Posted on
27
Jul 2021

## GMAT Calculator & Mental Math – All You Need To Know

Author: Apex GMAT
Contributor: Ilia Dobrev
Date: 27 July, 2021

Are you allowed to use a calculator on the GMAT? While this may seem as a pretty straightforward question to answer, it does deserve a separate blog post as it hides details that are vital for achieving a top GMAT score not only on the quantitative section, but on the exam as a whole.

Well, the answer is both Yes and No. This article aims to examine the different scenarios where you are allowed to use such a device and how you can make full use of its potential. But, if you are used to doing math with a calculator, do not worry as we have compiled a list of some mental math techniques that you can use to your advantage and even save much important time while still being spot-on with your answers.

#### Calculator on the GMAT | Explained

• You are not allowed to bring your own calculator to the GMAT exam.
According to the GMAC, no personal items are allowed in the exam room of any of the certified test centers.
• You cannot use a calculator on the Quantitative section of the GMAT.
Despite the fact that we are so used to using calculators to help us with arithmetic operations, you should not feel intimidated that you are not allowed to use any type of calculator on the GMAT Quantitative section. However, you will be provided with a blank canvas by the proctor of the exam where you will have plenty of space to practice to manually compute any calculations, if needed.

You should not worry as the GMAT exam is not designed to test you on complicated mathematical operations or complex calculations. Instead, the quant section draws from secondary-level math skills like basic algebra and geometry, which are mastered in high school, to test other kinds of abilities like critical thinking, logical reasoning, and problem-solving. In fact, the majority of the Quant questions can (and should) be answered without any calculations beyond estimation. A typical example of how you can use mental math to get to the right answer while saving precious time on the GMAT is the Movie Night combinatorics problem. Another type of common GMAT quant questions are data sufficiency problems, which are also more about reasoning than  calculations. You’ll only need to do basic calculations and can rely on estimation for anything more complicated. If you have to do the math, the GMAC usually keeps the numbers simple and avoids decimals. When you see large numbers or complex fractions, then it’s a good bet that there’s an easier solution path to embark on other than calculating.

Surprisingly or not, a calculator will be provided for use during the GMAT Integrated Reasoning section of the test. This GMAT calculator has the standard basic functions, CE (clear entry) button, C (clear) button, an sqrt function, a % (percentage) button, and a 1/x button that calculates the reciprocal of the entry currently on the screen. Also, there is a row with the standard memory functions

• MS (memory store) stores the current entry in the calculator’s memory for subsequent use.
• MR (memory recall) displays the latest number stored in the calculator’s memory so that it can be used for the next calculations.
• M+ (memory addition) adds the current entry to the value that is currently stored in the calculator’s memory. This button is helpful when you need to add a long series of numbers and don’t want to retype each one.
• MC (memory clear) erases whatever is in the current memory. You should click this button before every new calculation scenario.

### Survival Tips & Tricks

#### Do not overuse the IR calculator.

While you are provided with a basic GMAT calculator during the Integrated Reasoning section, you might not want to use it too often as you’ll waste more time than you’ll save. You can also apply the solution paths you are using in the Quant section to some problems in the Integrated Reasoning section.

#### Constantly practice Mental Math operations.

A huge morale boost is that mental math operations are easy to learn with some practice. You can add, multiply, subtract, and divide when you pay bills, check out at the grocery store, calculate a tip, etc. without using a calculator.

#### Make accurate estimations

The key to saving a considerable amount of time on the GMAT exam is efficiency in estimations. Transform numbers to less unwieldy figures like 0 or 5 for the purpose of calculations. You can then browse the answer choices to see which is closest to your preliminary estimate.

#### Do not use a calculator when you are prepping for the GMAT quant section.

This is a great way to practice mental math operations outside the daily life operations. The test setting and quant context will let you get used to this environment so that you know what to expect on test day.

#### Familiarize yourself with a basic GMAT calculator and do use its Memory functions.

As this will be your only technical aid during the GMAT Integrated Reasoning section, you’d better spend some time making the most out of it. Especially when you are pressed by time, memorizing calculated values for further operations in the calculator’s memory can be crucial for staying on track with a healthy exam pace.

#### Guide yourself by looking at the answer choices.

Looking at the answer choices can immediately permit you to eliminate a couple of options. Even if you are pushed by time, you can easily make a more educated decision depending on your reasoning that will boost the chances of picking the correct answer.

