Weighted Averages in Mixtures and Grouped Data

Hello! In our previous lesson, we mastered calculating simple averages and using shortcuts like the deviation method to handle problems where elements are added or removed. This foundational skill is crucial for today's topic.

This lesson builds directly on that knowledge. We'll explore weighted averages, a concept that is essential for your SSC CGL preparation. The learning outcome for this session is to apply the weighted average to solve problems on mixtures and grouped data. We'll see that a weighted average is what you use when some items in a set are more "important" or frequent than others. This idea frequently appears in questions about combining groups with different stats or mixing ingredients with different prices.

1. Beyond Simple Averages: The Concept of "Weight"

In the last lesson, every student or item we considered had an equal impact on the average. But what happens when that's not the case?

Imagine a shopkeeper mixes two types of rice. One is cheap, and the other is expensive. If he mixes a large amount of the cheap rice with a small amount of the expensive one, the final price should be closer to the cheap price, right? A simple average of the two prices wouldn't be accurate. This is where the "weight"—in this case, the quantity of each type of rice—comes into play.

Let's look at a classic example:

This image illustrates how to calculate the net cost of a mixture. Notice that the quantities (2kg and 3kg) act as "weights" for their respective prices (Rs. 10 and Rs. 40).

The calculation shown is:

\frac{( 2 \times 10 ) + ( 3 \times 40 )}{2 + 3} = \frac{20 + 120}{5} = \frac{140}{5} = 28

The resulting price per kg is Rs. 28. A simple average would have been $(10 + 40) /2 = 25$ , which is incorrect because it ignores the different quantities being mixed.

This leads us to the formal definition of a weighted average.

2. The Weighted Average Formula

The weighted average is calculated by multiplying each value by its corresponding weight, summing these products, and then dividing by the sum of the weights.

The general formula is:

Weighted Average = \frac{w _{1} x _{1} + w _{2} x _{2} + \dots + w _{n} x _{n}}{w _{1} + w _{2} + \dots + w _{n}} = \frac{\sum w _{i} x _{i}}{\sum w _{i}}

Here:

$x_{i}$ represents the individual values (like price, score, age).
$w_{i}$ represents the corresponding weights (like quantity, frequency, number of students).

This concept is explained very clearly from the basics in the following video. It covers several real-world examples, such as calculating average test scores, GPA, and solving a mixture problem.

How To Find The Weighted Mean and Weighted Average In Statistics

This video from The Organic Chemistry Tutor provides a thorough introduction to the weighted average formula and its applications.

Please watch the following segments: The basic formula: This part introduces the formula using the frequency of numbers as weights. Class test scores: A classic example of finding the average score for a class made of different groups of students. This is a common "grouped data" problem. Final semester grade: This shows how your final school/college grade is a weighted average, which is a great real-world analogy. Mixture problem: This directly applies the concept to an antifreeze mixture, explaining the intuition that the final average will be closer to the component with more "weight".

3. Shortcut for Exam Problems: The Deviation Method for Groups

While the formula works perfectly, for competitive exams like the SSC CGL, we need faster methods. The good news is that the deviation method we learned in the last lesson can be extended to handle weighted averages. This is a powerful shortcut.

The instructor in the next video (speaking in Hindi, as is common in SSC coaching) brilliantly explains how to apply this method to grouped data. The key trick he introduces is to use the ratio of the weights to simplify the numbers.

Average (औसत) के Basic Concepts + 3 IMP Types 🔥 by Aditya Ranjan Sir Maths @rankersgurukullive

This video by Aditya Ranjan Sir demonstrates the deviation shortcut for weighted average problems, a crucial technique for exam speed.

