Statistics – Mean and Variance – JEE Main 23 Jan 2026 Shift 1

Solution:

We are given the mean of 8 numbers is $\frac{7}{2}$.
$$ \frac{-10 – 7 – 1 + x + y + 9 + 2 + 16}{8} = \frac{7}{2} $$

$$ \frac{9 + x + y}{8} = \frac{7}{2} $$

$$ 9 + x + y = 28 \Rightarrow x + y = 19 $$

We are given the variance is $\frac{293}{4}$. Using the formula $\sigma^2 = \frac{\sum x_i^2}{n} – (\bar{x})^2$:
$$ \frac{(-10)^2 + (-7)^2 + (-1)^2 + x^2 + y^2 + 9^2 + 2^2 + 16^2}{8} – \left(\frac{7}{2}\right)^2 = \frac{293}{4} $$

$$ \frac{100 + 49 + 1 + x^2 + y^2 + 81 + 4 + 256}{8} – \frac{49}{4} = \frac{293}{4} $$

$$ \frac{491 + x^2 + y^2}{8} = \frac{293}{4} + \frac{49}{4} $$

$$ \frac{491 + x^2 + y^2}{8} = \frac{342}{4} = \frac{171}{2} $$

$$ 491 + x^2 + y^2 = 684 $$

$$ x^2 + y^2 = 193 $$

Now we have a system of equations:
1) $x + y = 19$
2) $x^2 + y^2 = 193$

We know that $(x+y)^2 = x^2 + y^2 + 2xy$.
$$ (19)^2 = 193 + 2xy $$

$$ 361 = 193 + 2xy \Rightarrow 2xy = 168 \Rightarrow xy = 84 $$

The numbers $x$ and $y$ are roots of the quadratic equation $t^2 – (x+y)t + xy = 0$.
$$ t^2 – 19t + 84 = 0 $$

$$ (t-12)(t-7) = 0 \Rightarrow \{x, y\} = \{12, 7\} $$

We need to find the mean of the 4 numbers: $x, y, x+y+1, |x-y|$.
Substituting the values $x=12, y=7$:
1. $x = 12$
2. $y = 7$
3. $x+y+1 = 12+7+1 = 20$
4. $|x-y| = |12-7| = 5$

$$ \text{Mean} = \frac{12 + 7 + 20 + 5}{4} $$

$$ \text{Mean} = \frac{44}{4} = 11 $$

Ans. (1)