Let the mean and variance of 8 numbers -10, -7, -1, x, y, 9, 2, 16 be 7/2 and 293/4, respectively. T

Q: Wait, do I really need to solve for $x$ and $y$ individually? That seems like it would take forever and involve a quadratic equation.

That is the exact trap the examiner wants you to fall into! In JEE Advanced, time is your most precious resource. Notice that the question asks for the mean of a set that is defined by the sum and difference of $x$ and $y$. By focusing on the aggregate properties—the sum and the sum of squares—you bypass the need to find $x$ and $y$ individually. It is not just faster; it is mathematically more elegant.

Q: Why did we use the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ instead of the definition $\frac{\sum (x_i - \bar{x})^2}{n}$?

Great observation! While both formulas are mathematically equivalent, the one we used is a 'computational shortcut.' If you use the definition $\frac{\sum (x_i - \bar{x})^2}{n}$, you would have to subtract the mean from every single term, including the variables $x$ and $y$, which creates messy expressions. The formula $\frac{\sum x_i^2}{n} - (\bar{x})^2$ allows you to handle the known constants and the variables $x^2$ and $y^2$ separately, making the algebra much cleaner.

Q: Is it possible that $|x-y|$ could be negative?

No, never! The absolute value operator $| \cdot |$ is defined to return a non-negative value. Geometrically, $|x-y|$ represents the distance between $x$ and $y$ on the number line. Distance is always a positive quantity. If you ever find yourself with a negative value inside an absolute sign, pause and recheck your algebraic steps—you likely missed a square or a sign convention.

Question

Let the mean and variance of 8 numbers $-10, -7, -1, x, y, 9, 2, 16$ be $\frac{7}{2}$ and $\frac{293}{4}$, respectively. Then the mean of 4 numbers $x, y, x + y + 1, |x - y|$ is :

Options:

(A) 11

(B) 12

(C) 10

(D) 9

Sigma Percentile · Accepted Answer

### The Art of Statistical Elegance

Welcome, future engineer. Today, we are going to dissect a problem that looks like a standard statistics question but is actually a masterclass in algebraic efficiency.

In the high-stakes environment of the JEE Advanced, the difference between a top rank and a missed opportunity often comes down to one thing: recognizing when to stop calculating and start thinking.

### Phase 1: The Mean Strategy

We are given a set of eight numbers: $-10, -7, -1, x, y, 9, 2, 16$. We know the mean is $\bar{x} = \frac{7}{2}$.

The definition of the mean is the sum of all observations divided by the total count, $n=8$. So, we write:

$$\frac{-10 - 7 - 1 + x + y + 9 + 2 + 16}{8} = \frac{7}{2}$$

Instead of panicking about the variables $x$ and $y$, let's group the knowns. The sum of the negative numbers is $-18$, and the sum of the positive numbers is $27$.

Thus, $-18 + 27 = 9$. Our equation simplifies beautifully to $\frac{9 + x + y}{8} = \frac{7}{2}$.

Multiplying by $8$, we get $9 + x + y = 28$, which leads us to our first vital piece of information: $x + y = 19$. We have successfully captured the sum of our variables without ever needing to know their individual identities.

### Phase 2: The Variance Challenge

Next, we tackle the variance, $\sigma^2 = \frac{293}{4}$. The most efficient tool here is the computational formula: $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.

Let's calculate the sum of the squares of our known constants: $(-10)^2 + (-7)^2 + (-1)^2 + 9^2 + 2^2 + 16^2 = 100 + 49 + 1 + 81 + 4 + 256 = 491$.

Now, we plug this into our variance equation:

$$\frac{491 + x^2 + y^2}{8} - \left(\frac{7}{2}ight)^2 = \frac{293}{4}$$

This looks daunting, but watch how it collapses. We have $\frac{491 + x^2 + y^2}{8} - \frac{49}{4} = \frac{293}{4}$.

Adding $\frac{49}{4}$ to both sides gives $\frac{491 + x^2 + y^2}{8} = \frac{342}{4}$. Multiplying by $8$ yields $491 + x^2 + y^2 = 684$.

Subtracting $491$, we find $x^2 + y^2 = 193$. We now have the sum and the sum of squares—the two building blocks of any algebraic relationship.

### Phase 3: The Algebraic Bridge

We need the mean of $x, y, x + y + 1, |x - y|$. We already know $x+y=19$ and $x+y+1=20$.

The only missing piece is $|x-y|$. This is where we invoke the power of the identity $(x+y)^2 + (x-y)^2 = 2(x^2+y^2)$.

Substituting our known values:

$$19^2 + (x-y)^2 = 2(193)$$

$$361 + (x-y)^2 = 386$$

$$(x-y)^2 = 25$$

Taking the square root, we find $|x-y| = 5$. The mystery is solved.

### Phase 4: The Final Calculation

The mean of our four target numbers is simply the sum of these numbers divided by $4$:

$$	ext{Mean} = \frac{x + y + (x + y + 1) + |x - y|}{4}$$

Substituting our values:

$$	ext{Mean} = \frac{19 + 20 + 5}{4} = \frac{44}{4} = 11$$

There it is. By refusing to solve for $x$ and $y$ individually, we saved precious minutes and avoided the potential for error.

Remember, in physics and mathematics, the most elegant path is often the one that respects the structure of the problem rather than brute-forcing the variables. Keep this mindset, and you will conquer any problem the JEE throws your way. The final answer is **11**.

Animated Solution for Mathematics - Statistics: Let the mean and variance of 8 numbers $- 10, - 7, - 1, x, y, 9, 2, 16$ be $\frac{7}{2}$ and $\frac{293}{4}$ , respectively. Then the mean of 4 numbers $x, y, x + y + 1, ∣ x - y ∣$ is :

Select Answer:

Visualized Solution

Identify the Data Set

Apply the Mean Formula

Solve for $x + y$

Apply the Variance Formula

Calculate Known Squares

Solve for $x^{2} + y^{2}$

Find $∣ x - y ∣$

Identify the Target Numbers

Calculate the Final Mean

Conclusion and Key Takeaways

The Sigma Insight: Measures of Dispersion

The Art of Statistical Elegance

Phase 1

Phase 2

Phase 3

Phase 4

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