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JEE Main 202623 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2026 · 23 January · Morning Shift · Mathematics · Statistics

Let the mean and variance of 8 numbers -10,-7,-1, x, y, 9,2,16 be...

Let the mean and variance of 8 numbers −10,−7,−1,x,y,9,2,16-10,-7,-1, x, y, 9,2,16 be 72\frac{7}{2} and 2934\frac{293}{4}, respectively. Then the mean of 4 numbers x,y,x+y+1,∣x−y∣x, y, x+y+1,|x-y| is :

Options

  1. A11
  2. B12
  3. C10
  4. D9

Correct answer

A. 11

Step-by-step solution

Mean =72= \frac{7}{2}: −10−7−1+x+y+9+2+16=28⇒x+y=19-10-7-1+x+y+9+2+16 = 28 \Rightarrow x+y = 19.
Variance =2934= \frac{293}{4}: ∑xi28−494=2934⇒∑xi2=684\frac{\sum x_i^2}{8} - \frac{49}{4} = \frac{293}{4} \Rightarrow \sum x_i^2 = 684.
Known squares: 100+49+1+81+4+256=491100+49+1+81+4+256 = 491, so x2+y2=193x^2+y^2 = 193.
(x−y)2=x2+y2−2xy=193−(361−193)=25⇒∣x−y∣=5(x-y)^2 = x^2+y^2-2xy = 193 - (361-193) = 25 \Rightarrow |x-y| = 5.
x+y+1=20x+y+1 = 20.
Mean of x,y,x+y+1,∣x−y∣=19+20+54=444=11x, y, x+y+1, |x-y| = \frac{19 + 20 + 5}{4} = \frac{44}{4} = 11.
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