JEE Main 2022 · 25 July · evening · Single correct · +4 / −1

If the mean deviation about median for the numbers 3, 5, 7, 2k, 12, 16, 21, 24, arranged in the ascending order, is 6 then the median is :

  1. (a) 11.5
  2. (b) 10.5
  3. (c) 12
  4. (d) 11
Answer and solution

Answer (d) 11

Solution

Maths sir · 4 steps

With eight numbers the median sits between 2k and 12, so the deviations split neatly into two groups.

  1. Find the median. There are 8 numbers, so the median is the average of the 4th and 5th values: M = (2k + 12)/2 = k + 6.
  2. Remove the modulus. The four values below M give deviations M - x, and the four values above give x - M. The four copies of M cancel, so the sum is (sum of upper four) minus (sum of lower four).

    Σ |xᵢ - M| = (12 + 16 + 21 + 24) - (3 + 5 + 7 + 2k) = 58 - 2k

  3. Use the given value. Mean deviation = (58 - 2k)/8 = 6, so 58 - 2k = 48 and 2k = 10. The list 3, 5, 7, 10, 12, … is still in ascending order.
  4. So. Median = (10 + 12)/2 = 11. Check: the deviations 8, 6, 4, 1, 1, 5, 10, 13 add to 48, and 48 ÷ 8 = 6. The answer is (d).

The idea to keep. For mean deviation about the median, the sum of |xᵢ - M| is just (sum of upper half) minus (sum of lower half). No casework with the modulus is needed.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 100 times in JEE Main since 2020. This question is from 2022.