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Posted on
21
Jan 2021

Rope Problem – Graphic Solution Path

Hi guys! Today we’re going to look at the rope problem. And this is a fairly straight forward problem with an excellent graphic solution path. But there are some obstacles in our way to that graphic solution path.

Obstacles To Avoid

The first thing to watch out for here is the phrasing of the problem. You’ll notice it is phrased in an awkward way: rather than telling us where the rope is cut, it tells us one length relative to the other. The other obstacle is that we immediately want to jump into the math. Either setting up an algebraic equation or, otherwise, not visualizing the rope.

And this is an error not because it’s that much more difficult to do it mathematically, but because it’ll take you a bit more time and it will be less clear. You won’t be as confident in your answer choice relative to actually being able to see it.

Visualize the Problem

So, what you want to do is visualize the actual rope. And we’ve got one right here. So, if this is 40 feet long, and one side is 18 feet longer than the other then we wanna take the 18 and make that the longer piece, and then the other two pieces are distributed among the short side and the rest of the long side. Once we have that we can say, well, if this long part here is 18, then these two pieces must be 22 they also must be equal. And this is much quicker and clearer than setting up an equation 2x+18 = 40

We’re doing the same thing but here it’s easy to say: okay, 11; 11+18 is 29, that gets us our 40. And we’re there, we’re confident, we move on.

This is a great example of a straightforward problem that can be done in 15 seconds and if you’re doing it in a minute you’re spending too much time. Hope this helps, and we’ll see you guys next time!

For other problem related to this, try out the Test Averages Problem.

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Posted on
15
Jan 2021

Counting Primes Exponent Inequality – GMAT Problem

Today, we are looking at counting a primes exponent inequality problem. Despite all those scary terms, this one is actually fairly straightforward once you master the ability to count prime factors.

Counting primes is all about understanding how many versions of each prime are necessary to construct the entire prime factorization of an integer. In this problem, we are comparing 25s and 5s and we are being asked how many 25s versus how many 5s there are.

Notice how we are not diving into the math immediately. We are first putting this in terms of counting only. 5 to the 12th means that we are actually multiplying 5 by itself 12 times. Like this: 5x5x5x5x5x5x5x5x5x5x5x5. We can now say we have 12 fives. The question then becomes: how many 25s is this equivalent to?

We are now looking for inequality by forming a baseline of equivalents. We now understand how much too many or too few would be. The key question here is how many 5s make up 25? The answer is not 5: we are not dividing or multiplying. 2 prime factors of 5 make 25. 5×5. That is 25=5 square. We wouldn’t know how many 25 it takes to hate more than 12 5s. Where each 25 is the equivalent of 2 5s, 6 25s is the same as 12 5s. So, we need now a 7th 25 in order to have more 5s than the 12 5s on the other side.

And that is our answer: 7. Answer choice B.

For additional problems like this, especially counting primes and number theory problems, check out these videos. 

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The Area of an Equilateral Triangle
Posted on
14
Jan 2021

The Area of an Equilateral Triangle

By: Rich Zwelling (Apex GMAT Instructor)
Date: Jan 14, 2021

As promised, we will now connect the 30-60-90 triangle to the equilateral triangle, specifically its area. There is a formula for the area of an equilateral triangle as it relates to the length of its side s, and it is as follows:

Equalateral triangles GMAT picture 1

But more likely than not for the GMAT, you’ll need to understand how this formula is derived. And the √3 term in the area is a big clue.

First, it helps to remember that an equilateral triangle has all equal angles as well as all equal sides. And given that the angles in a triangle must sum to 180 degrees, each angle must be 60 degrees:

Equalateral triangles GMAT picture 2

Now, what happens when we take such a triangle and split it down the middle?

Equilateral triangles GMAT picture 3This should look familiar. Because the line segment down the middle acts as an angle bisector, the 60 degree angle at the top vertex becomes two 30-degree angles. Take a moment to consider what this produces and what the implications are.

