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The new Official Guide books have landed! In part 1, we talked about the changes to the Quant portions of the guides.
Already read that part? Then let’s talk about those interesting problems I mentioned last time! Note: I can’t reprint the problems here for copyright reasons. You will actually have to have a copy of the book in order to follow along. If you don’t have a copy, skip ahead to part 3, where we’ll talk about verbal.
Note: I strongly recommend that you not read this until you have worked on these problems yourself! If you can figure out what’s going on yourself, the lesson will stick much better in the end. 🙂
Big Book #116 and #140
Guiding principle: Do NOT do textbook math. Work backwards.
My colleague Whitney Garner and I both snorted when we saw the first one of these. #116. The math is truly ridiculous…unless you approach it as a real-world problem, not a textbook math problem. The problem talks about 1/3 of the total, but we don’t have a real number for the total (we only know it’s 37 + x + 32). So the math could get crazy messy…unless you work backwards from the answer choices. Even then, the math is messy unless you also estimate at one point.
So let’s take answer (B) 9. (When working backwards, start with B or D.) If x = 9, then there are a total of 78 marbles. One-third are blue, so 78/3 = 26. Now we have to test these numbers with the percentages listed in the final column to see whether we have a match. Don’t calculate 10.8% of anything—estimate! (By the way…it makes sense that you can do this…because you shouldn’t have only part of a marble…)
A little more than 10% of 37 is 4. 50% of 32 is 16. And 66.7% of x = 9 is 6. Does that add up right? Yep. 26 blue ones. We’ve found the answer!
#140 is equally weird until you treat the answer choices as your starting point. The question asks for the least and greatest possible values of n. Look at those answers.
There are only three possibilities for the least value: 0, 1, or 2. Since they’re asking for the least, try the smallest one (0) first to see whether it’s possible.
Once you figure out whether the lower bound is 0, 1, or 2, that leaves you with either one or two remaining answers. At most, you’ll need to test one of the upper bound numbers and you’re done. (This time, test the larger number first, since you’re looking for the greatest possibility.)
Big Book #194 and #283:
Guiding principle: know how to think about probability.
#194 is a PS problem and the answers range from really small to just under half to just over half to decently bigger than half. So you might be able to narrow down by estimating.
Pretend you’re the one going camping. How’s it going to work out?
Saturday: no rain
Sunday: no rain
The probability of rain is 0.2, so the probability of no-rain is 0.8. You’ve got a really good chance to stay on Saturday. On Sunday, there’s a decent chance that you can still stay but it’s a bit worse than it was solely on Saturday. Still more than 50% though.
It does rain Monday, however, so you’ve got to factor in that low probability of 0.2. That should drop your odds below half. Answers (D) and (E) are out and even (C) is probably too close to half. And (A) is vanishingly small, so it’s probably (B).
The actual math would be (0.8)(0.8)(0.2) for the scenario described (no rain, no rain, rain). Ignore the decimals and multiply out the numbers: (8)(8)(2) = 128. Only one answer matches that digit sequence: (B)!
The second problem here, #283, is a DS. Let’s call a job offer a “Yes” answer (ie, yes the company wants Jill). In order to calculate the probability, we need to know the probability of earning a Yes from each, or (Yes)(Yes).
Notice that the statements combine the info about the two companies. Statement (1) is the equivalent of (No)(No). What are the other possible outcomes?
So statement (1) doesn’t give us enough to get to (Yes)(Yes). What about statement (2)?
Interesting! This one is (Yes)(No) + (No)(Yes). She gets an offer from the first one but not the second one or she gets an offer from the second one but not the first one.
By itself, that’s not enough, but it you put the two statements together, you have the probabilities for three out of the four possible scenarios.
All of the possible scenarios must add up to a probability of 1, so you can subtract the other three scenarios to find the probability of the one desired scenario (Yes)(Yes). Sufficient! The answer is (C).
In both cases, these two problems are medium to higher-level, but they don’t require crazy calculations. They do require, though, that you know how to think about probability in general. For instance, if you know the probability of all possible scenarios except for one, then you can always find the probability of the one missing scenario.
Big Book #310 and #?
Guiding principle: try to use what you’ve already figured out to help give you a boost on harder problems.
