### The GMAT Official Guide 2019 Edition, Part 2

Here we go, here we go! Welcome to part 2 of a little series on the GMAT Official Guide 2019 edition, hot off the presses; if you’d like, you can start with the first installment of this article series. Today’s post focuses on Data Sufficiency.

#### What’s New in Data Sufficiency for the GMAT Official Guide 2019 Edition?

There are 26 new Data Sufficiency (DS) problems (and 26 were dropped).

As we saw for Problem Solving, most of the new problems are on the easier to medium side. There are still some quite hard older problems, but the fact that there aren’t many new hard ones for the second year in a row leads me to believe that we should expect to hear an announcement in the not-too-distant future about some new hard-question product (or perhaps that an online product has had more hard questions added to it?). If so, I’m looking forward to it!

Six of the new problems are about statistics—some easier ones around average and median, including one that had some really clever wording (#326) and another that tested standard deviation in a way that I don’t recall seeing before on published official questions (#290). More on both of those below.

Ratios are also covered in 6 new problems, most of which are on the easier side. And we can toss in a 7th, harder problem that’s about probability at a high level…but you really need to understand ratios in order to be able to solve.

Anecdotally, I’ve noticed that students in my classes tend to struggle with ratios—it’s not something we tend to see in obvious ways in the real world, even in quant-focused jobs. But understanding how ratios work can actually be really valuable in the business world, so I recommend spending some of your precious study time making sure you understand how the fundamentals work here.

Again anecdotally speaking, my students who have struggled with ratios have told me they really like the “ratio table” we introduce in our Foundations of Math strategy guide—so check that out if you need a little work here. (See the Fractions, Decimals, Percents, & Ratios chapter—look for the stuff labeled Ratios at the end of the chapter.)

I didn’t spot any other big trends in the remaining new problems, but I did quite like #301.

Want to try the three that I’ve mentioned? Try them on your own first: #290, #301, and #326. I’ll give some hints below, then talk about how I solved each one.

As before, I can’t repeat the full text of the problems for copyright reasons—you’ll need to get the book…or at least go view these problems in the “Look Inside” feature on Amazon. (Note that you can only look at a small number of pages in any one book…so you can’t just use that feature to do the whole GMAT Official Guide 2019 edition without paying for it. Besides, they do work hard to produce these problems—so we should pay for something that’s of good value. )

If you want to use the Amazon search function, look up this book, click on the Look Inside link, and then use the following search terms (do use the quotes for the ones that have more than one word):

#290 “history class”

#301 Shana

#326 “9 of the 20 houses”

#### If you’re stuck, here are some hints. But promise me: Don’t look until you really need to!

#290: What is standard deviation a measure of? What about mean—what does that mean? (Pun intended… 😁)

#301: “Less than” 48. That’s interesting.

#326: “More than 9 of the 20”?? That’s super-interesting wording.

#### Need another hint? Here you go.

#290: The question asks for the maximum. Can you come up with multiple scenarios that fit the facts in the statements but provide different possible maximum scores?

