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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 Explained | Estimation & Scenario Solution Path

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
31
Mar 2021

Ace GMAT Data Sufficiency Questions with this Science Fair Problem

Data Sufficiency Problem Video Transcript

Introduction to Data Sufficiency

Hey guys! Today we’re looking at the Science Fair Problem. In this Data Sufficiency, we’re being asked how many, discrete number, of the 900 students at the school attended all three days. And we can surmise that they’re going to come at us by giving us different breakdowns of how different groups of students behaved and so most likely we’re going to need more than one piece of information to come together in order to give us the precise amount. The only way, typically, that we would have a single piece of information be sufficient is if they gave us the inverse and told us how many, or what percentage, or what fraction of students didn’t attend on all three days. Where we could then compute the opposite.

Statement 1

Let’s take a look: Number 1 is telling us that 30% or 270 of the students attended two or more days. If we break this up into a chart, we see this block that’s undefined but we know that 270 attended either two days or three days. Some mix of them, but we don’t know that mix. Therefore, this doesn’t give us what we need from the box and it’s insufficient. However, we could use it possibly with other information that distinguishes between the two day visitors and the three day visitors.

Statement 2

Number 2 gives us relative information based upon some other number: 10% of those that attended at least one day. That means of all those that attended at all, for one day, for two days, for three days, 10% of those belong in the three-day box. However, we don’t know how many students that is. So 2 is insufficient. When we try and combine them notice that the information from 2 slices and dices a piece of information that 1 doesn’t give us. There’s no way to reconcile the 10% from that big group into the group that just attended two days or three days. Therefore, we don’t have enough information.

Answer

The answer choice is E: both together are still insufficient. Hope this helped. Guys thanks for watching! For other examples of DS problems where you can make charts to fill in the blanks and find the square you need check out the links below and we’ll see you again soon.

If you enjoyed this Data Sufficiency problem video try this Standard Deviation Problem

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GMAT Prime numbers article with questions
Posted on
29
Mar 2021

A Primer on Primes

By: Rich Zwelling (Apex GMAT Instructor)
Date: 30th March 2021

As I said in my previous post, GMAT Prime Numbers are my favorite topic. This is because not only are they inherently interesting mathematically but they show up in unexpected circumstances on GMAT problems, even when the term “prime” is not explicitly mentioned.

But before we get to that, I thought it would help to review a basic definition:

If you’ve gone through school, you’ve likely heard the definition of a prime as “any number that can be divided only by 1 and itself.” Or put differently, “any number that has only 1 and itself as factors.”  For example, 3 is a prime number, because 1 and 3 are the only numbers that are factors of 3.

However, there is something slightly problematic here. I always then ask my students: “Okay, well then, is 1 prime? 1 is divisible by only 1 and itself.” Many people are under the misconception that 1 is a prime number, but in truth 1 is not prime

There is a better way to think about prime number definitionally:

*A prime number is any number that has EXACTLY TWO FACTORS*

By that definition, 1 is not prime, as it has only one factor

But then, what is the smallest prime number? Prime numbers are also by definition always positive, so we need not worry about negative numbers. It’s tempting to then consider 3, but don’t overlook 2. 

Even though 2 is even, it has exactly two factors, namely 1 and 2, and it is therefore prime. It is also the only even prime number. Take a moment to think critically about why that is before reading the next paragraph…

Any other even number must have more than two factors, because apart from 1 and the number itself, 2 must also be a factor. For example, the number 4 will have 1 and 4 as factors, of course, but it will also have 2, since it is even. No even number besides 2, therefore, will have exactly two factors. 

Another way to read this, then, is that every prime number other than 2 is odd

You can see already how prime numbers feed into other number properties so readily, and we’ll talk much more about that going forward. But another question people often ask is about memorization: do I have to memorize a certain number of prime values? 

It’s good to know up to a certain value. but unnecessary to go beyond that into conspicuously larger numbers, because the GMAT as a test is less interested in your ability to memorize large and weird primes and more interested in your reasoning skills and your ability to draw conclusions about novel problems on the fly. If you know the following, you should be set (with some optional values thrown in at the end):

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, (41, 43)

Thankfully, you’ll notice the list is actually pretty manageable. 

(And an interesting note that many people forget that 27 is actually not prime. But don’t beat yourself up if this happens to you: Terence Tao, one of the world’s leading mathematicians and an expert on prime numbers, actually slipped briefly on national television once and said 27 was prime before catching himself. And he’s one of the best in the world. So even the best of the best make these mistakes.)

