### The Importance of Getting to No on the GMAT — Part 2

Last time, we talked about how crucial it is to develop the instinct to go for the “No” when taking the GMAT. If you haven’t read the first installment, do so right now, then come back here to learn more.

I left you with this GMATPrep® problem from the free exams.

“*If 0 <

r< 1 <s< 2, which of the following must be less than 1?“I.

“II.

rs“III.

s–r“(A) I only

“(B) II only

“(C) III only

“(D) I and II

“(E) I and III”

Let’s talk about it now!

### Everything you need to know about the New Official Guides, Part 3

I have now done every last one of the new quant problems in both new books—and there are some really neat ones! I’ve also got some interesting observations for you. (If you haven’t yet read my earlier installments, start here.)

In this installment, I’ll discuss my overall conclusions for quant and I’ll also give you all of the problem numbers for the new problems in both the big OG and the smaller quant-only OG.

## What’s new in Quant?

Now that I’ve seen everything, I’ve been able to spot some trends across all of the added and dropped questions. For example, across both The Official Guide for GMAT® Review (aka the big book) and The Official Guide for GMAT® Quant Review (aka quant-only or the quant supplement), Linear Equation problems dropped by a count of 13. This is the differential: new questions minus dropped questions.

That’s a pretty big number; the next closest categories, Inequalities and Rates & Work, dropped by 5 questions each. I’m not convinced that a drop of 5 is at all significant, but I decided that was a safe place to stop the “Hmm, that’s interesting!” count.

Now, a caveat: there are sometimes judgment calls to make in classifying problems. Certain problems cross multiple content areas, so we do our best to pick the topic area that is most essential in solving that problem. But that 13 still stands out.

The biggest jump came from Formulas, with 10 added questions across both sources. This category includes sequences and functions; just straight translation or linear equations would go into those respective categories, not formulas. Positive & Negative questions jumped by 7, weighted average jumped by 6, and coordinate plane jumped by 5.

Given that Linear Equations dropped and Formulas jumped, could it be the case that they are going after somewhat more complex algebra now? That’s certainly possible. I didn’t feel as though the new formula questions were super hard though. It felt more as though they were testing whether you could *follow directions*. If I give you a weird formula with specific definitions and instructions, can you interpret correctly and manipulate accordingly?

If you think about it, work is a lot more like this than “Oh, here are two linear equations; can you solve for *x*?” So it makes sense that they would want to emphasize questions of a more practical nature.

### GMAT Problem Solving Strategy: Test Cases

If you’re going to do a great job on the GMAT, then you’ve got to know how to Test Cases. This strategy will help you on countless quant problems.

This technique is especially useful for Data Sufficiency problems, but you can also use it on some Problem Solving problems, like the GMATPrep® problem below. Give yourself about 2 minutes. Go!

* “For which of the following functions *f* is *f*(*x*) = *f*(1 – *x*) for all *x*?

(A) | f(x) = 1 – x |

(B) | f(x) = 1 – x^{2} |

(C) | f(x) = x^{2} – (1 – x)^{2} |

(D) | f(x) = x^{2}(1 – x)^{2} |

(E) | f(x) = x / (1 – x)” |

Testing Cases is mostly what it sounds like: you will test various possible scenarios in order to narrow down the answer choices until you get to the one right answer. What’s the common characteristic that signals you can use this technique on problem solving?

The most common language will be something like “Which of the following must be true?” (or “could be true”).

The above problem doesn’t have that language, but it does have a variation: you need to find the answer choice for which the given equation is true “for all *x*,” which is the equivalent of asking for which answer choice the given equation is always, or must be, true.

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### Tackling Max/Min Statistics on the GMAT (part 3)

Welcome to our third and final installment dedicated to those pesky maximize / minimize quant problems. If you haven’t yet reviewed the earlier installments, start with part 1 and work your way back up to this post.

I’d originally intended to do just a two-part series, but I found another GMATPrep® problem (from the free tests) covering this topic, so here you go:

“A set of 15 different integers has a median of 25 and a range of 25. What is the greatest possible integer that could be in this set?

“(A) 32

“(B) 37

“(C) 40

“(D) 43

“(E) 50”

Here’s the general process for answering quant questions—a process designed to make sure that you *understand* what’s going on and come up with the best *plan* before you dive in and *solve*:

Fifteen integers…that’s a little annoying because I don’t literally want to draw 15 blanks for 15 numbers. How can I shortcut this while still making sure that I’m not missing anything or causing myself to make a careless mistake?

Hmm. I could just work backwards: start from the answers and see what works. In this case, I’d want to start with answer (E), 50, since the problem asks for the greatest possible integer.

