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Posted on
28
Jul 2021

GMAT Trade Show Problem – Data Sufficiency

GMAT Trade Show Problem Introduction

Today we’re going to take a look at the Trade Show Problem and this is a GMAT Data Sufficiency problem with averages as the focal point. But really the concept of average is distracting from this problem. So, if we take a look at the question stimulus, we want to figure out what we need, but we need to synthesize some of the information there to understand what we know.

We’re being asked whether or not it gets above a certain threshold an average of 90, and over six days that’s going to be over a total of 540 points. Notice how I did it mathematically, you can represent it graphically as a rectangle, but 90 times 6 is that 540 points. We know though that all of our days at a minimum are 80 which means we can build up from that piece of knowledge. We have 80 x 6 = 480 points and we want to know if we have more or less than 60 points above that minimum that we’re already working with that’s what we need.

Solving the GMAT Problem

Ways we might get it include any number of slices and dices for the performance of the rest of the days and the difficulty of this problem in large part will be dependent on how convoluted the GMAT gives us the introduced information on number one and two.

When we look at number one, we’re told that the final four days average out to a hundred. Once again, like with other average problems, each of the individual four days the performance doesn’t matter. We can just say each is exactly 100 and make that assumption, which means each is 20 over – we’re 80 points over the mean. Because we want to know whether we are more or less than 60 points, this knowledge that we’re 100 points tells us “Yes, definitively. We are over that average of 90, we’re over that surplus of 60 points.” So, number one is sufficient.

Number two gives us the opposite information, it talks about the minimum, and, in aggregate, that doesn’t let us know directly whether or not we make those 60 points. That is it’s possible but it’s also possible that we don’t, because we’re dealing with a minimum rather than a maximum or rather we’re dealing with information that can lie on either side of what we need. Therefore 2 is insufficient. Our answer here is A.

I hope that was useful. GMAT nation stay strong, keep averaging. You guys got this! I believe in you. If you want to test your GMAT Data Sufficiency skills, check out the Science Fair problem.

 

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Posted on
21
Jul 2021

GMAT 3D Geometry Problem – GMAT Math – Quant Section

GMAT 3D Geometry Problem 

In this problem we’re going to take a look at 3D objects and in particular a special problem type on the GMAT that measures the longest distance within a three-dimensional object. Typically, they give you rectangular solids, but they can also give you cylinders and other such objects. The key thing to remember about problems like this one is that effectively we’re stacking Pythagorean theorems to solve it – we’re finding triangles and then triangles within triangles that define the longest distance.

This type of problem is testing your spatial skills and a graphic or visual aid is often helpful though strictly not necessary. Let’s take a look at how to solve this problem and because it’s testing these skills the approach is generally mathematical that is there is some processing because it’s secondary to what they’re actually testing.

gmat 3d geometry question

GMAT 3D Geometry Problem Introduction

So, we have this rectangular solid and it doesn’t matter which way we turn it – the longest distance is going to be between any two opposite corners and you can take that to the bank as a rule: On a rectangular solid the opposite corners will always be the longest distance. Here we don’t have any way to process this central distance so, what we need to do is make a triangle out of it.

Notice that the distance that we’re looking for along with the height of 5 and the hypotenuse of the 10 by 10 base will give us a right triangle. We can apply Pythagoras here if we have the hypotenuse of the base. We’re working backwards from what we need to what we can make rather than building up. Once you’re comfortable with this you can do it in either direction.

Solving the Problem

In this case we’ve got a 10 by 10 base. It’s a 45-45-90 because any square cut in half is a 45-45-90 which means we can immediately engage the identity of times root two. So, 10, 10, 10 root 2. 10 root 2 and 5 makes the two sides. We apply Pythagoras again. Here it’s a little more complicated mathematically and because you’re going in and out of taking square roots and adding and multiplying, you want to be very careful not to make a processing error here.

Careless errors abound particularly when we’re distracted from the math and yet we need to do some processing. So, this is a point where you just want to say “Okay, I’ve got all the pieces, let me make sure I do this right.” 10 root 2 squared is 200 (10 times 10 is 100, root 2 times root 2 is 2, 2 times 100 is 200). 5 squared is 25. Add them together 225. And then take the square root and that’s going to give us our answer. The square root of 225 is one of those numbers we should know. It’s 15, answer choice A.

Okay guys for another 3D and Geometry problem check out GMAT 680 Level Geometry Problem – No Math Needed! We will see you next time.