As you might have guessed, this line segment produces two 30-60-90 right triangles:

Equilateral triangles GMAT picture 4

Not only that, but we can then use s to denote the side length of the equilateral triangle and map out each segment of the 30-60-90 right triangles. Before viewing the diagram below, take a moment to consider what the height of the triangle would be.

Remember that the ratio of side lengths is 1 : √3 : 2. If we fill in all of the appropriate lengths, we would get the following:

Equilateral triangles GMAT picture 5Now, we’re very close to deriving the area of the triangle, which is simply base*height/2. In this case, the base is s, while the height is s√3/2.

This is how we finally get the universal formula for an equilateral triangle:

Area = base * height / 2
Area = (s) * (s√3/2) / 2
Area = (s) * (s√3/4)
Area = (s2√3) / 4.

Now that we’ve seen the relationship between equilateral and 30-60-90 triangles, let’s see how it plays out in an official GMAT problem:

The figure shown above consists of three identical circles that are tangent to each other. If the area of the shaded region is 64√3 – 32π, what is the radius of each circle?

Equilateral triangles GMAT picture 6

A. 4
B. 8
C. 16
D. 24
E. 32

Using signals

This is a complex problem that seems intimidating at first. However, if we use signals the problem is giving us, we can get to the answer more quickly than we might initially think. What signals does the area of the shaded region give us? Think about it before reading on…

If we look closely at the diagram, we see that an equilateral triangle is involved. We know this because each side of the triangle consists of two radii of each circle (i.e. the distance from the center to the outer edge of the circle), thus each side of the triangle must be equal. That’s a big hint that the √3 term is linked to the area formula we’ve been talking about.

Likewise, although it is not the subject of this post, the term using π is associated with circles in this case, the areas of the identical circles. (For reference, the area of a circle is πr2, and the circumference of a circle is 2πr.)

Conceptually, we should be able to see that 64√3 – 32π represents the area of the equilateral triangle minus the area of the three small sectors from the circles. 

Now, rather than do any unnecessarily complicated math, we should take notice that the question asks for the radius of each circle, and each side of the equilateral triangle is 2r:

Equilateral triangles GMAT picture 7

We already know that the area of the equilateral triangle is 64√3, and we have the formula for that area, so we are just a few steps away from solving for the radius.

Remember the formula, where s is the length of the side of the equilateral triangle:
Area = (s2√3) / 4

Substitute:
64√3 = (s2√3) / 4

Since √3 is common to both sides, you can divide it out:

64 = s2 / 4
256 = s2

Now, normally, you would say that s could be 16 or -16, but since this is a geometric quantity, we only deal in nonnegative quantities. Therefore:

s = 16, giving us the length of each side of the equilateral triangle.

Be careful, however. This could trap you into picking answer choice C. Remember to check exactly what the question asks for. We were asked for the radius of the circle, which as we see in the above diagram is half the length of s. The correct answer is B.

Again, it’s very important to notice that we didn’t do anything with the circles. The 64√3 term and the equilateral triangle were enough to get us the length of each side and thus the radius. Look for signals to help short-circuit problems and avoid lengthy solution paths.

Now that we’ve reviewed all of the basic triangles, we’ll do a little more next time on how triangles can appear in other shapes, such as circles and rectangles. We got a little taste today, so hopefully that will give you a good idea.

Find more articles in our triangle series here:
30-60-90 Triangles
Triangle Overview
Isosceles Triangles
Triangles within other shapes
Pythagorean Identities
45-45-90 Triangles

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Posted on
14
Jan 2021

Averages Problem No.1 : Test Averages

Hey guys, today we’re going to take a look at the test averages problem. This is a very straightforward mathematically oriented average problem or at least it can be. But there are very strong graphic solution paths here and there’s also a really strong sort of intuitive running tally counting solution path here. We’re going to start out with the math though, just because that’s how a lot of people are familiar with this problem. Before we jump into the heavier duty quicker sort of stuff. 

