x varies inversely to the square of y. Given that y = 5 when x = 6, find the value of x when y = 4.

Correct option is C

Given:
x varies inversely to the square of y.
When y = 5, x = 6.
Find x when y = 4.
Formula Used:
x ∝ 1 / y²
x = k / y²
Solution:
​$$\text{When } y = 5,\, x = 6 \\6 = \frac{k}{5^2} \\6 = \frac{k}{25} \\k = 6 \times 25 = 150 \\[6pt]\text{Now for } y = 4: \\x = \frac{150}{4^2} \\x = \frac{150}{16} \\x = 9.375$$​​