**Allegation and Mixture :**

In this section, we are going to learn, how to solve problems on allegation and mixture step by step.

**1. Allegation : **

This is a rule which we can use to find the ratio in which we have to mix two or more types of ingredients to produce the mixture in a desired price. **2. Mean price :**

The cost price of the mixture per given unit is called mean price. **3. Rule to find the ratio for producing mixtures :**

To produce the mixture, the ratio in which the cheaper and dearer to be mixed is

= (d - m) : (m - c)

4. Say, a bucket has "x" units of liquid. From which "y" units of liquid taken out and replaced by water. After repeating this process "n" times,

Let us look at some practice problems on "Allegation and Mixture".

**Problem 1 :**

What is the ratio in which the two types of wheat mixed where the price of the first type is $9.30 per kg and the second type is $10.80 per kg so the mixture is having worth $10 per kg ?

**Solution :**

From the given information, we have

Cost price of the cheaper (c) = $9.30

Cost price of the dearer (d) = $10.80

Cost price of the mixture (m) = $10

Rule to find the ratio for producing mixture is

= (d - m) : (m - c)

So, we have

(d - m) : (m - c) = (10.8 - 10) : (10 - 9.3)

(d - m) : (m - c) = 0.8 : 0.7

(d - m) : (m - c) = 8 : 7

Hence the ratio in which the first kind and second kind to be mixed is 8 : 7.

**Problem 2 :**

Find the ratio in which, water to be mixed with milk to gain 20% by selling the mixture at cost price.

**Solution :**

Let us consider the water as cheaper and milk as dearer.

Let the cost price of 1 ltr of pure milk be $1 and water be $0.

Now we take some quantity of milk (less than 1 ltr), add some water and make it to be 1 ltr milk-water mixture.

Let "x" be the cost price in 1 ltr of the mixture.

Because the gain is 20%, selling price of 1 ltr of the mixture is

= x + 20% of x

= x + (20x / 100)

= (100x / 100) + (20x / 100)

= (100x + 20x) / 100

= 120x / 100

= 6x / 5

**Given :** The mixture is sold at the cost price of pure milk.

That is,

S.P of 1 ltr of mixture = C.P of 1 ltr of pure milk

6x / 5 = 1

Multiply both sides by 5 / 6.

x = 5 / 6

So, the cost price of 1 ltr of the mixture is $5/6.

Now, we can consider the following points.

**Cheaper : **Cost price of 1 ltr water (c) = $0

**Dearer :** Cost price of 1 ltr pure milk (d) = $1

**Mean price : **Cost price of 1 ltr mixture (m) = $5/6

Rule to find the ratio for producing mixture is

= (d - m) : (m - c)

Then, we have

(d - m) : (m - c) = (1 - 5/6) : (5/6 - 0)

(d - m) : (m - c) = 1/6 : 5/6

(d - m) : (m - c) = 1 : 5

Hence, water and milk have to be mixed in the ratio to gain 20% is 1 : 5.

**Problem 3 :**

The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio, the liquids in vessels A and B be mixed to obtain a new mixture in vessel C consisting half milk and half water ?

**Solution :**

Let the cost price of 1 liter pure milk be $1.

Amount of milk in 1 liter of mixture in A = 4/7 liter

Amount of milk in 1 liter of mixture in B = 2/5 liter

Amount of milk in 1 liter of mixture in C = 1/2 liter

Let us assume the cost price of the liquid mixture in A as cheaper, B as dearer and C as mean price.

C.P of 1 liter mixture in A (c) = $4/7

C.P of 1 liter mixture in B (d) = $2/5

C.P of 1 liter mixture in C (m) = $1/2

Rule to find the ratio for producing mixture is

= (d - m) : (m - c)

So, we have

(d - m) : (m - c) = (2/5 - 1/2) : (1/2 - 4/7)

(d - m) : (m - c) = (4 - 5) / 10 : (7 - 8) / 14

(d - m) : (m - c) = (-1) / 10 : (-1) / 14

(d - m) : (m - c) = 1 / 10 : 1 / 14

L.C.M of (10, 14) is 70.

