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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.

Reviewing the Answer Signals

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.

Clustered Answer Choices

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.

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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 Speed Distance problem.

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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.

1. 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.

2. 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. 

3. 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.

4. 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.

5. 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: Combinatorics & Algebra

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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. Let’s take a look at it: 

At a wedding, the bride’s side has 143 guests and the groom’s side has 77 guests. What is the largest number of identical tables that can be created if each table has to have an equal number of guests from the bride’s side and the groom’s side? An identical table is one where the number of guests from the bride’s side is the same at every table and the number of guests from the groom’s side is the same at every table.

A. 3
B. 5
C. 7
D. 11
E. 13

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, we’re 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.

This 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 are 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 the 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 broken 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.

Number Theory 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.

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