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← JEE Main 2026 · 4 Apr · Shift 1
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JEE Main 2026 · 4 Apr · Shift 1, Mathematics question 10

Suppose that the mean and median of the non-negative numbers 21, 8, 17, aa, 51, 103, bb, 13, 67, (a>b)(a>b), are 40 and 21, respectively. If the mean deviation about the median is 26, then 2a2a is equal to:

Choose one option
Answer and worked solution

Official answer: D

Mean 40. The total is 360; the known seven numbers sum to 280, so a+b=80a+b=80.

Mean deviation 26. ∑∣x−21∣=234\sum|x-21|=234, and the known numbers contribute 183, so ∣a−21∣+∣b−21∣=51|a-21|+|b-21|=51.

Solve. Both ≥ 21 would give 38. So b<21<ab<21<a: a−b=51a-b=51, a=65.5a=65.5, b=14.5b=14.5. The median is still 21 ✓.

Answer. 2a=1312a=131.

(D)