Mixtures and Alligations: Solved Examples | CSAT Preparation - UPSC PDF Download

Mixtures & Alligation

• As the dictionary meaning of Alligation (mixing), we will deal with problems related to mixing of different compounds or quantities. The concept of alligation and weighted average are the same. 
• When two or more quantities are mixed together in different ratios to form a mixture, then ratio of the quantities of the two constituents is given by the following formulae:
  
  
• Gives us the ratio of quantities in which the two ingredients should be mixed to get the mixture.

Example.1: A sum of Rs 39 was divided among 45 boys and girls. Each girl gets 50 paise, whereas a boy gets one rupee. Find the number of boys and girls.

• Average amount of money received by each =
• Amount received by each girl = 50 paise = Rs
• Amount received by each boy = Re. 1
  
  
  
  Number of girls = 45 – 33 = 12.

Important Funda

• Always identify the ingredients as cheaper & dearer to apply the alligation rule.
  In the alligation rule, the variables c, d & m may be expressed in terms of percentages (e.g. A 20% mixture of salt in water), fractions (e.g. two-fifth of the solution contains salt) or proportions (e.g. A solution of milk and water is such that milk : Water = 2 : 3). 
• The important point is to remember is that c & d may represent pure ingredients or mixtures.

Mixing a pure component into a solution

Example.2: A jar contains a mixture of two liquids A and B in the ratio 4 : 1. When 10 litres of the liquid B is poured into the jar, the ratio becomes 2: 3. How many litres of liquid A were contained in the jar?

Method 1: (weighted average or equation method)

• Let the quantities of A & B in the original mixture be 4x and x litres.
• According to the question  
  12x = 2x + 20
  ⇒ 10x = 20
  ⇒ x = 2
• The quantity of A in the original mixture = 4x = 4 × 2 = 8 litres.

Method 2: (Alligation with composition of B)

• The average composition of B in the first mixture is 1/5.
• The average composition of B in the second mixture = 1
• The average composition of B in the resultant mixture = 3/5
• Hence applying the rule of Alligation we have
  [1 – (3/5)]/[(3/5) – (1/5)] = (2/5)/(2/5) = 1
  So, initial quantity of mixture in the jar = 10 litres.
  And, quantity of A in the jar = (10 × 4)/5 = 8 litres.
  

 Method 3: (Alligation with percentage of B)

• The percentage of B in 1st mixture = 20%
• The percentage of B in 2nd mixture = 100%
• The percentage of B in Final Mixture = 60%
• By rule of allegation we have
  Volume 1^st : Volume 2^nd = (100% - 60%) : (60% - 20%)
  V₁ : V₂ = 1 : 1
  Volume of mixture 1^st = 10 litres
  Volume of A in mixture 1^st = 80% of 10 litres = 8 litres. Answer
  

Removal and Replacement

Example.3: Nine litres of solution are drawn from a cask containing water. It is replaced with a similar quantity of pure milk. This operation is done twice. The ratio of water to milk in the cask now is 16 : 9. How much does the cask hold?

• Let there be x litres in the cask
• After n operations:
  
  (1 – 9/x)² = 16/25
  ∴ x = 45 litres.

Example.4: There are two containers A and B of milk solution. The ratio of milk and water in container A is 3 : 1 and in container B, it is 4 : 1. How many liters of container B solution has to be added to 20. lts of container ‘A’ solution such that in the resulting solution; the ratio of milk to water should be 19 : 6?

• In container A, the part of milk =
  
• It is given 20 lts of container A is added. So, the quantity of container B should be 5 lts.

 Example.5: There are two alloys A and B. Alloy A contains zinc, copper and silver, as 80% 15% and 5% respectively, whereas alloy B also contains the same metals with percentages 70%, 20%, 10% respectively. If these two alloys are mixed such that the resultant will contain 8% silver, what is the ratio of these three metals in the resultant alloy?

• Since the resultant alloy contains 8% silver, first we will find, in what ratio these two alloys A and B were mixed to form the resultant.
• Then the resultant zinc percentage is
  
• So, copper percentage = 100 – (74 + 8) = 18
  ∴ The ratio of these metals = 74 : 18 : 8 = 37 : 9 : 4.
  

 Example.6: The cost of an apple is directly proportional to square of its weight in a fruit bazaar. Two friends A and B went there to purchase apples. A got exactly 5 apples per kg and each apple is of same weight. Where as B got exactly 4 apples per kg each weight is exactly same. If B paid Rs. 10 more than A per kg apples, what is the cost of an apple which weighs 1 kg?

• It is given cost ∝ (weight)2
  ⇒ c = k w².
• A got 5 apples per kg and each apple is of same weight. ⇒ Each apple is 200 gm. = 1/5 kg.