#### Do not freak out if you see large numbers.

Remember that the people who stand behind the GMAT are aware that they are designing questions that are supposed to be answered without using a calculator. This also keeps the arithmetic from being too difficult and gives you the opportunity to apply a more straightforward approach.

Posted on
13
May 2021

## GMAT Factors Problem

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

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

#### The Most Difficult Part of The GMAT Problem

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

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

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

#### GMAT Problem Form

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

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

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

Posted on
01
Apr 2021

## GMAT Prime Factorization (Anatomy of a Problem)

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

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

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

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

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

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

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

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

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

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

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

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

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

Mathematically, you can also see this as:

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

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

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

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

This rule applies for any factor, not just 3.

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

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

What about 11? Here you have two options:

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

→ 11*11 + 11*2 → 11*(11+2) → 11*13

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

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

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

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

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

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

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

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…

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.

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:

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

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:

Notice 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…

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…

Posted on
23
Feb 2021

## An Intro to Combination Math

Last time, we looked at the following GMAT combinatorics practice problem, which gives itself away as a PERMUTATION problem because it’s concerned with “orderings,” and thus we care about the order in which items appear:

At a cheese tasting, a chef is to present some of his best creations to the event’s head judge. Due to the event’s very bizarre restrictions, he must present exactly three or four cheeses. He has brought his best cheddar, brie, gouda, roquefort, gruyere, and camembert. How many potential orderings of cheeses can the chef create to present to the judge?

A) 120
B) 240
C) 360
D) 480
E) 600

(Review the previous post if you’d like an explanation of the answer.)

Now, let’s see how a slight frame change switches this to a COMBINATION problem:

At a farmers market, a chef is to sell some of his best cheeses. Due to the market’s very bizarre restrictions, he can sell exactly two or three cheeses. He has brought his best cheddar, brie, gouda, roquefort, gruyere, and camembert. How many potential groupings of cheeses can he create for display to customers?

A) 6
B) 15
C)
20
D) 35
E) 120

Did you catch why this is a COMBINATION problem instead of a PERMUTATION problem? The problem asked about “groupings.” This implies that we care only about the items involved, not the sequence in which they appear. Cheddar followed by brie followed by gouda is not considered distinct from brie followed by gouda followed by cheddar, because the same three cheeses are involved, thus producing the same grouping

So how does the math work? Well, it turns out there’s a quick combinatorics formula you can use, and it looks like this:

Let’s demystify it. The left side is simply notational, with the ‘C’ standing for “combination.” The ‘n’ and the ‘k’ indicate larger and smaller groups, respectively. So if I have a group of 10 paintings, and I want to know how many groups of 4 I can create, that would mean n=10 and k=4. Notationally, that would look like this:

Now remember, the exclamation point indicates a factorial. As a simple example, 4! = 4*3*2*1. You simply multiply every positive integer from the one given with the factorial down to one.

So, how does this work for our problem? Let’s take a look:

At a farmers market, a chef is to sell some of his best cheeses. Due to the market’s very bizarre restrictions, he can sell exactly two or three cheeses. He has brought his best cheddar, brie, gouda, roquefort, gruyere, and camembert. How many potential groupings of cheeses can he create for display to customers?

A) 6
B) 15
C)
20
D) 35
E) 120

The process of considering the two cases independently will remain the same. It cannot be both two and three cheeses. So let’s examine the two-cheese case first. There are six cheese to choose from, and we are choosing a subgroup of two. That means n=6 and k=2:

Now, let’s actually dig in and do the math:

From here, you’ll notice that 4*3*2*1 cancels from top and bottom, leaving you with 6*5 = 30 in the numerator and 2*1 in the denominator:

That leaves us with:

6C2 = 15 combinations of two cheeses

Now, how about the three-cheese case? Similarly, there are six cheeses to choose from, but now we are choosing a subgroup of three. That means n=6 and k=3:

From here, you’ll notice that the 3*2*1 in the bottom cancels with the 6 in the top, leaving you with 5*4 = 20 in the numerator:

That leaves us with:

6C3 = 20 combinations of three cheeses

With 15 cases in the first situation and 20 in the second, the total is 35 cases, and our final answer is D.

Next time, we’ll talk about what happens when we have permutations with repeat elements.

In the meantime, as an exercise, scroll back up and return to the 10-painting problem I presented earlier and see if you can find the answer. Bonus question: redo the problem with a subgroup of 6 paintings instead of 4 paintings. Try to anticipate: do you imagine we’ll have more combinations in this new case or fewer?

