Solving Mixture Problems

1. Blending Ingredients: The Rule of Alligation

Imagine mixing two types of rice, one cheap and one expensive, to create a blend with a medium price. How do you figure out the final price? This is a classic weighted average problem. The final price is the weighted average of the prices of the two types of rice, where the "weights" are the quantities you mix.

The Rule of Alligation is simply a visual and faster way to solve these weighted average problems.

From Weighted Average to Alligation

For someone with your background in computer science, seeing how a shortcut is derived from first principles can be very helpful. The Rule of Alligation isn't magic; it's just a clever rearrangement of the weighted average formula you already know.

The Alligation Method in Action

The derivation shows us that the ratio of the quantities mixed is inversely proportional to the difference of their prices from the mean price. The "criss-cross" or "butterfly" method provides a visual way to remember and apply this.

Here’s the structure:

Let's apply this to a simple problem.

Example: In what ratio must a grocer mix tea at ₹60/kg and tea at ₹80/kg so that the mixture is worth ₹70/kg?

1. Set up the diagram:

   • Price of Cheaper (C): 60
   • Price of Dearer (D): 80
   • Mean Price (M): 70

2. Apply the criss-cross rule:

   • Find the difference between the Dearer price and Mean price: $D-M=80-70=10$. Write this under the Cheaper price.
   • Find the difference between the Mean price and Cheaper price: $M-C=70-60=10$. Write this under the Dearer price.

   Cheaper(60)       Dearer(80)
        \           /
         \         /
          Mean(70)
         /         \
        /           \
   (80-70)=10      (70-60)=10
   

3. The Resulting Ratio: The ratio of the quantity of Cheaper tea to Dearer tea is 10 : 10, which simplifies to 1 : 1.

This method is incredibly fast once you get the hang of it. You can use it for prices, percentages (like alcohol concentration), or any other measurable attribute.

2. Mixture Replacement

Replacement problems are a different but equally common type. Here, a part of the mixture is removed and replaced by one of the pure components (or a different mixture entirely).

The Core Logic of Replacement

The most crucial concept to remember is this: When you remove a quantity of the mixture, you remove each component in its current proportion.

For example, if you have a 100-litre mixture of milk and water in a 3:2 ratio (60L milk, 40L water) and you remove 10 litres of the mixture, you are not removing 10L of just milk or just water. You are removing:

• Milk: $\frac{3}{5}\times 10L=6L$
• Water: $\frac{2}{5}\times 10L=4L$

This is the most common point of confusion, and mastering this logic is key.

The Replacement Formula (for Repeated, Equal Replacements)

When a problem involves removing a certain quantity and replacing it with the same quantity of another liquid, and this process is repeated multiple times, using a formula is the fastest approach.

If you start with a total quantity $Q$ of a solution containing an initial quantity $I$ of a pure liquid, and in each step you replace a quantity $q$ with another liquid, the final quantity of the pure liquid ($F$) after $n$ operations is:

$F=I\times {\left(1-\frac{q}{Q}\right)}^n$

Note: Often, you start with a pure liquid, so the initial quantity $I$ is the same as the total quantity $Q$. In that case, the formula becomes:

$\text{Final Quantity of original liquid}=Q\times {\left(1-\frac{q}{Q}\right)}^n$

Let's see how to apply this.

A More Complex Replacement Problem

CAT problems often involve an extra twist. Let's look at a typical example where you need to work backward to find the original quantity.

Problem: A vessel contains a mixture of milk and water in the ratio 3:2. If 10 litres of the mixture is removed and replaced with water, the ratio of milk to water becomes 2:3. Find the original quantity of the mixture.

This type of problem is best solved with a clear, step-by-step algebraic approach, tracking the quantity of each component.

Let's see this solved in the resource we just looked at. The algebraic method shown is a reliable way to tackle such questions when the formula doesn't directly apply.

The method in Example 8 is powerful because it relies on fundamentals:

1. Assume the initial quantities based on the ratio (e.g., 3x milk, 2x water).
2. Calculate the amount of each component removed (based on the ratio).
3. Calculate the new quantities after removal and addition.
4. Set up an equation based on the new ratio and solve for x.

Conclusion

In this lesson, we tackled the two main types of mixture problems you'll face in the CAT exam.

Key Takeaways:

• Blending Problems: These are weighted average problems in disguise. The Rule of Alligation is your go-to shortcut for finding the ratio in which components are mixed. Remember the criss-cross method: (Dearer - Mean) and (Mean - Cheaper).
• Replacement Problems: The core logic is that components are removed in their existing proportions.
• For repeated, equal-volume replacements from an initial quantity $Q$, the formula $\text{Final}=\text{Initial}\times (1-q/Q{)}^n$ is the fastest method.
• For more complex scenarios (e.g., finding the initial quantity), a step-by-step algebraic approach where you track each component is the most reliable strategy.

In our next and final lesson for this module, we will expand on the Rule of Alligation and see how it can be applied to a wider range of weighted average problems, not just mixtures. This will solidify your understanding of this versatile and time-saving technique.