1. The Philosophy of “Equal Distribution” and Deviations
In competitive examinations like the UPSC Civil Services Aptitude Test (CSAT), SSC CGL, and CGPSC, questions on averages have evolved. They are no longer simple computational exercises of adding numbers and dividing them by the count. Instead, modern exams test your conceptual grasp of average as a “balancing point” or the concept of Equal Distribution.
Imagine five individuals with different amounts of money: ₹10, ₹20, ₹30, ₹40, and ₹50. The sum of their money is ₹150. If we pool this money together and distribute it equally among all five, each person receives ₹30. This ₹30 is the average. Thus, the average acts as a leveler. If an entity has a value higher than the average, it possesses a “surplus”; if it has a value lower, it has a “deficit.”
This brings us to the core logical principle of averages, known as the Deviation Method:
“The sum of deviations of all observations from their arithmetic mean is always equal to zero.”
To understand this, let us look at the deviations of our previous values (10, 20, 30, 40, 50) from their average (30):
- Deviation of 10: \(10 – 30 = -20\) (Deficit of 20)
- Deviation of 20: \(20 – 30 = -10\) (Deficit of 10)
- Deviation of 30: \(30 – 30 = 0\) (No deviation)
- Deviation of 40: \(40 – 30 = +10\) (Surplus of 10)
- Deviation of 50: \(50 – 30 = +20\) (Surplus of 20)
If you sum these deviations: \((-20) + (-10) + 0 + (+10) + (+20) = 0\). This principle allows us to solve complex calculation-heavy problems mentally. For instance, if you need to find the average of 1008, 1012, 996, 992, and 1002, you can assume a baseline mean of 1000. The deviations are \(+8, +12, -4, -8, +2\). The sum of these deviations is \(+10\). Distribute this net surplus of 10 equally among the 5 numbers (\(10 \div 5 = +2\)). Add this to your assumed mean: \(1000 + 2 = 1002\). The actual average is 1002.
2. Weighted Average: Combining Groups
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