What happens when Permutations have repeat elements?
Posted on
25
Feb 2021

What happens when Permutations have repeat elements?

By: Rich Zwelling (Apex GMAT Instructor)
Date: 25 February 2021

Permutations With Repeat Elements

As promised in the last post, today we’ll discuss what happens when we have a PERMUTATIONS situation with repeat elements. What does this mean exactly? Well, let’s return to the basic example in our intro post on GMAT combinatorics:

If we have five distinct paintings, and we want to know how many arrangements can be created from those five, we simply use the factorial to find the answer (i.e. 5! = 5*4*3*2*1 = 120). Let’s say those paintings were labeled A, B, C, D, and E. 

Now, each re-arrangement of those five is a different PERMUTATION, because the order is different:

ABCDE
EBADC
CADBE


etc

Remember, there are 120 permutations because if we use the blank (or slot) method, we would have five choices for the first blank, and once that painting is in place, there would be four left for the second blank, etc…

_5_  _4_  _3_  _2_  _1_ 

…and we would multiply these results to get 5! or 120.

However, what if, say we suddenly changed the situation such that some of the paintings were identical? Let’s replace painting C with another B and E with another D:

ABBDD

Suddenly, the number of permutations decreases, because some paintings are no longer distinct. And believe it or not, there’s a formulaic way to handle the exact number of permutations. All you have to do is take the original factorial, and divide it by the factorials of each repeat. In this case, we have 5! for our original five elements, and we now must divide by 2! for the two B’s and another 2! for the two D’s:

  5!
——
2! 2!     

= 5*4*3*2*1
   ————-
  (2*1)(2*1)

= 5*2*3
= 30 permutations

As another example, try to figure out how many permutations you can make out of the letters in the word BOOKKEEPER? Give it a shot before reading the next paragraph.

In the case of BOOKKEEPER, there are 10 letters total, so we start with a base of 10! 

We then have two O’s, two K’s and three E’s for repeats, so our math will look like this:

   10!
———
2! 2! 3! 

Definitely don’t calculate this, though, as GMAT math stays simple and likes to come clean. Remember, we’ll have to divide out the repeats. You are extremely unlikely to have to do this calculation for a GMAT problem, however, since it relies heavily on busy-work mechanics. The correct answer choice would thus look like the term above. 

Let’s now take a look at an Official Guide question in which this principle has practical use. I’ll leave it to you to discover how. As usual, give the problem a shot before reading on:

A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16

Quick Probability Review

Remember from our post of GMAT Probability that, no matter how complicated the problem, probability always boils down to the basic concept of:

    Desired Outcomes
———————————–
Total Possible Outcomes

In this case, each child has two equally likely outcomes: boy and girl. And since there are four children, we can use are blank method to realize that we’ll be multiplying two 4 times:

_2_  _2_  _2_  _2_   =  16 total possible outcomes (denominator)

This may give you the premature notion that C or E must be correct, simply because you see a 16 in the denominator, but remember, fractions can reduce! We could have 4 in the numerator, giving us a fraction of 4/16, which would reduce to 1/4. And every denominator in the answer choices contains a factor of 16, so we can’t eliminate any answers based on this. 

Now, for the Desired Outcomes component, we must figure out how many outcomes consist of exactly two boys and two girls. The trick here is to recognize that it could be in any order. You could have the two girls followed by the two boys, vice versa, or have them interspersed. Now, you could brute-force this and simply try writing out every possibility. However, you must be accurate, and there’s a chance you’ll forget some examples. 

What if we instead write out an example as GGBB for two girls and two boys? Does this look familiar? Well, this should recall PERMUATIONS, as we are looking for every possible ordering in which the couple could have two girls and two boys. And yes, we have two G’s and two B’s as repeats. Here’s the perfect opportunity to put our principle into play:

We have four children, so we use 4! for our numerator, then we divide by 2! twice for each repeat:

  4!
——
2! 2! 

This math is much simpler, as the numerator is 24, while the denominator is 4. (Remember, memorize those factorials up to 6!)

This yields 6 desired outcomes of two boys and two girls. 

With 6 desired outcomes of 16 total possible outcomes, our final probability fraction is 6/16, which reduces to 3/8. The correct answer is A.

Next time, we’ll look into combinatorics problems that involve restrictions, which can present interesting conceptual challenges. 

Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements

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An Introduction to combination math
Posted on
23
Feb 2021

An Intro to Combination Math

By: Rich Zwelling (Apex GMAT Instructor)
Date: 23 Feb 2021

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: 

combinations problem

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:

combinatorics and permutations on the GMAT, combination math on the gmat

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:

combinations and permutation on the gmat, combination math on the gmat

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

combinatorics and permutations on the GMAT, combination math on the gmat

combinatorics and permutations on the GMAT, combination math on the gmat

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:

combinatorics and permutations on the GMAT, combination math on the gmat 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:

solving a combinatorics problem

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:

combination problem on the gmat answer

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?

Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements

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A Continuation of GMAT Permutation Math
Posted on
18
Feb 2021

A Continuation of Permutation Math

By: Rich Zwelling (Apex GMAT Instructor)
Date: 16 Feb 2021

Review of example from last post

Last time, when we started our discussion of GMAT Combinatorics, we gave a brief example of GMAT permutations in which we had five paintings and asked how many arrangements could be made on a wall with those paintings. As it turns out, no complicated combinatorics formula is necessary. You can create an easy graph with dashes and list five options for the first slot, leaving four for the second slot, and so on:

_5_  _4_ _3_ _2_ _1_

Then multiply 5*4*3*2*1 to get 120 arrangements of the five paintings. Remember you could see this notationally as 5!, or 5 factorial. (It’s helpful to memorize factorials up to 6!)

More permutation math

But there could be fewer slots then items. Take the following combinatorics practice problem:

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

First, as a review, how do we know this is a PERMUTATION and not a COMBINATION? Because order matters. In the previous problem, the word “arrangements” gave away that we care about the order in which items appear. In this problem, we’re told that we’re interested in the “orderings” of cheeses. Cheddar followed by gouda would be considered distinct from gouda followed by cheddar. (Look for signal words like “arrangements” or “orderings” to indicate a PERMUTATION problem.)

In this case, we must consider the options of three or four cheeses separately, as they are independent (i.e. they cannot both happen). But for each case, the process is actually no different from what we discussed last time. We can simply consider each case separately and create dashes (slots) for each option. In the first case (three cheeses), there are six options for the first slot, five for the second, and four for the third:

_6_  _5_  _4_

We multiply those together to give us 6*5*4 = 120 possible ways to present three cheeses. We do likewise for the four-cheese case:

_6_  _5_  _4_  _3_

We multiply those together to give us 6*5*4*3 = 360 possible ways to present four cheeses.

Since these two situations (three cheeses and four cheeses) are independent, we simply add them up to get a final answer of 120+360 = 480 possible orderings of cheeses, and the correct answer is D. 

You might have also noticed that there’s a sneaky arithmetic shortcut. You’ll notice that you have to add 6*5*4 + 6*5*4*3. Instead of multiplying each case separately, you can factor out 6*5*4 from the sum, as follows:

6*5*4 + 6*5*4*3

= 6*5*4 ( 1 + 3)

= 6*5*4*4

= 30*16 OR 20*24

= 480

Develop the habit of looking for quick, efficient ways of doing basic arithmetic to bank time. It will pay off when you have to do more difficult questions in the latter part of the test. 

Now that we have been through GMAT permutations, next time, I’ll give this problem a little twist and show you how to make it a COMBINATION problem. Until then…

Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements

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Posted on
17
Feb 2021

Data Sufficiency: Area of a Triangle Problem

Hey guys! Today we’re checking out a geometry Data Sufficiency problem asking for the area of a triangle, and while the ask might seem straightforward, it’s very easy to get caught up in the introduced information on this problem. And this is a great example of a way that the GMAT can really dictate your thought processes via suggestion if you’re not really really clear on what it is you’re looking for on DS. So here we’re looking for area but area specifically is a discrete measurement; that is we’re going to need some sort of number to anchor this down: whether it’s the length of sides, or the area of a smaller piece, we need some value!

Begin with Statement #2

Jumping into the introduced information, if we look at number 2, because it seems simpler, we have x = 45 degrees. Now you might be jumping in and saying, well, if x = 45 and we got the 90 degree then we have 45, STOP. Because if you’re doing that you missed what I just said. We need a discrete anchor point. The number of degrees is both relative in the sense that the triangle could be really huge or really small, and doesn’t give us what we need. So immediately we want to say: number 2 is insufficient. Rather than dive in deeply and try and figure out how we can use it, let’s begin just by recognizing its insufficiency. Know that we can go deeper if we need to but not get ourselves worked up and not invest the time until it’s appropriate, until number 1 isn’t sufficient and we need to look at them together.

Consider Statement #1

Number 1 gives us this product BD x AC = 20. Well here, we’re given a discrete value, which is a step in the right direction. Now, what do we need for area? You might say we need a base and a height but that’s not entirely accurate. Our equation, area is 1/2 x base x height, means that we don’t need to know the base and the height individually but rather their product. The key to this problem is noticing in number 1 that they give us this B x H product of 20, which means if we want to plug it into our equation, 1/2 x 20 is 10. Area is 10. Number 1 alone is sufficient. Answer choice A.

