JEE Main202623 January 2026Morning ShiftMathematicsStatisticsActual

Let the mean and variance of 8 numbers -10,-7,-1, x, y, 9,2,16 be 7 2 and 293 4 , respectively. Then the mean of 4 numbers x, y, x+y+1,|x-y| is :

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Correct answer

A. 11

Step-by-step solution

Mean = 7 2 : -10-7-1+x+y+9+2+16 = 28 x+y = 19 . Variance = 293 4 : x_i^2 8 - 49 4 = 293 4 x_i^2 = 684 . Known squares: 100+49+1+81+4+256 = 491 , so x^2+y^2 = 193 . (x-y)^2 = x^2+y^2-2xy = 193 - (361-193) = 25 |x-y| = 5 . x+y+1 = 20 . Mean of x, y, x+y+1, |x-y| = 19 + 20 + 5 4 = 44 4 = 11 .

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