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Posted on
14
Jul 2021

GMAT 680 Level Geometry Problem – No Math Needed!

GMAT Geometry Problem

Hey guys, top level geometry problems are characterized typically by stringing a whole bunch of different rules together and understanding how one thing relates to the next thing, to the next thing. Until you get from the piece of information you started with to the conclusion. We’re going to start out by taking a look at this problem using the z equals 50° and seeing how that information goes down the line.

top level geometry problem

But afterwards we’re going to see a super simple logical pathway utilizing a graphic scenario that makes the z equals 50° irrelevant. To begin with we’re being asked for the sum of x and y and this will come into play on the logical side. We need the sum not the individual amounts but let’s begin with the y. We have a quadrilateral and it has parallel sides which means the two angles z and y must equal 180°. That’s one of our geometric rules. If z is 50° that means y is 130° and we’re halfway there.

Next we need to figure out how x relates and there are several pathways to this. One way we can do it is drop. By visualizing or dropping a third parallel line down, intersecting x, so on the one hand we’ll have 90 degrees. We’ll have that right angle and on the other we’ll have that piece. Notice that the parallel line we dropped and the parallel line next to z are both being intersected by the diagonal line going through which means that that part of x equals z. So we have 50° plus 90° is 140°. 130° from the y, 140° from the x, gives us 270°.

Another way we can do this is by taking a look at the right triangle that’s already built in z is 50° so y is 1 30°. now the top angle in the triangle must then be 180° minus the 130° that is 50°. it must match the z again we have the parallel lines with the diagonal coming through then the other angle the one opposite x is the 180° degrees that are in the triangle minus the 90° from the right triangle brings us to 90° minus the 50° from the angle we just figured out means that it’s 40° which means angle x is 180° flat line supplementary angles minus the 40° gives us 140° plus the 130° we have from y again we get to 270°.

Graphic Solution Path

Now here’s where it gets really fun and really interesting. We can run a graphic scenario here by noticing that as long as we keep all the lines oriented in the same way we can actually shift the angle x up. We can take the line that extends from this big triangle and just shift it right up the line until it matches with the y. What’s going to happen there, is we’re going to see that we have 270° degrees in that combination of x and y and that it leaves a right triangle of 90°, that we can take away from 360° again to reach the 270°.

Here the 50° is irrelevant and watch these two graphic scenarios to understand why no matter how steep or how flat this picture becomes we can always move that x right up and get to the 270°. That is the x and y change in conjunction with one another as z changes. You can’t change one without the other while maintaining all these parallel lines and right angles. Seeing this is challenging to say the least, it requires a very deep understanding of the rules and this is one of those circumstances that really points to weaknesses in understanding most of what we learn in math class in middle school, in high school. Even when we’re prepping only scratches the surface of some of the more subtle things that we’re either allowed to do or the subtle characteristics of rules and how they work with one another and so a true understanding yields this very rapid graphic solution path.

Logical Solution Path

The logical solution path where immediately we say x and y has to be 270° no matter what z is and as you progress into the 80th, 90th percentile into the 700 level on the quant side this is what you want to look for during your self prep. You want to notice when there’s a clever solution path that you’ll overlook because of the rules. Understand why it works and then backtrack to understand how that new mechanism that you discovered fits into the framework of the rules that we all know and love. Maybe? I don’t know if we love them! But they’re there, we know them, we’re familiar with them, we want to become intimate. So get intimate with your geometry guys put on some al green light some candles and I’ll see you next time.

If you enjoyed this problem, try other geometry problems here: GMAT Geometry.

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19
May 2021

GMAT Algebra Problem – Parts – Hotdogs & Donuts

GMAT Algebra Problem Introduction

Hi guys. Today I’m here with a classic GMAT Algebra problem, what we call a parts problem. And if you take a look at this problem you’re going to realize that it just looks like a bunch of algebra. But the key here is in how you frame it. We’ve got this diner or whatnot selling hot dogs and then after that point, so imagine like a timeline, they start selling donuts. Then they give us a piece of information about hot dogs to donuts over that course of secondary time but then give us this overarching total number of food products sold.

Distill The Ration

So what we need to do are two steps: the first one is fairly straightforward. We see that we have to get rid of the hot dogs that were sold in advance in order to distill the ratio but then the ratio can seem very, very complex, especially because it just tells us seven times and a lot of times the GMAT will do this as a way to throw us off the scent. So when we have seven times, what that means is we have eight parts. That is it’s saying for every one of these we have one, two, three, four, five, six, seven of these. Meaning in total there are eight. So while seven is kind of a scary number, eight is a number we can divide by easily. You always want to look for that when you’re given a ratio of one thing to another especially when they say something times as many.

