Maths Q8 – JEE-MAIN 2025

If the mean and the variance of

6, 4, a, 8, b, 12, 10, 13

are

9

and

9.25

respectively, then

a + b + a b

is equal to :

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

Step 1: Calculate the sum of observations using the mean✦ Active

The given data set is $6, 4, a, 8, b, 12, 10, 13$ . The number of observations is $n = 8$ . The mean is $\bar{x} = 9$ .

\bar{x} = \frac{\sum x_{i}}{n} ⟹ 9 = \frac{6 + 4 + a + 8 + b + 12 + 10 + 13}{8}

⟹ 72 = 53 + a + b ⟹ a + b = 19 (Equation 1)

Step 2: Calculate the sum of squares of observations using the variance○ Expand

The variance is $σ^{2} = 9.25$ . Using the formula for variance:

σ^{2} = \frac{\sum x_{i}^{2}}{n} - (\bar{x})^{2}

9.25 = \frac{6^{2} + 4^{2} + a^{2} + 8^{2} + b^{2} + 12^{2} + 10^{2} + 13^{2}}{8} - 9^{2}

9.25 = \frac{36 + 16 + a^{2} + 64 + b^{2} + 144 + 100 + 169}{8} - 81

9.25 = \frac{529 + a^{2} + b^{2}}{8} - 81

90.25 = \frac{529 + a^{2} + b^{2}}{8}

722 = 529 + a^{2} + b^{2} ⟹ a^{2} + b^{2} = 193 (Equation 2)

Step 3: Find

a b

and the required expression○ Expand

Using the algebraic identity $(a + b)^{2} = a^{2} + b^{2} + 2 a b$ and substituting values from Equation 1 and Equation 2:

(19)^{2} = 193 + 2 a b

361 = 193 + 2 a b

2 a b = 361 - 193 = 168

a b = 84

The required expression is $a + b + a b$ :

a + b + a b = 19 + 84 = 103

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Maths Question 8 – JEE-MAIN 2025