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

Read more
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

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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
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
Permutations with Restrictions GMAT Article
Posted on
02
Mar 2021

Permutations with Restrictions

By: Rich Zwelling, Apex GMAT Instructor
Date: 2nd March, 2021

So far, we’ve covered the basics of GMAT combinatorics, the difference between permutations and combinations, some basic permutation and combination math, and permutations with repeat elements. Now, we’ll see what happens when permutation problems involve conceptual restrictions that can obscure how to approach the math.

To illustrate this directly, let’s take a look at the following Official Guide problem:

The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A) 12
B) 18
C) 24
D) 36
E) 48

Did you catch the restriction? Up until the end, this is a standard permutation with repeats combinatorics problem, since there are five letters and two repeats of the letter ‘I’. However, we’re suddenly told that the two I’s must be separated by at least one other letter. Put differently, they are not allowed to be adjacent.

So how do we handle this? Well, in many cases, it’s helpful to set aside what we want and instead consider what we don’t want. It seems counterintuitive at first, but if we consider the number of ways in which the two I’s can appear together (i.e. what is not allowed) and then subtract that number from the total number of permutations without any restrictions, wouldn’t we then be left with the number of ways in which the two I’s would not appear together (i.e. what is allowed)? 

Let’s demonstrate: 

In this case, we’ll pretend this problem has no restrictions. In the word “DIGIT,” there are five letters and two I’s. Using the principle discussed in our Permutations with Restrictions post, this would produce 5! / 2! = 60 permutations. 

However, we now want to subtract out the permutations that involve the two I’s side by side, since this condition is prohibited by the problem. This is where things become less about math and more about logic and conceptual understanding. Situationally, how would I outline every possible way the two I’s could be adjacent? Well, if I imagine the two I’s grouped together as one unit, there are four possible ways for this to happen:

II DGT

D II GT

DG II T

DGT II

For each one of these four situations, however, the three remaining letters can be arranged in 3*2*1 = 6 ways. 

That produces a total of 6*4 = 24 permutations in which the two I’s appear side by side.

Subtract that from the original 60, and we have: 60 – 24 = 36. The correct answer is D

As you can see, this is not about a formula or rote memorization but instead about logic and analytical skills. This is why tougher combinatorics questions are more likely to involve restrictions.

Here’s another Official Guide example. As always, give it a shot before reading on:

Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36

Explanation

This is a classic example of a problem that will tie you up in knots if you try to brute force it. You could try writing up examples that fit the description, such as 717, 882, 939, or 772, trying to find some kind of pattern based on what does work. But as with the previous problem, what if we examine conceptually what doesn’t work?

This will be very akin to how we handle some GMAT probability questions. The situation desired is 2 digits equal and 1 different. What other situations are there (i.e. the ones not desired)?  Well, if you take a little time to think about it, there are only two other possibilities: 

  1. The digits are all the same
  2. The digits are all different

If we can figure out the total number of permutations without restrictions and subtract out the number of permutations in the two situations just listed, we will have our answer. 

First, let’s get the total number of permutations without restrictions. In this case, that’s just all the numbers from 701 up to 999. (Be careful of the language. Since it says “greater than 700”, we will not include 700.)

To get the total number of terms, we must subtract the two numbers then add one to account for the end point. So there are (999-701)+1 = 299 numbers in total without restrictions.

(Another way to see this is that the range between 701 and 999 is the same as the range between 001 and 299, since we simply subtracted 700 from each number, keeping the range identical. It’s much easier to see that there are 299 numbers in the latter case.)

Now for the restrictions. How many of these permutations involve all the digits being the same? Well, this is straightforward enough to brute force: there are only 3 cases, namely 777, 888, and 999. 

How about all the digits being different? Here’s where we have to use our blank (or slot) method for each digit:

___ ___ ___

How many choices do we have for the first digit? The only choices we have are 7, 8, and 9. That’s three choices:

_3_  ___ ___

Once that first digit is in place, how many choices do we have left for the second slot? Well, there are 10 digits, but we have to remove the one already used in the first slot from consideration, as every digit must be different. That means we have nine left:

_3_  _9_  ___

Using the same logic, that leaves us eight for the final slot:

_3_  _9_  _8_

Multiplying them together, we have 3*9*8 = 216 permutations in which the digits are different.

So there are 216+3 = 219 restrictions, or permutations that we do not want. We can now subtract that from the total of 299 total permutations without restrictions to get our final answer of 299-219 = 80. The correct answer is C.

