# ALLEGATION AND MIXTURE

Allegation and Mixture :

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

## Allegation and Mixture - Basic Concept

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".

## Allegation and Mixture - Practice Problems

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)

=  10 ⋅ (10/10 - 1/10)

=  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"

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