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JEE Mains Maths · Statistics

Finding Unknown Observations

Recovering missing observations from a given mean and variance: two unknowns give their sum and the sum of their squares, and a part of the data is found by subtracting known totals.

Why this matters

Nineteen PYQs, fourteen of them multiple choice, and three from 2026. Fourteen give a data set with unknown values, usually a pair a and b, and ask for them or for something built from them; five split the data into a known part and the rest — the first four of five, the values left after removing some, or the last value. Two ideas cover the page.

Concept 1 of 2: Two unknowns from the mean and variance

The mean fixes ∑xi\sum x_i, so it gives a+ba+b. The variance fixes ∑xi2\sum x_i^2, so it gives a2+b2a^2+b^2. From these two, abab and a−ba-b follow at once. Many questions ask for abab, ∣a−b∣|a-b| or a+b+aba+b+ab, so you often never need aa and bb one by one.

Definition

  • Mean: a+b=nxˉ−(sum of known values)a+b=n\bar x-(\text{sum of known values}).
  • Variance: a2+b2=n(σ2+xˉ2)−(sum of known squares)a^2+b^2=n(\sigma^2+\bar x^2)-(\text{sum of known squares}).
  • ab=(a+b)2−(a2+b2)2ab=\frac{(a+b)^2-(a^2+b^2)}{2}.
  • (a−b)2=2(a2+b2)−(a+b)2(a-b)^2=2(a^2+b^2)-(a+b)^2; the order (a>ba>b) picks the sign.

The pair from its sum and sum of squares

(a−b)2=2(a2+b2)−(a+b)2(a-b)^2=2\left(a^2+b^2\right)-(a+b)^2

Worked example

The mean and variance of 3,5,a,b,93,5,a,b,9 are 6 and 4, with a>ba>b. Find aa and bb.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 2 · Q58Moderate

Example 1 · Statistics · Finding Unknown Observations

Let the mean and the variance of seven observations 2,4,α,8,β,12,14,α<β2,4,\alpha,8,\beta,12,14,\alpha<\beta, be 8 and 16 respectively. Then the quadratic equation whose roots are 3α+23\alpha+ 2 and 2β+12\beta+ 1 is :

Keep the square of the mean

The variance gives ∑xi2=n(σ2+xˉ2)\sum x_i^2=n(\sigma^2+\bar x^2), not nσ2n\sigma^2. Dropping the xˉ2\bar x^2 term gives an a2+b2a^2+b^2 that is too small, often negative.

Concept 2 of 2: A part of the data

Totals add. The whole data has a ∑x\sum x and a ∑x2\sum x^2; a part of it takes away its own share of each. So find the two totals of the whole, subtract the known part, and work out the mean and variance of what is left with its own count. The same step run backwards finds the last value, or even the count nn.

Definition

  • Whole: ∑x=nxˉ\sum x=n\bar x, ∑x2=n(σ2+xˉ2)\sum x^2=n(\sigma^2+\bar x^2).
  • The rest: subtract the known part's ∑x\sum x and ∑x2\sum x^2.
  • Mean and variance of the rest use its own count mm.

Totals of the whole

∑xi=nxˉ,∑xi2=n(σ2+xˉ2)\sum x_i=n\bar x,\qquad \sum x_i^2=n\left(\sigma^2+\bar x^2\right)

Worked example

Six observations have mean 5 and variance 4. The value 9 is removed. Find the mean and variance of the other five.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q59Moderate

Example 2 · Statistics · Finding Unknown Observations

The mean and variance of nn observations are 8 and 16, respectively. If the sum of the first ( n−1n- 1 ) observations is 48 and the sum of squares of the first ( n−1n - 1 ) observations is 496 , then the value of n is :

Divide by the new count

After removing values, the mean and variance of the rest use the new count mm, not nn. Only the totals carry over.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Two unknowns from the mean and variance

    The pair from its sum and sum of squares

    (a−b)2=2(a2+b2)−(a+b)2(a-b)^2=2\left(a^2+b^2\right)-(a+b)^2
  • A part of the data

    Totals of the whole

    ∑xi=nxˉ,∑xi2=n(σ2+xˉ2)\sum x_i=n\bar x,\qquad \sum x_i^2=n\left(\sigma^2+\bar x^2\right)

Watch out for (2)

Test yourself on Statistics

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.