What if we were in a situation where we were told to buy a rice mixing two different quantities of rice of different prices. What would be the price of the final nixed rice ? Today we are going to learn how can we approach these situations rather sums with the objective of minimum time consumption. This process ,we call it alligation.

When two different mixtures of price P1 and P2 are mixed together in quantities Q1 and Q2 respectively , then according to the rule of allegation we can say that :

Q1/Q2 = P2- Mean/ Mean – P1

Where the mean price (price of the mixture) is denoted as : Mean

A descriptive image is given below :

Example:

Two quantities of milk costing 10 rs and 20 rs are mixed in a ratio such that the cost of the mixture is 18rs. What is the ratio of the quantities mixed ?

Solution : By the formula of alligation :

10 20

18

X Y

So, X/Y = 20-18 / 18-10 = 2/8 = 1:4

Hence they must be mixed in the ratio : 1:4

You can also do a cross-check to verify. 20*4 + 10*1/ 4+1 = 90/5 = 18

Hence Verified!

Suppose we are put in another situation where we are having a bucket filled with milk. We need to take a portion of it and replace with water. And again, so this process can be continued for a number of times. So after that what is the amount of milk/water that is present in the bucket. These sums are very important from CAT’s point of view. So, we will learn how to tackle these problems with the help of replacement rule.

Let us take an example to illustrate :

Roy has a container of 100L filled with milk. He takes out 10L from it, replaces with water. He repeats the process again. What is the amount of milk in the container after the second replacement is done ?

Solution: We will make a table for better understanding :

Milk Water

100 0

Taking 10 L from the above solution we will be left with 90L milk and 10 L water (after we add 10L water to it)

90 10

Now, taking 10L from this solution, Milk taken = (10/Total Mixture) *Amount of milk in the mixture

= 10/100 *90 = 9L

So we will have 90-9 L = 81L milk. And 10+9 = 19L of water.

You see, since same quantity of mixture is replaced, the total amount always remains SAME, i.e, 100 L in this case.

So final Ratio = 81:19

It will be lengthy, so why not use a formula ?

We will use this in all replacement sums :

If a vessel contains “x” amount of L1 and if “y” amounts to be withdrawn and replaced by liquid L2, and the operation is repeated ‘n’ times together, then, L1 left in vessel after “n” replacements = [(x-y)/x]^{n}

Lets do the above sum by this formula :

X=100, y=10, n=2

So, (100-10/100)^2= 81/100

So milk = 81L

Water = 100-81 = 19L

So, it took us only 5 seconds to do this by the formula.

We have cleared our basic. Practice these kind of sums and you should be in full flow. Happy replacing ;)

Alligation rule for 3 quantities:

Mixtures as the name suggests contain certain different kinds of liquids in their composition in definite amounts. With the help of this lesson we will try to do sums related to Mixtures. We have already learnt alligation. Mixtures is another important concept.

Suppose we have a mixture where there are 2 components, A and B in some ratio. The total amount of mixture = M. Applying the concept of ratio, we get:

Amount of A = (A/(A+B ) ) * M

Amount of B = (B/(A+B ) ) * M

If there are 3 different type of products that are needed to be mixed in a definite ratio to get the Mean mixture, we will consider the following:

Let the cost price of 1^{st} product = P1

Let the cost price of 2^{nd} product = P2

Let the Cost price of 3^{rd} product = P3

Cost of the Mean price of the Mixture = M

Then the ratio in which the three items should be mixed is =

(P2-M)*(P3-M) : (M-P1)*(P3-M) : (P2-M)*(M-P1)

Concept:

When two mixtures containing the same items in it and are added in definite amounts, then the ratio is calculated as follows.

Let us consider a mixture of M1 Kg containing A1 and B1 amounts of Ingredients X and Y and a mixture of M2 Kg containing A2 and B2 amounts of Ingredients X and Y.

Quantity of X/ Quantity of Y = (Quantity of X in M1 + M2) / (Quantity of Y in M1 + M2), we get the ratio as

[ ( (A1/A1+B1)*M1) + ( (A2/A2+B2)*M2) ] / [ ( (B1/A1+B1)*M1) + ( (B2/A2+B2)*M2) ] |

What if we were in a situation where we were told to buy a rice mixing two different quantities of rice of different prices. What would be the price of the final nixed rice ? Today we are going to learn how can we approach these situations rather sums with the objective of minimum time consumption. This process ,we call it alligation.

