Posted on
19
May 2021

## GMAT Algebra Problem – Parts – Hotdogs & Donuts

#### GMAT Algebra Problem Introduction

Hi guys. Today I’m here with a classic GMAT Algebra problem, what we call a parts problem. And if you take a look at this problem you’re going to realize that it just looks like a bunch of algebra. But the key here is in how you frame it. We’ve got this diner or whatnot selling hot dogs and then after that point, so imagine like a timeline, they start selling donuts. Then they give us a piece of information about hot dogs to donuts over that course of secondary time but then give us this overarching total number of food products sold.

#### Distill The Ration

So what we need to do are two steps: the first one is fairly straightforward. We see that we have to get rid of the hot dogs that were sold in advance in order to distill the ratio but then the ratio can seem very, very complex, especially because it just tells us seven times and a lot of times the GMAT will do this as a way to throw us off the scent. So when we have seven times, what that means is we have eight parts. That is it’s saying for every one of these we have one, two, three, four, five, six, seven of these. Meaning in total there are eight. So while seven is kind of a scary number, eight is a number we can divide by easily. You always want to look for that when you’re given a ratio of one thing to another especially when they say something times as many.

#### Solving the GMAT Problem

We take that thirty thousand two hundred knock off the fifty four hundred and get to twenty four thousand eight hundred and lo and behold that’s divisible by eight meaning each part is going to be 3100. Notice there’s no complex division there, 24 divided by 8, 800 divided by 8 and that’s the sort of mental math we can expect from the GMAT always. Which as you’ve seen before: if you’re doing that you’re doing something wrong.

Each part is 3100 and we’re concerned with the seven parts so we can either scale that 3100 up by seven into 21700, again the math works out super smoothly or we can take the 24800 knock off 3100 and get to that 21700. Notice in the answer choices there’s a few things to address sort of common errors that might be made.

On one of the answer choices what you’re looking at is dividing the total, the 30 200 by eight and multiplying by seven that is seven eighths of it without getting rid of those first 5400. Another answer is close to our 21700 correct answer and this is also a fairly reliable signal from the GMAT.

When they give you a range of answers but two of them are kind of tightly clustered together a lot of times it’s going to be one of the two and that second one there is to prevent you from too roughly estimating. But at the same time if you’re short on time or just in general you want to hone down and understand what you’re supposed to do that serves as a really strong signal. And then one of the answer choices is the 1/8 of it rather than the 7/8.

I want to speak a little more deeply about that signal about those two tightly clustered answer choices because as I said it can help you narrow to a very quick 50/50 when you’re constrained for time or this problem is just one that’s really not up your alley but it also can be leveraged in a really, really neat way.

If we assume that one or the other is the answer choice we can differentiate these two different answer choices by what they’re divisible by and so notice the 21700 is very clearly, with strong mental math is divisible by seven. Where the other one is not. Also neither of them are divisible by eight. We can look at these two say okay one of them is probably right, one of them is divisible by seven, the other one is not, so there’s our right answer and we can move on to the next problem. So I hope this helps. Write your comments and questions below. Subscribe to our channel at Apex GMAT here and give us a call if we can ever help you.

To work on similar GMAT algebra problem/s see this link: Work Rate Problem.

Posted on
17
Sep 2020

## Which Is The Greatest – GMAT Problem

Today we’re going to look at a GMAT problem that screams for estimation but can really tie you in knots if you don’t have the right pivot question, the right perspective. Of the following which is greatest? And on its surface this would seem like a straightforward question except of course the GMAT being the GMAT they’re going to give you a bunch of numbers that are going to be hard to interpret. One part of this problem is simply training. The square root of 2, the square root of 3, the square root of 5. These are common, especially root 2 and root 3 because we see them a lot on triangle problems.