So, multiply both the terms of the ratio (1/10 : 1/14) by 70.

(d - m) : (m - c) = 70 / 10 : 70 / 14

(d - m) : (m - c) = 7 : 5

Hence, the ratio in which the liquids in vessels A and B be mixed is 7 : 5.

**Problem 4 :**

A dealer mixes tea costing $6.92 per kg with tea costing $7.77 per kg and sells the mixture at $8.80 per kg and earns a profit of 17.5 % on his sale price. In what ratio does he mix them ?

**Solution :**

**Given :** The mixture is sold at $8.80 per kg and 17.5% profit earned on the sale price.

Profit = 17.5% of 8.80

Profit = 0.175 ⋅ 8.80

Profit = 1.54

Then the cost price of of the mixture is

= Selling price - Profit

= 8.80 - 1.54

= 7.26

Let the cost price of tea costing $6.92 per kg as cheaper and $7.77 as dearer.

Then the mean price is $7.26.

More clearly,

Cheaper (c) = 6.92

Dearer (d) = 7.77

Mean price (m) = 7.26

Rule to find the ratio for producing mixture is

= (d - m) : (m - c)

So, we have

(d - m) : (m - c) = (7.77 - 7.26) : (7.26 - 6.92)

(d - m) : (m - c) = 0.51 : 0.34

(d - m) : (m - c) = 51 : 34

(d - m) : (m - c) = 3 : 2

Hence, the required ratio is 3 : 2.

**Problem 5 :**

A container has 10 liters of milk. 1 liter of liquid is taken out and replaced by water. If this processed is repeated thrice, find the amount of pure milk in the container.

**Solution :**

The formula to find the amount of pure milk is,

= x ⋅ (1 - y/x)^{n}

Here,

x = 10 ltrs

y = 1 ltr

n = 3

Then, we have

= 10 ⋅ (1 - 1/10)^{3 }

= 10 ⋅ (10/10 - 1/10)^{3 }

= 10 ⋅ (9 / 10)^{3}

= 10 ⋅ (0.9)^{3}

= 10 ⋅ 0.729

= 7.29

Hence, the amount of pure milk in the container will be 7.29 liters.

After having gone through the stuff given above, we hope that the students would have understood "Allegation and mixture"

Apart from the stuff given above, If you want to know more about "Allegation and mixture",please click here

Apart from the stuff given on "Allegation and mixture", if you need any other stuff in math, please use our google custom search here.

Widget is loading comments...

You can also visit our following web pages on different stuff in math.

**WORD PROBLEMS**

**Word problems on simple equations **

**Word problems on linear equations **

**Word problems on quadratic equations**

**Area and perimeter word problems**

**Word problems on direct variation and inverse variation **

**Word problems on comparing rates**

**Converting customary units word problems **

**Converting metric units word problems**

**Word problems on simple interest**

**Word problems on compound interest**

**Word problems on types of angles **

**Complementary and supplementary angles word problems**

**Markup and markdown word problems **

**Word problems on mixed fractrions**

**One step equation word problems**

**Linear inequalities word problems**

**Ratio and proportion word problems**

**Word problems on sets and venn diagrams**

**Pythagorean theorem word problems**

**Percent of a number word problems**

**Word problems on constant speed**

**Word problems on average speed **

**Word problems on sum of the angles of a triangle is 180 degree**

**OTHER TOPICS **

**Time, speed and distance shortcuts**

**Ratio and proportion shortcuts**

**Domain and range of rational functions**

**Domain and range of rational functions with holes**

**Graphing rational functions with holes**

**Converting repeating decimals in to fractions**

**Decimal representation of rational numbers**

**Finding square root using long division**

**L.C.M method to solve time and work problems**

**Translating the word problems in to algebraic expressions**

**Remainder when 2 power 256 is divided by 17**

**Remainder when 17 power 23 is divided by 16**

**Sum of all three digit numbers divisible by 6**

**Sum of all three digit numbers divisible by 7**

**Sum of all three digit numbers divisible by 8**

**Sum of all three digit numbers formed using 1, 3, 4**

**Sum of all three four digit numbers formed with non zero digits**