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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gmat consecutive integers article
Posted on
23
Mar 2021

Consecutive Integers (plus more on Odds and Evens)

By: Rich Zwelling, Apex GMAT Instructor
Date: 23rd March, 2021

In our last post, we left you with a GMAT Official Guide Data Sufficiency problem to tackle regarding Odd/Even Number Theory. Here it is, if you didn’t get a chance to do it before:

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

Solution:

The title of today’s post gives a little hint. We discussed last time that for a product of integers to be even, all you need is a single even integer in the set. (Conversely, for the entire product to be odd, every integer must be odd.) 

How does that affect how we interpret Statement (1)? Well, this is where taking a purely Algebraic approach can get you into trouble. Why not take a NARRATIVE APPROACH? What is the equation really telling us narratively about the relationship between x and y? Develop the  habit of thinking about numbers narratively instead of purely algebraically, as this can make numerical relationships easier to understand.

Statement (1), in essence, is really telling us that y and x are consecutive integers. If I take y and add one to it to get x, they must be consecutive. So what are the implications for the number property of the question (even/odd)? Well, between any pair of consecutive integers, one must be odd and one must be even. I don’t know which is which, but it doesn’t matter in this case, because I care only about the product. Whether it’s Odd*Even or Even*Odd, the final product of xy will always be Even. 

Even if you don’t see this narratively, you could use a scenario-driven approach and test simple numbers using the equation. If x = y + 1, try y = 2 and x = 3 to get xy = 6. Now, that’s just one instance of xy being even, so that doesn’t prove it’s always even. But then if I use y = 3 and x = 4 to get xy = 12, I again get an even result for xy. Using y = 4 and x = 5 would again yield an even xy result. Hopefully what I will realize, at this point, is that I am switching between x=odd, y=even and x=even, y=odd, and yet I still end up with xy=even

Statement (1) is SUFFICIENT

And remember: the GMAT is very fond of testing interesting properties of consecutive integers.

For Statement (2), we discussed that division is less amenable to hard-fast rules of odd/even properties. For that reason, you could definitely use a scenario-driven approach. But hopefully, this approach would lead you towards a consideration of our previously discussed undesired possibility. Here’s what I mean:

If x/y is even, does that guarantee that xy is even? If you’re picking numbers, you must pick ones that fit the statement. It’s tempting to pick ones that contradict the statement (i.e. proving the statement wrong), but remember that the question is up for debate, not the statement. The statement is given to you as iron-clad fact. 

So what if, for example, x = 4 and y = 2? That works for Statement (2), because x/y would certainly be even (4/2 = 2). And that would lead you to xy = 8, which is of course even. 

You could pick many such examples. But here’s where the undesired part comes into play. Is it possible for us to pick numbers here such that xy becomes odd? Well, for xy to be odd, both x and y would have to be odd. But if x and y were both odd, could x/y be even as Statement (2) says? Even if you can’t see the answer right away, try some numbers here, knowing that x and y must be integers:

x=5, y=35/3 (not even)
x=3, y=5 → 3/5  (not even)
x=3, y=7 → 3/7 (not even)
x=9, y=3 → 3 (not even)

You’ll see that no matter what numbers you try, you’ll never get an even result for x/y. From a number theory perspective, this is because to get an even result, you must retain a factor of 2 in the numerator of the fraction. But we don’t even start with a factor of 2 if we have only odd numbers to begin with. 

In conclusion, there’s no way that x and y can both be odd in Statement (2), meaning xy is not odd, and that guarantees xy is even. Statement (2) is also SUFFICIENT.

The correct answer is D. Each statement is SUFFICIENT on its own.

For next time, give the following Official Guide problem a shot, and use it as a chance to practice a NARRATIVE APPROACH:

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.

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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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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When Probability Meets Combinatorics: One Problem, Two Approaches article
Posted on
16
Mar 2021

When Probability Meets Combinatorics: One Problem, Two Approaches

By: Rich Zwelling, Apex GMAT Instructor
Date: 16th March, 2021

Now, we’d like to take a look at an Official GMAT Probability problem to pull everything together. The following is a good example for two reasons:

 1. It illustrates a quirky case that is difficult more conceptually than mathematically, and thus is better for the GMAT.

 2. It can be tackled either through straight probability or through a combination of probability and combinatorics.

Here’s the question:

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

A) 1/24
B)
1/8
C) 1/4
D) 1/3
E) 3/8

First, as always, give the problem a shot before reading on for the explanation. If possible, see if you can tackle it with both methods (pure probability and probability w/ combinatorics). 

Explanation #1:

First, we’ll tackle pure probability. Let’s label the letters A, B, C, and D, and let’s say that A is the letter we wish to match with its correct envelope. The other three will be matched with incorrect envelopes. We now must examine the individual probabilities of the following events happening (green for correct, red for incorrect):

_A_   _B_   _C_   _D_

For the above, each slot represents a letter matched with an envelope. There are four envelopes and only one is correct for letter A. That means Tanya has a 1/4 chance of placing letter A in its correct envelope:

_1/4__   _B_   _C_   _D_

We now desire letter B to be placed in an incorrect envelope. Two of the remaining three envelopes display incorrect addresses, so there is a 2/3 chance of that happening:

_1/4__   _2/3_   _C_   _D_

We then desire letter C to also be placed in an incorrect envelope. Only one of the remaining two envelopes displays an incorrect address, so there is a 1/2 chance of that happening:

_1/4__   _2/3_   _1/2_   _D_

At that point, the only remaining option is to place the last remaining letter in the last remaining envelope (i.e. a 100% chance, so we place a 1 in the final slot):

_1/4__   _2/3_   _1/2_   _1_

Multiplying the fractions, we can hopefully see that some cancelling will occur:

¼ x ⅔ x ½ x 1

= 1 x 2 x 1
   ———–
   4 x 3 x 2

= 1/12

But lo and behold, 1/12 is not in our answer choices. Did you figure out why?

