Problems based on mixtures, more commonly known as alligation is a fairly frequent concept tested on the exam. Since a major portion of this article is devoted to mathematics, let us begin with a little bit of English. ‘Alligation’ which means ‘a method of solving mixture problems’ sounds very similar to ‘allegation’ which means ‘a claim or assertion that someone has done something illegal or wrong, typically one made without proof.’ Carefully note the ‘e’ instead of the ‘i’. Words which sound the same but are spelt differently are called ‘Homophones’.

Now back to mathematics,

Your knowledge of this concept can be tested in multiple ways. In this article we are going to discuss two types of problems on mixtures, each with a different difficulty level.

The first type.

This is straightforward and the easiest to solve. You have two mixtures with their ingredients given. They are mixed together to give you a new mixture. Of the three mixtures, you could be asked to find out the composition of any one. An example would be:

A 100 gm mixture is 30% salt. Some salt is added to it to make it into a 50% salt mixture, find the amount of salt added.

Now before we show you how to do it the right way, let us understand where most people tend to go wrong. The first mixture has 30gms of salt, the final mixture apparently has 50gms of salt so there has been an addition of 20gms. This is flawed reasoning.

Let us understand what we are doing here. The original mixture weighed 100 gms and had 30 gms of salt. When I added 20 gms of salt, the quantity of the original mixture became 120 gms and the amount of salt became 50gms. It is very clear that 50gms is not 50% of 120gms.

When I say the solution has 50gms of salt, I am assuming that even after addition of salt, the amount of the mixture has remained the same i.e. 100gms. The amount of the final mixture will be 100 + x where x is the amount of salt that has been added. The total amount of salt that is present in the mixture would be 30gms that was originally present, plus the ‘x’ gms that has been added. so 30 + x is 50% of 100 + x. now if you solve for x, you will arrive at 40.

If you add 40gms to the current mixture, the new mixture weighs 140gms and the amount of salt becomes 30 + 40 = 70. 70gms is 50% of 140gms.

Now let us complicate this for the second type.

Let us say that we have two barrels. One barrel contains 30% salt, the other barrel contains 60% salt. If you mix these in the ratio 2:7, what is the percentage of salt in the resulting mixture?

Now, this is an example where we are combining concepts from both percentages and ratios, inorder to solve these, we need to be able to convert ratios into percents and vice versa.

Let us assume that the remaining part of the mixture is water (you could assume it to be anything!). So, the ratio of salt to water becomes 3:7. For the mixture in the other barrel, it would be 3:2 (6:4 reduced to 3:2).

I have to mix these in the ratio, 2:7 i.e. if I take two litres from the first barrel, I need to take 7 litres from the second barrel.

The first step would be to find the amount of salt in the new mixture.

Contribution from the first barrel would be 2 × (3/10) …..I

Each unit of mixture from the first barrel would be (3 /10)th salt (3 parts salt, 7 parts water so a total of 10 parts). Since we are taking two units, we multiply it by 2.

Contribution from the second barrel would be 7 × (3/5)…..II

Each unit of mixture from the second barrel would be (3/5)th salt. Since we are taking, seven units, we multiply this by 7.

Similarly the amount of water would be 2 × (7/10) …..III and 7 × (2/5) …..IV from barrels 1 and 2 respectively.

The ratio of salt to water would be (I+II)/(III+IV) = (3/5 + 21/5)/(7/5 + 14/5) = 24/21 = 8/7 .

So the percentage of salt in the mixture would be (8/15) × 100 = 53.33%

A more abstract way to solve this would be:

If we mixed both in equal amounts, the percentage of salt in the resulting mix would be 45%, the average of 30 and 60.

Since, we are mixing them in the ratio, 2:7, we divide the range from 30 to 60 into 9 equal parts. Each part would be 30/9 = 3.33. So, two parts would be 6.66. Contribution from barrel two is greater, so the percentage of salt will be closer to 60 as compared to 30, and the exact percentage would be 60 – 6.66 = 53.33%

 

The abstract way is faster but more difficult to understand and apply while the previous method is easier to understand and feels more logical. When you start preparing for the test, stick to the old lengthy methods because they offer clarity. As you get into advanced stages of preparation, you can switch to abstract methods because you shall be more receptive to them once you have mastered the basics.