JEE Main 2023 · 30 January · evening · Single correct · +4 / −1

Let S be the set of all values of a₁ for which the mean deviation about the mean of 100 consecutive positive integers a₁, a₂, a₃, … ., a₁₀₀ is 25 . Then S is :

  1. (a) {9}
  2. (b) φ
  3. (c) {99}
  4. (d) N
Answer and solution

Answer (d) N

Solution

Maths sir · 4 steps

Mean deviation does not care where the numbers start, only how they are spread, so a₁ drops out.

  1. Write the terms. Call a₁ = a. The 100 numbers are a, a+1, …, a+99, so their mean is a + 49.5.
  2. Find the deviations. Subtract the mean from each term. The term a + k has deviation |k - 49.5|, and a cancels completely. The deviations are 49.5, 48.5, …, 0.5 on one side and 0.5, 1.5, …, 49.5 on the other.

    MD = 2(0.5 + 1.5 + … + 49.5)/100 = (2 × 1250)/100 = 25

  3. Add them up. The bracket has 50 terms with average 25, so its sum is 1250. The mean deviation is 25 whatever the value of a.
  4. So. Every positive integer a₁ gives mean deviation 25, so S = ℕ, option (d).

The idea to keep. Adding the same number to every observation shifts the mean by that number too, so mean deviation, variance and standard deviation do not change. Check this before solving for a starting value.

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

Study Statistics

Asked about 100 times in JEE Main since 2020. This question is from 2023.