JEE MAIN - Mathematics (2020 - 7th January Evening Slot - No. 7)
If the mean and variance of eight numbers 3, 7, 9, 12, 13, 20, x and y be 10 and 25 respectively,
then x.y is equal to _______.
Answer
54
Explanation
Mean = $${{3 + 7 + 9 + 12 + 13 + 20 + x + y} \over 8}$$ = 10
16 = x + y ....(1)
Variance ($${\sigma ^2}$$) = 25
$$ \Rightarrow $$ $${{{3^2} + {7^2} + {9^2} + {{12}^2} + {{13}^2} + {{20}^2} + {x^2} + {y^2}} \over 8}$$ - 100 = 25
$$ \Rightarrow $$ 125 × 8 = 9 + 49 + 81 + 144 + 169 + 400 + x2 + y2 - 800
$$ \Rightarrow $$ x2 + y2 = 148
We know, (x + y)2 = x2 + y2 + 2xy
$$ \Rightarrow $$ 256 = 148 + 2xy
$$ \Rightarrow $$ x.y = 54
16 = x + y ....(1)
Variance ($${\sigma ^2}$$) = 25
$$ \Rightarrow $$ $${{{3^2} + {7^2} + {9^2} + {{12}^2} + {{13}^2} + {{20}^2} + {x^2} + {y^2}} \over 8}$$ - 100 = 25
$$ \Rightarrow $$ 125 × 8 = 9 + 49 + 81 + 144 + 169 + 400 + x2 + y2 - 800
$$ \Rightarrow $$ x2 + y2 = 148
We know, (x + y)2 = x2 + y2 + 2xy
$$ \Rightarrow $$ 256 = 148 + 2xy
$$ \Rightarrow $$ x.y = 54
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