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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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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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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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odds and evens on the gmat
Posted on
18
Mar 2021

Odds and Ends (…or Evens)

By: Rich Zwelling, Apex GMAT Instructor
Date: 18th March, 2021

Last time, we signed off with an Official Guide GMAT problem that provided a nice segue into Number Theory, specifically today’s topic of GMAT Odds and Evens. Now we’ll discuss the solution. Here’s the problem, in case you missed it and want to try it now:

If x is to be chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7}, what is the probability that xy will be even?

A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 5/6

Method #1 (Certainly passable, but not preferable)

Since there are so few numbers involved here, you could certainly take a brute-force approach if pressed for time and unsure of a faster strategy. It doesn’t take long to map out each individual product of x*y systematically and then tally up which ones are even. Here I’ll use red to indicate not even and green to indicate even:

1*5 = 5
1*6 = 6
1*7 = 7

2*5 = 10
2*6 = 12
2*7 = 14

3*5 = 15
3*6 = 18
3*7 = 21

4*5 = 20
4*6 = 24
4*7 = 28

Since we know that all probability is Desired Outcomes / Total Possible Outcomes, and since we have 8 even results out of 12 total possible outcomes, our final answer would be 8/12 or 2/3.

However, this is an opportune time to introduce something about odd and even number properties and combine it with the method from our “undesired” probability post…

Method #2 (Far preferable)

First, some number theory to help explain:

You might have seen that there are rules governing how even and odd numbers behave when added, subtracted, or multiplied (they get a little weirder with division). They are as follows for addition and subtraction:

Even ± Odd = Odd

Odd ± Odd = Even

And with multiplication, the operative thing is that, when multiplying integers, just a single even number will make the entire product even. So for example, the following is true:

Odd * Odd * Odd * Odd * Odd * Odd * Odd * Odd * Odd * … * Odd = Odd

But introduce just a single even number into the above product, and the entire product becomes even:

Even * Odd * Odd * Odd * Odd * Odd * Odd * Odd * Odd * Odd * … * Odd = Even

This makes sense when you think about it, because you are introducing a factor of 2 to the product, so the entire product must be even

When considering the above rules, you could memorize them, or you could turn to SCENARIO examples with simple numbers to illustrate the general pattern. For example, if you forget what Odd * Odd is, just multiply 3 * 5 to get 15, which is odd, and that will help you remember. 

This can also help you see that introducing just a single even number makes an entire product even: 

3 * 3 * 3 = 27

2 * 3 * 3 * 3 = 54

So how does all of that help us get the answer to this problem faster?

Well, the question asks for the desired outcome of xy being even. That presents us with three possibilities:

 1. x could be even AND y could be odd
 2. x could be odd AND y could be even
 3. x could be even AND y could be even

This is why I, personally, find it less helpful to think of individual multiplication rules for even numbers and more helpful to think in terms of: “The only way a product of integers is odd is if every integer in the set is odd.”

Because now, we can just think about our undesired outcome, the only outcome left:

4. x is odd AND y is odd, making the product xy odd

And how many such outcomes are there in this problem? Well at this point, we can treat it like a simple combinatorics problem. There are two odd numbers in the x group (1 and 3) and two odd numbers in the y group (5 and 7):

_2_ * _2_ = 4 possible odd xy products 

And for the total, there are four possible x values and three possible y values:

_4_ * _3_ = 12 total possible xy products 

That gives us a 4/12 or 1/3 probability of getting our undesired odd xy product. And as discussed in the previous post, we can now simply subtract that 1/3 from 1 to get:

1 – 1/3  = 2/3 probability that we get our desired even xy product.

For a little “homework,” try the following Official Guide GMAT problem. It has an underlying topic that we’ll discuss next time:

If x and y are integers, is xy even?
(1) x = y + 1.
(2) x/y is an even integer.

If you enjoyed this GMAT odds and evens article watch Mike solve this Number Theory problem with multiple solution paths.