Doing the Math

So to solve this problem we want to take an average. But one of the components of our average is missing. So we have four things with an average of 78, and a fifth unknown. That means we can assume that each of the first four exams were 78. So we’ve got 4 times 78 plus X over 5. The total number of exams is going to give us our average of 80. Then through algebra, algebraic manipulation we multiply the 5 over, we get 400 equals 4 times 78 plus x. The 4 times 78 is 312. We subtract that off the 4 and that brings you to 88. Answer choice E.

Graphic Solution Path: Poker Chips

Let’s take a look at this a little differently. One of the ways I like looking at averages is imagining stacks of poker chips and you can have stacks of anything. I like poker chips because they fit together and you can make two stacks equal very easily so what we’re being told here is we have four stacks of 78 a fifth unknown stack but if we equalize them all that is if we take chips off of the unknown stack and distribute them all the stacks will be 80. That means that the fifth stack needs to be 80 and then it needs two poker chips for each of the other four stacks to bring those 78’s up. We can also envision this as just a rectangle our goal is 80 but we have 78, and our goal is five tests but we have four so we have 78 by four here. And then 80 by 5 here what’s missing is the full 80 and then 2 on each of four stacks of 48.

Running Tally Method: Intuitive Approach

The most powerful way to handle this problem though is probably by doing a running tally. Don’t even worry about the visualization but rather notice that, I’ve got 47 8s each of those are too short so I’m two, four, six. eight points short on the last test. I need to get the 80 that I want plus those eight points that I’m short bringing us to 88. And anybody who’s sweated like A+, B+, A- or a C+, B- has done this math. So if you characterize it like that a lot of times it becomes much more intuitive and once again allows you to cultivate confidence for a deeper treatment and more complex averages problems and mean problems check out the snack shop problem, check out the company production problem and there’s a great ds problem that we do the trade show problem you’ll find links to all of them just below and I hope this helped. 

 

Enjoyed this Averages Problem ? Try another type of GMAT problem to get familiar with all question types on the exam: Remainder Number Theory Problem.

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The 5-12-13 and 7-24-25 Right Triangles
Posted on
12
Jan 2021

The 5-12-13 and 7-24-25 Right Triangles

By: Rich Zwelling (Apex GMAT Instructor)
Date: Jan 12, 2021

The 5-12-13 and 7-24-25 Right Triangles

Although the 3-4-5 right triangle is by far the most common of the so-called “Pythagorean triples” tested on the GMAT, there are a few others worth knowing. First, a little review: 

You’ll recall that the Pythagorean Theorem ( + b² = c²) holds for any right triangle where a and b are the two legs and c is the hypotenuse, and that the 3-4-5 triangle represents the smallest such triangle with all integer side lengths:

5-12-13 and -7-24-25 Triangle Identities problem 1

This works not only for 3-4-5 but also for 6-8-10, 9-12-15, or any other multiples of each side length.

5-12-13 and -7-24-25 Triangle Identities problem 2No matter what positive integer n you choose for the figure above, you will produce a valid right triangle.

So now we come to the main topic: what are some other common “Pythagorean triples” the GMAT may test? The next base triples that fit the Pythagorean Theorem are 5-12-13 and 7-24-25. These work because if you check the arithmetic, 5² + 12² = 13² and 7² + 24² = 25²:

5-12-13 and -7-24-25 Triangle Identities problem 3

As we’ve continually discussed, however, your success on more difficult GMAT problems will require you to go beyond mere rote memorization. Let’s take a look at an Official Guide Data Sufficiency problem that illustrates how the test can force you to engage some higher-level reasoning skills:

5-12-13 and -7-24-25 Triangle Identities problem 4

If A is the area of a triangle with sides of lengths x, y, and z as shown above, what is the value of A?

(1) z = 13

(2) A = 5y/2

Give it a try on your own before reading any further.

As with any Data Sufficiency question, let’s identify what we’re asked to find. A represents the area of the triangle, which is found by multiplying base by height and dividing by 2. That means A = xy/2, since x and y represent the height and base, respectively. 