Now, here’s an Official Guide problem that takes the basics of Prime Numbers and forces you to do a little reasoning. As usual, give it shot before reading the explanation:

The product of all the prime numbers less than 20 is closest to which of the following powers of 10 ?

A) 109
B) 108
C) 107
D) 106
E) 105

Explanation

For this one, you have a little hint going in, as we’ve provided you with the necessary list of primes you’ll use to find the product.

And the language given (“closest to”) is a huge hint that you can estimate:

2*3*5*7*11*13*17*19 ~= ??

Since powers of 10 are involved, let’s try to group the numbers to get 10s as much as possible. The following is just one of many ways you could do this, but the universal easiest place to start is the 2 and the 5, so let’s multiply those. We’ll mark numbers we’ve accounted for in red:

(2*5)*3*7*11*13*17*19 ~= ??

10*3*7*11*13*17*19 ~= ??

Next, we can look at the 19 and label it as roughly 20, or 2*10:

10*3*7*11*13*17*19 ~= ??

10*3*7*11*13*17*20 ~= ??

10*3*7*11*13*17*2*10 ~= ??

We could also take the 11 and estimate it as another 10:

10*3*7*11*13*17*2*10 ~= ??

10*3*7*10*13*17*2*10 ~= ??

At this point, we should be able to eyeball this. Remember, it’s estimation. We may not know 17*3 and 13*7 offhand. But we know that they’re both around or less than 100 or 102. And a look at the answer choices lets us know that each answer is a factor of 10 apart, so the range is huge. (In other words, estimation error is not likely to play a factor.)

So it’s not unreasonable in the context of this problem to label those remaining products as two values of 102:

10*3*7*10*13*17*2*10 ~= ??

10*(102)*10*(102)*2*10 ~= ??

And at this point, the 2 is negligible, since that won’t be enough to raise the entire number to a higher power of 10. What do we have left?

101*(102)*101*(102)*101 ~= 107 

The correct answer is C. 

Next time, we’ll get into Prime Factorizations, which you can do with any positive integer.

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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Consecutive Integers and Data Sufficiency (Avoiding Algebra) Article
Posted on
25
Mar 2021

Consecutive Integers and Data Sufficiency (Avoiding Algebra)

By: Rich Zwelling (Apex GMAT Instructor)
Date: 25 March 2021

Last time, we left off with the following GMAT Official Guide problem, which tackles the Number Theory property of consecutive integers. Try the problem out, if you haven’t already, then we’ll get into the explanation:

The sum of 4 different odd integers is 64. What is the value of the greatest of these integers?
(1) The integers are consecutive odd numbers
(2) Of these integers, the greatest is 6 more than the least.

Explanation (NARRATIVE or GRAPHIC APPROACHES):

Remember that we talked about avoiding algebra if possible, and instead taking a narrative approach or graphic approach if possible. By that we meant to look at the relationships between the numbers and think critically about them, rather than simply defaulting to mechanically setting up equations.

(This is especially helpful on GMAT Data Sufficiency questions, on which you are more interested in the ability to solve than in actually solving. In this case, once you’ve determined that it’s possible to determine the greatest of the four integers, you don’t have to actually figure out what that integer is. You know you have sufficiency.)

Statement (1) tells us that the integers are consecutive odd numbers. Again, it may be tempting to assign variables or something similarly algebraic (e.g. x, x+2, x+4, etc). But instead, how about we take a NARRATIVE and/or GRAPHIC approach? Paint a visual, not unlike the slot method we were using for GMAT combinatorics problems:

___ + ___ +  ___ + ___  =  64

Because these four integers are consecutive odd numbers, we know they are equally spaced. They also add up to a definite sum.

This is where the NARRATIVE approach pays off: if we think about it, there’s only one set of numbers that could fit that description. We don’t even need to calculate them to know this is the case.

You can use a scenario-driven approach with simple numbers to see this. Suppose we use the first four positive odd integers and find the sum:

_1_ + _3_ +  _5_ + _7_  =  16

This will be the only set of four consecutive odd integers that adds up to 16. 

Likewise, let’s consider the next example:

_3_ + _5_ +  _7_ + _9_  =  24

This will be the only set of four consecutive odd integers that adds up to 24. 

