JEE Main 2026 — Apr 4, Shift 1 — Mathematics solution

JEE Main2026Apr 4, Shift 1Mathematics

Suppose that the mean and median of the non-negative numbers $21, 8, 17, a, 51, 103, b, 13, 67$ , $(a > b)$ , are $40$ and $21$ , respectively. If the mean deviation about the median is $26$ , then $2 a$ is equal to :

A
$109$
B
$117$
C
$161$
D
$131$

Solution

Mean condition: $\frac{8 + 13 + 17 + 21 + 51 + 67 + 103 + a + b}{9} = 40 \Rightarrow a + b = 80$ .

The median is the $5$ th order statistic, which equals $21$ . With $a > b$ , ordering forces $b$ to sit among the small values and $a$ among the large, with $21$ in the middle. So $b \leq 17$ and $a \geq 51$ .

Mean deviation about median $21$ :

$\frac{∣8 - 21∣ + ∣13 - 21∣ + ∣ b - 21∣ + ∣17 - 21∣ + 0 + ∣51 - 21∣ + ∣ a - 21∣ + ∣67 - 21∣ + ∣103 - 21∣}{9} = 26.$

$\Rightarrow 13 + 8 + (21 - b) + 4 + 30 + (a - 21) + 46 + 82 = 234.$

$\Rightarrow 183 + (a - b) = 234 \Rightarrow a - b = 51.$

Combined with $a + b = 80$ : $a = \frac{131}{2}$ , so $2 a = 131$ .

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