Solving the GMAT Problem

We take that thirty thousand two hundred knock off the fifty four hundred and get to twenty four thousand eight hundred and lo and behold that’s divisible by eight meaning each part is going to be 3100. Notice there’s no complex division there, 24 divided by 8, 800 divided by 8 and that’s the sort of mental math we can expect from the GMAT always. Which as you’ve seen before: if you’re doing that you’re doing something wrong.

Each part is 3100 and we’re concerned with the seven parts so we can either scale that 3100 up by seven into 21700, again the math works out super smoothly or we can take the 24800 knock off 3100 and get to that 21700. Notice in the answer choices there’s a few things to address sort of common errors that might be made.

Reviewing the Answer Signals

On one of the answer choices what you’re looking at is dividing the total, the 30 200 by eight and multiplying by seven that is seven eighths of it without getting rid of those first 5400. Another answer is close to our 21700 correct answer and this is also a fairly reliable signal from the GMAT.

When they give you a range of answers but two of them are kind of tightly clustered together a lot of times it’s going to be one of the two and that second one there is to prevent you from too roughly estimating. But at the same time if you’re short on time or just in general you want to hone down and understand what you’re supposed to do that serves as a really strong signal. And then one of the answer choices is the 1/8 of it rather than the 7/8.

Clustered Answer Choices

I want to speak a little more deeply about that signal about those two tightly clustered answer choices because as I said it can help you narrow to a very quick 50/50 when you’re constrained for time or this problem is just one that’s really not up your alley but it also can be leveraged in a really, really neat way.

If we assume that one or the other is the answer choice we can differentiate these two different answer choices by what they’re divisible by and so notice the 21700 is very clearly, with strong mental math is divisible by seven. Where the other one is not. Also neither of them are divisible by eight. We can look at these two say okay one of them is probably right, one of them is divisible by seven, the other one is not, so there’s our right answer and we can move on to the next problem. So I hope this helps. Write your comments and questions below. Subscribe to our channel at Apex GMAT here and give us a call if we can ever help you.

To work on similar GMAT algebra problem/s see this link: Work Rate Problem.

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13
May 2021

GMAT Factors Problem – 700 Level GMAT Question

GMAT Factors Problem

Hey guys! Today we’re going to take a look at one of my favorite problems. It’s abstract, it’s oddly phrased and in fact the hardest part for many folks on this problem is simply understanding what’s being asked for. The difficulty is that it’s written in math speak. It’s written in that very abstract, clinical language that if you haven’t studied advanced math might be new to you.

How this breaks down is they’re giving us this product from 1 to 30, which is the same as 30!. 30*29*28 all the way down the line. Or you can build it up 1*2*3*……*29*30.

The Most Difficult Part of The GMAT Problem

And then they’re asking this crazy thing about how many k such that three to the k. What they’re asking here is how many factors of three are embedded in this massive product. That’s the hard part! Figuring out how many there are once you have an algorithm or system for it is fairly straightforward. If we lay out all our numbers from 1 to 30. And we don’t want to sit there and write them all, but just imagine that number line in your head. 1 is not divisible by 3. 2 is not divisible by 3, 3 is. 4 isn’t. 5 isn’t. 6 is. In fact, the only numbers in this product that concern us are those divisible by 3. 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.

Important Notes About Factors

Here it’s important to note that each of these components except the three alone has multiple prime factors. The three is just a three. The six is three and a two. The nine notice has a second factor of three. Three times three is nine and because we’re looking at the prime factors it has two. It’s difficult to get your head around but there are not three factors of three in nine when you’re counting prime factors.

Three factors of three would be 3 by 3 by 3 = 27. So notice that 3 and 6 have a single factor. 9 has a double factor. Every number divisible by 3 has one factor. Those divisible by 9 like 9, 18 and 27 are going to have a second factor and those divisible by 27, that is 3 cubed, are going to have a third factor. If we lay it out like this we see ten numbers have a single factor. Another of those three provide a second bringing us to thirteen. Finally, one has a third bringing us to fourteen. Answer choice: C.