#### Get Familiar With Identities

And knowing these identities by heart as an estimate is really, really valuable just for being able to get a bearing whether you’re on a geometry problem and you’re trying to navigate or make sure that your answer seems correct or if you’re in a problem like this knowing these identities root 2 is 1.4, root 3 is 1.7, root 5 is 2.2 is useful as a touchstone.

#### Break Down The Problem

But this problem in general and the greater problem can be broken down not by saying oh well this is 1.4, this is 1.7, but by asking ourselves well logically which is bigger which is smaller. Remember it’s a multiple choice exam and they’re asking for the biggest or the smallest or whatever it is but these are opportunities to compare not nail down knowledge and this attitude is exceptionally vital for the data sufficiency but it crops up in problem solving a lot more than people might care to admit.

Especially if you’ve been there just trying to study and study and study and get to a precise answer on a lot of these things. So, let’s start just by taking a look at a few things. First square root of three square, root of two which one’s larger? If you said root three you are correct. How much larger? That might be a little bit more difficult to ascertain but if you say 1.7 versus 1.4 maybe 20 percent larger 3 is 50% larger than 2 so root 3 is going to be some smaller percentage larger than root two. But either way we know that root three is the bigger one it’s going to be the dominant value so the question becomes how much larger? Or which part of the answer drives the answer choice?

#### What Do We Know?

So we know that the integers 2 and 3 are more meaningful, larger than the square roots because the square roots are components of those integers. So between A and B, a drives the question that is the three drives the root two more than the two drives the root three. We can take a look at the following two and notice that both of them are around root three.

That is if we take apart the ugly part, which is the square root and take a look at the rest of it – four over five, five over four, these numbers are about one and compared to the two root three we have and the three root two which we’ve already decided is even stronger we don’t really need to entertain C and D all that much. Just to understand that oh they’re about a root three and that’s not going to be enough.

#### Looking At Answer Choice E

Finally, we have E. E is a little funky but we can ask ourselves how many times will root 3, will this 1.7 go into 7 and we get this answer that it’s a bit below 4. Compared with 3 root 2 which is 4.2 (3 times 1.4), we still have a driving the answer. You guys see how this is a marriage of doing a little bit of estimation but also really keeping your framing as is this greater or less than. Now we’ve included a bunch of other different answer choices here for you to take a look at play around with it and see if you can get yourself familiar with comparing these things because the GMAT is only going to come at you with things like square roots that are unfamiliar.

So it’s a fairly defined GMAT problem in that sense. I hope this helps, questions below, like us, subscribe, keep checking in and we’ll see you again real soon.

If you enjoyed this GMAT problem, try these problems next: Probability problem, and the

Posted on
06
Aug 2020

## Probability GMAT Problem

Probability GMAT Problems can be super complex if you don’t frame it correctly. One of the keys to looking at probability problems, particularly conditional probability and independent probability problems, is breaking each part up into its own entity, and a lot of times this clarifies the problem.

#### Introduction To The XYZ Probability Problem

Let’s take a look at this ‘XYZ’ probability problem. Xavier, Yvonne, and Zelda are solving problems. We’re given the 3 probabilities for correct answers and we’re being asked what’s the probability of X being right and solving it, Y solving it, and Z not solving it.

The first thing we can look at is, say: “Well what’s the probability of Zelda not solving it?” And it’s just going to be the flip, the other side of 5/8 to bring us up to 1. If she solves it 5 out of 8 times, she’s not going to solve it the other 3 out of 8 times. So, we’re dealing with 1/4, 1/2, and 3/8.

#### Doing The Math May Seem Simple

The math here is straightforward, multiply them together. But that might not be readily apparent, or at the very least, just plugging it into that formula can get you into trouble. So, here’s where owning it conceptually and mapping it out with a visualization helps you take command of this problem.

#### Xavier Getting It Correct

Since each probability is independent of the others we can look at them independently. What’s the probability of Xavier getting this correct? 1 out of 4 times. So, we can say in general, for every four attempts, he gets it correct once or 25%. If, and only if Xavier gets it correct can we move on to the next part – Yvonne.