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### Tackling Max/Min Statistics on the GMAT (Part 2)

Last time, we discussed two GMATPrep® problems that simultaneously tested statistics and the concept of maximizing or minimizing a value. The GMAT could ask you to maximize or minimize just about anything, so the latter skill crosses many topics. Learn how to handle the nuances on these statistics problems and you’ll learn how to handle any max/min problem they might throw at you.

Feel comfortable with the two problems from the first part of this article? Then let’s kick it up a notch! The problem below was written by us (Manhattan Prep) and it’s complicated—possibly harder than anything you’ll see on the real GMAT. This problem, then, is for those who are looking for a really high quant score—or who subscribe to the philosophy that mastery includes trying stuff that’s harder than what you might see on the real test, so that you’re ready for anything.

Ready? Here you go:

“Both the average (arithmetic mean) and the median of a set of 7 numbers equal 20. If the smallest number in the set is 5 less than half the largest number, what is the largest possible number in the set?

“(A) 40

“(B) 38

“(C) 33

“(D) 32

“(E) 30”

Out of the letters A through E, which one is your favorite?

You may be thinking, “Huh? What a weird question. I don’t have a favorite.”

I don’t have one in the real world either, but I do for the GMAT, and you should, too. When you get stuck, you’re going to need to be able to let go, guess, and move on. If you haven’t been able to narrow down the answers at all, then you’ll have to make a random guess—in which case, you want to have your favorite letter ready to go.

If you have to think about what your favorite letter is, then you don’t have one yet. Pick it right now.

I’m serious. I’m not going to continue until you pick your favorite letter. Got it?

From now on, when you realize that you’re lost and you need to let go, pick your favorite letter *immediately* and move on. Don’t even think about it.

Read more

### Tackling Max/Min Statistics on the GMAT (Part 1)

Blast from the past! I first discussed the problems in this series way back in 2009. I’m reviving the series now because too many people just aren’t comfortable handling the weird maximize / minimize problem variations that the GMAT sometimes tosses at us.

In this installment, we’re going to tackle two GMATPrep® questions. Next time, I’ll give you a super hard one from our own archives—just to see whether you learned the material as well as you thought you did.

Here’s your first GMATPrep problem. Go for it!

“*Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight, in kilograms, of the lightest box?

“(A) 1

“(B) 2

“(C) 3

“(D) 4

“(E) 5”

When you see the word *maximum *(or a synonym), sit up and take notice. This one word is going to be the determining factor in setting up this problem efficiently right from the beginning. (The word *minimum* or a synonym would also apply.)

When you’re asked to maximize (or minimize) one thing, you are going to have one or more decision points throughout the problem in which you are going to have to maximize or minimize some other variables. Good decisions at these points will ultimately lead to the desired maximum (or minimum) quantity.

This time, they want to maximize the lightest box. Step back from the problem a sec and picture three boxes sitting in front of you. You’re about to ship them off to a friend. Wrap your head around the dilemma: if you want to maximize the *lightest* box, what should you do to the other two boxes?

Note also that the problem provides some constraints. There are three boxes and the median weight is 9 kg. No variability there: the middle box must weigh 9 kg.

The three items also have an average weight of 7. The total weight, then, must be (7)(3) = 21 kg.

Subtract the middle box from the total to get the combined weight of the heaviest and lightest boxes: 21 – 9 = 12 kg.

The heaviest box has to be equal to or greater than 9 (because it is to the right of the median). Likewise, the lightest box has to be equal to or *smaller* than 9. In order to maximize the weight of the lightest box, what should you do to the heaviest box?

Minimize the weight of the heaviest box in order to maximize the weight of the lightest box. The smallest possible weight for the heaviest box is 9.

If the heaviest box is minimized to 9, and the heaviest and lightest must add up to 12, then the maximum weight for the lightest box is 3.

The correct answer is (C).

Make sense? If you’ve got it, try this harder GMATPrep problem. Set your timer for 2 minutes!

“*A certain city with a population of 132,000 is to be divided into 11 voting districts, and no district is to have a population that is more than 10 percent greater than the population of any other district. What is the minimum possible population that the least populated district could have?

“(A) 10,700

“(B) 10,800

“(C) 10,900

“(D) 11,000

“(E) 11,100”

Hmm. There are 11 voting districts, each with some number of people. We’re asked to find the *minimum* possible population in the *least* populated district—that is, the smallest population that any one district could possibly have.

Let’s say that District 1 has the minimum population. Because all 11 districts have to add up to 132,000 people, you’d need to *maximize* the population in Districts 2 through 10. How? Now, you need more information from the problem:

“no district is to have a population that is *more than 10 percent greater* than the population of any other district”

So, if the smallest district has 100 people, then the largest district could have up to 10% more, or 110 people, but it can’t have any more than that. If the smallest district has 500 people, then the largest district could have up to 550 people but that’s it.