GMAT Problem Form

So let’s take a look at this problem by writing a new one just to reinforce the algorithm. For the number 100 factorial. How many factors of seven are there? So first we ask ourselves out of the 100 numbers which ones even play? 7, 14… 21 so on and so forth. 100 divided by 7 equals 13. So there are 13 numbers divisible by 7 from 1 to 100. Of those how many have more than one factor of 7? Well we know that 7 squared is 49. So only those numbers divisible by 49 have a second factor. 49 and 98. There are none that have three factors of 7 because 7 cubed is 343. If you don’t know it that’s an identity you should know. So here our answer is 13 plus 2 = 15.

Try a few more on your own. This one’s great to do as a problem form and take a look at the links below for other abstract number theory, counting prime type problems as well as a selection of other really fun ones. Thanks for watching guys and we’ll see you soon.

If you enjoyed this GMAT factors problem, here is an additional number theory type problem to try next: Wedding Guest Problem.

 

 

 

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Posted on
23
Apr 2021

Standard Deviation – Clustering (Birds) Problem

Hey guys! Today we’re going to take a look at a DS problem that is a skills problem, focused on GMAT standard deviation.

Standard Deviation & Variance

What they’re asking here is do we have enough information to compute a standard deviation? It’s useful to think of standard deviation as clustering. If we have a whole series of points we can define how clustered or un-clustered the group of points is. That’s all that’s standard deviation, that’s all that variance is. So if we have all the points that works. What we should be on the lookout here for are parametric measurements. Especially things like the average number is, because while the average can be used to compute standard deviation, we need to know how each of the points differs from the average. But if we have each of the points we always get the average. That is, we can compute the average. So the average is a nice looking piece of information that actually has little to no value here. So let’s jump into the introduced information.

Statement 1

Number 1 BOOM – tells us that the average number of eggs is 4 and that’s great except that it doesn’t tell us about the clustering. If we run some scenarios here we could have every nest have 4 eggs or we could have 5 nests have 0, 5 nests have 8, or 9 nests have 0, 1 nest has 40. These are all different clusterings and we could end up with anything in between those extremes as well. So number 1 is insufficient.

Statement 2

Number 2: tells us that each of the 10 bird’s nests has exactly 4 eggs. What does this mean? We have all 10 points. They happen to all be on the average, which means the standard deviation is 0. that is there’s no clustering whatsoever. But 2 gives us all the information we need so B – 2 alone is sufficient is the answer here.

Hope this was useful guys, check out the links below for a video about how to compute standard deviation as a refresher, as well as other problems related to this one. Thanks for watching we’ll see you again real soon

If you enjoyed this GMAT problem, try another one next: Normative Distribution

 

 

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Posted on
14
Apr 2021

GMAT Percentage Problem – Unemployment Rate – Multiple Solution Paths

GMAT Percentage Problems

Hey guys, GMAT Percentage problem/s are commonplace on the GMAT and today we’re going to take a look at one that is straightforward but could very easily get you caught up with the math. In this problem, notice that there’s the word “approximately.” That always means there’s an Estimation Solution Path. We’ll take a look at that first but then we’re going to look at a Scenario Solution Path, which for many people is a lot more natural. In addition to seeing that word approximately you can see that there’s this massive spread within the answer choices. Once again pushing us towards an Estimation Solution Path.

Estimation Solution Path

So let’s dive in: The unemployment rate is dropping from 16% to 9% and your quick synthesis there should be: okay it’s being cut about in half or a little less than half. And monitoring that directionality is important. Additionally, the number of workers is increasing. So we have lower unemployment but a greater number of workers. So we have two things, two forces working against one another. If the number of workers were remaining equal then our answer would be about a 50% decrease or just under a 50% decrease, so like 45% or something like that. But because we’re increasing the number of workers, our decrease in unemployment is lower. That is we have more workers, so we have a larger number of unemployed so we’re not losing as many actual unemployed people and therefore our answer is B: 30% decrease.

Scenario Solution Path

If we want to take a look at this via Scenario, we can always throw up an easy number like 100. We begin with 100 workers and 16% are unemployed so 16 are unemployed. Our workers go from 100 to 120. 9% of 120 is 9 plus 0.9 plus 0.9 = 10.8% or 11%. What’s the percentage decrease from 16 to 11? Well it’s not 50, that’s too big. It’s not 15, that’s too small. It’s about 30 and the math will bear us out there.

So thanks for watching guys! Check out the links below for other GMAT percentage problem/s and we look forward to seeing you again real real soon.