Don’t Get Caught Up With “X”

If we don’t recognize this then we get caught up with taking a look at x and what that means and trying to slice and dice this, which is complicated to say the least. And I want you to observe that if we get ourselves worked up about x, then immediately when we look at this BD x AC product, our minds are already in the framework of how to incorporate these two together. Whereas, if we dismiss the x is insufficient and look at this solo, the BD times AC, then we’re much more likely to strike upon that identity. Ideally though, of course, before we jump into the introduced information, we want to say, well, the area of a triangle is 1/2 x base x height. So, if I have not B and H individually, although that’ll be useful, B x H is enough. And then it’s a question of just saying, well, one’s got what we need – check. One is sufficient. Two doesn’t have what we need – isn’t sufficient, and we’re there. So,

I hope this helped. Look for links to other geometry and fun DS problems below and I’ll see you guys soon. Read this article about Data sufficiency problems and triangles next to get more familiar with this type of GMAT question.

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Posted on
16
Feb 2021

Triangles With Other shapes

By: Rich Zwelling (Apex GMAT Instructor)
Date: 16 Feb 2021

As discussed before, now that we’ve talked about the basic triangles, we can start looking at how the GMAT can make problems difficult by embedding triangles in other figures, or vice versa. 

Here are just a few examples, which include triangles within and outside of squares, rectangles, and circles:

triangles in other shapes GMAT article

Today, we’ll talk about some crucial connections that are often made between triangles and other figures, starting with the 45-45-90 triangle, also known as the isosceles right triangle.

You’ve probably seen a rectangle split in two along one of its diagonals to produce two right triangles:

triangles in other shapes gmat article gmat probelm

But one of the oft-overlooked basic geometric truths is that when that rectangle is a square (and yes, remember a square is a type of rectangle), the diagonal splits the square into two isosceles right triangles. This makes sense when you think about it, because the diagonal bisects two 90-degree angles to give you two 45-degree angles:

triangles in other shapes gmat article, 45 45 90 degree angle

(For clarification, the diagonal of a rectangle is a bisector when the rectangle is a square, but it is not a bisector in any other case.)

Another very common combination of shapes in more difficult GMAT Geometry problems is triangles with circles. This can manifest itself in three common ways:

  1. Triangles created using the central angle of a circle

triangle in a circle, gmat geometry article

In this case, notice that two of the sides of the triangle are radii (remember, a radius is any line segment from the center of the circle to its circumference). What does that guarantee about the triangle?

Since two side are of equal length, the triangle is automatically isosceles. Remember that the two angles opposite those two sides are also of equal measure. So any triangle with the center of the circle as one vertex and points along the circumference as the other two vertices will automatically be an isosceles triangle.

2. Inscribed triangles

triangle inscribed in circle, gmat problem

An inscribed triangle is any triangle with a circle’s diameter as one of its sides and a vertex along the circumference. And a key thing to note: an inscribed triangle will ALWAYS be a right triangle. So even if you don’t see the right angle marked, you can rest assured the inscribed angle at that third vertex is 90 degrees.

3. Squares and rectangles inscribed in circles

rectangle in circle, gmat geometry

What’s important to note here is that the diagonal of the rectangle (or square) is equivalent to the diameter of the circle.

Now that we’ve seen a few common relationships between triangles and other figures, let’s take a look at an example Official Guide problem:

A small, rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?

A) 19,200
B) 19,600
C) 20,000
D) 20,400
E) 20,800

Explanation

The diagonal splits the rectangular park into two similar triangles:

triangle in other shapes gmat problem

Use SIGNALS to avoid algebra

It can be tempting to then jump straight into algebra. The formulas for perimeter and diagonal are P = 2L + 2W an D2 = L2 + W2, respectively, where L and W are the length and width of the rectangle. The second formula, you’ll notice, arises out of the Pythagorean Theorem, since we now have two right triangles. We are trying to find area, which is LW, so we could set out on a cumbersome algebraic journey.

However, let’s try to use some SIGNALS the problem gives us and our knowledge of how the GMAT operates to see if we can short-circuit this problem.

We know the GMAT is fond of both clean numerical solutions and common Pythagorean triples. The large numbers of 200 for the diagonal and 560 for the perimeter don’t change that we now have a very specific rectangle (and pair of triangles). Thus, we should suspect that one of our basic Pythagorean triples (3-4-5, 5-12-13, 7-24-25) is involved.

Could it be that our diagonal of 200 is the hypotenuse of a 3-4-5 triangle multiple? If so, the 200 would correspond to the 5, and the multiplying factor would be 40. That would also mean that the legs would be 3*40 and 4*40, or 120 and 160.

Does this check out? Well, we’re already told the perimeter is 560. Adding 160 and 120 gives us 280, which is one length and one width, or half the perimeter of the rectangle. We can then just double the 280 to get 560 and confirm that we do indeed have the correct numbers. The length and width of the park must be 120 and 160. No algebra necessary.

Now, to get the area, we just multiply 120 by 160 to get 19,200 and the final answer of A.

