“I’m just not a math person” is probably the most common thing GMAT students tell me about themselves on the first day of class. I’m here to tell you the same thing I tell each of those students: Odds are, you’re totally wrong.

Read more

Recently, we took a look at a story problem dealing with ratios, and I finished up by giving you a second problem to test your skills. How did you do?

If you haven’t already, try the GMATPrep® problem below and then we’ll talk about it. Give yourself about 2 minutes. Go! Read more

How did it go last time with the rate problem? I’ve got another story problem for you, but this time we’re going to cover a different math area.

Just a reminder: here’s a link to the first (and long ago) article in this series: making story problems real. When the test gives you a story problem, do what you would do in the real world if your boss asked you a similar question: a back-of-the-envelope calculation to get a “close enough” answer.

If you haven’t yet read the earlier articles, go do that first. Learn how to use this method, then come back here and test your new skills on the problem below.

This is a GMATPrep® problem from the free exams. Give yourself about 2 minutes. Go!

* “Jack and Mark both received hourly wage increases of 6 percent. After the wage increases, Jack’s hourly wage was how many dollars per hour more than Mark’s?

“(1) Before the wage increases, Jack’s hourly wage was $5.00 per hour more than Mark’s.

“(2) Before the wage increases, the ratio of Jack’s hourly wage to Mark’s hourly wage was 4 to 3.”

Data sufficiency! On the one hand, awesome: we don’t have to do all the math. On the other hand, be careful: DS can get quite tricky.

Okay, you and your (colleague, friend, sister…pick a real person!) work together and you both just got hourly wage increases of 6%. (You’re Jack and your friend is Mark.) Now, the two of you are trying to figure out how much more you make.

Hmm. If you both made the same amount before, then a 6% increase would keep you both at the same level, so you’d make $0 more. If you made $100 an hour before, then you’d make $106 now, and if your colleague (I’m going to use my co-worker Whit) made $90 an hour before, then she’d be making…er, that calculation is annoying.

Actually, 6% is pretty annoying to calculate in general. Is there any way around that?

There are two broad ways; see whether you can figure either one out before you keep reading.

First, you could make sure to choose “easy” numbers. For example, if you choose $100 for your wage and half of that, $50 an hour, for Whit’s wage, the calculations become fairly easy. After you calculate the increase for you based on the easier number of $100, you know that her increase is half of yours.

Oh, wait…read statement (1). That approach isn’t going to work, since this choice limits what you can choose, and that’s going to make calculating 6% annoying.

Second, you may be able to substitute in a different percentage. Depending on the details of the problem, the specific percentage may not matter, as long as both hourly wages are increased by the same percentage.

Does that apply in this case? First, the problem asks for a relative amount: the difference in the two wages. It’s not always necessary to know the exact numbers in order to figure out a difference.

Second, the two statements continue down this path: they give relative values but not absolute values. (Yes, $5 is a real value, but it represents the difference in wages, not the actual level of wages.) As a result, you can use any percentage you want. How about 50%? That’s much easier to calculate.

Okay, back to the problem. The wages increase by 50%. They want to know the difference between your rate and Whit’s rate: Y – W = ?

“(1) Before the wage increases, Jack’s hourly wage was $5.00 per hour more than Mark’s.”

Okay, test some real numbers.

Case #1: If your wage was $10, then your new wage would be $10 + $5 = $15. In this case, Whit’s original wage had to have been $10 – $5 = $5 and so her new wage would be $5 + $2.50 = $7.50. The difference between the two new wages is $7.50.

Case #2: If your wage was $25, then your new wage would be $25 + $12.50 = $37.50. Whit’s original wage had to have been $25 – $5 = $20, so her new wage would be $20 + $10 = $30. The difference between the two new wages is…$7.50!

Wait, seriously? I was expecting the answer to be different. How can they be the same?

At this point, you have two choices: you can try one more set of numbers to see what you get or you can try to figure out whether there really is some rule that would make the difference always $7.50 no matter what.

If you try a third case, you will discover that the difference is once again $7.50. It turns out that this statement is sufficient to answer the question. Can you articulate why it must always work?

The question asks for the difference between their new hourly wages. The statement gives you the difference between their old hourly wages. If you increase the two wages by the same percentage, then you are also increasing the difference between the two wages by that exact same percentage. Since the original difference was $5, the new difference is going to be 50% greater: $5 + $2.50 = $7.50.

(Note: this would work exactly the same way if you used the original 6% given in the problem. It would just be a little more annoying to do the math, that’s all.)

Okay, statement (1) is sufficient. Cross off answers BCE and check out statement (2):

“(2) Before the wage increases, the ratio of Jack’s hourly wage to Mark’s hourly wage was 4 to 3.”

Hmm. A ratio. Maybe this one will work, too, since it also gives us something about the difference? Test a couple of cases to see. (You can still use 50% here instead of 6% in order to make the math easier.)