Another GMAT percentage problem

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Posted on
08
Apr 2021

GMAT Factorial Problem: Estimation & Scenario Solution

GMAT Factorial Introduction

Factorials and divisibility, together. Two mathematical kids from opposite sides of the tracks, they come together and fall in love and they create this problem. Here we’re asked what numbers might divide some new number 20 factorial plus 17. As a refresher, a factorial is simply the number times each integer below it. So in this case, 20! is equal to 20 x 19 x 18 …. x 3 x 2 x 1. It’s a huge number. And it’s not at all possible to process in GMAT time. What we want to notice about any factorial is that it has as factors every number that it contains. So 20! is divisible by 17, it’s divisible by 15, it’s divisible by 13, 9, 2, what have you and any combination of them as well.

What The GMAT is Counting On You Not Knowing

When we’re adding the 17 though, the GMAT is counting on the idea that we don’t know what to do with it and in fact that’s the entire difficulty of this problem. So I want you to imagine 20! as a level and we’re going to take a look at this graphically. So 20! can be comprised by stacking a whole bunch of 15’s up. Blocks of 15. How many will there be? Well 20 x 19 x 18 x 17 x 16 x 14 times all the way down the line. There will be that many 15’s. But 20! will be divisible by 15. Similarly, by 17, by 19, by any number. They will all stack and they all stack up precisely to 20! because 20! is divisible by any of them.

Answer

So when we’re adding 17 to our number all we need to see is that, hey, 15 doesn’t go into 17, it’s not going to get all the way up there. 17 fits perfectly. 19? guess what? It’s too big and we’re going to have a remainder. So our answer here is B, only 17.

For other problems like this, other factorials, and what have you, please check out the links below and we will see you next time. If you enjoyed this GMAT problem, try your hand at this Science Fair Problem.

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GMAT Prime Factorization Article
Posted on
01
Apr 2021

GMAT Prime Factorization (Anatomy of a Problem)

By: Rich Zwelling (Apex GMAT Instructor)
Date: 1st April 2021

First, if you’d prefer to go straight to the explanation for the solution to the problem given in the last post, continue to the end of this post. But we’ll start with the following Official Guide GMAT problem as a way to talk about GMAT Prime Factorization. Give the problem a shot, if you can:

How many prime numbers between 1 and 100 are factors of 7,150?

A) One
B) Two
C) Three
D) Four
E) Five

One of the things you’ll notice is the linguistic setup of the problem, which is designed to confuse you immediately (a common theme on GMAT problems). They get you panicking right away with mention of a large range (between 1 and 100), and then they compound your frustration by giving you a rather large value (7,150). 

Don’t let that convince you that you can’t do the problem. Because remember, the GMAT is not interested in large calculations, memorization involving large numbers, or weird arcana. Chances are, if you find yourself thinking about a complicated way to do a problem, you’re taking the wrong approach, and there’s a simpler way.

Try to pick out the most operative signal words, which let you know how to address the problem. We are dealing with prime numbers and also the topic of finding factors. The language of the problem may make you nervous, thinking that we must consider a slew of prime numbers up to 100. But the only primes we are really interested in are those that are actually factors of 7,150. 

So let’s focus our attention there. And we can do so with a prime factor tree. Does this bring back memories? 

Now, in the case of 7,150, we don’t have to break it down into prime numbers immediately. Split the number up into factors that are easy to recognize. In this case, the number ends in a zero, which means it is a multiple of 10, so we can start our tree like this:

prime factorization on the GMATNotice that the advantage here is two-fold: It’s easier to divide by 10 and the two resulting numbers are both much more manageable. 

Splitting up 10 into it’s prime factorization is straightforward enough (2 and 5). However, how do we approach 715? Well, it’s since it ends in a 5, we know it must be divisible by 5. At that point, you could divide 715 by 5 using long or short division… 

…or you could get sneaky and use a NARRATIVE approach with nearby multiples:

750 is nearby, and since 75/5 = 15, that must mean that 750/5 = 150. Now, 750 is 35 greater than 715. And since 35/5 = 7, that means that 715 is seven multiples of 5 away from 750. So we can take the 150, subtract 7, and get 143

Mathematically, you can also see this as: 

715/5 = (750-35)/5 = 750/5 – 35/5 = 150 – 7 = 143

So as stands, here’s our GMAT prime factorization:

prime factorization GMAT articleNow, there’s just the 143 to deal with, and this is where things get a bit interesting. There are divisibility rules that help make factoring easier, but an alternative you can always use is finding nearby multiples of the factor in question. 