Check out the following links for our other articles on triangles and their properties:

A Short Meditation on Triangles
The 30-60-90 Right Triangle
The 45-45-90 Right Triangle
The Area of an Equilateral Triangle
Isosceles Triangles and Data Sufficiency
Similar Triangles
3-4-5 Right Triangle
5-12-13 and 7-24-25 Right Triangles

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Combinatorics: Permutations and Combinations Intro
Posted on
11
Feb 2021

Combinatorics: Permutations and Combinations Intro

By: Rich Zwelling (Apex GMAT Instructor)
Date: 11 Feb 2021

GMAT Combinatorics. It’s a phrase that’s stricken fear in the hearts of many of my students. And it makes sense, because so few of us are taught anything about it growing up. But the good news is that, despite the scary title, what you need to know for GMAT combinatorics problems is actually not terribly complex.

To start, let’s look at one of the most commonly asked questions related to GMAT combinatorics, namely the difference between combinatorics and permutations

Does Order Matter?

It’s important to understand conceptually what makes permutations and combinations differ from one another. Quite simply, it’s whether we care about the order of the elements involved. Let’s look at these concrete examples to make things a little clearer:

Permutations example

Suppose we have five paintings to hang on a wall, and we want to know in how many different ways we can arrange the paintings. It’s the word “arrange” that often gives away that we care about the order in which the paintings appear. Let’s call the paintings A, B, C, D, and E:

ABCDE
ACDEB
BDCEA

Each of the above three is considered distinct in this problem, because the order, and thus the arrangement, changes. This is what defines this situation as a PERMUTATION problem. 

Mathematically, how would we answer this question? Well, quite simply, we would consider the number of options we have for each “slot” on the wall. We have five options at the start for the first slot:

_5_  ___ ___ ___ ___

After that painting is in place, there are four remaining that are available for the next slot:

_5_  _4_ ___ ___ ___

From there, the pattern continues until all slots are filled:

_5_  _4_ _3_ _2_ _1_

The final step is to simply multiply these numbers to get 5*4*3*2*1 = 120 arrangements of the five paintings. The quantity 5*4*3*2*1 is also often represented by the exclamation point notation 5!, or 5 factorial. (It’s helpful to memorize factorials up to 6!)

Combinations example

So, what about COMBINATIONS? Obviously if we care about order for permutations, that implies we do NOT care about order for combinations. But what does such a situation look like?

Suppose there’s a local food competition, and I’m told that a group of judges will taste 50 dishes at the competition. A first, a second, and a third prize will be given to the top three dishes, which will then have the honor of competing at the state competition in a few months. I want to know how many possible groups of three dishes out of the original 50 could potentially be selected by the judges to move on to the state competition.

The math here is a little more complicated without a combinatorics formula, but we’re just going to focus on the conceptual element for the moment. How do we know this is a COMBINATION situation instead of a permutation question? 

It’s a little tricky, because at first glance, you might consider the first, second, and third prizes and believe that order matters. Suppose that Dish A wins first prize, Dish B wins second prize, and Dish C wins third prize. Call that ABC. Isn’t that a distinct situation from BAC? Or CAB? 

Well, that’s where you have to pay very close attention to exactly what the question asks. If we were asking about distinct arrangements of prize winnings, then yes, this would be a permutation question, and we would have to consider ABC apart from BAC apart from CAB, etc. 

However, what does the question ask about specifically? It asks about which dishes advance to the state competition? Also notice that the question specifically uses the word “group,” which is often a huge signal for combinations questions. This implies that the total is more important than the individual parts. If we take ABC and switch it to BAC or BCA or ACB, do we end up with a different group of three dishes that advances to the state competition? No. It’s the same COMBINATION of dishes. 

Quantitative connection

It’s interesting to note that there will always be fewer combinations than permutations, given a common set of elements. Why? Let’s use the above simple scenario of three elements as an illustration and write out all the possible permutations of ABC. It’s straightforward enough to brute-force this by including two each starting with A, two each starting with B, etc:

ABC
ACB
BAC
BCA
CAB
CBA

But you could also see that there are 3*2*1 = 3! = 6 permutations by using the same method we used for the painting example above. Now, how many combinations does this constitute? Notice they all consist of the same group of three letters, and thus this is actually just one combination. We had to divide the original 6 permutations by 3! to get the correct number of permutations.

Next time, we’ll continue our discussion of permutation math and begin a discussion of the mechanics of combination math. 

Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements

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triangle inequality rule
Posted on
09
Feb 2021

Triangle Inequality Rule

By: Rich Zwelling (Apex GMAT Instructor)
Date: 9 Feb 2021

One of the less-common but still need-to-know rules tested on the GMAT is the “triangle inequality” rule, which allows you to draw conclusions about the length of the third side of a triangle given information about the lengths of the other two sides.

