The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80 . Then which one of the following gives possible values of a and b ? - Mathematics MCQ AIEEE 2026 | ExamTest.live

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The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80 . Then which one of the following gives possible values of a and b ?

Solution & Step-by-step Explanation

Mean = \frac{a + b + 8 + 5 + 10}{5} = 6 ⟹ a + b + 23 = 30 ⟹ a + b = 7 .Variance = \frac{\sum x _{i}^{2}}{n} - (mean)^{2} = 6.80 . \frac{a ^{2} + b ^{2} + 8 ^{2} + 5 ^{2} + 1 0 ^{2}}{5} - 6^{2} = 6.80 \frac{a ^{2} + b ^{2} + 64 + 25 + 100}{5} = 42.80 a^{2} + b^{2} + 189 = 214 a^{2} + b^{2} = 25 We have a + b = 7 and a^{2} + b^{2} = 25 .Substituting b = 7 - a : a^{2} + (7 - a)^{2} = 25 ⟹ a^{2} + 49 + a^{2} - 14 a = 25 2 a^{2} - 14 a + 24 = 0 ⟹ a^{2} - 7 a + 12 = 0 (a - 3) (a - 4) = 0 ⟹ a = 3 or a = 4 If a = 3, b = 4 . If a = 4, b = 3 .

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The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80 . Then which one of the following gives possible values of a and b ?

a = 0, b = 7

a = 5, b = 2

a = 1, b = 6

a = 3, b = 4

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