Watch these two sections to learn the shortcut: Applying deviation to groups: The instructor shows how to calculate the weighted average of groups by assuming a "base" average and then calculating the weighted deviations. He compares the long formula method to this fast shortcut. Using ratios to simplify: This is the most important part. Watch how he solves problems by first converting the number of items in each group into a simple ratio (e.g., 20 matches and 25 matches become a 4:5 ratio). This makes the multiplication much easier. The steps are: Simplify the weights (quantities) into their simplest ratio. Assume a convenient "base" average (usually one of the given averages). For each group, calculate: (Group Average - Base Average) × (Group's Ratio Weight). Sum these results to get the Total Deviation. Calculate the Correction: Total Deviation / (Sum of Ratio Weights). The Final Average is Base Average + Correction.

Let's apply this to an example from the nerdsjobportal article.

Problem: The average marks of 21 students in a class is 42. If 15 students with an average marks of 78 join them, find the average marks of all students.

Weights: 21 students and 15 students.
Ratio of Weights: The ratio of 21 to 15 is $7 : 5$ .
Base Average: Let's assume the base average is 42.
Deviations:
- Group 1 (7 parts): Average is 42. Deviation from base is $42 - 42 = 0$ . Total deviation = $0 \times 7 = 0$ .
- Group 2 (5 parts): Average is 78. Deviation from base is $78 - 42 = + 36$ . Total deviation = $+ 36 \times 5 = + 180$ .
Total Deviation = $0 + 180 = + 180$ .
Sum of Ratio Weights = $7 + 5 = 12$ .
Correction = $+ 180/12 = + 15$ .
Final Average = Base Average + Correction = $42 + 15 = 57$ .

This method avoids large multiplications like $21 \times 42$ and $15 \times 78$ , making it much faster and less error-prone.

4. Advanced Application: The Reverse Problem

A common twist in SSC exams is the "reverse" problem, where the overall weighted average is given, and you need to find a missing value for one of the subgroups.

The underlying principle here is that for the true average, the net weighted deviation must be zero. Think of it like a balanced scale or "तराजू" as the instructor in the video calls it. If one side goes down (negative deviation), the other must go up (positive deviation) by the same amount to stay balanced.

Average (औसत) के Basic Concepts + 3 IMP Types 🔥 by Aditya Ranjan Sir Maths @rankersgurukullive

This segment explains how to solve "reverse" weighted average problems where you need to find a missing value.

Watch from this section. The instructor explains that the sum of all weighted deviations from the true average must be zero. If you calculate the deviation for the known groups and find a total of, say, -42, the unknown group(s) must contribute a total deviation of +42 to balance it out.

This "balancing" technique is extremely effective for solving complex problems quickly.

5. Connecting to Mixtures and Alligation

As we've seen, problems involving mixtures are a direct application of weighted averages. The concept you may have heard of as "Alligation Medial" is simply the process of calculating a weighted average.

Mixture and Alligation, Short Tricks, Concepts, and Usage

This article from Oliveboard helps connect the language of "Mixtures and Alligation" to what we've just learned.

Read the sections titled "Terms Used in Mixture and Alligation Questions" and "Alligation Medial". Pay attention to the definition of 'Mean Price / Mean Value (M)' and notice how the article explicitly states that Alligation Medial works just like a weighted average. This confirms that the methods we've learned today are the foundation for mixture problems.

Conclusion

Today, we've built a strong understanding of weighted averages, moving from the basic formula to the powerful deviation shortcut essential for your exam.

Key Takeaways:

A weighted average is used when different items or groups contribute differently to the overall average. The formula is $\frac{\sum w _{i} x _{i}}{\sum w _{i}}$ .
Grouped data problems (e.g., averages of boys and girls) and mixture problems (e.g., mixing rice of different prices) are both solved using weighted averages.
The deviation method is a high-speed shortcut. By assuming a base average and using the ratio of weights, you can avoid large calculations.
For the true average, the net weighted deviation is always zero. This "balancing" principle is the key to solving reverse problems where a value is missing.

In our next lesson, we will introduce a visual technique derived from weighted averages: the Alligation Cross Method. This method is a fantastic shortcut for a specific type of mixture problem: when you know the individual values and the final mean value, and you need to find the ratio in which the items were mixed. You'll see it's just a different way of looking at the concepts we learned today.