Question 1098164
<br>Formally, you can represent this inverse variation with either
{{{y = k/x^2}}}  or  {{{x^2y = k}}}<br>
I think inverse variation problems are easier to solve informally by thinking of the second of these forms as saying that the product of x^2 and y is a constant.  If the product of two numbers is a constant, then if one goes up by a factor of x, the other has to go down by a factor of x.<br>
In your example, the value of x went up by a factor of 8 (4/0.5 = 8); so the value of x^2 went up by a factor of 8^2=64.  That means y has to go down by a factor of 64.  So the new value of y is {{{10/64 = 5/32)}}}