Allegation and mixture tricks play a major role in quantitative aptitude test. It is bit difficult to score marks in competitive exams without knowing the shortcuts related to time and distance. We might have already learned this topic in our lower classes.Even though we have been already taught this topic in our lower classes, we need to learn some more short cuts which are being used to solve the problems in the above topic.

The only thing we have to do is, we need to apply the appropriate short cut and solve the problems in a limited time. This limited time will be one minute or less than one minute in most of the competitive exams.

**Basic Concepts:**

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.

The ratio of the two types of ingredients to be mixed = (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

Students who are preparing to improve their aptitude skills and those who are preparing for this type of competitive test must prepare this topic in order to have better score. Because, today it is bit difficult to score marks in competitive exams without knowing time and distance shortcuts. Whether a person is going to write placement exam to get placed or a student is going to write a competitive exam in order to get admission in university, they must be time and distance shortcuts. This is the reason for why people must study this topic.

Students will learn, how and when they have to apply shortcuts to solve the problems which are related to time and distance. Apart from the regular shortcuts, students can learn some additional tricks in this topic allegation and mixture tricks. Already we are much clear with the four basic operations which we often use in math. They are addition, subtraction, multiplication and division. Even though we are much clear with these four basic operations, we have to be knowing some more stuff to solve the problems which are being asked from the topic time and distance in competitive exams. The stuff which I have mentioned above is nothing but the tricks and shortcuts which need to solve the problems in a very short time.

Short cut is nothing but the easiest way to solve problems related to allegation and misture. In competitive exams, we will have very limited time to solve each problem. Then only we will be able to attend all the questions. If we do problems in competitive exams in perfect manner with all the steps, it will definitely take much time and we may not able to attend the other questions. So we need some other way in which the problems can be solved in a very short time. The way we need to solve the problem quickly is called as shortcut.

**Here,
we are going to have some problems such that how allegation and mixture tricks can be used. You can check your answer online and see step by
step solution.**

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?

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)

(d-m):(m-c) = (10.8-10):(10-9.3) = 0.8:0.7 =8:7

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

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)

(d-m):(m-c) = (10.8-10):(10-9.3) = 0.8:0.7 =8:7

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

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

Let the cost price of 1 ltr pure milk be $1

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

Let "x" be the money we invest for milk in 1 ltr of milk-water mixture

Since the gain is 20%, selling price = x + 20%of x

Selling price = 120% of x = (120/100)x = (6/5)x

But the mixture is sold at the cost price of pure milk

So, we have (6/5)x = 1

x = 5/6

Therefore cost price of the milk in the mixture = $(5/6

Cost price of the water = $0

Rule to find the ratio for producing mixture = (d-m):(m-c)

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

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

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

Let "x" be the money we invest for milk in 1 ltr of milk-water mixture

Since the gain is 20%, selling price = x + 20%of x

Selling price = 120% of x = (120/100)x = (6/5)x

But the mixture is sold at the cost price of pure milk

So, we have (6/5)x = 1

x = 5/6

Therefore cost price of the milk in the mixture = $(5/6

Cost price of the water = $0

Rule to find the ratio for producing mixture = (d-m):(m-c)

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

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

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 both the vessels be mixed to obtain a new mixture in vessel c consisting half milk and half water?

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

Milk in 1 liter of mixture in A = 4/7 liter

Milk in 1 liter of mixture in B = 2/5 liter

Milk in 1 liter of mixture in C = 1/2 liter

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

(cost of water is $0)

Rule to find the ratio for producing mixture = (d-m):(m-c) (d-m):(m-c) = (2/5-1/2):(1/2-4/7) = 1/10:1/14 = 7:5

Hence the required ratio is 7:5

Milk in 1 liter of mixture in A = 4/7 liter

Milk in 1 liter of mixture in B = 2/5 liter

Milk in 1 liter of mixture in C = 1/2 liter

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

(cost of water is $0)

Rule to find the ratio for producing mixture = (d-m):(m-c) (d-m):(m-c) = (2/5-1/2):(1/2-4/7) = 1/10:1/14 = 7:5

Hence the required ratio is 7:5

4. A milk vender has two cans of milk. The first contains 25% water and the rest milk. The second contains 50% water and the rest milk. How many liters of milk should he mix from each of the cans so as to get 12 liters of milk such that that the ratio of water to milk is 3:5?

