Alligation Made EasyAlligation is a very strong tool/method that can be used to reduce a lot of calculations while solving questions in GMAT. However, many students have this common doubt that where can they use it and where they cannot. Let us understand this through this article.
Let us start this discussion with a question that is very common in GMAT mixture problems.
Question 1: 10 % acidic solution (by volume) is mixed with the 15 % acidic solution by volume to form a 13% acidic solution (by volume). What is the ratio of the quantity of the 13% solution to the 15% solution in the final solution? Solution:- There are two ways to go about this problem
- First is the orthodox way of taking variables:
- Let us assume the volume of 1st acid be \(x\) ml
- The volume of acid in this = 10% of \(x = \frac{10}{100}\times x=\frac{x}{10}\)
- And the volume of the 2nd acid be \(y\) ml
- The volume of acid in this = 15% of \(y = \frac{15}{100}\times y=\frac{3y}{20}\)
- Total acid mixed \(=\frac{x}{10}+\frac{3y}{20}=\frac{2x+3y}{20}\)
- This \(\frac{2x+3y}{20}\) ml acid happens to be 13% of the total solution which is \((x+y)\) ml
- Thus, we can write \(\frac{\frac{2x+3y}{20}}{x+y}=13\%\)
\(⇒\frac{2x+3y}{20(x+y)}=\frac{13}{100}\)
\(⇒\frac{2x+3y}{x+y}=\frac{13}{5}\)
\(⇒\frac{10x+15y}{x+y}=13\) - Does this look familiar? Yes, this is exactly what we do in the weighted average method
\(⇒10x+15y=13x+13y\)\(\)
\(⇒2y=3x\)\(\)
\(⇒\frac{x}{y}=\frac{2}{3}\) - So, the required answer is 2:3
- The second way is the Alligation Method
- The concentration of the two solutions is 10% and 15%
- Whereas the concentration of the resulting mixture is 13%
- So, we can use the alligation method and say:
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- The ratio of the quantity of 10% acetic solution to the 15% acetic acid = 2∶3
- So, you see how the alligation method made the solution to the problem so easy and quick
Always Alligation?It brings us to the question of that can we use this method in every scenario.
Let us take another mixture problem:
Question 2: Two types of rice costing $60 per kg and $40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice? Solution: - Can we apply alligation to this question?
- Can we assume the price of the mixed rice to be $x per kg and make this diagram?
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- Now does this mean \(x-40=2\) and \(60-x=3?\)
- We can see that we are getting 2 different values of x above
- Thus, we cannot use alligation in this way here
- Well, we can still use these numbers by using proportions here as follows,
- \(\frac{x-40}{60-x}=\frac{2}{3}\)
\(⇒3x-120=120-2x\)
\(⇒5x=240\)
\(⇒x=48\)
- So, the final concentration or the price of the mix will be $48 per kg
- The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?Alligation is generally associated with mixtures of questions
- Now is this method only limited to topics such as mixtures?
- Let us look at a question based on ‘interests’ and see how the method of alligation can be applied there:
Question 3: Mike invested a total of $25,000 in two accounts at simple interests of 8% and 12% respectively. The total interest he earned after a year is $2,600. What fraction of the total money was invested at the higher rate?Solution: - Now, we can get assume the money invested at 8% is x and money invested at 12% is 25000 – x and form an equation and solve it to get our answer
- However, there is a quicker way using alligation. Let us see
- The total interest earned is \($2,600\) which is \(\frac{2600}{25000}=10.4\)% of the total investment
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- Fraction of total money that was invested at a higher rate i.e., 12% \(=\frac{2.4}{2.4+1.6}=\frac{2.4}{4}=\frac{3}{5}\)
- Hence the right answer is \(\frac{3}{3+5}=\frac{3}{8}\)
- From this, we can understand that the alligation method does not only need to be associated with mixture problems but can also be used to solve a lot of other questions where the proportionality works in a straightforward way
Having said that, one must have good clarity on the method and apply this method to different types of questions before trying them in the exam.