a type of Chemical Reaction in which two compounds exchange ions to form two new compounds.
  17. A neutralization reaction is a type of chemical reaction in which an acid and a base react to form a salt and water.
  18. A Precipitation reaction is a type of chemical reaction in which two solutions are mixed and a solid precipitate forms.
  19. An oxidation-reduction reaction is a type of chemical reaction in which one substance loses electrons and another substance gains electrons.
  20. A combustion reaction is a type of chemical reaction in which a substance reacts with Oxygen to produce heat and Light.
  21. A combination reaction is a type of chemical reaction in which two or more substances combine to form a new substance.
  22. A decomposition reaction is a type of chemical reaction in which a substance breaks down into two or more simpler substances.
  23. A displacement reaction is a type of chemical reaction in which one element displaces another element from a compound.
  24. A disproportionation reaction is a type of chemical reaction in which an element is both oxidized and reduced.
  25. A redox reaction is a type of chemical reaction in which electrons are transferred between two or more substances.
  26. A Catalyst is a substance that speeds up The Rate of a Chemical Reaction without itself being consumed.
  27. An inhibitor is a substance that slows down the rate of a chemical reaction.
  28. A homogeneous catalyst is a catalyst that is in the same phase as the reactants.
  29. A heterogeneous catalyst is a catalyst that is in a different phase than the reactants.
  30. An enzyme is a biological catalyst.
  31. A chemical equilibrium is a state of balance in which the rate of the forward reaction is equal to the rate of the reverse reaction.
  32. The law of mass action states that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants raised to their respective powers.
  33. The equilibrium constant is a constant that expresses the relationship between the concentrations of the reactants and products at equilibrium.
  34. Le Châtelier’s principle states that if a Stress is applied to a system at equilibrium, the system will shift in a way that relieves the stress.
  35. A buffer solution is a solution that resists changes in pH when acids or bases are added.
  36. A buffer solution contains a weak acid and its conjugate base, or a weak base and its conjugate acid.
  37. The pH scale is a scale that measures the acidity or basicity of a solution.
  38. A pH of 7 is neutral.
  39. A pH below 7 is acidic.
  40. A pH above 7 is basic.
  41. A strong acid is an acid that completely dissociates in water.
  42. A weak acid is an acid that does not completely dissociate in water.
  43. A strong base is a base that completely dissociates in water.
  44. A weak base is a base that does not completely dissociate in water.
  45. An indicator is a substance that changes color