#### Yvonne Getting It Correct

Xavier gets a correct 1 out 4 times then what are the chances that Yvonne gets a correct? 1 out of 2. So to have Xavier get it correct and then Yvonne get it correct it’s going to be 1 out of 8 times – 1/4 times 1/2.

It’s not that we can’t look at a Yvonne when Xavier gets it incorrect, it’s that it doesn’t matter. From a framing perspective, this is all about only looking at the probability for the outcome that we want and ignoring the rest.

#### Zelda Getting It Incorrect

Xavier: 1 out of 4, Yvonne: 1 out of 2, gets us to 1 out of 8. Then and only then, what are the chances that Zelda gets it incorrect? 1 out of 8 trials brings us to X and Y are correct, then we multiply it by the 3/8 that Zelda gets it incorrect. That gets us to 3/64. 3 out of every 64 attempts will end in ‘correct’, ‘correct’, ‘incorrect’.

This is one of those problems that may have to go through a few times but once you attach the explanation to it, you can’t mess up the math.

If you enjoyed this GMAT probability problem, try your hand at these other types of challenging problems: &

Posted on
07
Jul 2020

## GMAT Problem – Speed Distance Problem

Speed and distance problems are among the most complained about problems on the GMAT. Numerous clients come to us and say they have difficulty with speed and distance problems, word problems, or work rate problems. So we’re going to look at a particularly difficult one and see just how easy it can be with the right approach.

#### The Two Cars Problem

In this problem we have two cars – car ‘A’ and ‘B’. Car ‘A’ begins 20 miles behind car ‘B’ and needs to catch up. Our immediate DSM (Default Solving Mechanism) is to dive in and create an equation for this and that’s exactly what we don’t want to do.

These types of problems are notorious for being algebraically complex, while conceptually simple. If you hold on to the algebra, rather than getting rid of it, you’re going to have a hard time.

#### Solution Paths

In this problem we’re going to build up solution paths. We’re gonna skip the algebra entirely. We’re going to take a look at an iterative way to get to the answer and then do a conceptual scenario, where we literally put ourselves in the driver’s seat to understand how this problem works. So if we want to take the iterative process we can simply drive the process hour-by-hour until we get to the answer.

#### Iterative solution path

We can imagine this on a number line or just do it in a chart with numbers. ‘A’ starts 20 miles behind ‘B’ so let’s say ‘A’ starts at mile marker zero. ‘B’ starts at 20. After one hour ‘A’ is at 58, ‘B’ is at 70 and the differential is now -12 and not -20. After the second hour ‘A’ is at 116, ‘B’ is at 120. ‘A’ is just four behind ‘B’. After the third hour ‘A’ has caught up! Now it’s 4 miles ahead. At the fourth hour it’s not only caught up but it’s actually +12, so we’ve gone too far. We can see that the correct answer is between three and four and our answer is three and a half.

Now let’s take a look at this at a higher level. If we take a look at what we’ve just done we can notice a pattern with the catching up: -20 to -12 to -4 to +4. We’re catching up by 8 miles per hour. And if you’re self-prepping and don’t know what to do with this information, this is exactly the pattern that you want to hinge on in order to find a better solution path.

You can also observe (and this is how you want to do it on the exam) that if ‘A’ is going 8 miles an hour faster than ‘B’, then it’s catching up by 8 miles per hour. What we care about here is the rate of catching up, not the actual speed. The 50 and 58 are no different than 20 and 28 or a million and a million and eight. That is, the speed doesn’t matter. Only the relative distance between the cars and that it changes at 8 miles per hour.

Now the question becomes starkly simple. We want to catch up 20 miles and then exceed 8 miles, so we want to have a 28 mile shift and we’re doing so at 8 miles an hour. 28 divided by 8 is 3.5.

