Suppose a population A has 100 observations 101, 102, ..., 200 and another population B has 100 obse

Q: Wait, why can't I just calculate the variance using the standard formula $\frac{1}{n} \sum (x_i - \bar{x})^2$?

You absolutely can, and you would get the correct answer! But in a JEE Advanced exam, time is your most precious currency. Calculating the mean and then the sum of squares for 100 terms is a massive trap. It is designed to test if you know the *properties* of variance rather than just your ability to do arithmetic. Always look for the shortcut first.

Q: Does the number of observations ($n=100$) matter here?

Great question! In this specific problem, the number of observations is the same for both sets, which makes the comparison straightforward. If the number of observations were different, the variance might still be the same because variance depends on the *spread* of the data, not the count. However, always check if the sets are identical in structure.

Q: What if the problem asked for the ratio of standard deviations instead of variances?

The logic remains identical! Since the variance is the square of the standard deviation ($V = \sigma^2$), if $V_A = V_B$, then $\sigma_A^2 = \sigma_B^2$, which implies $\sigma_A = \sigma_B$. The ratio would still be 1. The beauty of these properties is that they hold true regardless of the specific metric you are looking at.

Question

Suppose a population A has 100 observations 101, 102, ..., 200 and another population B has 100 observations 151, 152, ..., 250. If $V_A$ and $V_B$ represent the variances of the two populations, respectively then $\frac{V_A}{V_B}$ is

Options:

(A) 1

(B) $\frac{9}{4}$

(C) $\frac{4}{9}$

(D) $\frac{2}{3}$

Sigma Percentile · Accepted Answer

### The Trap of Brute Force: A Lesson in Elegance

My dear student, welcome to a problem that serves as a perfect litmus test for the JEE Advanced mindset. When you first look at this question, your instinct—the one honed by years of school-level exams—is to grab your pen, calculate the mean of Population A, calculate the mean of Population B, and then start squaring differences.

Stop. Take a breath. If you start calculating, you have already fallen into the trap.

The JEE Advanced is not a test of your ability to perform arithmetic; it is a test of your ability to see the underlying structure of the universe. Let us peel back the layers of this problem together.

### Phase 1: Visualizing the Data

Imagine you are standing on a long, infinite number line. Population A is a block of $100$ soldiers standing in a line, starting at $101$ and ending at $200$. They are shoulder-to-shoulder, a solid, unbroken chain of integers.

Now, look at Population B. They are also $100$ soldiers, but they start at $151$ and end at $250$.

If you were to take a photograph of Population A and then a photograph of Population B, what would you notice? You would see that the 'shape' of the crowd is identical. The distance between the first person and the last person is the same, and the density of the crowd is the same. The only difference is that the entire group has marched forward by exactly $50$ units.

### Phase 2: The Definition of Variance

Now, let us talk about the soul of this problem: Variance. What is variance? It is not just a formula involving $\sum (x_i - \bar{x})^2$.

Conceptually, variance is a measure of the 'spread' or 'dispersion' of data. It asks: 'How far, on average, are these points from their own center?'

If I take a rigid object—like a ruler—and I slide it across the table, does the length of the ruler change? Of course not. The ruler is the same object; it has just changed its position. Variance is exactly like that ruler. It measures the length of the data set, not its location.

### Phase 3: The Mathematical Insight

Let us formalize this. Let $x_i$ be an observation in Population A, taking values from $101$ to $200$. Now, let $y_i$ be an observation in Population B, taking values from $151$ to $250$.

Notice the relationship: $151 = 101 + 50$, $152 = 102 + 50$, and so on. Mathematically, we can write this as $y_i = x_i + 50$.

This is a classic 'Change of Origin' scenario. There is a fundamental theorem in statistics that states: if you add a constant $c$ to every observation in a dataset, the variance remains unchanged.

Why? Because variance is defined as:

$$Var(X) = E[(X - E[X])^2]$$

If we replace $X$ with $X + c$, the new mean becomes $E[X + c] = E[X] + c$. When we calculate the new variance, the term $(X + c) - (E[X] + c)$ simplifies to $(X - E[X])$.

The constant $c$ cancels out perfectly! It vanishes into thin air.

### Phase 4: The Elegant Conclusion

We have established that $V_B = Var(Y) = Var(X + 50)$. Because of the property we just discussed, $Var(X + 50) = Var(X) = V_A$.

Therefore, $V_A = V_B$. When the question asks for the ratio $\frac{V_A}{V_B}$, we are simply asking for the ratio of a number to itself.

The answer is **$1$**.

Do you see the beauty here? We did not need to calculate a single square. We did not need to find a single mean. We simply understood the geometry of the data.

This is the hallmark of a true physicist and mathematician: finding the path of least resistance by understanding the fundamental laws of the system. Keep this perspective in your toolkit. Whenever you see a problem involving variance, ask yourself: 'Is this a shift, or is this a scale?'

If it is just a shift, the variance is invariant. You have mastered this concept. Now, go forth and apply this intuition to more complex problems.

Animated Solution for Mathematics - Statistics: Suppose a population A has 100 observations 101, 102, ..., 200 and another population B has 100 observations 151, 152, ..., 250. If $V_{A}$ and $V_{B}$ represent the variances of the two populations, respectively then $\frac{V _{A}}{V _{B}}$ is

Select Answer:

Visualized Solution

Population $A$

Population $B$

Concept of Variance

Visualizing the Spread

Mapping $A$ to $B$

Change of Origin Property

Equating $V_{A}$ and $V_{B}$

Calculating $\frac{V _{A}}{V _{B}}$

The Sigma Insight: Measures of Dispersion

The Trap of Brute Force

Phase 1

Phase 2

Phase 3

Phase 4

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