Remember, it helps to frame Data Sufficiency questions in terms of what information you need to get to the answer. We need to know the individual values of x and y. Or, as a matter of fact, we could have sufficiency if we knew xy as a product, even if we didn’t know the values of x and y, individually. For example, on a different problem with the same question, if the test had said that the product of the base and height were 30, that would have been sufficient, as that would be enough for us to deduce that the area is 15.  

You can save yourself much time and mental energy by having a solid idea of what information you need from the statements for sufficiency before you actually view the statements. 

Now that we know what information we need for sufficiency, let’s examine each statement on its own. Statement (1) should get you thinking about the 5-12-13 right triangle, as it tells us that the hypotenuse is 13. But be careful: this is where rote memorization only goes so far (and may actually get in the way). 

Does knowing that the hypotenuse is 13 guarantee that the other sides are 5 and 12? For all we know, they could be non-integers that fit + b² = 13². In fact, a and b could be equal — remember that we can’t assume that the figures are drawn to scale. Without a clear idea of what the base and height are, we cannot get a consistent product for xy. Statement (1) is INSUFFICIENT on its own.

Statement (2) is more complicated, as we have two variables, one of which is the area. But we already discussed that A = xy/2, so we can do a substitution:

A = 5y/2
xy/2 = 5y/2

At this point, we can see that the sides are identical, except that the x on the left has been replaced by a 5 on the right. Therefore, x must be 5. Again, this should get us thinking about the 5-12-13 triangle. But we should again remember that this alone does not guarantee that the other sides are 12 and 13. Even though x is 5, there could be multiple values for y, and that means multiple values for the product xy. Statement (2) is also INSUFFICIENT on its own.

This narrows the answer choices down to C (statements sufficient together) and E (statements insufficient together).

This is where previous knowledge of the 5-12-13 triangle helps. Ideally, once you see that the statements together tell you that x=5 and z=13, you will know without much thought that y must be 12. You won’t bother using the Pythagorean theorem and you certainly won’t wonder if y could have multiple values.

Without knowledge of the 5-12-13, one trap a test-taker could possibly fall into is viewing the two statements and noticing that there are 3 variables and only 2 equations. We need a full 3 equations with 3 variables if we’re going to solve for all 3 variables, and that may lead some to prematurely conclude that the answer is E. 

However, why is that a false conclusion?

Well, we’re not trying to solve for all variables. We’re only solving for one. It’s possible to solve for one variable, even if there are fewer equations than variables. 

In this case, now that we know that x=5 and y=12, we have our base and height, and we can solve for A, the area of the triangle. Note that I’m not going to bother solving, because for sufficiency, I don’t need to. I only care that I CAN solve. The final answer is (C).

We’ve now talked about the various Pythagorean triples and special right triangles. Next time, we’ll talk about how triangles can appear within OTHER shapes. And to tide yourself over, you can also link to our other article about triangles:

Triangle Overview
Equilateral Triangles
Isosceles Triangles
Triangles within other shapes
Pythagorean Identities
45-45-90 Triangles
30-60-90 Triangles

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Posted on
08
Jan 2021

The 30-60-90 Right Triangle

By: Rich Zwelling (Apex GMAT Instructor)
Date: Jan 7 2021

30-60-90 Right Triangle

In a previous piece, we covered the 45-45-90 right triangle, also known as the isosceles right triangle. There is another so-called “special right triangle” commonly tested on the GMAT, namely the 30-60-90 right triangle.

Like the isosceles right, its sides always fit a specific ratio, as seen in the above diagram (1 : √3 : 2). And it’s worth noting, as with all triangles, that the shortest side is opposite the smallest angle, while the longest side is opposite the largest angle, etc. 

Now, it’s easy enough to memorize this ratio and deduce what each side length will be, given that we are dealing with a 30-60-90 triangle. For example, Suppose we are given the following information:

This is low-level memorization, and we can deduce that the side opposite the 60-degree angle will be length 5√3, while the hypotenuse will be length 10.

But let’s look to this GMAT Official Guide problem to see something a little more high-level. Give it a shot before reading further:

In the figure above, V represents an observation point at one end of a pool. From V, an object that is actually located on the bottom of the pool at point R appears to be at point S. If VR = 10 feet, what is the distance RS, in feet, between the actual position and the perceived position of the object?

