How To Solve Alligation And Mixtures Questions Quickly

How to solve Alligation and Mixtures Questions Quickly

How To Solve Alligation And Mixtures Problems Quickly via latest methods, formulae is discussed on this page along with few variety of questions.

What is Alligation?

The rule of alligation help to find the ratio in which two or more variety of ingredients of a given price must be mixed to produce a mixture of desired price.It is a rule for the solution of problems concerning the compounding or mixing of ingredients

What is Mixture?

A mixture contains two or more commodities of certain quantity mixed together to get the desired quantity.

What is called Mean Price?

The cost of a unit quantity of the mixture is called the mean price.

Type 1 : How to Solve Alligation and Mixtures Questions Quickly (When two quantities are mixed then)

\left ( \frac{ \text{ Quantity of Cheaper }}{\text{ Quantity of dearer }} \right ) = \left ( \frac{ \text{ C.P of Dearer (d) – Mean price (m)}}{ \text{ Mean price(m) – C.P. of cheaper(c) }} \right)


Therefore, (Cheaper Quantity) : (Dearer Quantity) = (d – m) : (m-c)

How to solve Alligation  and Mixtures Questions quickly Ques 1

Calculate the average price of the resulting mixture when two variety of sugar at ₹12 per Kg and ₹15 per Kg are mixed in the ratio of 2:3. 

Solution:

Let average price be Aw

\frac{2}{3} = \frac{(15-Aw)}{ (Aw-12)}

2Aw-24 = 45-3Aw

5Aw = 69

Aw = 69/5 = ₹13.8 per Kg

How to solve Alligation  and Mixtures Questions quickly Ques 1

Two varieties of rice are in the ratio of 3:2 such that the average price of the resulting mixture is ₹15 per Kg. The price of one of the varieties is ₹10 per Kg. Find the price of the other variety of rice. 

Solution:

Let Price of another variety is A_{2}

\frac{3}{2}= \frac{(A2-15)}{(15-10)}
15 = 2A_{2}-30
\frac{45}{2} = A_{2}
A2 = ₹22.5 per Kg

How to solve Alligation  and Mixtures Questions quickly Ques 1

Two varieties of tea are mixed in some ratio. The cost of the first variety is ₹20 per Kg and that of second variety is ₹30 per Kg. If the average cost of the resulting mixture is ₹25 per Kg, find the ratio. 

Solution:

\frac{x}{y}= \frac{(30-25)}{(25-20)}

\frac{x}{y} =  \frac{5}{5} = 1:1

 

Type 2 : How to Solve Alligation and Mixtures Questions Quickly (when there is process of replacement)

If a container has x unit of liquid A from which y units are taken out and replaced by water. This process is repeated n number of time, then the quantity of pure liquid will be given by:

After n operations, the quantity of pure liquid = \left ( x \left( 1 – \frac{y}{x} \right)n \right)

Question 1 

A 420L of the mixture contains pure milk and water in the ratio of 2:5. Now 80 L of water is added to the mixture. Calculate the ratio of milk and water in the resulting mixture. 

Solution:

The quantity of milk in initial mixture =

\frac{2}{7} \times 420 = 120 \text{ L}

Quantity of water = 420-120 = 300L

Water added = 80L

Concentration of water in the resulting mixture = 300 + 80 = 380L

The ratio of the resulting mixture 120/380 = 6:19.

Question 2

A milkman mixes 30L of water in 90L of milk. He then sells 1/4th of this mixture. Now he adds water to replenish the quantity of milk sold. Find the current proportion of milk and water. 

Solution:

The initial ratio of milk and water = 90:30 = 3:1

Now, 1/4th of the mixture is sold, that is the total volume of the mixture is reduced by 25%.

In other words, both water and milk are reduced by 25%.

So, the volume of milk and water is 67.5L and 22.5 respectively.

Now, 30L (25% of the total mixture volume) of water is added to the mixture.

The volume of milk = 67.5L

Volume of water = 22.5+30= 52.5L

Current ratio = 67.5/52.5 = 9:7

Question 3

A 20 litres mixture of milk and water contains milk and water in the ratio 3 : 2. 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more. At the end of the two removal and replacement, what is the ratio of milk and water in the resultant mixture? 

Solution:

In 20 liters of mixture

Milk =

\frac{3}{5} \times 20 = 12 \text{ L}

Water = 8 litres

In 10 liters of mixture.

Milk = 6 liters

Water = 4 litres

On adding 10 liters of milk,

Milk = 12 – 6 + 10 = 16 L

Water = 8 – 4 = 4 L

Again. in 10 liters of mixture,

Milk = 4/5 * 10 = 8 L

Water = 2 litres
On adding 10 litres of milk,

Milk =16 – 8 + 10 = 18 litres

Water = 2 litres

Therefore, Required ratio = 18 : 2 = 9 : 1

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7 comments on “How To Solve Alligation And Mixtures Questions Quickly”


  • Sumaiya

    Please explain question 1 again. how the quantity of water in initial mixture is calculated?


    • Lavanya Chowdary

      In second question we have to calculate the ratio of water and milk in the given mixture.
      90:30=3:1
      hence,the person can sold the 1/4th of the mixture (1/4*100)=25%
      25% of the mixture can contain both milk and water .
      After he sold the 25% of mixture from that we have to calculate the what percent water and milk can contain the 90ml of mixture.
      (25/100)*90=22.5 of water can contain in that mixture.
      (75/100)*90=67.5 of milk can contain in that mixture.
      Hence , he added the 30L of water (the quantity of milk he sold 25% =30L) into the that.
      22.5+30=52.5
      then the volume of milk =67.5
      Then the ratio is =67.5:52.5
      =9:7