#### Conceptual scenario solution path

You might ask yourself what to do if you are unable to see those details. The hallmark of good scenarios is making them personal. Imagine you’re driving and your friend is in the car in front of you. He’s 20 miles away. You guys are both driving and you’re trying to catch up. If you drive at the same speed as him you’re never going to get there. If you drive one mile per hour faster than him you’ll catch up by a mile each hour. It would take you 20 hours to catch up. This framework of imagining yourself driving and your friend in the other car, or even two people walking down the street, is all it takes to demystify this problem. Make it personal and the scenarios will take you there.

Thanks for the time! For other solutions to GMAT problems and general advice for the exam check out the links below. Hope this helped and good luck!

Found it helpful? Try your hand at some other GMAT problems: Profit & Loss Problem.

Posted on
11
Jun 2020

## Snack Shop GMAT Problem

The Snack shop GMAT problem is an average or a mean problem. A characteristic of many average problems is that one big takeaway right at the outset is that the answer choices are clustered tightly together. We want to refrain from making any calculations.

The problem is below:

#### Selecting A Solution Path

If they’re looking for a level of precision, the estimation solution path isn’t available to us. If we dive into the problem, right from the first sentence we have sort of a conclusion that we can create via either a graphic or accounting solution path.

If you were the business owner immediately you’d say to yourself: Well for 10 days and an average of \$400 a day I made \$4000.

This is how we want to think about averages. Many times they’ll tell us a parameter about a length of time or over a certain universe of instances and here we want to treat them all as equal.

#### Solving the Problem

It doesn’t matter if one day we made 420 and another day we made 380. We can treat them in aggregate as all equal and start out with that assumption. That’s a very useful assumption to make on average problems. So, we start out knowing that we made 4,000.

What I want us to do is do a little pivot and notice from a running count standpoint how much above or below we are on a given day. So we’re told that for the first six days we averaged \$360 which means each of those six days we’re short \$40 from our average. That means in aggregate we’re short \$240. 6 days times \$40 –  and this has to be made up in the last 4 days.

Notice how we’re driving this problem with the story rather than with an equation. In the last four days we need to outperform our 400 by 240. 240 divided by 4 is 60. 60 on top of the 400 target

#### Graphical Solution Path

If we are more comfortable with graphic solution paths, imagine this in terms of 10 bars each representing \$400. Lowering six of those bars down by 40 and taking the amount that we push those first six down and distributing it among the last four bars gives us our \$460 total per day.

If you enjoyed this Snack Shop GMAT Problem, watch next.

Posted on
28
May 2020

## Gas Mileage GMAT Problem

The Gas Mileage problem is a classic example of the GMAT triggering one of our DSM’s: Our Default Solving Mechanisms for applied math. Yet there are three higher level solution paths that we can engage instead. So we are going to skip the math entirely on this one. In reading the question stimulus, there’s a signal that estimation is going to be a very strong and viable solution path and in fact for most folks estimation is the dominant solution path for this problem.

#### What to Take Note Of

Notice in the first sentence here that we are given the relationship between the efficiency for Car X and the efficiency for Car Y. When comparing 25 to 11.9, 11.9 is a little bit less than half. Whenever we have a relationship that is a little less or a little more than a factor, that’s a clear signal that the GMAT wants us to estimate.

Now, we have an inverse relationship here, between the efficiency of Cars X and Y and the amount of gas they use. So if Car Y is using a little half or rather if Car Y has a little less than half efficiency it’s going to use a little more than double the amount of gas. Managing the directionality of estimation is essential to make full use of this solution path.

#### Estimation Solution Path

Right off the bat, we have a sense that Car Y is going to use a little bit more than double the amount of gas. Now, all we need to do is figure out how much Car X will use. This is an exercise in mental math. Instead of dividing the 12,000 miles by 25 we want to build up from the 25 to 12,000.

