#1:Check my answer

[Yunie]

(a) y is inversely proportional to x^2. It is known that y = 10 for a particular value of x. Find the value of y when the value of x is halved.

My answer is 20, correct?

If not, can show me the workings, thanks?

[krtxmrtz]

(a) y is inversely proportional to x^2. It is known that y = 10 for a particular value of x. Find the value of y when the value of x is halved.

My answer is 20, correct?

If not, can show me the workings, thanks?

Nope...

You have y = k/x² where k is any constant number. If x is halved:

y(x/2) = k/(x/2)² = 4k/x²

This tells you that when x is halved y is multiplied by 4, consequently the value of y you want is 40.

[eranga262154]

y is inversely proportional to x^2.

So, there should be a constant. krtxmrtz use it as 'k' and he clearly solved it out, normally called as proportional constant. In mathematics by word you can say it as,

Two quantities are proportional, if they vary in a way that one quantities is a constant multiple of the other quantities.

Depend on that, logically solve the problem as follows. When the x goes to halved of the original the constant going to be halved. On you question it is change quadratically, so the constant going to be one-fourth. If it is going to inversely proportional, constant going to be 4, think in other way round. So to get the new value multiply the original value by 4.

When you much familiar with this, so easy to solve these type of problems.