Verified Solution • 2025–26 Session moderate

QJEEM20Madf00: JEE Main 2020 Mathematics - Class 11 JEEM20Madf00 Solution: The mean and variance of 7 observations are 8 and 16,…

The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is :

Multiple Choice Options

  • A 2
  • B 3
  • C 1
  • D 4

Step-by-Step Verified Solution

x‾=2+4++10+12+14+x+y7=8\overline x = {{2 + 4 + + 10 + 12 + 14 + x + y} \over 7} = 8

x + y = 14 ....(i)

(σ)2=∑(xi)2n−(∑xin)2{(\sigma )^2} = {{\sum {{{({x_i})}^2}} } \over n} - {\left( {{{\sum {{x_i}} } \over n}} \right)^2}

⇒\Rightarrow 16=4+16+100+144+196+x2+y22−8216 = {{4 + 16 + 100 + 144 + 196 + {x^2} + {y^2}} \over 2} - {8^2}

⇒\Rightarrow 16+64=460+x2+y2716 + 64 = {{460 + {x^2} + {y^2}} \over 7}

⇒\Rightarrow 560 = 460 + x2 + y2

⇒\Rightarrow x2 + y2 = 100 ......(ii)

Clearly by (i) and (ii),

(x + y)2 - 2xy = 100

⇒\Rightarrow (14)2 - 2xy = 100

⇒\Rightarrow 2xy = 96

⇒\Rightarrow xy = 48

Now, |x - y| = (x+y)2−4xy\sqrt {{{\left( {x + y} \right)}^2} - 4xy}

= 196−192\sqrt {196 - 192}

= 2

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