Ask ourselves, in a scenario type of way, how many 25’s go into 100 – The answer is 4. 4 quarters to a \$1. Then we can scale it up just by throwing some zeros on. So, 40 25’s are 1,000. How do we get from 1,000 to 12,000? We multiply by 12. So 40 times 12, 480 25’s gives us our 12,000 miles. Car X uses 480 gallons.

Therefore, Car Y is going to use a little more than double this and we point to answer C because we just need to answer the amount Y uses in addition to X. SO there is a bit of verbal play there that we also have to recognize. That’s the estimation solution path.

#### Graphical Solution Path

We can see this via the graphic solution path by imaging a rectangle, where we have the efficiency of the engine on one side and the amount of gallons on the other. With Car X, 25 miles per gallon time 480 gallons is going to give us the area of 12,000 miles. That is we’ve driven the 12,000 miles in that rectangle. If we are cutting it in half on efficiency, or a little more than half, we end up with two strips and if we lay them side by side we see that we’re doubling of going a little more than double on the amount of gas that we use to maintain that 12,000 mile area.

#### Logical Solution Path

Finally, we can look at this from a logical solution path which overlaps a bit with the estimation. But the moment we know that Y uses a little more than double the amount of gas of X, we can also look at and not manage that directionality and just say it uses about double. The only answer choice among our answer choices that is close but not exactly, is C – 520. 480 is our exact number and the A answer is way too low. It’s not close enough to 480 to be viable. So here is an example where, while best practices have us managing the directionality, we don’t even need to do that.

#### Similar Problems

For similar problems like this take a look at the Wholesale Tool problem, The Glucose Solution Problem and for a really good treatment of the graphic solution path check out Don’s Repair Job. There should be links to all three right below and I hope that this helps you guys on your way to achieving success on the GMAT.

If you enjoyed this Gas Mileage Problem but would like to watch more videos about Meta strategy, try “How coffee affects your GMAT performance“.

Posted on
14
Mar 2020

## Remainder Number Theory Problem

Today we’re going to be looking at this problem and our big question is that originally we’re given this unknown number N and we know we just have a remainder 3. So the problem is presenting us information in a way that we’re not used to seeing it and what we need to do is work backwards from this to drive the core insights.

#### Sorting Through the Information

So if we have a remainder of 3 on 23 this means that the chunk that isn’t remainder is 20. So what can our n be in those cases that will allow us to divide out by 20 and leave this remainder 3.

Well first we know that n has to be greater than 3 because in order to have a remainder the amount we’re dividing by has to be something greater. The moment the remainder equalizes the thing we’re dividing by of course we get one more tick in the dividing by box and the remainder goes back down to zero.

#### Solving

So with 23 and a remainder of 3 our key number to look at is 20. Our factors of 20, that is the things that divide evenly into 20, are 1, 2, 4, 5, 10 and 20. Of course 1 and 2 are below 3 and so they’re not contenders. So we end up with n being 4, 5, 10 or 20.

#### Check Against the Statements

So for number 1: Is N even? If N can be 4 but can also be 5 then we’re not assured that it’s even. Notice the data sufficiency problem type embedded here. So N is not necessarily even.

Is N a multiple of 5? Once again N is not because N could be 4 or 5. Finally, is in a factor of 20? And in this case it is because 4, 5, 10 and 20 as we just said are all factors of that 20 that we’re looking for. So our answer here is 3 alone, answer choice A.

#### More Practice

Now here’s a more challenging problem at the same form, see if you can do it and we’re going to come back and in the next video talk about the solution and give you another problem.

So if 67 is divided by some integer N the remainder is 7. Our three things to look at are whether:

• N is even?
• If N is a multiple of 10?
• Or N is a factor of 120?

So give this one a try and see if you can use the principles from the easier problem on this more challenging one to make sure that you actually understand what’s going on. If not, re-watch this video and see if a review might allow you to answer this question.

If you enjoyed using this video for practice, try this one next: Wedding Guest Problem.