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
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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An “Undesired” Approach to GMAT Probability gmat article
Posted on
11
Mar 2021

An “Undesired” Approach to GMAT Probability

By: Rich Zwelling, Apex GMAT Instructor
Date: 11th March, 2021

In our last post, we discussed a solution for the following question, which is a twist on an Official Guide GMAT probability problem:

Xavier, Yvonne, and Zelda individual probabilities for success on a certain problem are 1/4, 1/2 and 5/8, respectively. Xavier will attempt the problem first. If he solves it, Yvonne and Zelda will not attempt it. If Xavier cannot solve it, Yvonne will attempt it next. If she solves it, Zelda will not attempt it. If Yvonne cannot solve it, Zelda will then attempt it. What is the probability that Zelda does not get to attempt the problem?

A) 3/16
B)
5/8
C) 3/8
D) 5/64
E) 3/64

We also mentioned that there was an alternate way to solve it. Did you find it? In truth, it relates to something we discussed in a previous post we did on GMAT Combinatorics, specifically Combinations with Restrictions. In that post, we discussed the idea of considering combinations in which you’re not interested. It might seem counterintuitive, but if you subtract those out from the total number of combinations possible, you’re left with the number of combinations in which you are interested:

You can actually do something similar with probability. Take the following basic example:

Suppose I told you to flip a fair coin five times, “fair” meaning that it has an equal chance of landing heads-up or tails-up. I then wanted to know the probability that I flip at least one head. Now, when you think about it, the language “at least one” involves so many desired possibilities here. It could be 1 head, 2 heads, …, all the way up to 5 heads. I’d have to calculate each of those probabilities individually and add them up.

Or…

I could consider what is not desired, since the possibilities are so much fewer:

0 heads   |   1 head      2 heads      3 heads      4 heads      5 heads

All of the above must add to 100% or 1, meaning all possible outcomes. So why not figure out the probability that I get 0 heads (or all tails), and then subtract it from 100% or 1 (depending on whether I’m using a percentage or decimal/fraction)? I’ll then be left with all the possibilities in which I’m actually interested, without the need to do any more calculations.

Each time I flip the coin, there is a ½ chance that I flip a tail. This is the same each of the five times I flip the coin. I then multiply all of the probabilities together:

½ x ½ x ½ x ½ x ½ = 1 / 25  = 1 / 32

Another way to view this is through combinatorics. Remember, probability is always Desired outcomes / Total possible outcomes. If we start with the denominator, there are two outcomes each time we flip the coin. That means for five flips, we have 25 or 32 possible outcomes, as illustrated here with our slot method:

_2_  _2_  _2_  _2_  _2_ = 32

Out of those 32 outcomes, how many involve our (not) desired outcome of all tails? Well, there’s only one possible way to do that: 

_T_  _T_  _T_  _T_  _T_    ← Only 1 outcome possible

It really is that straightforward: with one outcome possible out of 32 total, the probability is 1/32 that we flip all tails. 

Now remember, that is our, not desired. Our desired is the probability of getting at least one head

0 heads   |   1 head      2 heads      3 heads      4 heads      5 heads

So, since the probability of getting 0 heads (all tails) is 1/32, we simply need to subtract that from 1 (or 32/32) to get our final result. The probability that we flip at least one head if we flip a fair coin five times is 31/32.

Application to problem from previous post

So now, how do we work that into the problem we did last time? Well, in the previous post, we took a more straightforward approach in which we considered the outcomes we desired. But can we use the above example and consider not desired instead? Think about it and give it a shot before reading the explanation:

Xavier, Yvonne, and Zelda individual probabilities for success on a certain problem are 1/4, 1/2 and 5/8, respectively. Xavier will attempt the problem first. If he solves it, Yvonne and Zelda will not attempt it. If Xavier cannot solve it, Yvonne will attempt it next. If she solves it, Zelda will not attempt it. If Yvonne cannot solve it, Zelda will then attempt it. What is the probability that Zelda does not get to attempt the problem?

