In this Blog we will begin with the Mixture and Alligation

What is a mixture? When we take two or more different ingredients or solutions and mix them together then the final product or solution is called the mixture.

Let’s get familiar with some terms that we are going to use in this chapter:

**Dearer Price:**The cost price of the dearer solution.**Cheaper Price:**The cost price of the cheaper solution.**Mean Price:**The cost price of the mixture. Mean Price will always lie between the cheaper and dearer price i.e. Cheaper Price < Mean Price < Dearer Price

** Rule of Mixture and Alligation:** To understand the formula of the Alligation and the rule of Alligation, we have to understand the concept of weighted average. For example, let us say that we buy 50 packs of chips at the cost of 10 rupees each and 30 packets at the cost of 20 rupees each. What is the average cost? Will the average will be determined by the 20 rupees packs or the 10 rupees packs? This is when we come across the concept of weighted average.

The weighted average here will be: [50×10]+[30×20]/80

In Alligation, we will use the same concept. Let us first define the few following terms:

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

Mean Price: The cost price of a unit quantity of the mixture is called the mean price. Now let us define the rule of Alligation.

We have a rule to for the mixture or Alligation to find the ratio of the amount of solutions mixed in it or to get the desired price of a mixture by adding two solutions of different prices. Let’s understand it.

According to this rule,

**Amount of cheaper: Amount of dearer (n**_{1}**:n**_{2}**) = A**_{2 }**– A**_{w}**/A**_{w }**– A**_{1}

Where, n_{1}/n_{2} is the ratio in which two quantities should be mixed and A_{1}, A_{2} and A_{w} are cheaper price, dearer price and mean price.

We will use the pictorial representation (Alligation) of this rule to solve the problems on Mixture as it makes it easier for us.

So, from here, we can get the ratio n_{1}:n_{2}.

Let’s solve some examples and see how this pictorial representation is more helpful than the formula.

**E.g.1: ** In what proportion, must wheat at Rs. 6.20 per kg be mixed with wheat at Rs. 7.20 per kg, so that the mixture be worth Rs. 6.50 per kg?

Given: A_{1 }= Rs. 6.20, A_{2 }= Rs. 7.20, A_{w }= Rs. 6.50

By formula:

Amount of cheaper: Amount of dearer (n_{1}:n_{2}) = A_{2 }– A_{w}/A_{w }– A_{1}

= (7.20 – 6.50)/(6.50 – 6.20)

= 0.70/0.30 = 7/3 or 7:3

Hence, required ratio = 7:3 (Ans).

**Rule Of Alligation**

Let us suppose that two ingredients of concentrations ‘a’ and ‘b’ respectively have been mixed in some proportion. Let ‘a’ be the cheaper component and ‘b’ be the dearer or the costlier component. Then the rule of alligation states that:

[{Quantity of Cheaper substance}/{Quantity of dearer substance}] = [(C.P. of dearer substance) – (Mean price)/(Mean price) – (C.P. of cheaper substance)]

Let ‘c’ be the cost price or C.P. of a unit quantity of a cheaper substance, ‘m’ be the mean price, ‘d’ be the cost price of a unit quantity of the dearer substance, then we can write:

(Quantity of the Cheaper Substance) : (Quantity of the Dearer Substance) = (d – m) : (m – c). Let us get a clearer picture with some an example.

Example 1: In what ratio must rice at Rs 9.30 per kg be mixed with rice sold at Rs. 10.80 per kg, so that the mixture be worth Rs. 10 per kg?

A) 2:1 B) 4:3 C) 6:5 D) 8:7

Answer: Using the rule of Alligation, we have:

C.P. of 1 kg of rice (in paise) = 1080 paise = d

Also the C.P. of 1 kg rice of 2nd kind (in paise) = 930 paise = c

Also, mean price of the mixture (per kg in paise) or m = 1000 paise. So from the rule of alligation, we have:

(Quantity of Cheaper rice) : (Quantity of dearer rice) = (1080 – 1000)/(1000 – 930)

Therefore the required ratio = 80 : 70 or 8:7 and hence the correct option is D) 8: 7

∴n_{1}:n_{2} = 0.70:0.30 = 7:3 (Ans).

**Note: **On observation, you will see that there is not much difference between the formula method and alligation but we are still practicing alligation as it takes away the burden to learn the formula.

**E.g. 2:**The amount of alcohol in two different medicines is 1.5% and 3.5% respectively. In what ratio they should be mixed such that amount of alcohol in new mixture is 2.5% ?

Here, we don’t have to deal with the price but percentage.

∴Required ratio = 1:1 (Ans).

**Imp: **In such question where we don’t have volumes, we will use the share (part) concept of ratios to solve the problems.

Let’s solve more example to understand it.

**E.g. 3:** Two vessels contain milk and water in the ratio 2;3 and 7:3. Find the ratio in which the contents of the two vessels have to be mixed so that the ratio of milk and water in the new mixture is equal.

Ratio of milk and water in 1st vessel = 2:3

Share of milk = 2/(2 + 3) = 2/5 [Using the share(part) concept of Ratio]

Ratio of milk and water in 2nd vessel = 7:3

Share of milk= 7/(7 + 3) = 7/10

Ratio of milk and water in the mixture is equal.