#301: Don’t try to write equations. Pretend you’re Shana, in the bookstore, and you’ve got $47.99 in your pocket. You’re trying to figure out whether you can afford these three books. How are you going to figure that out? #326: “More than 9” is the same thing as “at least 10.” Out of 20. And statement 2 mentions median. #### Solutions #290: The question stem doesn’t give us much—there are 10 students, so there are 10 scores in the mix. What’s the maximum score? No idea, but the fact that they’re asking for a maximum already signals to me that I’m likely going to need to test cases on this problem. Let’s go take a look at the statements. Knowing that the mean (average) score was 75 leaves open various possibilities. For instance, all 10 students could have scored 75. Or half of the students could have scored 70 and the other half could have scored 80. See what we just did there? We tested two possible cases. Each case used the given information (10 students total, mean score of 75) but we came up with two different possible maximum scores, so the first statement is not sufficient to answer the question. Eliminate answers (A) and (D). What about statement (2)? Standard deviation is a measure of the spread of the numbers. The closer together the numbers are, the smaller the standard deviation; the more spread out the numbers, the broader the standard deviation. First, remember that you’re supposed to forget statement (1) right now. If I only know the standard deviation but nothing at all about the scoring scale used (is it 1 to 10? 0 to 100??), I can’t say what the maximum score was. By itself, statement (2) is not sufficient; eliminate answer (B). What if we put the two statements together? Maybe this means the maximum score is 75 + 5 = 80? It doesn’t, actually. The calculation for standard deviation is super complicated, and the official explanation—which tests cases, too—gets into some…crazy calculations. So I’m just going to tell you something that you’ll know forever. If I tell you (1) the number of data points in a set, (2) the average value of those data points, and (3) the standard deviation of those points…then you cannot figure out the range of possible values, or the smallest or largest value, of the data points in that set. Lots of combinations are possible. Well, there is one exception: if the standard deviation is 0, then you can—because when the SD is 0, then every data point in the set has to be the same (and, therefore, the average is also equal to the value of each data point). The correct answer is (E). #301: Look at the statements. They never give us real values for the prices of the books—it’s always either “less than” or one price relative to another price. So test cases here. The most expensive book costs “less than$17.” For the sake of not driving yourself crazy, pretend the thing cost exactly $17. If you bought 3 books at that price, how much would you spend? (17)(3) = 51, which is over your$48 budget. Now, the book was “less than 17” but that means your three books could’ve been $16.99,$16.98, and $16.97, which is around$51 (i.e., over budget). Or the books could have been $4,$5, and $6, in which case you have enough money left over to take me out to lunch. Yay! 😎 I can’t tell whether Shana spent less than$48, so statement (1) is not sufficient; eliminate answers (A) and (D).

What about statement (2)? Here’s where my second hint comes in. Do not start to assign variables and write equations and ugh. Whip out an envelope and a pen—we’re going to do some back-of-the-envelope checks on these numbers.

The cheapest one was $3 less than the second most expensive. So I could have spent, say,$3, $6, and$7—coming in under my $48 budget. Or I could have spent, say,$50, $53, and$100—totally blowing my budget. Nope, this statement’s not sufficient either; eliminate answer (B).

Put them together. The most expensive has to be under $17. (Which is annoying, so let’s just call it$17 and remember to take off 1 penny at the end.)

The second most expensive has to be at least a penny less than the most expensive, which is annoying, so let’s just call it $17 and remember to take off 2 pennies at the end. And then the cheapest one is$3 less than that, so…$14 minus another two pennies. Oh, wait. This is interesting. So in my first test, I did (17)(3) = 51. Which is the same as 17 + 17 + 17 = 51. Now, I’ve got 17 + 17 + 14…which is 3 less, or 48. Oh…and then I need to take off a few more pennies. Wow! Shana did indeed spend less than$48 on her three books.

Statements (1) and (2) together are sufficient to answer the question. The correct answer is (C).

#326:

The question stem already mentions the concept of average and then asks that really weird “more than 9 of the 20” thing. More than 9 is the same thing as “at least 10” and they’ve already brought up statistics, so now I’m wondering whether they’re getting at something around median and I’m scanning the statements and…boom! Yes, statement (2) says something about median.

If I have 20 numbers in a set, which ones do I use to calculate median? 20 numbers = annoying, so let’s make this easier. If I have 6 numbers in the set, then I need to know (counting on my fingers here) numbers 3 and 4 in the set—those are the middle two. Number 3 is half of number 6 and 4 is just the next one up. So apply that pattern to the number 20: the number 10 is half and then I need to go one up to 11. I need 10 and 11…haha, that’s why they specified “more than 9.” Because 10 and 11 are the relevant numbers for the median.

Okay, what else do I know? The average is $160k and there are 20 houses total, so I can find the sum if I want to (since average = sum / # of terms). And what is the question really asking? Was house price #10 (if I were to write the numbers in increasing order) less than the average? Or was at least one half of the calculation for the median less than the average? Let’s find out! Statement (1) tells me the price of the most expensive house, or #20 on the list if I write them in increasing order…and that is totally useless when I’m trying to figure out something about the #10 house. This one isn’t sufficient; eliminate answers (A) and (D). Okay, statement (2)…the median was$150k. How do I calculate median again? Oh, right, find the average of houses #10 and #11. And remember that #10 has to be less than #11—since we’re doing median, which means they have to be written in increasing order. So the median is the average of just these two houses…and that median is only $150k…so house #10’s price had to be less than$150k.

And that’s less than the overall overage of $160k, so yes, at least 10 of the houses had prices that were less than$160k. Sufficient!