Case #1: If your initial wage was $4, then your new wage would be $4 + $2 = $6. Whit’s initial wage would have been $3, so her new wage would be $3 + $1.5 = $4.50. The difference between the new wages is $1.5.

Case #2: If your initial wage was $8, then your new wage would be $8 + $4 = $12. Whit’s initial wage would have been $6, so her new wage would be $6 + $3 = $9. The difference is now $3!

Statement (2) is not sufficient. The correct answer is (A).

Now, look back over the work for both statements. Are there any takeaways that could get you there faster, without having to test so many cases?

In general, if you have this set-up:

– The starting numbers both increase or decrease by the same percentage, AND

– you know the numerical difference between those two starting numbers

? Then you know that the difference will change by that same percentage. If the numbers go up by 5% each, then the difference also goes up by 5%. If you’re only asked for the difference, that number can be calculated.

If, on the other hand, the starting difference can change, then the new difference will also change. Notice that in the cases for the second statement, the difference between the old wages went from $1 in the first case to $2 in the second. If that difference is not one consistent number, then the new difference also won’t be one consistent number.

Key Takeaways: Make Stories Real

(1) Put yourself in the problem. Plug in some real numbers and test it out. Data Sufficiency problems that don’t offer real numbers for some key part of the problem are great candidates for this technique.

(2) In the problem above, the key to knowing you could test cases was the fact that they kept talking about the hourly wages but they never provided real numbers for those hourly wages. The only real number they provided represented a relative difference between the two numbers; that relative difference, however, didn’t establish what the actual wages were.

* GMATPrep® questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.

In the past, we’ve talked about making story problems real. In other words, when the test gives you a story problem, don’t start making tables and writing equations and figuring out the algebraic solution. Rather, do what you would do in the real world if someone asked you this question: a back-of-the-envelope calculation (involving some math, sure, but not multiple equations with variables).

If you haven’t yet read the article linked in the last paragraph, go do that first. Learn how to use this method, then come back here and test your new skills on the problem below.

This is a GMATPrep® problem from the free exams. Give yourself about 2 minutes. Go!

* “Machines X and Y work at their respective constant rates. How many more hours does it take machine Y, working alone, to fill a production order of a certain size than it takes machine X, working alone?

“(1) Machines X and Y, working together, fill a production order of this size in two-thirds the time that machine X, working alone, does.

“(2) Machine Y, working alone, fills a production order of this size in twice the time that machine X, working alone, does.”

You work in a factory. Your boss just came up to you and asked you this question. What do you do?

In the real world, you’d never whip out a piece of paper and start writing equations. Instead, you’d do something like this:

I need to figure out the difference between how long it takes X alone and how long it takes Y alone.

Okay, statement (1) gives me some info. Hmm, so if machine X takes 1 hour to do the job by itself, then the two machines together would take two-thirds…let’s see, that’s 40 minutes…

Wait, that number is annoying. Let’s say machine X takes 3 hours to do the job alone, so the two machines take 2 hours to do it together.

What next? Oh, right, how long does Y take? If they can do it together in 2 hours, and X takes 3 hours to do the job by itself, then X is doing 2/3 of the job in just 2 hours. So Y has to do the other 1/3 of the job in 2 hours. Read more

A couple of months ago, we talked about what to do when a geometry problem pops up on the screen. Do you remember the basic steps? Try to implement them on the below GMATPrep® problem from the free tests.

* ”In the xy-plane, what is the y-intercept of line L?

“(1) The slope of line L is 3 times its y-intercept
“(2) The x-intercept of line L is – 1/3”

My title (3 Steps to Better Geometry) is doing double-duty. First, here’s the general 3-step process for any quant problem, geometry included:

All geometry problems also have three standard strategies that fit into that process.

First, pick up your pen and start drawing! If they give you a diagram, redraw it on your scrap paper. If they don’t (as in the above problem), draw yourself a diagram anyway. This is part of your Glance-Read-Jot step.

Second, identify the “wanted” element and mark this element on your diagram. You’ll do this as part of the Glance-Read-Jot step, but do it last so that it leads you into the Reflect-Organize stage. Where am I trying to go? How can I get there?

Third, start Working! Infer from the given information. Geometry on the GMAT can be a bit like the proofs that we learned to do in high school. You’re given a couple of pieces of info to start and you have to figure out the 4 or 5 steps that will get you over to the answer, or what you’re trying to “prove.”

Let’s dive into this problem. They’re talking about a coordinate plane, so you know the first step: draw a coordinate plane on your scrap paper. The question indicates that there’s a line L, but you don’t know anything else about it, so you can’t actually draw it. You do know, though, that they want to know the y-intercept. What does that mean?

They want to know where line L crosses the y-axis. What are the possibilities?

Infinite, really. The line could slant up or down or it could be horizontal. In any of those cases, it could cross anywhere. In fact, the line could even be vertical, in which case it would either be right on the y-axis or it wouldn’t cross the y-axis at all. Hmm.
Read more