JEE Main 2021 · 20 July · morning · Single correct · +4 / −1

The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6 observations are 2, 4, 5 and 7, then the remaining two observations are :

  1. (a) 10, 11
  2. (b) 3, 18
  3. (c) 8, 13
  4. (d) 1, 20
Answer and solution

Answer (a) 10, 11

Solution

Maths sir · 3 steps

The mean fixes the sum of the two numbers, and the variance fixes the sum of their squares. Use both.

  1. Use the mean. The total of all six is 6 × 6.5 = 39. The four known values add to 2 + 4 + 5 + 7 = 18, so the missing two satisfy a + b = 21.

    σ² = (Σ xᵢ²)/n - x̄²

  2. Use the variance. (Σ xᵢ²)/6 = 10.25 + 6.5² = 10.25 + 42.25 = 52.5, so Σ xᵢ² = 315. The known squares add to 4 + 16 + 25 + 49 = 94, so a² + b² = 221.
    Optiona + ba² + b²Needs 21 and 221
    (a) 10, 1121100 + 121 = 221fits
    (b) 3, 18219 + 324 = 333fails
    (c) 8, 132164 + 169 = 233fails
    (d) 1, 20211 + 400 = 401fails
  3. So. From (a+b)² = 441 we get 2ab = 441 - 221 = 220, so ab = 110. The two numbers are 10 and 11, which is option (a).

The idea to keep. Mean gives you Σ x, variance gives you Σ x². With two unknowns, these two equations are exactly enough. Always check both, never just the sum.

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 2021.