(A) 10−53

(B) 10−52

(C) 2

(D) 2 1/2

(E) 4

 

(For starters, notice that the question they’re asking for — the distance between the actual position and the perceived position — is just line segment RS. Remember that the GMAT is very good at using complicated wording to frame a simple concept. Always simplify the question as quickly as possible.)

To understand this problem, let’s first talk about one of the higher-level ways the GMAT could test 30-60-90 triangles. Take this example:

Notice we are given no angles except the right angle. But we do have 2 sides and 1 angle in total, which is sufficient to form a unique triangle. Furthermore, did you identify anything that gives this away as a 30-60-90? 

The hypotenuse is twice the length of one of the sides, giving them a 2:1 ratio. That guarantees that the third side fits the √3 component of our ratio, giving that side a length of 5√3. So even without labeled angles:

A right triangle with a hypotenuse twice the length of one of its legs must be a 30-60-90 triangle.

That’s much more the kind of critical thinking the GMAT is interested in testing. 

Similarly, in this Official Guide problem, we are told that VR is length 10:

Notice that at this point, it’s up to you to make the deduction that we have a 30-60-90 triangle, and thus the distance from the right angle marker to point R must be 5√3:

From there, it’s straightforward to see that RS is simply the marked length of 10 minus the length of 5√3 we just deduced, thus leading us to answer choice A.

In terms of strategy, another point: a brief look at the answer choices at the start of the problem gives a strong hint that either a 30-60-90 or 45-45-90 triangle is involved. Notice that the first two answers feature a √3 and a √2 term, and this is clearly a geometry question. This gives you the opportunity to be preemptive and use the test’s patterns against itself. 

In our next post, we’ll talk about how 30-60-90 triangles can be used directly to calculate the area of equilateral triangles. You can also link to our other article about triangles:

 

A review of Triangles
Right Triangles
Equilateral Triangles
Isosceles Triangles
Triangles within other shapes
Pythagorean Identities

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45-45-90 triangle on the GMAT
Posted on
06
Jan 2021

45-45-90 Right Triangle – GMAT Geometry Guide

By: Rich Zwelling (Apex GMAT Instructor)
Date: Jan 6 2021

45-45-90 Right Triangle

Another of the commonly tested triangles on the GMAT is the 45-45-90, also known as the isosceles right triangle. Know that term, as it could appear by name in a question.

As shown in the above diagram, the side lengths of this triangle always fit the same ratio (1 : 1 : √2) , where the legs are the same length and the hypotenuse length is √2 times the leg length. For example, if the leg lengths were 3 instead of 1, then the hypotenuse would be 3√2 instead of simply √2.

But likewise, don’t forget that you can go backwards and divide the hypotenuse length by √2 to get to the leg length. It may seem obvious, but it presents an important point: what’s more important than simply memorizing the ratio is understanding the mathematical relationship between the side lengths. This will help you avoid trouble if the GMAT happens to give you a problem that doesn’t conform to expectations.

For example, the following problem fits expectations quite nicely:

A yard in the shape of an isosceles right triangle has a hypotenuse of length 10√2. What is the area of this yard?

From this information, it’s easy enough to deduce that the leg length is 10, and we can draw a diagram that looks roughly like this:


From there, we can easily calculate the area, which is base*height / 2, or in this case 10*10/2 = 50.

But what happens if we give the problem a little twist:

A yard in the shape of an isosceles right triangle has a hypotenuse of length 10. What is the area of this yard?

Did you catch the twist? We’re used to the hypotenuse including a √2. This is what the GMAT will do. They’ll throw you off-center, and you’ll have to adjust. But this is also why we said earlier that what matters more than memorizing the ratio of sides is understanding the relationships between the sides of an isosceles right triangle…

Remember we said that, just as we multiply the leg length by √2 to get to the hypotenuse length, so we must divide the hypotenuse length by √2 to get to the leg length. That must mean each leg has length 10/√2. 