This is the one where I wouldn’t tell you the second question number before—I suggested you try to find something similar to 310 yourself. I’ll be honest: this was a really challenging assignment. If you can do this, you have a good shot to get a really high quant score on the GMAT.
To me, these two were the most intriguing of all of the new quant problems. I haven’t seen prior problems test these principles in quite this way. Let’s start with the first one, 310. When I first saw it, I wanted to throw in the towel. Even just Testing Cases was going to be really tedious and annoying on this one.
I knew, though, that this test doesn’t ever require us to do super-tedious math. So I took another look at what they gave me and realized there was a really interesting hidden pattern.
The question stem asks me to add the same value, n, to each of 3 numbers. That’s a really weird request, so it caught my attention. I noticed that, since I’m adding n to each one, the differences between those three numbers never change. 94 is 25 larger than 69 and 94 + n is 25 larger than 69 + n. I had no idea what that meant; I just noticed it.
Statement (1) brings in what seems like crazy info: 69 + n and 94 + n both represent the squares of two consecutive integers.
Ah! That’s what did it for me. I already knew another principle that I’ll explain to you right now. Square some consecutive positive integers. Notice anything?
1, 4, 9, 16, 25, 36, 49, …
As you get bigger, the difference between the numbers always gets larger. The difference between 1 and 4 is 3, the difference between 4 and 9 is 5, and the difference between 9 and 16 is 7. And that keeps going. The key detail: each pair of consecutive integers-squared has its own unique difference.
So 69 + n and 94 + n represent the squares of two consecutive integers. Whatever those two squares are, they are 25 apart. Their difference is 25. There’s only one pair of positive consecutive integers-squared that are 25 apart. (I’m not going to go figure out what they are, of course. This is DS. But I know from the pattern that there can be only one pairing with a difference of 25.) Done! Sufficient.
What about statement (2)? Same deal. This time, we’ve got two different numbers (and a difference of 121 – 94 = 27), but the same principle: there’s only one pairing of consecutive integers-squared that has a difference of 27. Sufficient!
Which other problem later in this same chapter touches on similar principles? Drum-roll please: it’s 327. Check it out.
In this case, though, the problem cannot be solved via the given information. The two statements each tell you that one of the terms is a perfect square. The problem mentions nothing about the two terms being consecutive, though, nor could they be, since the difference between consecutive squares is always odd and the difference between the two terms here is 24, an even number.
Each statement can’t be sufficient by itself because the other term could be a non-integer; there are infinite possibilities. That knocks out answers (A), (B), and (D). The real question is whether putting the two statements together will narrow things down to just one possible value.
At this point, it’s a judgment call whether you want to take the time to see whether you really can find two different cases (using real numbers). It’s tricky on this problem, since both x and x + 24 have to be perfect squares—but that right there is your clue. Test out small perfect squares to find ones that are exactly 24 apart…and don’t bother trying to test consecutive pairings, since you know those will be an odd number apart, not an even number. Try numbers that are 2 apart, 4 apart, 6 apart…
1 and 3 are too close for their squares to be 24 apart. What about 1 and 5? Bingo! That pairing actually works.
What about 2 and 4 or 3 and 5? Too close. 4 and 6? Almost… 5 and 7? Bingo! At least two pairings work, so even together, these guys aren’t sufficient. The answer is (E).
You could also just gamble that, without the consecutive constraint, chances are that more than one combo works—since it seems as though the constraint is now so specific that only one number could work. (ie, it’s a trap.)
The key takeaway: having thought through the first one already, I actually had a shot at figuring out the more complicated second one in a reasonable amount of time. If I’d seen the second one first, I likely would have decided the problem wasn’t worth it once I got to (C) and (E).
What about Verbal?
Next time, we’ll dive into the verbal sections of the Big OG and the verbal review. In some later articles, I’ll also provide you with lists of the new questions in each book.
Happy studying! 📝
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Stacey Koprince is a Manhattan Prep instructor based in Montreal, Canada and Los Angeles, California. Stacey has been teaching the GMAT, GRE, and LSAT for more than 15 years and is one of the most well-known instructors in the industry. Stacey loves to teach and is absolutely fascinated by standardized tests. Check out Stacey’s upcoming GMAT courses here.