JEE Main 2026 — Jan 24, Shift 2 — Mathematics solution

JEE Main2026Jan 24, Shift 2Mathematics

Let $X = {x \in N : 1 \leq x \leq 19}$ and for some $a, b \in R$ , $Y = {a x + b : x \in X}$ . If the mean and variance of the elements of Y are 30 and 750, respectively, then the sum of all possible values of $b$ is

A
20
B
80
C
100
D
60

Solution

For a linear transformation $Y = a X + b$ : $μ_{Y} = a μ_{X} + b$ and $σ_{Y}^{2} = a^{2} σ_{X}^{2}$ .

For $X = {1, 2, \dots, 19}$ : $μ_{X} = 10$ and $σ_{X}^{2} = \frac{n ^{2} - 1}{12} = \frac{361 - 1}{12} = 30$ .

From variance: $a^{2} \cdot 30 = 750 \Rightarrow a^{2} = 25 \Rightarrow a = \pm 5$ .

From mean: $30 = 10 a + b$ . So $b = 30 - 50 = - 20$ when $a = 5$ , and $b = 30 + 50 = 80$ when $a = - 5$ .

Sum of values of $b = - 20 + 80 = 60$ .

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