You can then take 10/√2 and multiply it by √2/√2 to de-radicalize the denominator and get (10√2) / 2, or a leg length of 5√2:

Notice again that we have a more unfamiliar form, with the √2 terms in the legs and an integer in the hypotenuse. We can’t count on the GMAT to give us what we’re used to. 

Now we can calculate the area:

Area = (base*height)/2 = (5√2)(5√2)/2 = (5*5)(√2*√2)/2 = (25)*(2) / 2 = 25

 

Problem #1

Now, to try this on your own, take a look at this Official Guide problem:

If a square mirror has a 20-inch diagonal, what is the approximate perimeter of the mirror, in inches?

(A)   40
(B)   60
(C)   80
(D)   100
(E)   120

Explanation:

This is a nice change-up, because it involves another shape. Did you notice that splitting a square along its diagonal creates two isosceles right triangles

Once you realize this, you can divide 20 by √2 to get 20/√2, then multiply top and bottom by √2 to get x=10√2.

Since the question asks for perimeter, we can multiply this by four to get 40√2. 

The final step is to realize that √2 is approximately 1.4. If we multiply 40 by 1.4, the only answer choice that possibly makes sense is 60, and thus the correct answer is B

 

After reviewing the 45-45-90 triangle identity, these further articles in the triangle geometry series will take you through more identities, each of the specific triangles and how the GMAT uses them to test your critical and creative solving skills:
Triangle Overview
Right Triangles
Equilateral Triangles
Isosceles Triangles
Triangles within other shapes
Pythagorean Identities

Read more
3-4-5 Triangle
Posted on
15
Dec 2020

The 3-4-5 Right Triangle – GMAT Geometry Guide

By: Rich Zwelling (Apex GMAT Instructor)
Date: 17 Dec 2020

Right Triangle Identities: 3-4-5

Right triangles always adhere to the same basic relationship, reflected by the Pythagorean Theorem, or + b² = c², where a, b, and c match the triangle sides as pictured above. c always represents the longest side, called the hypotenuse.

But rather than use the formula directly, the most common way the GMAT will test knowledge of the formula is through the simplest integer values that fit this relationship. The most common is + 4² = 5² → 9 + 16 = 25, as pictured below:

What’s important to remember is that this relationship works not only for 3-4-5, but also for any corresponding multiples, such as 6-8-10 or 9-12-15 or any other multiples of the original numbers.

6-8-10 triangle on the gmat

GMAT Triangle Problem #1

If you rely solely on the formula, you could certainly get the job done, but it will take you a lot longer. Here’s an Official Guide problem that drives this point home:

The figure above shows a path around a triangular piece of land. Mary walked the distance of 8 miles from P to Q and then walked the distance of 6 miles from Q to R. If Ted walked directly from P to R, by what percent did the distance that Mary walked exceed the distance that Ted walked?

(A)   30%
(B)   40%
(C)   50%
(D)   60%
(E)   80%

If you know your so-called “Pythagorean Triples” from memory (e.g. 3-4-5, 6-8-10), this problem moves along much faster. By test day, you should know within seconds that segment PR is length 10, no calculations involved. 

After that, the bulk of your time should be spent calculating the percent difference between Mary’s total distance (14) and Ted’s total distance (10). (Answer: Since Mary walked 4 more miles more than Ted’s original 10, and 4 is 40% of 10, this makes B the correct answer.)

GMAT Triangle Problem #2

Also, it’s much more likely that the GMAT will test your knowledge of completeness of information with respect to Right Triangles, especially on Data Sufficiency. Give this problem a shot before reading on:

What is the area of triangle ABC pictured above?

  1. The length of segment AB is 5
  2. The perimeter of triangle ABC is 12

A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D) EACH statement ALONE is sufficient to answer the question asked.
E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

If you know your 3-4-5 Pythagorean triple really well, you may prematurely view Statement (1) as sufficient, since you may believe the hypotenuse of 5 automatically guarantees the legs of the triangle must be 3 and 4. But this assumes the legs must have integer lengths. In reality, the legs could be any non-integer lengths that satisfy + b² = 5²

This is a classic way the GMAT could throw you off your guard. And it’s another way to test that only one side of a triangle is not enough to give you complete information about the entire triangle. Statement (1) is actually INSUFFICIENT, because we do not have information about a unique triangle, and thus could not possibly know the area.

