Standardized Test Math Mechanic: An Introduction to SAT Prep

SAT

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I am happy to introduce Standardized Test Math Mechanic, a new blog series at Cambridge Coaching dedicated to the process of approaching mathematics problems on standardized tests.

Hearing the words “math on standardized tests” may be enough to send you into a cold sweat.  But it doesn’t need to be that way! 

The good news about the math problems that you will encounter on tests like the SAT, GRE, and GMAT is that the concepts are ones that most of you will have covered already in your high school math classes.  The bad news is – you may have forgotten how to do Algebra!  But with a little practice, I think that you will find solving Geometry problems to be as painless as tying your shoelaces.  

So how do I start preparing for math on the SAT?

To help you get your bearings, over the next few weeks, I will be looking at different types of math problems and discussing methods to tackle them.  For many challenging math problems, it is uncommon that there is one single way to arrive at the solution, so we will explore different approaches to solve the problems.

Are we going to talk about math or actually solve problems?

While we could start with math theories, it is much more fun to start with a problem. 

Here is one that I found in an old SAT test:

If x2/y is an integer, but x/y is not an integer, which of the following could be the values of x and y ?

(A) x = 1, y = 1
(B) x = 3, y = 2
(C) x = 4, y = 2
(D) x = 6, y = 3
(E) x = 9, y = 3

What is the SAT saying?

Before we dive in, let’s review a bit of nomenclature and what the question is asking.  Integers are whole numbers like 1, -3, 5, and 0; 3.2 and 3/2 are not integers.  The question makes a distinction between a number x squared divided by a number y and simply x divided by y

Now let’s solve the problem!

With a question like this, there are at least two methods to approach the answer.  One way is to look at the answer choices and begin testing each choice.  The advantage to this strategy is that it’s fairly straightforward; the disadvantage is that it may be time-consuming and cause you to do a lot of arithmetic and, in the process, make a careless calculation error.

That doesn’t sound like a great strategy.  Do you have other tips?

A second way to approach the problem is using factors.  Factors of a number are positive integers that can be divided evenly into the number – that is, without a remainder.  For example, the factors of 8 are 1, 2, 4, and 8.  If x/y is not an integer, we know that y is not a factor of x.  In answer choices A, C, and E, y is a factor of x, so these choices can be eliminated.  This leaves B and D; B does not satisfy the condition that  is an integer, while D satisfies both conditions listed in the question.  The correct answer is D.  

Tune in next week for the second installment of the Math Mechanic.

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