The biggest challenge that we hear from students is not necessarily how to factor algebra questions or solve for the area of a circle, but rather, even being able to figure out the first step in solving a data sufficiency or problem solving question.

Once that first step is determined, the next steps fall into place, and for some test takers, from there it is just a matter of efficiency. But, how really do we figure out that very first step?

In today’s post we’ll start with one huge helpful tip:

• Start by writing out what you know.

Dissecting a word problem, line-by-line, or simplifying an equation often gives us the confidence we need to push the envelope further. Take this question, for example:

There are 8 books in a shelf that consist of 2 paperback books and 6 hardback books. If 4 books are selected at random without replacement, how many different combinations are there that at least one paperback book is selected?

A. 40
B. 45
C. 50
D. 55
E. 60

This is a classic “at least” combination problem that spins many test takers into a frenzy. How in the world are you supposed to figure out such a complex combination problem in two short minutes?

Breaking up the question, piece by piece, can help us get organized to how we are going to tackle this problem.

Total number of books = 8

Total number of paperback books = 2

Total number of hardback books = 6

What is the question being asked?

• To pick 4 books
• To make sure one of those 4 books is paperback

Ah, we feel better already – don’t you? In looking at our set-up, hopefully we’ve realized that there are only two paperback books, so the amount of combination calculations we have to make is ½ of what we probably anticipated at first read.

From there, we should know that there are likely a couple different ways we can sold this problem.

The route most test takers will select is:

Total numbers of combinations that have 2 paperback + total number of combinations that have 1 paperback = the total combinations that have at least one paperback

= 2C2 * 6C2 + 2C1 * 6C3
= 15 + 40
= 55 ways

Super easy, right? An even easier way to setup this problem is to say:

Number of combinations which have at least one paperback is found by finding the total number of combinations minus total number of combinations in which there is no paperback:

= 8C4 – 6C4
= 55 ways

Most “at least” problems are easier found by looking at the Total – None to find the number of combinations.

But, the key thing to recognize here is that by writing out what we know and organizing our information, the right answer becomes much easier to calculate that we projected.

So, starting by writing out what you know is an excellent strategy for “just getting started.” Check-in next where we tackle our next tip – having a “running checklist.”