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

Milk in 1 liter mixture in the 1st can = 3/4 (that is 75%)

Milk in 1 liter mixture in the 2nd can = 1/2 (that is 50%)

Milk in 1 liter mixture in the final mix = 5/8

(from the given ratio w:m = 3:5)

C.P of 1 liter mixture in the 1st can (c) = $3/4

C.P of 1 liter mixture in the 2nd can (d) = $1/2

C.P of 1 liter mixture in the final mix (m) = $5/8

Rule to find the ratio for producing mixture = (d-m):(m-c)

(d-m):(m-c) = 1/2-5/8:5/8-3/4 = 1/8:1/8 = 1:1

The above found ratio 1:1 says that equal quantity of milk should be taken from each of the cans.

Since he wants to get 12 liters of milk, he should take 6 liters of milk from each of the cans.

The correct answer is option (B) (6,6).

Milk in 1 liter mixture in the 1st can = 3/4 (that is 75%)

Milk in 1 liter mixture in the 2nd can = 1/2 (that is 50%)

Milk in 1 liter mixture in the final mix = 5/8

(from the given ratio w:m = 3:5)

C.P of 1 liter mixture in the 1st can (c) = $3/4

C.P of 1 liter mixture in the 2nd can (d) = $1/2

C.P of 1 liter mixture in the final mix (m) = $5/8

Rule to find the ratio for producing mixture = (d-m):(m-c)

(d-m):(m-c) = 1/2-5/8:5/8-3/4 = 1/8:1/8 = 1:1

The above found ratio 1:1 says that equal quantity of milk should be taken from each of the cans.

Since he wants to get 12 liters of milk, he should take 6 liters of milk from each of the cans.

The correct answer is option (B) (6,6).

5. A merchant has 1000 kg of sugar, the part of which he sells at 8% profit and the rest at 18% profit. He gets 14% profit on the whole.The quantity sold at 18% profit is

From the given information, we have

Profit on the first part (c) = 8%

Profit on the second part (d) = 18%

Profit on the whole (mixture) (m) = 14%

Rule to find the ratio for producing mixture = (d-m):(m-c)

(d-m):(m-c) = (18-14):(14-8) = 4:6 = 2:3

Quantity of the 2nd kind = 1000X3/5 = 600 kg

The quantity sold at 18% profit is 600 kg

Profit on the first part (c) = 8%

Profit on the second part (d) = 18%

Profit on the whole (mixture) (m) = 14%

Rule to find the ratio for producing mixture = (d-m):(m-c)

(d-m):(m-c) = (18-14):(14-8) = 4:6 = 2:3

Quantity of the 2nd kind = 1000X3/5 = 600 kg

The quantity sold at 18% profit is 600 kg

6. 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 have to mix them?

From the given information, we have

Cost price of the cheaper (c) = 6.92

Cost price of the dearer (d) = 7.77

Cost price of the mixture = selling price - profit

= 8.80-17.5%of8.8 = 8.80-1.54 = 7.26

Therefore, cost price of the mixture (m) = 7.26

Rule to find the ratio for producing mixture = (d-m):(m-c)

(d-m):(m-c) = (7.77-7.26):(7.26-6.92) = 0.51:0.34 = 51:34

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

Hence the required ratio is 3:2

Cost price of the cheaper (c) = 6.92

Cost price of the dearer (d) = 7.77

Cost price of the mixture = selling price - profit

= 8.80-17.5%of8.8 = 8.80-1.54 = 7.26

Therefore, cost price of the mixture (m) = 7.26

Rule to find the ratio for producing mixture = (d-m):(m-c)

(d-m):(m-c) = (7.77-7.26):(7.26-6.92) = 0.51:0.34 = 51:34

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

Hence the required ratio is 3:2

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