How can you use that to figure out how to split up the 132,000 people?

In the given problem, the number of people in the smallest district is unknown, so let’s call that *x*. If the smallest district is *x*, then calculate 10% and add that figure to *x*: *x* + 0.1*x* = 1.1*x*. The largest district could be 1.1*x* but can’t be any larger than that.

Since you need to maximize the 10 remaining districts, set all 10 districts equal to 1.1*x*. As a result, there are (1.1*x*)(10) = 11*x* people in the 10 maximized districts (Districts 2 through 10), as well as the original *x *people in the minimized district (District 1).

The problem indicated that all 11 districts add up to 132,000, so write that out mathematically:

11*x* + *x* = 132,000

12*x* = 132,000

*x* = 11,000

The correct answer is (D).

Practice this process with any max/min problems you’ve seen recently and join me next time, when we’ll tackle a super hard problem.

### Key Takeaways for Max/Min Problems:

(1) Figure out what variables are “in play”: what can you manipulate in the problem? Some of those variables will need to be maximized and some minimized in order to get to the desired answer. Figure out which is which at each step along the way.

(2) Did you make a mistake—maximize when you should have minimized or vice versa? Go through the logic again, step by step, to figure out where you were led astray and why you should have done the opposite of what you did. (This is a good process in general whenever you make a mistake: figure out why you made the mistake you made, as well as how to do the work correctly next time.)

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

### GMAT Prep Story Problem: Make It Real Part 2

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.

### GMAT Prep Story Problem: Make It Real

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*

### GMAT Quant: Reflect before you Work

Stop! Before you dive in and start calculating on a math problem, reflect for a moment. How can you set up the work to minimize the number of annoying calculations?

Try the below Percent problem from the free question set that comes with your GMATPrep® software. The problem itself isn’t super hard but the calculations can become time-consuming. If you find the problem easy, don’t dismiss it. Instead, ask yourself: how can you get to the answer with an absolute minimum of annoying calculations?

District |
Number of Votes |
Percent of Votes for Candidate P |
Percent of Votes for Candidate Q |

1 |
800 |
60 |
40 |

2 |
1,000 |
50 |
50 |

3 |
1,500 |
50 |
50 |

4 |
1,800 |
40 |
60 |

5 |
1,200 |
30 |
70 |

* ” The table above shows the results of a recent school board election in which the candidate with the higher total number of votes from the five districts was declared the winner. Which district had the greatest number of votes for the winner?

“(A) 1

“(B) 2

“(C) 3

“(D) 4

“(E) 5”

Ugh. We have to figure out what they’re talking about in the first place!

The first sentence of the problem describes the table. It shows 5 different districts with a number of votes, a percentage of votes for one candidate and a percentage of votes for a different candidate.

Hmm. So there were two candidates, P and Q, and the one who won the election received the most votes *overall*. The problem doesn’t say who that was. I could calculate that from the given data, but I’m not going to do so now! I’m only going to do that if I have to.

Let’s see. The problem then asks which district had the greatest number of votes for the winner. Ugh. I am going to have to figure out whether P or Q won. Let your annoyance guide you: is there a way to tell who won without actually calculating all the votes?

### The 4 Math Strategies Everyone Must Master, Part 1

We need to know a lot of different facts, rules, formulas, and techniques for the quant portion of the test, but there are 4 math strategies that can be used over and over again, across any type of math—algebra, geometry, word problems, and so on.

Do you know what they are?

Try this GMATPrep® problem and then we’ll talk about the first of the four strategies.

* ” If

mv<pv< 0, isv> 0?“(1)

m<p“(2)

m< 0”

All set?

How did you do the problem? Most quant questions have more than one possible approach and this one is no exception—but I want to use this problem to talk about a particular technique called *Testing Cases*.

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. When we have these kinds of questions, we can use theory to solve—but that can get very confusing very quickly. Instead, try testing real numbers to “prove” the theory to yourself.

(Note: I chose a particularly tough question for this exercise; testing cases can also be useful and fast on easier questions!)

This problem gives one inequality:

“mv<pv< 0″

The test writers are hoping that you’ll say, “Oh, let’s just divide by *v* to get rid of it, so the equation is really *m* < *p* < 0.” But that’s a trap! Why?

When you divide an inequality by a negative, you have to flip the signs. But you don’t know whether *v* is positive or negative, so you don’t know whether to flip the signs! Never divide an inequality by a variable if you don’t know the sign of the variable.

The question itself contains a clue (two, actually!) pointing to this trap. The given inequality asks about “< 0” and the question also asks whether *v* > 0? *Less than zero* and *greater than zero* are code for “I’m testing you on positives and negatives.”

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