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

Posted on
26
Nov 2020

## GMAT Calculator & Mental Math – All You Need To Know

Are calculators allowed on the GMAT? It seems like a pretty straightforward question, but the details are a bit more complicated.

The short answer is: yes and no. In fact, the calculator question holds the key for a strong performance on the exam as a whole. This article explains when calculator use is permitted and, more importantly, when using a calculator isn’t the best approach to solving a given problem.

So if you’re used to using a calculator on math tests, don’t worry! We’ve provided a list of some handy mental math techniques and time saving strategies that will enhance your performance on the Quant section and beyond.

## Calculators on the GMAT | Explained

• You are not allowed to bring your own calculator to the GMAT exam.

According to the GMAC, no personal items are allowed in the exam room at any certified test center.

However, the proctor will provide a blank canvas with plenty of space to perform any necessary calculations by hand.

• ###### You cannot use a calculator on the Quantitative section.

There’s no reason to be intimidated by the restriction on calculators. Although most of us are used to using calculators for arithmetic, the GMAT is not designed to test your ability to perform complex mathematical operations. The Quant section draws from secondary-level math and basic algebra and geometry to test other skill sets, such as critical thinking, logical reasoning, and problem solving.

In fact, the majority of the Quant questions can and should be answered without any calculations beyond estimation.

For example, data sufficiency problems, which are more geared towards reasoning than math skills, typically only call for basic calculations and estimation. If you do need to do math, keep in mind that the GMAC designers usually keep numbers simple and avoid decimals. If you see large numbers or complex fractions, it’s a good bet that there’s an easier solution path.

For another example of how mental math can save you time, see our explanation of the movie night combinatorics problem

• ###### You can use an on-screen calculator on the Integrated Reasoning section.

Surprisingly or not, a calculator will be provided for the Integrated Reasoning section. This GMAT calculator has the standard basic functions, CE (clear entry) button, C (clear) button, an sqrt function, a % (percentage) button, and a 1/x button that calculates the reciprocal of the entry currently on the screen. There is also a row with the standard memory functions

• MS (memory store) stores the current entry in the calculator’s memory.
• MR (memory recall) displays the last number stored in the calculator’s memory.
• M+ (memory addition) adds the current entry to the value stored in the calculator’s memory. This button is helpful when you need to add a long series of numbers, but don’t have time to retype each one.
• MC (memory clear) erases whatever is in the current memory. Use it before every new calculation set.

## Improve your Mental Math and Reduce Calculator Dependence

### Survival Tips & Tricks

• #### Do not overuse the IR calculator.

Although the GMAT provides a basic calculator for the Integrated Reasoning section, don’t use it too often. You’ll waste more time than you save. However, you can apply some of the same solution paths used in the Quant section to problems in Integrated Reasoning.

• #### Practice mental math operations regularly.

Mental math operations are easy to learn with some practice, and mastering mental math can provide a significant morale boost leading up to your test date. You can add, multiply, subtract, and divide when you pay bills, check out at the grocery store, calculate a tip, etc. without using a calculator.

Try putting away the calculator and practicing mental math in your daily life to save time and, ultimately, enhance your GMAT score.

• #### Make accurate estimations

Learning to estimate efficiently is the key to saving considerable amounts of time on the GMAT. Convert unwieldy numbers to more manageable figures, like 0 or 5, for the quicker calculations. Then, you can browse the answer choices and select the answer that’s closest to your preliminary estimate.

• #### Don’t use a calculator when prepping for the Quant section.

Preparing without a calculator is a great way to practice mental math operations outside of your daily life. The test setting and Quant context will help accustom you to the environment. You’ll feel more prepared if you know exactly what to expect on test day.

• #### Familiarize yourself with a basic GMAT calculator and practice using its memory functions.

Since the on-screen calculator will be your only technical aid during the Integrated Reasoning section, it’s smart to spend some time getting used to it. When you’re pressed for time, the calculator’s memory function can be a crucial tool for staying on track with a healthy exam pace.

Sometimes, you can eliminate a couple of answer choices immediately.

Even when time is in short supply, you can make educated guesses and use your reasoning skills to boost your chance of arriving at the correct answer.

• #### Don’t panic if you see big numbers.

Keep in mind that the people behind the GMAT are aware that they’re designing questions to be answered without calculators. This limits the difficulty of the arithmetic and encourages test-takers to look for the more straightforward approach.

Contributor: Ilia Dobrev
Date: 26th November, 2020