We can’t treat letter A as the only possible correct letter. Any of the four letters could possibly be the correct one. However, the good news is that in any of the four cases, the math will be exactly the same. So all we have to do is take the original 1/12 we just calculated and multiply it by 4 to get the final answer: 4 x 1/12 = 4/12 = 1/3. The correct answer is D.

Explanation #2:

So what about a combinatorics approach?

As we’ve discussed in our previous GMAT probability posts, all probability can be boiled down to Desired Outcomes / Total Possible Outcomes. And as we discussed in our posts on GMAT combinatorics, we can use factorials to figure out the total possible outcomes in a situation such as this, which is actually a simple PERMUTATION. There are four envelopes, so for the denominator of our fraction (total possible outcomes), we can create a slot for each envelope and place a number representing the letters in each slot to get:

_4_  _3_  _2_  _1_  =  4! = 24  possible outcomes

This lets us know that if we were to put the four letters into the four envelopes at random, as the problem says, there would be 24 permutations, giving us the denominator of our fraction (total possible outcomes). 

So what about the desired outcomes? How many of those 24 involve exactly one correctly placed letter? Well, let’s again treat letter A as the correctly placed letter. Once it’s placed, there are three slots (envelopes) left: 

___  ___  ___ 

But the catch is: the next envelope has only two letters that could go into it. Remember, one of the letters correctly matches the envelope in address, and we want a mismatch:

_2_  ___  ___ 

Likewise, that would leave two letters available for the next envelope, but only one of them would have the wrong address:

_2_  _1_  ___ 

And finally, there would be only one choice left for the final envelope:

_2_  _1_  _1_ 

That would mean for the correctly-placed A letter, there are only two permutations in which each of the other letters is placed incorrectly:

_2_ x  _1_ x  _1_ = 2 possible outcomes.

But as before, we must consider that any of the four letters could be the correct letter, not just letter A. So we must multiply the 2 possible outcomes by four to get 8 desired outcomes involving exactly one letter being placed in its correct envelope. That gives us our numerator of Desired Outcomes. Our denominator, remember, was 24 total possible outcomes. So our final answer, once again, is 8/24 = 1/3.

This is a great example of how GMAT combinatorics can intersect with probability.

To tide you over until next time, give this Official GMAT problem a try. It will also give a nice segue into Number Theory, which we’ll begin to talk more about going forward. Explanation next time…

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

 

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
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
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Posted on
10
Mar 2021

GMAT Ratio Problem – Mr. Smiths Class

GMAT Ratio DS Problem

Expressing Different Notations

Hey guys!

Expressing different notations is often challenging when you’re first starting out on the GMAT and by different notations mean percentages fractions decimals ratios. We learn all these separately and we tend to of them as separate systems of math when in fact they’re all different expressions of the same math. One half is no different from 0.5 is no different from 50 percent there are different ways of the same thing.

Breaking Down The Problem

In this problem all their testing is our ability to shift notations. We’re being asked what the ratio, keyword ratio, is between boys and girls in the or what do we need is just that a ratio it’s fairly straightforward. So they’re probably going to come to us with weird information that doesn’t quite look like a ratio. The big thing to note before we dive in is that when we’re being asked for a ratio. In fact, when we’re being asked for any sort of relative notation, fractions, percentages, anything that needs a base that is compared to a whole. We don’t need precise numbers.

Possible ways to solve this problem

So this leaves us open either to run scenarios if we want to or to deal entirely in the relative. So we’re looking for an expression of that ratio in a non-ratio sort of language. Number one tells us there are three times as many boys and girls. We can run a scenario with 3 boys, 1 girl, 75 boys, 25 girls, but we’re being given that ratio. It’s being expressed in language rather than with the term ratio or with the two dots : in between but it’s still a ratio. So it’s sufficient!

What Did You Miss?

Correction!! Number one states there are three times as many girls as there are boys. Why do we leave that error in? To point out that here it doesn’t matter. We’re not looking to determine whether the ratio is 1 boy to 3 girls or 3 girls to 1 boy or 3 boys to 1 girl. The only thing that matters, the threshold issue on this problem, is getting to a single specific ratio. What that is or in this case even reversing the boys and girls doesn’t matter because it’s a referendum on the type of information that we have. The moment we have a quantitative comparison of boys and girls coming from number one we know that number one is sufficient. Being able to have flexibility and even focus on the more abstract thing you’re looking for sometimes leads to careless errors on the details though and this is important. Many times those careless errors don’t matter, freeing yourself up to make those and understanding that you don’t have to manage the nitty-gritty once you have the big abstract understanding is very important.

Looking at Statement No. 2

Number two goes fractional, telling us that 1/4 of the total class is boys. We can break that into a ratio by understanding that a ratio compares parts to parts whereas a fraction is part of a whole so one out of four has a ratio of one to three. If this isn’t immediately obvious, imagine a pizza and cut it into four slices. One slice is one quarter of the total pizza the comparison of the one slice to the other three slices is the ratio one to three so if you get one slice and your friends get the other ones. The ratio of your slice to the others is 1:3. You have 1/4 of the total so two is also sufficient. Therefore, the answer choice here is D.

Hope this helped guys! Practice this skill of going in between these different notations because it’s one that pays off in dividends. Check out the links below for other problems and we’ll see you again real soon.

If you enjoyed this GMAT Ratio DS Problem, try your hand at this Triangle DS Problem.

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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