Posted on
08
Jul 2019

## GMAT Confidence

Mike Diamond, Head of instruction and Jaymes Kine one of the lead instructors here are going to talk about some of the questions that we’ve gotten on our social media platforms.

We asked for you guys to give us some questions you have asked the questions and I guess kind of talked ourselves into a corner because now we have to answer them. Do you have the answers? I have the answer guide!

#### Prep Materials

What materials to use? But I want to talk especially about the quality materials out there. There’s a wide variety of materials. Wide variety of quality but even the good materials in terms of the problems often do a great disservice to our clients by saying this is how you do it.

You look in the back and you see this especially, it’s more apparent with the quantitative. Here’s all the algebra you need to do, and so many times it’s not about the answer, it’s about thinking about other solution paths and getting around the answer. Yeah, and this is actually something that happens a lot, it’s mainly one of the major pitfalls for self-prep.

When you’re studying on your own we read through the official guide or whatever books you may have. By the way, we recommend always getting the official guide, other books to supplement are great, but at least the official guide.

We have our own materials that we would we would recommend obviously, but let’s go to the official guide on the answers there. When people are self-prepping they will go through the official guide, then they’ll be reading the official answers which they start with here’s the algebra problem and work you all the way through every algebraic step.

#### GMAT Confidence

I don’t know if you guys can tell we haven’t rehearsed and this is actually a GMAT moment. Yeah, we maybe should have rehearsed, you know failing to plan is planning to fail I think is what we say. Yes and no though, but we have all the answers.

We don’t need to go over the answers in advance we don’t need a script because we live and breathe the GMAT, this is all we do. And in that way this is a strong parallel for the GMAT. When you’re sitting the exam you don’t need to know the answers in advance ABCD whatever. If you have the tools and the knowledge, the experience, the expertise, then the answers are going to flow. And all that I think adds up to GMAT confidence to know you can go into that test and you can take it right.

Now we may be over confident here but it’s something we’re working on. But being confident when you get to the test is definitely key. And it’s often a characteristic of really high achieving test takers. To be sure none of our instructors went to the test saying, “Oh, I’m nervous I’m not going to do well,” and when we take follow-up tests from time to time that’s that’s not a question.

That’s characteristic of just about everyone I’ve met, clients, instructors, whatever that have scored above that 720/730 mark.

I think irrational confidence is probably my best my best characteristic as a test taker. That’s your best characteristic overall! Oh, thanks man!

If you enjoyed this video, see: Quant Versus Verbal.

Posted on
06
Jul 2019

## Quant Versus Verbal

It’s time for quant versus verbal, one of the most common questions we get. Where should I start?

#### Quant

It won’t surprise our lovely viewers that it all depends on the person, but let’s talk in some generalizations. One thing you might be surprised by, maybe not so surprised to learn, is that a distinct majority of the people we work with come to us for quantitative help versus verbal help.

At least that’s what they state upfront. Many of them end up only getting quant anyhow but a lot of people state that they only need quant and then they end up needing verbal help as well. Once your quant outstrips your verbal you want to bring them up to parity because that’s highly rewarded by the scoring algorithm.

We talk, we read, we write, we live, we’re immersed in a world of language, a verbal world. Where even math professors only math a few hours a day. Okay yeah there is a verb – to math! This is not a GMAT word but it’s an Apex word because we math frequently. Yes!

#### Fluency

So the issue there is fluency. If you’re already fluent in English, all the lessons you need to learn are much more easily attainable. Whereas with quantitative concepts even ones you think you know, often there’s more context. So you need a longer time period and more contact density with them in order to absorb all the stuff you need to then be flexible with them the same way you’re likely already flexible with the English language.

#### Verbal

A big part of that is that the verbal section is the verbal section but the math section is math in English. They’re not just equations. They’re not just giving you specific mathematics problems per se. They are giving you math problems wrapped up in words.