Often times, this rule is presented in two parts, but I find it is easiest to condense it into one, simple part that concerns a sum and a difference. Here’s what I mean, and we’ll use a SCENARIO:

Suppose we have a triangle that has two sides of length 3 and 5:

triangles inequalities 1

What can we say about the length of the third side? Of course, we can’t nail down a single definitive value for that length, but we can actually put a limit on its range. That range is simply the difference and the sum of the lengths of the other two sides, non-inclusive.

So, in this case, since the difference between the lengths of the other two sides is 2, and their sum is 8, we can say for sure that the third side of this triangle must have a length between 2 and 8, non-inclusive. [Algebraically, this reads as (5-3) < x < (5+3) OR 2 < x < 8.]

If you’d like to see that put into words:

**The length of any side of a triangle must be shorter than the sum of the other two side lengths and longer than the difference of the other two side lengths.**

It’s important to note that this works for any triangle. But why did we say non-inclusive? Well, let’s look at what would happen if we included the 8 in the above example. Imagine a “triangle” with lengths 3, 5, and 8. Can you see the problem? (Think about it before reading the next paragraph.)

Imagine a twig of length 3 inches and another of length 5 inches. How would you form a geometric figure of length 8 inches? You’d simply join the two twigs in a straight line to form a longer, single twig of 8 inches. It would be impossible to form a triangle with a side of 8 inches with the original two twigs.

triangle inequalities 2

 

If you wanted to form a triangle with the twigs of 3 and 5, you’d have to “break” the longer twig of 8 inches and bend the two twigs at an angle for an opportunity to have a third side, guaranteed to be shorter than 8 inches:

triangle inequalities 3

The same logic would hold for the other end of the range (we couldn’t have a triangle of 3, 5, and 2, as the only way to form a length of 5 from lengths of 2 and 3 would be to form a longer line segment of 5.)

Now that we’ve covered the basics, let’s dive into a few problems, starting with this Official Guide problem:

If k is an integer and 2 < k < 7, for how many different values of k is there a triangle with sides of lengths 2, 7, and k?
(A) one
(B) two
(C) three
(D) four
(E) five

Strategy: Eliminate Answers

As usual with the GMAT, it’s one thing to know the rule, but it’s another when you’re presented with a carefully worded question that tests your ability to pay close attention to detail. First, we are told that two of the lengths of the triangle are 2 and 7. What does that mean for the third side, given the triangle inequality rule? We know the third side must have a length between 5 (the difference between the two sides) and 9 (the sum of the two sides).

Here, you can actually use the answer choices to your advantage, at least to eliminate some answers. Notice that k is specified as an integer. How many integers do we know now are possible? Well, if k must be between 5 and 9 (and remember, it’s non-inclusive), the only options possibly available to us are 6, 7, and 8. That means a maximum of three possible values of k, thus eliminating answers D and E.

Since the GMAT is a time-intensive test, you might have to end up guessing now and then, so if you can strategically eliminate answers, it increases your chances of guessing correctly.

Now for this problem, there’s another condition given, namely that 2 < k < 7. We already determined that k must be 6, 7, or 8. However, of those numbers, only 6 fits in the given range 2 < k < 7. This means that 6 is the only legal value that fits for k. The correct answer is A.

Note:

It’s important to emphasize that the eliminate answers strategy is not a mandate. We’re simply presenting it as an option that works here because it is useful on many GMAT problems and should be explored and practiced as often as possible.

Check out the following links for our other articles on triangles and their properties:

A Short Meditation on Triangles
The 30-60-90 Right Triangle
The 45-45-90 Right Triangle
The Area of an Equilateral Triangle
Triangles with Other Shapes
Isosceles Triangles and Data Sufficiency
Similar Triangles
3-4-5 Right Triangle
5-12-13 and 7-24-25 Right Triangles

Read more
gmat in toronto
Posted on
02
Feb 2021

Taking the GMAT in Toronto

Table of Contents:

  1. Who administers the GMAT test?
  2. What does the test center look like?
  3. Where are the test centers located?
  4. Test center holidays
  5. Top MBA programs in the area
  6. Tips
  7. Test Day FAQs

About ¾ of the way through your extensive GMAT prep you should begin to start planning your test day, including scheduling the test, preparing your trip to the test center, and even pre-visiting the test center so that you know exactly where it is. This guide is here to offer you all the required information related to taking the GMAT in Toronto.

Who Administers The GMAT in Toronto: 

Pearson Professional Centers – administers the GMAT and EA exam on behalf of the GMAC. To find out more about the Pearson Professional test centers visit https://www.pearson.com/us/.

What Does The Test Center Look Like?

A Pearson Professional Center will include individual testing areas for each test taker with a separation screen between each test-taker.

 

Where are the Test Centers in Toronto?

