Question

X varies inversely as the square of y. If x = 6 when y = 2, then the value of x when y = 8 will be

Answer 1

According to the question X=1/y^2
when x=6 the value of y =2
so x and y vary with a difference 4
so
the value of x when y = 8 is 12

Answer 2

The value of x when y = 8 is 3/8.

• Given:

x varies inversely as the square of y

x $\propto$ $\frac{1}{y^2}$

∴ x = $\frac{k}{y^2}$

• Now, when x = 6 , y=2.

• Substituting in the equation, we find the value of k

6 = $\frac{k}{2^2}$

k = 24

• Now , the equation becomes x = $\frac{24}{y^2}$

We have to find the value of x when y = 8

• Now,

x = $\frac{24}{8^2}$

x = $\frac{3}{8}$

• The value of x is 3/8.