Likewise, Statement (2) is INSUFFICIENT, because there are many ways to generate a right triangle with a perimeter of 12.

When we combine the statements, however, it’s interesting to note that, as a rule, we know we have a unique triangle if we’re given both the perimeter and the hypotenuse. As such, we would be able to find the area (even though we don’t have to calculate it), and thus the answer is C. Don’t do any math!

Takeaways

The big takeaway here is that, rather than have you use the Pythagorean theorem directly, the GMAT will try to force you into false conclusions, such as believing a hypotenuse of 5 gives you all the information you need. Be on your toes! Make sure to thoroughly examine all information given to you!

 

The 3-4-5 triangle is not the only identity to review in this triangle geometry section, here are some other identities and triangle related topics to review:
GMAT Triangles
Right Triangles
Equilateral Triangles
Isosceles Triangles
Triangles within other shapes
Pythagorean Identities
45-45-90 Triangles

 

 

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Posted on
15
Dec 2020

A short meditation on GMAT triangles – Read this if you struggle with triangle problems on the GMAT!

By: Rich Zwelling (Apex GMAT Instructor)

Date: 15 Dec 2020

GMAT Triangles

Close your eyes, imagine a triangle, then open your eyes again.

Which of the following did you picture? Something different?

Triangle problems explained

Well, every one of the above is tested on the GMAT. And you can’t stick to just one conception of a single shape. You have to be familiar with different types of triangles and all of their characteristics. 

For example, suppose you need to find the two-dimensional area within a triangle. What’s the proper way to do this for a right triangle (2), which contains a 90-degree angle? Even if I know the proper formula, would applying it prove different if I needed to use it on triangle (4), an obtuse triangle, which has one angle greater than 90 degrees?

What about the perimeter, or the sum of the lengths of the three sides of the triangle? Are there simple, straightforward ways to calculate this in the case of the equilateral triangle (1), where each side is the same length. Or the isosceles triangle (3), where two of the three sides are the same length?

What about that isosceles triangle? Did you notice that the two sides that are the same length are opposite two angles of equal measure? What are the implications of this? 

How about the fact that all the angles in any triangle add up to the same degree measure, no matter how the triangle looks?

And then there’s something as simple as the base of the triangle. Take triangle (4):

gmat triangles

As shown, the shortest side is the base, or bottom side, of the triangle. But in truth, any side of a triangle could be considered the base. What if we took the same triangle and rotated it clockwise:

gmat triangle guide articleNow suddenly, the same triangle has the longest side as its base. This could completely change how you calculate, for example, the area of the triangle.

What about the relationships among the sides of a right triangle, for example?

It turns out that the sides a, b, and c, of all right triangles conform to a special mathematical relationship, no matter their lengths. And in its most elementary forms as well. For example, did you know that if a right triangle has short sides (legs) of lengths 3 and 4, the longest side (hypotenuse) will always be 5?

gmat triangle - right triangle

The GMAT is less interested in very complex math and much more interested in variations on basic facts like this one. 

Also, what happens when triangle types are combined? What happens when a triangle is both isosceles AND right? How does this affect the relationships among the sides?

Another way the test can give triangles (and other shapes) a twist is to combine them with other shapes. You may see questions that involve, for example, a triangle within a circles (inscribed):

triangle inscribed in a circle gmat triangle problemThis presents all sorts of opportunities for testing not only the measurements of triangles and of circles individually but also the relationships among those measurements and the ways in which they might overlap

This is of course meant to be a brief overview of the basic forms GMAT triangles can take. But the above poses critical thinking questions you should examine as you move forward with your preparation. Remember: don’t just think in terms of memorizing facts! There will be a handful of formulas to know, to be sure, but once those are out of the way, it will be up to you to be on the lookout for new ways that knowledge could be applied and tested. 