These are the test centers in Toronto where test-takers had the best experience:

Pearson Professional Centres-Toronto (Downtown) ON

1075 Bay Street
Suite 525
Toronto, Ontario M5S 2B1
Canada

By car:gmat test center in toronto 1

From Central Park Lodge (7 minutes):

  • Take Madison Ave and Huron St to Bloor St W
  • Turn left onto Bloor St W
  • Turn right onto Devonshire Pl
  • Take Wellesley St W to Bay St

Test-takers’ review:

This test center was rated 5.0 by Google reviewers.

Pearson Professional Centres-Toronto (West) ON

1 Eva Road
Suite 400to
Toronto, Ontario M9C 4Z5
Canada

By car:

From Central Park Lodge (25 minutes):

  • Get on Gardiner Expy W from Queen’s Park Cres W and University Ave
  • Continue on Gardiner Expy W to Etobicoke. Take the Burnhamthorpe Rd/Rathburn Rd exit from ON-427 N
  • Drive to your destination

Test-takers’ review:

This test center was rated 3.6 by Google reviewers. They mentioned that the facilities were clean, quiet, and organized.

Test Center Holidays: 

The most popular times for GMAT preparation and test-taking are during the holiday seasons. Be mindful of dates that you will not be able to take the GMAT or EA at any of the test centers mentioned above. Pearson test centers are closed during the following dates:

  • 1 Jan- New Year’s Day
  • 17 Feb- Family Day Canada
  • 10 Apr- Good Friday
  • 18 May- Victoria Day
  • 1 Jul- Canada Day
  • 7 Sep – Labour Day
  • 12 Oct- Thanksgiving
  • 25 Dec- Christmas Day
  • 28 Dec – Boxing Day
Top MBA Programs In Toronto

Tips:

  • During the test there will not be complete silence – you will be able to hear noise from other test takers so it is best to prepare for this by studying for the exam in similar scenarios. This can prepare you for any distractions (such as coughing, sneezing, or computer clicking sounds) that might occur while taking your exam.  
  • Try to spend some time actually prepping in the lobby of the test center weeks/days in advance of your exam date. Since the place will be familiar to you come test day this can help curb test anxiety should you have any.

Test Day FAQs

Here are the top 5 questions that clients ask us about exam day information:

  • Are you allowed to listen to music while taking the GMAT exam?

You are not allowed to listen to music while taking the GMAT exam and you are not allowed to wear earphones as well. 

  • What should I do if I fall sick on the exam day?

If you do not feel well come exam day you will have to make the decision as to whether or not you can take the test and perform at your best. Most people will not be able to do this so it will be best to cancel it. If you do so on the day of the exam you will incur a loss of your full $250 exam fee. If you cancel the exam 7 days in advance you will be charged a penalty of $50. If it is the first time that you will sit the exam and you are up for sitting through a 4 hour test, this may be a good opportunity to experience the test as you have the ability to cancel the score right afterwards if you are unhappy with it. Ultimately, it is best to take the GMAT when you are feeling your best as this will result in your optimum test performance. 

  • What can I bring with me to the test center?

You are allowed in the test center with the following:

  • GMAT approved identification
  • Appointment confirmation letter or email you received from Pearson VUE
  • Prescription eyeglasses
  • Light sweater or light non-outerwear jacket
  • Comfort items only if they were pre-approved as an accommodation received in advance

Any additional personal belongings that you bring with you such as your cell phone, bag, snacks, and earphones will need to be stored in one of the provided lockers. You may eat your snack during the breaks. Any cell phone use throughout the test time (including breaks) is prohibited. 

The test center will provide you with everything that you need in order to take the test including scratch paper and a pencil. 

  • Should I wear a mask during the exam?

At the test centers above they strongly recommend that you wear a face mask or some type of face-covering in the test center and for the duration of your test to protect yourself and others. Test centers do not provide face masks for candidates. 

Please note that if you have any flu-like symptoms upon arrival at the test center, you may be requested to reschedule your exam for another time when you are in full health.

  • What can I expect at the test center?

A usual test center is typically quite small. Once you arrive you will have to provide the administrator with the relevant documents and while these are being processed you will be asked to wait in the waiting area. In this area, you can still access all your personal belongings up until you are called into the testing room. 

Once in the room, you will be allocated an individual exam station where you will find a computer. 

Find more FAQs: HERE

For any questions or comments please reach out to us at www.apexgmat.com.
To speak to an Apex instructor about your GMAT prep, schedule a call HERE.

Read more
similar triangles on the gmat
Posted on
02
Feb 2021

Similar Triangles – GMAT Geometry

By: Rich Zwelling (Apex GMAT Instructor)
Date: 2 Feb 2021

One of the most important things to highlight here is that “similar” does not necessarily mean “identical.” Two triangles can be similar without being the same size. For example, take the following:

similar triangles on the GMAT 1

Even though the triangles are of different size, notice that the angles remain the same. This is what really defines the triangles as similar.