Next time, we’ll take a look at a few examples of combinatorics problems involving COMBINATIONS with restrictions.

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
5 Takeaways from a Successful EA Journey
Posted on
25
Feb 2021

5 Takeaways from a Successful EA Journey

By: Apex GMAT
Contributor: Uerda Muca
Date: 25th February, 2021

Each client that contacts us is in a different stage of their Executive Assessment prep, but a universal constant is that each is striving for a great 165+ EA score. The threshold difficulty standing in their way is the lack of a proper mindset, which in turn can lead to a poor performance, whether due attitude, inefficient solving mechanisms, misplaced focus, or myriad other issues. No matter what, mindset leads the way to performance.

To adjust one’s way of perceiving problems requires much more intricate work than cramming a bunch of material, facts, and figures. Taking the time to understand this and elevate your approach to the test is challenging but ultimately rewarding come test day. Here are five takeaways that anyone scoring 150 or better on the Executive Assessment comes to realize along their EA journey. These insights that help test takers thrive help top performers continue to excel in their EMBA/grad school programs and in their post-EMBA careers, long after the Executive Assessment is a distant memory. 

It is not what you know… it’s what you do with it. 

The EA is a psychometric exam. It expects you to be knowledgeable in a core group of secondary school concepts. It’s not a knowledge test, but it uses this universe of information as a baseline that everyone reasonably has been exposed to long before they thought about the Executive Assessment. The exam tests not so much your knowledge but your creative application of that knowledge.

In the process of preparing for your EA it is vital to maximize your performance, which necessitates deep understanding of seemingly straightforward concepts so that you can be flexible in how you navigate them. For instance, in the Integrated Reasoning section there is a high chance that you will come across an unfamiliar graph you need to use. In such a case the ability to draw conclusions from known graphs and apply them to the new situation is much more valuable than having seen the specific graph before.

This holds true well beyond the exam. The amount of information you will be exposed to within the 2 years of a top tier EMBA program is staggering. In order to thrive in this demanding environment you must be selective, actively deciding what information you take on to master, and use universal thinking tools (heuristics and mental models) to be adaptable as new concepts and information come your way.

For the Executive Assessment, the core concepts are indeed essential. But it is also important to notice what concepts and information you can derive from fundamental knowledge and how to do so, hence not needing to memorize it. Knowing how to successfully apply your knowledge will result in efficiency which will afford you the ability and time to excel in the EA, explore what your EMBA has to offer, and be a thought leader in your chosen career. 

Prioritization is crucial 

On the Executive Assessment there are harsh penalties for unanswered questions, so it is vital to complete each section in the time allotted. Therefore, proper time and process management is critical when sitting the exam. Essentially, each problem represents a decision where you must weigh the likelihood of obtaining a correct answer, your time commitment to that problem, ancillary considerations like stress and focus management, and how this problem fits into your larger strategy for the section and the exam. Ultimately, you must decide how much time it is worth expending on each problem as part of your core process.

This mental cost benefit analysis must be deeply embedded in your thought process to achieve an elite EA score. With the proper calibration, this sense will certainly be useful in business school and beyond. In the professional world, there will always be time constraints – be it stringent deadlines or time zone differences. Being able to prioritize focus and make decisions quickly and accurately while navigating uncertainty and incomplete information is a huge strength. Similarly, actively choosing to abandon a low value or less important task so that you can fully devote to solving an issue of importance is not a sign of weakness or incapability, but rather an asset in a world that will always ask more of you than you can give. Time is scarce in the workplace, and just like on the Executive Assessment, you should prioritize what adds the most value to your bottom line. 

Every problem has multiple solution paths

 A common theme in our client’s feedback is their fascination with a core principle that we teach; that every EA problem has multiple solution paths and that sensitivity to how you solve the problem is more important than simply arriving at the correct answer. Let’s take a means and averages problem from the Quant section as an example. Many would be tempted to solve this mathematically straightaway, but this problem can be solved more efficiently using a scenario or a graph rather than processing equations, delivering greater clarity and freeing up valuable time for other, more challenging problems.

Wresting yourself away from the paradigm that a problem has a single “correct” solution path is essential to conquering the Executive Assessment but is also valuable in life. Very few things are clear cut and unambiguous, and training yourself to recognize multiple ways to get to the same destination is important, especially if you can recognize them before committing to any specific path. Seeking answers beyond the ordinary and obvious will provide you with innovative ways of overcoming obstacles and drive progress, and make you a thought leader among your peers and in your graduate program and organization.

