X is directly proportional to A² and A is inversely proportional to Y. If X is 97 when Y is 6, find X when Y is 2.

Correct option is A

Given:

X $$\propto A^2$$​​

A $$\propto \dfrac{1}{Y}$$​

when Y = 6, X = 97; find X when Y = 2

Formula Used:

Since A $$\propto \dfrac{1}{Y}$$​, A$$= \dfrac{k}{Y}$$​ 

​$$X \propto A^2 \implies X = \dfrac{K}{Y^2}$$​​

Therefore, X Y$$^2 = \text{constant}$$​​

Solution:
At Y = 6: K$$= X Y^2 = 97 \times 6^2 = 97 \times 36 = 3492$$​​
At Y = 2: X$$= \dfrac{K}{Y^2} = \dfrac{3492}{2^2} = \dfrac{3492}{4} = 873$$​​