Maths Q7 – JEE-MAIN 2026

A set of four observations has mean 1 and variance 13. Another set of six observations has mean 2 and variance 1. Then, the variance of all these 10 observations is equal to :

Clue 1 of 3: Understand the definitions

Recall the definitions of mean and variance for a set of observations.

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Step 1: Calculate Sum of Squares for Each Set✦ Active

For the first set of $n_{1} = 4$ observations with mean ${\bar{x}}_{1} = 1$ and variance $σ_{1}^{2} = 13$ , the sum of squares is $\sum x_{i}^{2} = n_{1} (σ_{1}^{2} + {\bar{x}}_{1}^{2})$ . For the second set of $n_{2} = 6$ observations with mean ${\bar{x}}_{2} = 2$ and variance $σ_{2}^{2} = 1$ , the sum of squares is $\sum y_{i}^{2} = n_{2} (σ_{2}^{2} + {\bar{x}}_{2}^{2})$ .

\sum x_{i}^{2} = 4 (13 + 1^{2}) = 4 (14) = 56

\sum y_{i}^{2} = 6 (1 + 2^{2}) = 6 (5) = 30

Step 2: Calculate Combined Mean and Total Sum of Squares○ Expand

The total number of observations is $N = n_{1} + n_{2} = 4 + 6 = 10$ . The total sum of observations is $\sum X = n_{1} {\bar{x}}_{1} + n_{2} {\bar{x}}_{2}$ . The total sum of squares is $\sum X^{2} = \sum x_{i}^{2} + \sum y_{i}^{2}$ . The combined mean is $\bar{X} = \frac{\sum X}{N}$ .

\sum X = (4 \times 1) + (6 \times 2) = 4 + 12 = 16

\bar{X} = \frac{16}{10} = 1.6

\sum X^{2} = 56 + 30 = 86

Step 3: Calculate the Combined Variance○ Expand

The combined variance $σ^{2}$ is given by the formula $σ^{2} = \frac{\sum X^{2}}{N} - (\bar{X})^{2}$ . Substitute the calculated values to find the final variance.

σ^{2} = \frac{86}{10} - (1.6)^{2}

σ^{2} = 8.6 - 2.56

σ^{2} = 6.04

💡 Teacher's Secret Hint

Ensure to square the combined mean correctly before subtracting.

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Maths Question 7 – JEE-MAIN 2026

Clue 1 of 3: Understand the definitions

🧠 Full Solution Path