When two different mixtures of price P1 and P2 are mixed together in quantities Q1 and Q2 respectively , then according to the rule of allegation we can say that :

Q1/Q2 = P2- Mean/ Mean – P1

Where the mean price (price of the mixture) is denoted as : Mean

A descriptive image is given below :

Example:

Two quantities of milk costing 10 rs and 20 rs are mixed in a ratio such that the cost of the mixture is 18rs. What is the ratio of the quantities mixed ?

Solution : By the formula of alligation :

10 20

18

X Y

So, X/Y = 20-18 / 18-10 = 2/8 = 1:4

Hence they must be mixed in the ratio : 1:4

You can also do a cross-check to verify. 20*4 + 10*1/ 4+1 = 90/5 = 18

Hence Verified!

Suppose we are put in another situation where we are having a bucket filled with milk. We need to take a portion of it and replace with water. And again, so this process can be continued for a number of times. So after that what is the amount of milk/water that is present in the bucket. These sums are very important from CAT’s point of view. So, we will learn how to tackle these problems with the help of replacement rule.

Let us take an example to illustrate :

Roy has a container of 100L filled with milk. He takes out 10L from it, replaces with water. He repeats the process again. What is the amount of milk in the container after the second replacement is done ?

Solution: We will make a table for better understanding :

Milk Water

100 0

Taking 10 L from the above solution we will be left with 90L milk and 10 L water (after we add 10L water to it)

90 10

Now, taking 10L from this solution, Milk taken = (10/Total Mixture) *Amount of milk in the mixture

= 10/100 *90 = 9L

So we will have 90-9 L = 81L milk. And 10+9 = 19L of water.

You see, since same quantity of mixture is replaced, the total amount always remains SAME, i.e, 100 L in this case.

So final Ratio = 81:19

It will be lengthy, so why not use a formula ?

We will use this in all replacement sums :

If a vessel contains “x” amount of L1 and if “y” amounts to be withdrawn and replaced by liquid L2, and the operation is repeated ‘n’ times together, then, L1 left in vessel after “n” replacements = [(x-y)/x]^{n}

Lets do the above sum by this formula :

X=100, y=10, n=2

So, (100-10/100)^2= 81/100

So milk = 81L

Water = 100-81 = 19L

So, it took us only 5 seconds to do this by the formula.

We have cleared our basic. Practice these kind of sums and you should be in full flow. Happy replacing ;)

Alligation rule for 3 quantities:

Mixtures as the name suggests contain certain different kinds of liquids in their composition in definite amounts. With the help of this lesson we will try to do sums related to Mixtures. We have already learnt alligation. Mixtures is another important concept.

Suppose we have a mixture where there are 2 components, A and B in some ratio. The total amount of mixture = M. Applying the concept of ratio, we get:

Amount of A = (A/(A+B ) ) * M

Amount of B = (B/(A+B ) ) * M

If there are 3 different type of products that are needed to be mixed in a definite ratio to get the Mean mixture, we will consider the following:

Let the cost price of 1^{st} product = P1

Let the cost price of 2^{nd} product = P2

Let the Cost price of 3^{rd} product = P3

Cost of the Mean price of the Mixture = M

Then the ratio in which the three items should be mixed is =

(P2-M)*(P3-M) : (M-P1)*(P3-M) : (P2-M)*(M-P1)

Concept:

When two mixtures containing the same items in it and are added in definite amounts, then the ratio is calculated as follows.

Let us consider a mixture of M1 Kg containing A1 and B1 amounts of Ingredients X and Y and a mixture of M2 Kg containing A2 and B2 amounts of Ingredients X and Y.

Quantity of X/ Quantity of Y = (Quantity of X in M1 + M2) / (Quantity of Y in M1 + M2), we get the ratio as

[ ( (A1/A1+B1)*M1) + ( (A2/A2+B2)*M2) ] / [ ( (B1/A1+B1)*M1) + ( (B2/A2+B2)*M2) ] |