On nearly every GMAT, you’ll see at least one of the “Min/Max” variety of word problems, a category that’s difficult for even the brightest quant minds largely for one major reason: these aren’t your typical word problems, and they don’t lend themselves very well to algebra. They tend to be every bit as “situational” as “mathematical” and in fact are labeled “scenario-driven Min/Max problems” in the Veritas Prep Word Problems lesson. Why? Because they’re almost entirely driven by the situation, including:

**The figures almost always have to be integers.** The problems use situations like “the number of people” or “the number of trees,” a subtle clue that algebra won’t quite work because you’re not using all real numbers, but instead nonnegative integers. But be careful (as you’ll see below).

**The questions ask for a very specific value in a very specific way.** You’ll often see them ask “did at least three” (3 or more means “yes”) or “was the number sold greater than 50″ (50 itself means “no” – to get “yes” it has to be 51 or more, provided you’re dealing with integers).

**The rules of the game often dictate whether repeat numbers are allowed.** Quite often you’ll find a stipulation that “no two could be the same” (but make sure you see that stipulation before you act on it!).

**Some of the information in a Data Sufficiency version of a Min/Max is much more sufficient than it usually appears.** This is largely because of the scenario, numbers, and question stem they’ve carefully crafted to sneak sufficiency past you.

Let’s consider an example so that you can see how one of these works:

Five friends recently visited a famous chocolatier, and collectively purchased a total of 16 pounds of fudge. Did any one friend purchase more than 5 pounds of fudge?

(1) No two friends purchased the same amount of fudge.

(2) The minimum increment in which the chocolatier sells fudge is one pound.

Look at the familiar symptoms of a min/max problem:

*The question stem asks a yes/no question about a very specific value (5 pounds)

*Statement 1 provides the caveat “no two can be the same”

*While the problem itself doesn’t dictate “integers” via the scenario – “pounds of fudge” can certainly come in fractions – Statement 2 comes in to limit the values to integers

Now, if you’re looking at the information from the question stem and statement 1, you could try to set up some algebra:

The given information: a + b + c + d + e = 16

Statement 1: a > b > c > d > e

The question, then: Is a > 5?

You should immediately see that this isn’t sufficient; with nonintegers in play, a could be 15.9 and the other four could add up to 0.1 (“yes”) or they could each be right around the average of 3.2, just a hair off to satisfy the inequality (“no”). But you should also see what makes problems like this tricky with algebra – there are a lot of variables and there’s a lot of inequality. Min/Max problems tend to require a lot more trial and error, and live up to their name because the technique that works best on them is to minimize and maximize particular values to figure out the possible range of the value in question. Eschewing algebra, let’s look at statement 2:

Given Information: 16 total pounds were purchased.

Statement 2: The purchases had to be in integer increments.

The question: Was one of those integers 5 or higher?

Here, to find the maximum value you can minimize the other values. What if four friends didn’t buy anything (0, 0, 0, 0) and the fifth bought all 16 pounds? That’s a resounding “yes”. But they could have split things much more easily – you’d do this by maximizing the smallest value(s). 3, 3, 3, 3, 3 would give you 15, allowing that one final pound to go to the highest making the highest value 4. So there’s your “no” and statement 2 is not sufficient.

When you take the statements together, however, you should see what really makes these problems tick. With algebra it’s still awful:

a + b + c + d + e = 16

a > b > c > d > e

a, b, c, d, and e are integers

Is a > 5?

But with an intent to minimize the highest value (by maximizing the others, sucking as much value away as possible) and maximize the highest value (by minimizing the others to drive all value toward the highest), you have a blueprint for trial and error.

Maximize the highest value / Minimize the others. To make sure you can get a “yes”, minimize the smallest values to see how high the highest can go. That means 0, 1, 2, and 3 – a total of 6 pounds leaving 10 for the highest. It’s easy to get a “yes”.

Minimize the highest value / Maximize the others. Since highest = 5 gives you “no”, see if you can then minimize that highest (5) and maximize the others (4, 3, 2, and 1). But notice that that only gives you a total of 15, and you need to account for 16. And here you cannot give that extra pound to any of the lower values without matching a higher one (add it to 1 and you match 2; add it to 2 and you match 3; etc.). So this guarantees that the highest value is 6 or more, and the answer is sufficient, C.

More importantly, look at the technique – many great mathematical minds hate these problems because the “pure math” algebra is so ugly…but the GMAT loves these because they force you to think logically through a few situations. Since so many of these are Yes/No Data Sufficiency problems, keep in mind that your goals are to “prove insufficiency” looking for both a Yes and a No answer, by:

Minimizing the highest value by maximizing the others

Maximizing the highest value by minimizing the others

Minimizing the lowest value by maximizing the others

Maximizing the lowest value by minimizing the others

Essentially to ______ize one value, do the opposite to the others, and doing so will help you test the possible range. As you do so, make sure you consider:

-Can the values be nonintegers, negative numbers, or 0? (often the scenario dictates that the answer to a few of these is “no”)

-Can values repeat?

Min/Max Scenario problems can be a pain, as they maximize the amount of time you have to spend on them while minimizing your score. But if you know the game, you have an advantage – these problems are all about trial-and-error of Min/Max situations and about taking acute inventory of what is allowable for the values you do try. Play the game correctly, and you’ll be set up for maximal success with minimal (comparative) effort.

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*By Brian Galvin*