# How to work faster on the GRE.

I'm fixin' to go to grad school next year, so lately I've been studying for the mathematics subject GRE. What I have heard, from MSE and other places, is that the most important thing is to learn to work quickly, since the test is so long. I am not good at this. I've spent the last 2 years studying grad level material in algebra almost exclusively, so calculus/differential equations is not my strong point, and those problems (i.e. half the exam) take me way too long. I realize this is a pretty malleable question, but I was hoping somebody might have general tips on how to get faster at these types of computational problems.

An example of the type of thing that's giving me trouble:

Let $P$ be the tangent plane to the surface $$y^2z-2xz^2+3x^2y=2$$ at the point $Q=(1,1,1)$. Which of the following poitns also lies on $P$?

(a) $(4,5,-3)$

(b) $(6,-4,3)$

(c) $(3,-1,5)$

(d) $(5,3,-2)$

(e) $(-2,4,2)$

I got the expression for the plane, then went through the answers and plugged them in one by one, until finally (e) turned out to be the correct one. This turned out to be exactly how the answer key said to do it, but despite knowing precisely what I was doing going in, the calculation took forever due to messing up arithmetic and general slowness... something like $15$ minutes.

Is there any trick to cutting down on time with computational questions like this, other than not being inept at arithmetic?

• Standardized testing is a scam. Anyhow, good luck on your test! – rschwieb Oct 8 '12 at 12:28

For example, rather than finding the actual equation of the plane and plugging them in, I would have computed the gradient vector, which is $$( -2z^2+6xy, 3x^2+2zy, y^2-4xz)\mid_{(1,1,1)}= (4,5,-3).$$ Any point $P$ that lines on the plane in question must satisfy $(P- (1,1,1))\cdot (4,5,-3)=0$ so subtract 1 from each coordinate and just compute dot products until you find one that is $0.$ You wouldn't even have to compute (a) or (d) properly, you'll notice by signs that the dot product will be positive. It is quite similar to finding the equation of the plane first and plugging them in, but seems to save at least a little bit of time.