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

Memorize what? I’m not going to tell you yet. Try this problem from the GMATPrep® free practice tests first and see whether you can spot the most efficient solution.

All right, have you got an answer? How satisfied are you with your solution? If you did get an answer but you don’t feel as though you found an elegant solution, take some time to review the problem yourself before you keep reading.

Step 1: Glance Read Jot

Take a quick glance; what have you got? PS. A given equation, xy = 1. A seriously ugly-looking equation. Some fairly “nice” numbers in the answers. Hmm, maybe you should work backwards from the answers?

Jot the given info on the scrap paper.

Step 2: Reflect Organize

Oh, wait. Working backwards isn’t going to work—the answers don’t stand for just a simple variable.

Okay, what’s plan B? Does anything else jump out from the question stem?

Hey, those ugly exponents…there is one way in which they’re kind of nice. They’re both one of the three common special products. In general, when you see a special product, try rewriting the problem usually the other form of the special product.

Step 3: Work

Here’s the original expression again:

Let’s see.

Interesting. I like that for two reasons. First of all, a couple of those terms incorporate xy and the question stem told me that xy = 1, so maybe I’m heading in the right direction. Here’s what I’ve got now:

And that takes me to the second reason I like this: the two sets of exponents look awfully similar now, and they gave me a fraction to start. In general, we’re supposed to try to simplify fractions, and we do that by dividing stuff out.

How else can I write this to try to divide the similar stuff out? Wait, I’ve got it:

The numerator: 

The denominator: 

They’re almost identical! Both of the terms cancel out, as do the terms, leaving me with:

I like that a lot better than the crazy thing they started me with. Okay, how do I deal with this last step?

First, be really careful. Fractions + negative exponents = messy. In order to get rid of the negative exponent, take the reciprocal of the base:

Next, dividing by 1/2 is the same as multiplying by 2:

That multiplies to 16, so the correct answer is (D).

Key Takeaways: Special Products

(1) Your math skills have to be solid. If you don’t know how to manipulate exponents or how to simplify fractions, you’re going to get this problem wrong. If you struggle to remember any of the rules, start building and drilling flash cards. If you know the rules but make careless mistakes as you work, start writing down every step and pausing to think about where you’re going before you go there. Don’t just run through everything without thinking!

(2) You need to memorize the special products and you also need to know when and how to use them. The test writers LOVE to use special products to create a seemingly impossible question with a very elegant solution. Whenever you spot any form of a special product, write the problem down using both the original form and the other form. If you’re not sure which one will lead to the answer, try the other form first, the one they didn’t give you; this is more likely to lead to the correct answer (though not always).

(3) You may not see your way to the end after just the first step. That’s okay. Look for clues that indicate that you may be on the right track, such as xy being part of the other form. If you take a few steps and come up with something totally crazy or ridiculously hard, go back to the beginning and try the other path. Often, though, you’ll find the problem simplifying itself as you get several steps in.

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

A while back, we talked about the 4 GMAT math strategies that everyone needs to master. Today, I’ve got some additional practice for you with regard to one of those strategies: Testing Cases.

Try this GMATPrep® problem:

* ” If xy + z = x(y + z), which of the following must be true?

“(A) x = 0 and z = 0

“(B) x = 1 and y = 1

“(C) y = 1 and z = 0

“(D) x = 1 or y = 0

“(E) x = 1 or z = 0

How did it go?

This question is called a “theory” question: there are just variables, no real numbers, and the answer depends on some characteristic of a category of numbers, not a specific number or set of numbers. Problem solving theory questions also usually ask what must or could be true (or what must not be true). When we have these kinds of questions, we can use theory to solve—but that can get very confusing very quickly. Testing real numbers to “prove” the theory to yourself will make the work easier.

The question stem contains a given equation:

xy + z = x(y + z)

Whenever the problem gives you a complicated equation, make your life easier: try to simplify the equation before you do any more work.

xy + z = x(y + z)

xy + z = xy + xz

z =  xz

Very interesting! The y term subtracts completely out of the equation. What is the significance of that piece of info?

Nothing absolutely has to be true about the variable y. Glance at your answers. You can cross off (B), (C), and (D) right now!

Next, notice something. I stopped at z = xz. I didn’t divide both sides by z. Why?

In general, never divide by a variable unless you know that the variable does not equal zero. Dividing by zero is an “illegal” move in algebra—and it will cause you to lose a possible solution to the equation, increasing your chances of answering the problem incorrectly.

The best way to finish off this problem is to test possible cases. Notice a couple of things about the answers. First, they give you very specific possibilities to test; you don’t even have to come up with your own numbers to try. Second, answer (A) says that both pieces must be true (“and”) while answer (E) says “or.” Keep that in mind while working through the rest of the problem.

z =  xz

Let’s see. z = 0 would make this equation true, so that is one possibility. This shows up in both remaining answers.

If x = 0, then the right-hand side would become 0. In that case, z would also have to be 0 in order for the equation to be true. That matches answer (A).

If x = 1, then it doesn’t matter what z is; the equation will still be true. That matches answer (E).

Wait a second—what’s going on? Both answers can’t be correct.

Be careful about how you test cases. The question asks what MUST be true. Go back to the starting point that worked for both answers: z = 0.

It’s true that, for example, 0 = (3)(0).

Does z always have to equal 0? Can you come up with a case where z does not equal 0 but the equation is still true?

Try 2 = (1)(2). In this case, z = 2 and x = 1, and the equation is true. Here’s the key to the “and” vs. “or” language. If z = 0, then the equation is always 0 = 0, but if not, then x must be 1; in that case, the equation is z = z. In other words, either x = 1 OR z = 0.

The correct answer is (E).

The above reasoning also proves why answer (A) could be true but doesn’t always have to be true. If both variables are 0, then the equation works, but other combinations are also possible, such as z = 2 and x = 1.

Key Takeaways: Test Cases on Theory Problems

(1) If you didn’t simplify the original equation, and so didn’t know that y didn’t matter, then you still could’ve tested real numbers to narrow down the answers, but it would’ve taken longer. Whenever possible, simplify the given information to make your work easier.

(2) Must Be True problems are usually theory problems. Test some real numbers to help yourself understand the theory and knock out answers. Where possible, use the answer choices to help you decide what to test.

(3) Be careful about how you test those cases! On a must be true question, some or all of the wrong answers could be true some of the time; you’ll need to figure out how to test the cases in such a way that you figure out what must be true all the time, not just what could be true.

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