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QJEEM23Mae850: JEE Main 2023 Mathematics - Class 11 JEEM23Mae850 Solution: Let the mean and variance of 8 numbers be and respectively. If…

Let the mean and variance of 8 numbers x,y,10,12,6,12,4,8x, y, 10,12,6,12,4,8 be 99 and 9.259.25 respectively. If x > y , then 3x−2y3 x-2 y is equal to _____________.

Step-by-Step Verified Solution

xiamp;(xi−xˉ)amp;(xi−xˉ)2xamp;x−9amp;(x−9)2yamp;y−9amp;(y−9)210amp;1amp;112amp;3amp;96amp;−3amp;912amp;3amp;94amp;−5amp;258amp;−1amp;1x+y+92amp;amp;(x−9)2+(y−9)2+54\begin{array}{|c|c|c|}\hline x_i & (x_i-\bar{x}) & (x_i-\bar{x})^2 \\\hline x & x-9 & (x-9)^2 \\\hline y & y-9 & (y-9)^2 \\\hline 10 & 1 & 1 \\\hline 12 & 3 & 9 \\\hline 6 & -3 & 9 \\\hline 12 & 3 & 9 \\\hline 4 & -5 & 25 \\\hline 8 & -1 & 1 \\\hline x+y+92 & & (x-9)^2+(y-9)^2+54 \\\hline\end{array}

Now, mean (xˉ)=9(\bar{x})=9

amp;⇒x+y+528=9amp;⇒x+y=20\begin{aligned}& \Rightarrow \frac{x+y+52}{8}=9 \\\\& \Rightarrow x+y=20\end{aligned}

Also, variance =9.25=9.25

amp;⇒(x−9)2+(y−9)2+548=9.25amp;⇒x2+y2+81+81−2×9(x+y)=20amp;⇒x2+y2−18×20=−142amp;⇒x2+y2=218amp;⇒x2+(20−x)2=218amp;⇒x2+400+x2−40x=218amp;⇒2x2−40x+182=0amp;⇒x=40±124amp;⇒x=13 or x=7⇒y=7 or y=13amp; But xgt;yamp;∴x=13 and y=7amp; So, 3x−2y=39−14=25\begin{aligned}& \Rightarrow \frac{(x-9)^2+(y-9)^2+54}{8}=9.25 \\\\& \Rightarrow x^2+y^2+81+81-2 \times 9(x+y)=20 \\\\& \Rightarrow x^2+y^2-18 \times 20=-142 \\\\& \Rightarrow x^2+y^2=218 \\\\& \Rightarrow x^2+(20-x)^2=218 \\\\& \Rightarrow x^2+400+x^2-40 x=218 \\\\& \Rightarrow 2 x^2-40 x+182=0 \\\\& \Rightarrow x=\frac{40 \pm 12}{4} \\\\& \Rightarrow x=13 \text { or } x=7 \Rightarrow y=7 \text { or } y=13 \\\\& \text { But } x>y \\\\& \therefore x=13 \text { and } y=7 \\\\& \text { So, } 3 x-2 y=39-14=25\end{aligned}

Concept :

(a) Mean =Σxin=\frac{\Sigma x_i}{n}

(b) Variance =Σ(xi−xˉ)2n=\frac{\Sigma\left(x_i-\bar{x}\right)^2}{n}

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