A) 3/16
B) 5/8
C) 3/8
D) 5/64
E) 3/64

Explanation

In this question, our desired case is that Zelda does not attempt the problem. That means, quite simply, that our not desired case is that Zelda does get to attempt it. This requires us analytically to consider how such a case would arise. Let’s map out the possibilities with probabilities:

An “Undesired” Approach to GMAT Probability treeNotice that two complementary probabilities are presented for each box. For example, since there is a 1/4 chance Xavier solves the problem (left arrow), we include the 3/4 probability that he does not solve the problem (right arrow). 

If Zelda does get to attempt it, it’s clear from the above that first Xavier and Yvonne must each not solve it. There is a 3/4 and a 1/2 chance, respectively, of that happening. This is also a dependent situation. Xavier must not solve AND Yvonne must not solve. Therefore, we will multiply the two probabilities together to get ¾ x ½ = ⅜. So there is a 3/8 chance of getting our not desired outcome of Zelda attempting the problem.

So, we can finally subtract this number from 1 (or 8/8) and see that there is a 5/8 chance of Zelda not getting to attempt the problem. The correct answer is B.

Next time, we’ll discuss how GMAT Probability and Combinatorics can combine to form some interesting problems…

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

Read more
Independent vs Dependent Probability article for the GMAT
Posted on
09
Mar 2021

Independent vs. Dependent Probability

By: Rich Zwelling, Apex GMAT Instructor
Date: 8th March, 2021

Independent vs. Dependent Probability

As promised last time, we’ll return to some strict GMAT probability today. Specifically, we’ll discuss the difference between independent and dependent probability. This simply refers to whether or not the events involved are dependent on one another. For example, let’s take a look at the following Official Guide problem:

Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4, 1/2 and 5/8, respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem?

A) 11/8
B)
7/8
C) 9/64
D) 5/64
E) 3/64

In this case, we are dealing with independent events, because none of the probabilities affect the others. In other words, what Xavier does doesn’t affect Yvonne’s chances. We can treat each of the given probabilities as they are. 

So mathematically, we would multiply, the probabilities involved. (Incidentally, the word “and” is often a good indication that multiplication is involved. We want Xavier AND Yvonne AND not Zelda to solve the problem.) And if Zelda has a chance of solving the problem, that means she has a chance of not solving it. 

The answer would therefore be ¼ x ½ x ⅜  = 3/64 or answer choice E. 

What if, however, we changed the problem to look like this:

Xavier, Yvonne, and Zelda individual probabilities for success on a certain problem are 1/4, 1/2 and 5/8, respectively. Xavier will attempt the problem first. If he solves it, Yvonne and Zelda will not attempt it. If Xavier cannot solve it, Yvonne will attempt it next. If she solves it, Zelda will not attempt it. If Yvonne cannot solve it, Zelda will then attempt it. What is the probability that Zelda does not get to attempt the problem?

A) 3/16
B)
5/8
C) 3/8
D) 5/64
E) 3/64

As you can see, the problem got much more complicated much more quickly, because now, the question stem is dependent upon a very specific series of events. Now, the events affect one another. Xavier will attempt the problem, but what happens at this stage affects what happens next. If he solves it, everything stops. But if he doesn’t, the problem moves to Yvonne. So in effect, there’s a ¼ chance that he’s the only person to attempt the problem, and there’s a ¾ chance the problem moves to Yvonne.

This is most likely how the GMAT will force you to think about probability: not in terms of formulas or complicated mathematical concepts, but rather in terms of narrative within a new problem with straightforward numbers. 

That brings us to consideration of the question stem itself. What would have to happen for Zelda not to attempt the problem? Well, there are a couple of possibilities:

 1. Xavier solves the problem

If Xavier solves the problem, the sequence ends, and Zelda does not see the problem. This is one case we’re interested in, and there’s a ¼ chance of that happening. 