It’s straightforward from here to see that for any set of four consecutive odd integers, there will be a unique sum. (In truth, this principle holds for any set of equally spaced integers of any number.) This essentially tells us [for Statement (1)] that once we know that the sum is set at 64 and that the integers are equally spaced, we can figure out exactly what each integer is. Statement (1) is sufficient.

(And notice that I’m not even going to bother finding the integers. All I care about is that I can find them.)

Similarly, let’s take a graphic/narrative approach with Statement (2) by lining the integers up in ascending order:

_ + __ +  ___ + ____  =  64

But very important to note that we must not take Statement (1) into account when considering Statement (2) by itself initially, so we can’t say that the integers are consecutive. 

Here, we clearly represent the smallest integer by the smallest slot, and so forth. We’re also told the largest integer is six greater than the smallest. Now, again, try to resist the urge to go algebraic and instead think narratively. Create a number line with the smallest (S) and largest (L) integers six apart:

S—————|—————|—————|—————|—————|—————L

Narratively, where does that leave us? Well, we know that the other two numbers must be between these two numbers. We also know that each of the four numbers is odd. Every other integer is odd, so there are only two other integers on this line that are odd, and those must be our missing two integers (marked with X’s here):

S—————|—————X—————|—————X—————|—————L

Notice anything interesting? Visually, it’s straightforward to see now that we definitely have consecutive odd integers. Statement (2) actually gives us the same information as Statement (1). Therefore, Statement (2) is also sufficient. The correct answer is D

And again, notice how little actual math we did. Instead, we focused on graphic and narrative approaches to help us focus more on sufficiency, rather than actually solving anything, which isn’t necessary.

Next time, we’ll make a shift to my personal favorite GMAT Number Theory topic: Prime Numbers…

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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Posted on
17
Mar 2021

Sequence Problem on the GMAT

Key Information to Know About Sequences

Hey guys! When we see sequences on the GMAT whether in the problem solving or the data sufficiency section they have two important characteristics. One is how they work, the other is anchoring the sequence to a particular set of numbers. Let’s start by taking a look at the most basic sequence out there. Counting! The way counting works is that every time we go up a term in the sequence we add one and if we take number one as our first term then the term number and the value move in tandem. First term is 1. Second term is 2, 50th term is 50. We could also anchor it differently. Let’s say we wanted to say the first term is five, then the second term is six, third term is seven, fourth term is eight, so on and so forth.

Play around with this: do it with the even numbers or the odd numbers and try different anchor points. Sequences can seem more complicated than they are because we don’t think of them in this basic sort of way and because they’re expressed oftentimes with weird notation. So when we see some sequence with a little number below, it that’s called a Subscript. That tells you the number of term of the sequence that they’re talking about. So going back to our counting example, S1 the first term in the sequence equals 1. S2 equals 2, S sub 3 equals 3. If we were doing the even numbers starting at two S sub 1 equals 2, S sub 2 equals 4, S sub 3 equals 6, S sub 10 equals 20. So don’t get freaked out by the notation just because it looks like it comes out of some very crazy math book.

What We Need

The problem we’re going to look at today is asking us for the value of a specific term within a sequence and the what do we need comes in two parts. We need both how the sequence works and we need to know (not necessarily where it starts) but some anchor point to tell us what some term is relative to the sequence so we can figure out any other term above or below that. We’re going to say that again: we don’t need the beginning or ending term, just any term with a specific value that along with the rules allows us to get to any other place in the sequence.

Which Statement to Begin With

Generally, when we are looking for two pieces of information we should be attuned to looking for a (C) or an (E) answer choice but that’s not always the case. If we dive into the introduced information, we’ll start with number 2 and the reason is that it’s going to be easier to evaluate. At first glance it’s simpler and you always want to start out with the easier piece and work your way up. Number 2 gives us a term. It gives us the first term, but we don’t know how the sequence works therefore it’s insufficient. Number 1 gives us the 298th term and describes how the sequence works so we’re getting both pieces together in number 1.

Answer

Therefore (A) 1 alone is sufficient. Notice here that because we’re primed for (C) / (E) answer choice, looking for two pieces of information, the GMAT is betting that we think to ourselves hey I need the first term in this sequence and the 298th term doesn’t tell me anything. They’re looking for us to answer (C) that we need them both. But once you understand sequences you’ll never fall for it. Hope this helped guys, check out other sequence problems below and we’ll see you again soon.

If you enjoyed this GMAT sequences video, try your hand at this Ratio problem next.

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What happens when Permutations have repeat elements?
Posted on
25
Feb 2021

What happens when Permutations have repeat elements?