That goes both ways, there are quantitative problems particularly on the critical reasoning and a lot of times these aren’t: here are some numbers; figure it out. Rather, the cost-of-living index is growing more quickly than inflation, more than pensions or something like that. Where you have some sort of abstract inequality buried in a property – they require mathematical reasoning.

That’s how it goes, so anyway there’s a lot of overlap on the GMAT but especially on the quantitative side, a lot of the difficulty is puzzling out what you need to answer, not doing the equation but you’re saying: what that hell is this asking me for?

#### Non-Native English Speakers

This is something else that we feel like a lot of the other test prep factories don’t really do a good enough job in my opinion. Emphasizing what many of you may be thinking right now which is verbal help and mathematical help with verbal for non-native English speakers. There are plenty of students who come to us who are actually very good mathematicians as it were and it’s the English that they need a little bit of help with. Not as it pertains to the verbal section but actually it’s the English on the quant section that’s difficult.

Absolutely, there’s vocabulary, there’s context, but what’s really important here is that native speakers and non-native speakers pick up language differently. Even the way you learned English if you’re a non-native speaker affects how we approach working with you on the verbal. So if you’re a non-English speaker don’t be too concerned that that’s a disadvantage.

Something I’d like to point out to my students quite often is that the GMAT is actually created specifically for native English speakers and a lot of the test itself is meant to trick native English speakers. So coming at it actually from a non-native speaking background can actually help you kind of skip over all of the little traps that are set up for native speakers. So don’t despair, it’s not that you’re at a distinct disadvantage, you just have some different kind of work to do to prepare.

If you enjoyed this video, watch GMAT Confidence

Posted on
21
Jun 2019

## Number Theory Problem Form – Wedding Guest

Today we’re going to be looking at what at first seems to be an allocation problem. But on further reflection, actually turns out to be a much simpler number theory problem.

If we take a look at the problem stem these numbers 143 and 77. They stick out to us and they stick out not just because they don’t seem to have any relative association but also because they’re sort of odd-looking numbers, they don’t look like most the numbers were used to seeing. Say 48 or 24 or 36 something easily divisible clearly breakable into factors. Here were given these two disparate numbers and we’re being asked to formulate not what the tables are made up of but how many tables there are.

###### Solving the Problem

So we look at these two numbers and we examine first the 77 because it’s a simpler lower number. 77 breaks into factors of 7 and 11. This clues us in as to what to look for out of the 143. 143 must have a factor of 7 or 11. And in fact 143 is evenly divisible by 11 and it gives us 13.

Which means that the maximum number of tables is 11. Each one has 13 people from the bride’s side and 7 people from the groom’s side. 13 plus 7 there’s 20 people at each table. Times the 11 tables is 220.

And we can back check our math, 143 plus 77 is 220. We don’t need to go that far but that might help deliver some comfort to this method. So in reality this is a very creative clever way the GMAT is asking us for a greatest common factor.

###### Graphical Solution Path

Another way to think about this is that we need an equal number of groups from the bride and groom side. The number of people on the bride’s side doesn’t have to equal the number of people on the groom’s side. We just need them broke in into the same number of equal groups. Graphically, the illustration shows us how a certain number of different sized groups combined into this common table. So 13 and 7 and we have 11 groups of each.

###### Problem Forming

This is a great problem to problem form. It will give you some additional mental math or common result experience by forcing you to figure out numbers that you can present that at first don’t look like they match but in fact do have a common factor. You’ll notice that if they had given you 16 people and 36 people finding that common factor might be easier.

So as you problem form this try and do it in a way that sort of obscures the common factor. Either try it maybe where they have multiple common factors and you tweak things like what is the greatest number what is the fewest number of tables. Or even do a perspective shift and take a look at say how many guests are represented at each table. Or on the bride’s side or on the groom’s side. Of course there are conceptual shifts to this and you can make this story about anything. Once you control the story or rather the structure of the story this problem becomes very very straightforward.

Hope this helped, I look forward to seeing you guys soon.