Share of milk in mixture = ½

Using alligation:

Required ratio= 2/10 : 1/10 = 2:1 (Ans).

Let’s deal with some problems that involve multiple concepts.

**E.g. 4: **In what ratio should Darjeeling Tea costing Rs. 400 per kg be mixed with Assam Tea costing Rs. 300 per kg, so that by selling the mixture at Rs. 408, there is a gain of 20%?

We don’t have the Cost Price (CP) of the mixture so we cannot apply the formula as we have to keep the same quantities in the formula. For now I am finding the CP, we will learn how to find CP in Profit and Loss chapter in depth.

CP of Assam Tea, A_{1 }= 300 per kg

CP of Darjeeling Tea, A_{2 }= 400 per kg

Selling Price (SP) of Mixture = Rs. 408

CP of Mixture, A_{w }= ?

Profit = 20%

⇒ Multiplying Factor = 6/5

CP × (100 + Gain%)/100 = SP

⇒ CP × (100 + 20)/100 = 408

⇒ CP × 120/100 = 408

⇒ CP × 6/5 = 408

⇒ CP = 408 × 5/~~6~~

⇒ CP = 68 × 5 = Rs. 340

⇒ A_{w} = Rs. 340 per kg

By formula:

Amount of cheaper: Amount of dearer (n_{1}:n_{2}) = A_{2 }– A_{w}/A_{w }– A_{1}

= (400 – 340)/(340 – 300)

= 60/40 = 3/3 or 3:2

But the ratio asked in the problem is Darjeeling tea(Dearer) to Assam tea(cheaper)

Hence, required ratio = 2:3 (Ans).

By alligation:

∴n_{1}:n_{2} = 60:40 or 3:2

But the ratio asked in the problem is Darjeeling tea (dearer) to Assam tea (cheaper)

Hence, required ratio= 2:3 (Ans).

**Note:**

- Always take the ratio between similar quantities.
- Always make sure about the ratio asked in the problem. In this case if we haven’t checked there were chances that we marked the wrong answer.

**E.g. 5: **A trader bought a bag of 40 kg of basmati rice at Rs. 125 per kg and another bag of 60 kg at Rs. 150 per kg. He sold the entire stock at a profit of 20%. Find the selling price per kg.

Given: n_{1}=40 kg, n_{2}=60 kg, A_{1}=Rs. 125 per kg, A_{2}= Rs. 150

n_{1}:n_{2} = 40:60= 2:3

By Alligation:

n_{1}:n_{2} = 2:3 = (150 – A_{w}) : (A_{w }– 125)

Or, 2/3 = (150 – A_{w}) / (A_{w }– 125)

⇒ 2 × (A_{w }– 125) = 3 × (150 – A_{w})

⇒ 2A_{w }– 2 × 125 = 3 × 150 – 3A_{w}

⇒ 2A_{w }+ 3A_{w} = 3 × 150 + 2 × 125

⇒ 5A_{w} = 450 + 250 = 700

⇒ A_{w} = 700/5

⇒ A_{w} = Rs. 140 per kg.

Therefore, the CP of mixture = 140 per kg.

CP × (100 + Gain%)/100 = SP

⇒ 140 × (100 + 20)/100 = SP

⇒ CP × 120/100 = SP

⇒ ~~140 ~~× 6/~~5~~ = SP

⇒ SP = Rs. (28 × 6) = Rs. 168 per kg (Ans).

- If a container initially contains a units of liquid and b units of liquid is taken out and it is filled with b units of another liquid, then after n operations, the final quantity of the original liquid in the container is given as
**[a × (1 – b/a)**^{n}**]**units.

Let’s understand this with an example.

**E.g: **A container contains 40 L of milk. From this container, 4 L of milk was taken out and replaced by water. This process was further repeated 1 time. How much is the milk now there in the container.

Initial quantity of milk = 40L

4 L of milk was taken out and replaced by water,

Now, Quantity of milk = 36 L and Quantity of Water = 4 L.

When 4 L of this mixture is taken out,

Quantity of milk taken out = 4 × 36/40 = 3.6L

Quantity of milk left = 36 – 3.6 = 32.4 L Ans

By formula:

Here, a = 40L, b = 4 L and n = 3

Using formula,

Quantity of milk left= [a × (1 – b/a)n] L

= [40 × (1 – 4/40)2] L

= [40 × (1 – 1/10)2] L

= [40 × (9/10)2] L

= (40 × 81/100) L

= (2 × 81/5) L = 162/5 L = 32.4 L (Ans).

__Miscellaneous Examples__**:**

Let’s solve one miscellaneous question type with the help of multiplying factor in the ratios.

**E.g.:** In a mixture 25 litres, the ratio of milk and water is 4:1. How many litres of milk must be added to make the ratio 16:1?

In 25 litre mixture,

Ratio of milk and water = 4:1

∴Milk in the mixture = 4 × 25/5 = 20 L

And, water in the mixture = 5L

Here, multiplying factor for water is 5.

∴Milk required to make the ratio 16:1 = 16 × 5= 80 L

Since, initially we have 20 L.

∴Milk to be added = 80 – 20 = 60 L (Ans).