By: Rich Zwelling, Apex GMAT Instructor
Date: 25th February, 2021

Permutations With Repeat Elements

As promised in the last post, today we’ll discuss what happens when we have a PERMUTATIONS situation with repeat elements. What does this mean exactly? Well, let’s return to the basic example in our intro post on GMAT combinatorics:

If we have five distinct paintings, and we want to know how many arrangements can be created from those five, we simply use the factorial to find the answer (i.e. 5! = 5*4*3*2*1 = 120). Let’s say those paintings were labeled A, B, C, D, and E. 

Now, each re-arrangement of those five is a different PERMUTATION, because the order is different:

ABCDE
EBADC
CADBE


etc

Remember, there are 120 permutations because if we use the blank (or slot) method, we would have five choices for the first blank, and once that painting is in place, there would be four left for the second blank, etc…

_5_  _4_  _3_  _2_  _1_ 

…and we would multiply these results to get 5! or 120.

However, what if, say we suddenly changed the situation such that some of the paintings were identical? Let’s replace painting C with another B and E with another D:

ABBDD

Suddenly, the number of permutations decreases, because some paintings are no longer distinct. And believe it or not, there’s a formulaic way to handle the exact number of permutations. All you have to do is take the original factorial, and divide it by the factorials of each repeat. In this case, we have 5! for our original five elements, and we now must divide by 2! for the two B’s and another 2! for the two D’s:

  5!
——
2! 2!     

= 5*4*3*2*1
   ————-
  (2*1)(2*1)

= 5*2*3
= 30 permutations

As another example, try to figure out how many permutations you can make out of the letters in the word BOOKKEEPER? Give it a shot before reading the next paragraph.

In the case of BOOKKEEPER, there are 10 letters total, so we start with a base of 10! 

We then have two O’s, two K’s and three E’s for repeats, so our math will look like this:

   10!
———
2! 2! 3! 

Definitely don’t calculate this, though, as GMAT math stays simple and likes to come clean. Remember, we’ll have to divide out the repeats. You are extremely unlikely to have to do this calculation for a GMAT problem, however, since it relies heavily on busy-work mechanics. The correct answer choice would thus look like the term above. 

Let’s now take a look at an Official Guide question in which this principle has practical use. I’ll leave it to you to discover how. As usual, give the problem a shot before reading on:

A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16

Quick Probability Review

Remember from our post of GMAT Probability that, no matter how complicated the problem, probability always boils down to the basic concept of:

    Desired Outcomes
———————————–
Total Possible Outcomes

In this case, each child has two equally likely outcomes: boy and girl. And since there are four children, we can use are blank method to realize that we’ll be multiplying two 4 times:

_2_  _2_  _2_  _2_   =  16 total possible outcomes (denominator)

This may give you the premature notion that C or E must be correct, simply because you see a 16 in the denominator, but remember, fractions can reduce! We could have 4 in the numerator, giving us a fraction of 4/16, which would reduce to 1/4. And every denominator in the answer choices contains a factor of 16, so we can’t eliminate any answers based on this. 

Now, for the Desired Outcomes component, we must figure out how many outcomes consist of exactly two boys and two girls. The trick here is to recognize that it could be in any order. You could have the two girls followed by the two boys, vice versa, or have them interspersed. Now, you could brute-force this and simply try writing out every possibility. However, you must be accurate, and there’s a chance you’ll forget some examples. 

What if we instead write out an example as GGBB for two girls and two boys? Does this look familiar? Well, this should recall PERMUATIONS, as we are looking for every possible ordering in which the couple could have two girls and two boys. And yes, we have two G’s and two B’s as repeats. Here’s the perfect opportunity to put our principle into play:

We have four children, so we use 4! for our numerator, then we divide by 2! twice for each repeat:

  4!
——
2! 2! 

This math is much simpler, as the numerator is 24, while the denominator is 4. (Remember, memorize those factorials up to 6!)

This yields 6 desired outcomes of two boys and two girls. 

With 6 desired outcomes of 16 total possible outcomes, our final probability fraction is 6/16, which reduces to 3/8. The correct answer is A.

Next time, we’ll look into combinatorics problems that involve restrictions, which can present interesting conceptual challenges. 

Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements
Permutations With Restrictions
Combinations with Restrictions
Independent vs Dependent Probability
GMAT Probability Math – The Undesired Approach
GMAT Probability Meets Combinatorics: One Problem, Two Approaches

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