JEE Main 2022 · 27 Jun · Shift 2 · Mathematics Q16

MathematicsSingle correct+4 / -1Question 16 of 90

The mean and variance of the data 4, 5, 6, 6, 7, 8, $x$ , $y$ , where $x<y$ , are 6 and $\dfrac94$ respectively. Then $x^4+y^2$ is equal to

162

320

674

420

Answer and worked solution

Official answer: B

Mean. $36+x+y=48$ , $x+y=12$ .

Variance. $\frac{226+x^2+y^2}8-36=\frac94$ gives $x^2+y^2=80$ , so $xy=32$ : $x=4$ , $y=8$ .

Value. $256+64=320$ .

(B)

JEE Main 2022 · 27 Jun · Shift 2, Mathematics question 16