For example, is 143 divisible by 3? There is a rule for divisibility by 3, but you could also compare 143 against 150. 150 is a multiple of 3, and 143 is a distance of 7 away. 7 is not a multiple of 3, and therefore 143 is not a multiple of 3.

This rule applies for any factor, not just 3. 

Now we can test the other prime numbers. (Don’t test 4 and 6, for example. We know 143 is not even, so it’s not divisible by 2. And if it’s not divisible by 2, it can’t be divisible by 4. Likewise, it’s not divisible by 3, so it can’t be divisible by 6, which is a multiple of 3.)

143 is not divisible by 5, since it doesn’t end in a 5 or 0. It’s not divisible by 7, since 140 is divisible by 7, and 143 is only 3 away. 

What about 11? Here you have two options: 

  1. Think of 143 as 110+33, which is 11*10 + 11*3 → 11*(10+3) → 11*13
  2. If you know your perfect squares well, you could think of 143 as 121+22 

→ 11*11 + 11*2 → 11*(11+2) → 11*13

Either way, you should arrive at the same prime factorization:

GMAT prime factorization QuestionNotice that I’ve marked all prime numbers in blue. This result shouldn’t be a surprise, because notice that everything comes relatively clean: there are only a few prime numbers, they are relatively small, and there is just one slight complication in solving the problem (the factorization of 143). 

So what is the answer? Be very careful that you don’t do all the hard work and falter at the last second. There are five ends to branches in the above diagram, which could lead you prematurely to pick answer choice E. But two of these branches have the same number (5). There are actually only four distinct primes (2, 5, 11, 13). The correct answer is D.

And again, notice that the range given in the question stem (1 to 100) is really a linguistic distraction to throw you off track. We don’t even go beyond 13. 

Next time, we’ll talk about the fascinating topic of twin primes and how they connect to divisibility.

Find other GMAT Number Theory topics here:
Odds and Ends (…or Evens)
Consecutive Integers (plus more on Odds and Evens)
Consecutive Integers and Data Sufficiency (Avoiding Algebra)
GMAT Prime Factorization (Anatomy of a Problem)
A Primer on Primes

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Posted on
24
Mar 2021

Standard Deviation Problem On The GMAT (Normative Distribution)

Standard Deviation 700+ Tips

Hey guys! Today, we’re going to take a look at a standard deviation problem. And standard deviation is a concept that only comes up infrequently on the GMAT. So it’s more important to have a basic understanding of the concepts associated with it than to go really deep. This is true largely for much of the statistics, probability, and combinatorics problems. They show up infrequently until you get to the higher levels and even when you’re at the higher levels, relative to the algebra, arithmetic and even the geometry problems, they play such a small role.

And yet there’s so much math there that it’s very easy to get caught up in spending a lot of time prepping on problems or on these types of math that offer very little in return relative to spending your prep time really mastering the things that come up frequently. I’m not saying don’t learn this stuff I’m saying balance it according to its proportionality on the GMAT. As a general rule you can assume that stat, combinatorics and probability, all that advanced math, constitutes maybe 10 to 15% of what you’ll see on the GMAT. So keep that in mind as you prep.

Problem Language

In this problem the first step is to figure out what the heck we’re actually being asked for and it’s not entirely clear. This one’s written a little bit in math speak. So we have a normal distribution which doesn’t really matter for this problem but if you studied statistics it just means the typical distribution with a mean m in the middle and a standard deviation of d which they tell us is a single standard deviation. So they’re really just telling us one standard deviation but they’re saying it in a very tricky way and they’re using a letter d. If it helps you can represent this graphically.

Graphical Representation

Notice that they tell you something that you may already know: that one standard deviation to either side of the mean is 68 in a normal distribution. This breaks up to 34, 34. But they’re asking for everything below. The +1 side of the distribution. Since the m, the mean is the halfway point, we need to count the entire lower half and the 34 points that are in between the mean and the +d, the +1 standard deviation. This brings us to 84 which is answer choice D. This is primarily a skills problem, that is, you just need to know how this stuff works. There’s no hidden solution path and the differentiation done by the GMAT here is based upon your familiarity with the concept. Rather than heavy-duty creative lifting as we see on so many other problems that have more familiar math, that everyone kind of knows.

I hope this was helpful. Check out below for other stat and cool problem links and we’ll see you guys next time. If you enjoyed this GMAT Standard Deviation problem, try this Data Sufficiency problem next.

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