Now, what makes this interesting is that the measurements associated with the triangle increase proportionally. For example, if we were to present a triangle with lengths 3, 5, and 7, and we were to then tell you that a similar triangle existed that was twice as large, the corresponding side lengths of that similar triangle would have to be 6, 10, and 14. (This should be no surprise considering our lesson on multiples of Pythagorean triples, such as 3-4-5 leading to 6-8-10, 9-12-15, etc.)

You can also extend this to Perimeter, as perimeter is another one-dimensional measurement. So, if for example we ask:

similar triangles on the GMAT 2

A triangle has line segments XY = 6, YZ = 7, and XZ = 9. If Triangle PQR is similar to Triangle XYZ, and PQ = 18, as shown, then what is the perimeter of Triangle PQR?

Answer: Perimeter is a one-dimensional measurement, just as line segments are. As such, since PQ is three times the length of XY, that means the perimeter of Triangle PQR will be three times the perimeter of Triangle XYZ as well. The perimeter of Triangle XYZ is 6+7+9 = 22. We simply multiply that by 3 to get the perimeter of Triangle PQR, which is 66.

Things can get a little more difficult with area, however, as area is a two-dimensional measurement. If I double the length of each side of a triangle, for example, how does this affect the area? Think about it before reading on…

SCENARIO

Suppose we had a triangle that had a base of 20 and a height of 10:

similar triangles on the GMAT 3

The area would be 20*10 / 2 = 100.

Now, if we double each side of the triangle, what effect does that have on the height? Well, the height is still a one-dimensional measurement (i.e. a line segment), so it also doubles. So the new triangle would have a base of 40 and a height of 20. That would make the area 40*20 / 2 = 400.

Notice that since the original area was 100 and the new area is 400, the area actually quadrupled, even though each side doubled. If the base and height are each multiplied by 2, the area is multiplied by 22. (There’s a connection here to units, since units of area are in square measurements, such as square inches, square meters, or square feet.)

Now, let’s take a look at the following original problem:

Triangle ABC and Triangle DEF are two triangular pens enclosing two separate terrariums. Triangle ABC has side lengths 7 inches, 8 inches, and 10 inches. A beetle is placed along the outer edge of the other terrarium at point D and traverses the entire perimeter once without retracing its path. When finished, it was discovered that the beetle took three times as long as it did traversing the first terrarium traveling at the same average speed in the same manner. What is the total distance, in inches, that the beetle covered between the two terrariums?

A) 25
B) 50
C) 75
D) 100
E) 125

Explanation

This one has a few traps in store. Hopefully you figured out the significance of the beetle taking three times as long to traverse the second terrarium at the same average speed: it’s confirmation that the second terrarium has three times the perimeter of the first. At that point, you can deduce that, since the first terrarium has perimeter 7+8+10 = 25, the second one must have perimeter 25*3 = 75. However, it can be tempting to then choose C, if you don’t read the question closely. Notice the question effectively asks for the perimeters of BOTH terrariums. The correct answer is D.

GMAT Triangle Series Articles:

A Short Meditation on Triangles
The 30-60-90 Right Triangle
The 45-45-90 Right Triangle
The Area of an Equilateral Triangle
Triangles with Other Shapes
Isosceles Triangles and Data Sufficiency
Similar Triangles
3-4-5 Right Triangle
5-12-13 and 7-24-25 Right Triangles

Read more
GMAT in Los Angeles
Posted on
29
Jan 2021

Taking the GMAT in Los Angeles

Table of Contents:

  1. Who administers the GMAT test?
  2. What does the test center look like?
  3. Where are the test centers located?
  4. Test center holidays
  5. Top MBA programs in the area
  6. Tips
  7. Test Day FAQs

About ¾ of the way through your extensive GMAT prep you should begin to start planning your test day, including scheduling the test, preparing your trip to the test center, and even pre-visiting the test center so that you know exactly where it is. This guide is here to offer you all the required information related to taking the GMAT in Los Angeles.

Who Administers The GMAT in Los Angeles: 

Pearson Professional Centers – administers the GMAT and EA exam on behalf of the GMAC. To find out more about the Pearson Professional test centers visit https://www.pearson.com/us/.

What Does The Test Center Look Like?

A Pearson Professional Center will include individual testing areas for each test taker with a separation screen between each test-taker.

Where are the Test Centers in Los Angeles?

These are the test centers in Los Angeles where test-takers had the best experience:

Pearson Professional Centers-Pasadena (LA) CA

70 S. Lake Avenue
Suite 840
Union Bank Building
Pasadena, California 91101
United States

By car:gmat in la test center 1

From Central LA (24 minutes):

  • Get on US-101 S from S Wilton Pl and Melrose Ave
  • Take CA-110 N/Arroyo Seco Pkwy to E Glenarm St in Pasadena
  • Take S Los Robles Ave and Cordova St to S Lake Ave

Test-takers’ review:

This test center was rated 5.0 by Google reviewers. They mentioned that the staff was extremely helpful and professional and the test center was highly monitored.

