Question

when x varies inversly as y² and x=3 when y=4 find the x when y=2
class 8 chapter 11
plss solve this ​

Answer 1

Given,

x varies inversely with y².

x = 3, when y = 4.

To find,

Value of x when y = 2.

Solution,

We can solve this problem simply by following the below process.

It is given here, that x varies inversely with y².

⇒ $x\propto \frac{1}{y^2}$

Now removing the proportional sign. Let the constant of proportionality be k. So,

$x=k\frac{1}{y^2}$

⇒ $x=\frac{k}{y^{2} }$

To find the value of k, we need to use the given values, that is,

when y = 4, x = 3, and put in above equation.

⇒ $3=\frac{k}{4^2}$

⇒ k = 3×16

⇒ k = 48.

Putting this value in the previous equation, we get a relation between x and y.

$x=\frac{48}{y^2}$

Now, we have to find x for the value of y = 2.

So, putting y = 2 in the above equation, we get,

$x = \frac{48}{2^2}$

$x=\frac{48}{4}$

⇒ x = 12.

Therefore, for the given condition, when y = 2, x will be 12.