 2. Xavier does not solve, but then Yvonne solves

There’s a ½ chance of Yvonne solving, but her seeing the problem is dependent upon the ¾ chance that Xavier does not solve. So in reality, we must multiply the two numbers together to acknowledge that the situation we want is “Xavier does not solve AND Yvonne does solve.” This results in ¾ x ½ = ⅜ 

The two above cases constitute two independent situations that we now must add together. For Zelda not to see the problem, either Xavier must solve it OR Yvonne must solve it. (The word “or” is often a good indication that addition will be used).

This leads us to our final probability of ¼ + ⅜ = that Zelda does not get to attempt the problem.

There is an alternative way to solve this problem, which we’ll talk about next time. It will segue nicely into the next topic, which we’ve already hinted at in our posts on GMAT combinatorics. Until then…

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

Read more
GMAT Combinations with Restrictions Article
Posted on
04
Mar 2021

Combinations with Restrictions

By: Rich Zwelling, Apex GMAT Instructor
Date: 4th March, 2021

In our previous post, we discussed how GMAT combinatorics problems can involve subtracting out restrictions. However, we discussed only PERMUTATIONS and not COMBINATIONS.

Today, we’ll take a look at how the same technique can be applied to COMBINATION problems. This may be a bit more complicated, as you’ll have to use the formula for combinations, but the approach will be the same.

Let’s start with a basic example. Suppose I were to give you the following problem:

The board of a large oil company is tasked with selecting a committee of three people to head a certain project for the following year. It has a list of ten applicants to choose from. How many potential committees are possible?

This is a straightforward combination problem. (And we know it’s a COMBINATION situation, because we do not care about the order in which the three people appear. Even if we shift the order, the same three people will still comprise the same committee.)

We would simply use the combination math discussed in our Intro to Combination Math post:

                         10!
 10C3 =       ————-
                     3! (10-3!)

 

   10!
———
3! (7!)

 

10*9*8
———
3!

 

10*9*8
———
3*2*1

= 120 Combinations 

However, what if we shifted the problem slightly to look like the following? (As always, give the problem a shot before reading on…):

The board of a large oil company is tasked with selecting a committee of three people to head a certain project for the following year. It has a list of ten applicants to choose from, three of whom are women and the remainder of whom are men. How many potential committees are possible if the committee must contain at least one woman?

A) 60
B) 75
C) 85
D) 90
E) 95

In this case, there’s a very important SIGNAL. The language “at least one” is a huge giveaway. This means there could be 1 woman, 2 women, or 3 women which means we would have to examine three separate cases. That’s a lot of busy work. 

But as we discussed in the previous post, why not instead look at what we don’t want and subtract it from the total? In this case, that would be the case of 0 women. Then, we could subtract that from the total number of combinations without restrictions. This would leave behind the cases we do want (i.e. all the cases involving at least one woman). 

We already discussed what happens without restrictions: There are 10 people to choose from, and we’re selecting a subgroup of 3 people, leading to 10C3  or 120 combinations possible. 

But how do we consider the combinations we don’t want? Well, we want to eliminate every combination that involves 0 women. In other words, we want to eliminate every possible committee of three people that involves all men. So how do we find that?

Well, there are seven men to choose from, and since we are choosing a subgroup of 3, we can simply use 7C3 to find the number of committees involving all men:

                       7!
7C3 =       ————-
                 3! (7-3!)

 

  7!
———
3! (4!)

 

7*6*5
———
    3!

= 7*5 = 35 Combinations involving all men

So, out of the 120 committees available, 35 of them involve all men. That means 120-35 = 85 involve at least one woman. The correct answer is C. 

Next time, we’ll return to probability and talk about how the principle of subtracting out elements that we don’t want can aid us on certain questions. Then we’ll dovetail the two and talk about how probability and combinatorics can show up simultaneously on certain questions.

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

Read more