Focusing on the structure of the EA helps you compare solution paths and choose the best for the current challenge, resulting in not only a correct answer, but a timely one. Moreover, thinking of a problem from multiple perspectives means that you take into consideration unlikely or unnoticed features of a problem, and when applied to a business setting, this added vision can drive great insight into stakeholders interests and uncover innovative solutions to intractable problems.

In order to succeed first know yourself

The Executive Assessment is not an exam where you can get 100% of the problems correct. In fact, your score will not depend on the number of questions that you get correct, but rather by the difficulty level of the ones that you do get correct. Since the EA is computer adaptive, it increases in difficulty until it matches a candidate’s capabilities, and the aim as a test taker is to get to the most difficult problems that you can handle, and then get most of those correct. In this way, the Executive Assessment drives you to perform at your best rather than spending a lot of time testing fundamentals.

Ultimately, this means that you must decide how to allocate your time and energy to produce the best performance. This means understanding your strengths and weaknesses, evaluating each problem in light of those, and then deciding which problems make sense to handle, which make sense to invest extra time in, and which (few) problems you might want to walk away from right off the bat in order to preserve your valuable time for higher value problems. Don’t simply put your head down and try to get everything right – at least not at first.

During your EA preparation you should be conscious of how you perform and how you have progressed from where you began. If you struggle to finish a practice exam in a timely manner, this is a sign that your time management skills require polishing, and that you’re not making timing decisions well. If you perform well on Sentence Correction but not on the Reading Comprehension, then that means that you can spend less time on SC questions and reallocate that time towards Reading Comp, thereby increasing your score and building confidence in your Executive Assessment allocation decisions along the way.

Understanding your strengths and weaknesses, and the workload you handle best will help you excel in your career. Furthermore, knowing your capabilities can aid you when setting work boundaries and defining your professional skill set on the other side of business school. Successful professionals know how to focus on what they do best, and remove those tasks that impinge upon their productivity and value.

In this way, they don’t find themselves taking on too much, and are able to have work life balance, all while placing them in a position to continue to achieve because of that balance. Overworking is counterproductive because it drives burnout and reduces focus and efficiency. In much the same way that athletes require proper rest for peak performance, those working in intellectually rigorous fields requiring creativity need mental breaks for better focus, clarity and job performance. In this sense, being aware of your own limitations will guide you towards a healthy work-life balance and in turn increase productivity. 

5% talent and 95% hard work

Being naturally intuitive with numbers or extremely well-read can provide a great footing for your EA preparation. Without further development, however, natural talent can only take you so far. The Executive Assessment begins testing you the moment you can no longer trust your intuition and talent, and then need to rely upon knowing what you don’t know, and navigating towards deeper insights. The EA tests a range of skills such as critical assessment of data, ability to reason and analytical thinking. This means that being knowledgeable and skillful with fundamentals, or being a strong student only lays the foundation for success. It’s persistence, determination, and having a comprehensive study plan and clear understanding of this exam’s architecture that defines those who score 165 or better.

The good news is that the skills necessary to get a 165+ EA score can be cultivated and enhanced with hard work, perseverance and determination. Moreover, these same skills can help you get the most out of your EMBA program and career and enhance your skill set. For example, in business school you may come across an exceptional mathematician pursuing a concentration in Marketing because she has identified a weak point, and wants to focus on how to conduct research, to write and communicate clearly and effectively and to understand and implement data in the decision making process. Similarly, someone with average mathematical Mathematical ability might excel in Finance courses because of the skills he has developed – analytical thinking, problem solving, and constructing mental models.

Conclusion

The most important thing is to put in the hard work (effortful learning, not just a lot of prep time) to grow those top-level skills, regardless of how naturally gifted you are in a given subject. Marketing isn’t all creativity nor is finance all math, and in this way professional challenges are similar to the Executive Assessment itself, which is neither about Math nor English grammar. Compensating your weaknesses and enhancing your strengths in your chosen concentration will be a vital part of your EMBA experience, and it should start with your EA preparation.

Your Executive Assessment journey can be pleasant and enriching rather than an arduous, distasteful experience that you dread having to go through. With the proper mindset, guidance and support you can grow through your EA experience to acquire valuable skills that will help you for years to come.

Schedule a call with one of our experienced Executive Assessment consultants at +41 41 534 98 78, +44 (0) 79 4361 2406 or +1 (917) 819-5945  and get a head start on the road to achieving your goal.

If you enjoyed this EA journey article, learn more about the EA exam in this overview.

Read more