I am discussing direct and inverse variation on my blog nowadays.
http://www.veritasprep.com/blog/2013/01 ... -directly/
http://www.veritasprep.com/blog/2013/02 ... inversely/

Next post will discuss a similar question using both. I have tried to show every step here to avoid confusion but on the blog, I am going to be more direct.
By the way, the question is not very GMAT-like.

\(x = ky^2\) (x is directly proportional to y^2. k is any constant)
\(xz = k\) (I am assuming that indirectly is actually inversely proportional)

Both together, \(\frac{xz}{y^2} = k\) (Assume z to be constant, you get x is directly proportional to y^2. Assume y to be constant, you get x is inversely proportional to z)

\(\frac{x_1*z_1}{(y_1)^2} = \frac{x_2*z_2}{(y_2)^2}\)
Given that \(z_2 = (\frac{1}{2})z_1\) and that \(x_1\)should be equal to \(x_2\). What is the relation between \(y_1\) and \(y_2\)?

\(\frac{x_1*z_1}{(y_1)^2} = \frac{x_1*(1/2)z_1}{(y_2)^2}\)
\((y_2)^2 = \frac{(y_1)^2}{2}\)
\(y_2 = \frac{y_1}{\sqrt{2}}\)

\(y_2 = \frac{y_1*\sqrt{2}}{2}\)
\(y_2 = 0.7*y_1\)

(A decrease of 30%)

Answer (D)
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Original question from GMAT Prep:



Discussed here: the-rate-of-a-certain-chemical-reaction-is-directly-90119.html (I guess many Gmat Hacks questions are just reworded problems from GMAT Prep).
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Normally i plug numbers to see what happens if certain change applies, i am not sure whether i am doing correct but please comment if anything wrong in my method:

Let say the efficiency index y=10 and investment index=1, so, per question terms, production index x=y^2/x=10^2/1=100; now we change the investment index to half (i assumed that it has been decreased to half since it is not clear in the question) and we have the following: 100=y^2/0.5=(y^2)*2=100 --> y=\sqrt{50} --> equals to about 7, then we can see that efficiency index should be reduced to approximately 30% (from 10 to 7).

I hope that my way of thinking is correct.
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“Production index x is directly proportional to the square of efficiency index y, and indirectly proportional to investment index z” means x = k * y^2/z for some positive value k.

Now we are given that z becomes z/2 and we can let y become ty for some positive value t. So we have x = k * (ty)^2/(z/2). Since we are keeping x the same and recall that x = k * y^2/z, so we have:

k * y^2/z = k * (ty)^2/(z/2)

y^2/z = t^2 * y^2 * 2/z

1 = t^2 * 2

1/2 = t^2

t = √(1/2) ≈ 0.7

Thus we see that the efficiency index y has to become 0.7y, or a 30% decrease, in order to keep the production index x the same.

Answer: D
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