 

As the next step in your discovery and practice of GMAT triangles visit these posts:

Right Triangles
Equilateral Triangles
Isosceles Triangles
Triangles within other shapes
Pythagorean Identities
45-45-90 Triangles

 

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gmat probability problems
Posted on
20
Nov 2020

GMAT How-to: Probability Problems

GMAT probability questions, which test logical reasoning skills, tend to be quite daunting. The good news is that they don’t appear very frequently; the Quant section contains no more than three or four probability questions. However, since so many test-takers struggle with these questions, mastering probability can be an excellent way to boost your overall score. 

GMAT probability questions aren’t so hard once you’ve grasped the basic concepts. Like the majority of the Quant section, probability questions only cover high school level material. The principle challenge is the tricky wording. 

This article will cover some methods to simplify probability questions and boost your Quant score. 

What Is Probability?

The first step to mastering probability is to break down the basic idea:

Probability = the number of desired outcomes / the total number of outcomes

Or in other words, the chance of something happening is the quotient of the number of desired outcomes and the total number of possible outcomes.

A coin flip is one generic example that can help us understand probability.

There are two possible outcomes when we flip a coin: heads or tails. If we want the coin to land on heads, then we divide 1 (the chance that the coin will land on the desired outcome, heads) by 2 (all possible outcomes, heads and tails), and the result is ½ or 0.5 (50%), meaning that there is a 50% chance that the coin will land on heads.

Although this is an elementary example, it demonstrates the fundamental concept behind all probability problems–a ratio between a part and a whole expressed as a fraction or percentage.

Probability of Independent Events

The probability of x discrete events occurring is the product of all individual probabilities.

For example, imagine that we toss a coin twice. Each toss is independent of the other, meaning that each toss has an equal chance of landing on either heads or tails (0.5). If we want to calculate the chance of getting heads twice in a row, we need to multiply the probability of getting heads the first time by the probability of getting heads the second time. 

Or, represented as an equation:

 ½ x ½ = ¼ 

We get a 0.25 or 25% chance that the coin will land on heads twice. 

Probability of Getting Either A or B

Keep in mind that the sum of all possible events is equal to 1 (100%). 

If we continue with the coin toss example, we know that the probability of landing on heads is 0.5, and that the probability of landing on tails is also 0.5. Therefore:

0.5 + 0.5 = 1

The possibility of landing on either heads or tails is equal to 1, or 100%. In other words, every time we flip a coin, we can be certain that it will land on heads or tails.

Probability Of An Event Not Occurring

Following the concept that the sum of all possible events is 1, we can conclude that the probability of event A not happening (A’) is 1 – A, or equal to the probability of event B occurring.

The chance that the coin will not land on heads is equal to the chance that the coin will land on tails:

1 – 0.5 = 0.5

This method is most useful in situations with many favorable events and fewer unfavorable ones. Since time management is essential on the GMAT, it’s better to avoid solution paths that require more calculations. Subtracting the number of unfavorable events from the whole is quicker and simpler, and thus, less likely to result in mistakes.

Pay Attention to Keywords

Read each problem’s wording with great care to determine exactly which operations to use. 

For example, if the problem uses the word “and,” you need to find the product of the probabilities. If the question uses the word “or,” you need to solve for their sum.

If we flipped one coin and we wanted to know the chances of landing on either heads or tails, we would calculate it like this:

0.5 + 0.5 = 1

Similarly, if we were to toss two coins and we wanted to find the probability of landing on both heads and tails, we would use this equation:

0.5 x 0.5 = 0.25

Avoid Common Errors

Minor errors, such as missing possible events, can lead to incorrect answers.

These pointers will help you avoid some common mistakes on probability questions:

  • List all possible events before starting any calculations;
  • Sum up the probabilities of all possible events to make sure they add up to 1;
  • If there are several different arrangements possible (for example, picking different colored balls from a box), find the probability of one of the events and multiply it by the number of different possible arrangements.

If you enjoyed this article make sure to check out our other How To articles like: Efficient Learning & Verbal section.

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