Alternative Options Outside the City:

If you would like to get away from the hustle of the city for your exam or if test centers within the city do not have any openings, there are two other options for you in either Ontario or Lake Forest:

Pearson Professional Centers-Ontario (LA) CA

3401 Centrelake Drive
Suite 675
Centrelake Plaza
Ontario, California 91761
United States

By car:gmat in la test center 2

From Central LA (46 minutes):

  • Get on US-101 S from S Wilton Pl and Melrose Ave
  • Take I-10 E to N Haven Ave in Ontario. Take exit 56 from I-10 E
  • Continue on N Haven Ave to your destination

Test-takers’ review:

This test center was rated 4.6 by Google reviewers. They mentioned that the staff was courteous, helpful, and professional. The facility was clean and there is free parking.

Pearson Professional Centers-Lake Forest (LA) CA

23792 Rockfield Blvd.
Suite 200
Lake Forest, California 92630
United States

By car:gmat in la test center 3

From Central LA (49 minutes):

  • Get on US-101 S from S Wilton Pl and Melrose Ave
  • Take I-5 S to Lake Forest Dr in Orange County. Take exit 92 from I-5 S
  • Follow Lake Forest Dr and Rockfield Blvd to your destination in Lake Forest

Test-takers’ review:

This test center was rated 4.5 by Google reviewers. The reviewers mentioned that the staff was professional and helpful, and the facility was easy-to-find and quiet.

Test Center Holidays: 

The most popular times for GMAT preparation and test-taking are during the holiday seasons. Be mindful of dates that you will not be able to take the GMAT or EA at any of the test centers mentioned above. Pearson test centers are closed during the following dates:

  • 1 Jan – New Year’s Day 
  • 2 Apr – Good Friday   
  • 5 Apr – Easter Monday 
  • 3 May – May Day 
  • 31 May – Late May Bank Holiday   
  • 30 Aug – August Bank Holiday
  • 25 Dec – Christmas Day
  • 26 Dec – Boxing Day 
  • 27 Dec – Christmas Holiday
  • 28 Dec – Boxing Day Holiday
Top MBA Programs In Los Angeles

Tips:

  • During the test there will not be complete silence – you will be able to hear noise from other test takers so it is best to prepare for this by studying for the exam in similar scenarios. This can prepare you for any distractions (such as coughing, sneezing, or computer clicking sounds) that might occur while taking your exam.  
  • Try to spend some time actually prepping in the lobby of the test center weeks/days in advance of your exam date. Since the place will be familiar to you come test day this can help curb test anxiety should you have any.

Test Day FAQs

Here are the top 5 questions that clients ask us about exam day information:

  • Are you allowed to listen to music while taking the GMAT exam?

You are not allowed to listen to music while taking the GMAT exam and you are not allowed to wear earphones as well. 

  • What should I do if I fall sick on the exam day?

If you do not feel well come exam day you will have to make the decision as to whether or not you can take the test and perform at your best. Most people will not be able to do this so it will be best to cancel it. If you do so on the day of the exam you will incur a loss of your full $250 exam fee. If you cancel the exam 7 days in advance you will be charged a penalty of $50. If it is the first time that you will sit the exam and you are up for sitting through a 4 hour test, this may be a good opportunity to experience the test as you have the ability to cancel the score right afterwards if you are unhappy with it. Ultimately, it is best to take the GMAT when you are feeling your best as this will result in your optimum test performance. 

  • What can I bring with me to the test center?

You are allowed in the test center with the following:

  • GMAT approved identification
  • Appointment confirmation letter or email you received from Pearson VUE
  • Prescription eyeglasses
  • Light sweater or light non-outerwear jacket
  • Comfort items only if they were pre-approved as an accommodation received in advance

Any additional personal belongings that you bring with you such as your cell phone, bag, snacks, and earphones will need to be stored in one of the provided lockers. You may eat your snack during the breaks. Any cell phone use throughout the test time (including breaks) is prohibited. 

The test center will provide you with everything that you need in order to take the test including scratch paper and a pencil. 

  • Should I wear a mask during the exam?

At the test centers above they strongly recommend that you wear a face mask or some type of face-covering in the test center and for the duration of your test to protect yourself and others. Test centers do not provide face masks for candidates. 

Please note that if you have any flu-like symptoms upon arrival at the test center, you may be requested to reschedule your exam for another time when you are in full health.

  • What can I expect at the test center?

A usual test center is typically quite small. Once you arrive you will have to provide the administrator with the relevant documents and while these are being processed you will be asked to wait in the waiting area. In this area, you can still access all your personal belongings up until you are called into the testing room. 

Once in the room, you will be allocated an individual exam station where you will find a computer. 

Find more FAQs: HERE

For any questions or comments please reach out to us at www.apexgmat.com.
To speak to an Apex instructor about your GMAT prep, schedule a call HERE.

Read more