JEE Main 2026 — Jan 23, Shift 1 — Mathematics solution

JEE Main2026Jan 23, Shift 1Mathematics

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 :

A
$11$
B
$9$
C
$10$
D
$12$

Solution

Mean equation: sum of the 8 numbers $= 8 \times \frac{7}{2} = 28$ . The known six sum to $- 10 - 7 - 1 + 9 + 2 + 16 = 9$ , so $x + y = 19$ .

Variance equation. Using $σ^{2} = \frac{\sum x _{i}^{2}}{8} - μ^{2} = \frac{293}{4}$ , we get $\sum x_{i}^{2} = 8 (\frac{293}{4} + \frac{49}{4}) = 684$ . Sum of known squares $= 100 + 49 + 1 + 81 + 4 + 256 = 491$ , so $x^{2} + y^{2} = 193$ .

Solve. With $x + y = 19$ and $x^{2} + y^{2} = 193$ : $2 x y = 361 - 193 = 168$ , so $x y = 84$ . The pair is $(x, y) = (12, 7)$ (taking $x = 12, y = 7$ ).

Required mean. The four numbers are $12, 7, 20, 5$ ; mean $= \frac{44}{4} = 11$ .

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