4 techniques to ace gmat sentence correction questions
Posted on
29
Apr 2021

4 Techniques to Ace GMAT Sentence Correction Questions

By: Apex GMAT
Contributor: Andrej Ivanovski
Date: 29th April 2021

The GMAT Sentence Correction questions are one of the three question formats that comprise the Verbal section, with the other two being Reading Comprehension and Critical Reasoning. Test takers should expect to come across anything between 11 and 16 sentence correction questions on the exam. Each GMAT Sentence Correction question contains a part that is underlined, and you will be prompted to identify the mistake in the sentence and replace it with one of the five options provided. Even though this might sound like a piece of cake at first glanceglimpse, there is a catch. The reason that most GMAT test takers find the Sentence Correction questions challenging is the fact that the sentences provided are usually several lines long and the grammatical mistakes are not very apparent. If you follow these 4 GMAT Sentence Correction techniques you will find it a lot easier to spot the mistakes and ace the GMAT Sentence Correction questions.

Get rid of the extra information

The GMAC intentionally makes the GMAT Sentence Correction problems long by including a lot of fluff and descriptive information which very often covers up the error and makes it very difficult to spot. Therefore, getting rid of that extra information would not only make the sentence shorter and simpler, but it would also make it easier for you to uncover the mistake. But, how do you know which part of the sentence to get rid of?

  • Look for parts of the sentence set off by commas. Oftentimes, the part that is set off the comma only serves to better explain or give more details about the subject, and when removed it would not affect the meaning of the sentence. Here’s what extra information looks like in a sentence (note that there are no mistakes in the given example):

Maria, Stephen’s youngest and most talented daughter, moved to Sweden. 

Maria, Stephen’s youngest and most talented daughter, moved to Sweden. 

In the sentence above, the part set off by commas is not necessary to convey the meaning of the sentence. So, even if you get rid of that part, you would still be left with a complete sentence. However, one caveat to keep in mind is that the extra information does not necessarily have to be separated by two commas, as it can come at the beginning or the end of the sentence (a modifier), in which case it would only be set off with a single comma.

  • Look for adjectives and adverbial phrases. These could be a little more challenging to find, as they are not set off by commas and one needs to understand the meaning of the sentence in order to identify them.

A group of young men coming from Dubai held a conference in New York.

The sentence above can exist without the two underlined parts: of young men and coming from Dubai. Even though they make the sentence more descriptive, they do not convey the main meaning of the sentence, and can therefore be taken out of the sentence for the sake of simplicity and spotting the mistake more easily.

Pay attention to the meaning

We have already established that grammar is vital if you want to do well on the GMAT Sentence Correction problems. Is grammar necessary? Absolutely! Is grammar everything that you need? Definitely not! No matter how good you are at grammar, solely relying on it is guaranteed to get you stuck at one point or another.

It is often the case that GMAT Sentence Correction problems are free of grammatical errors, but contain logical ones. GMAT test-makers are actually hoping that test-takers will only rely upon grammar and would not pay attention to less formal errors, so if you want to do well on this type of question you absolutely need to pay attention to the meaning of the sentence.

In order to do so, you first need to read the sentence carefully and try to understand the meaning behind it. Oftentimes, it might seem that the sentence is perfectly correct and free of grammar mistakes, and you would not be able to find a logical gap or an inconsistency. In that case, you will want to look through the answers provided and try to assess the message that they are trying to convey. When doing that, you might get an idea of what could be wrong with the original sentence and in that way find the correct one.

Use “splits”

Another strategy which includes using the answer choices in order to successfully answer the GMAT Sentence Correction problems is the so-called “splits” strategy. This strategy involves trying to find similarities and dissimilarities, or any kind of patterns in the answer choices. In order to explain this strategy, we will use a GMAT Sentence Correction problem from the GMAT Official Guide.

The overall slackening of growth in productivity is influenced less by government regulation, although that is significant for specific industries like mining, than the coming to an end of a period of rapid growth in agricultural productivity.

  • the coming to an end of
  • the ending of
  • by the coming to an end of
  • by ending
  • by the end of

In a question like this, the mistake might not be apparent at first. Therefore, in order to get an idea of what the mistake could be, we will have a look at the answer choices. In there, we can see two patterns: C, D and E all contain “by”, whereas A and B do not. If we look at the sentence, we can see that the first part of it says “is influenced less by”, which implies that the second part of the questions has to begin with “…than by”. Therefore, the split AB, and we continue looking for the answer in the CDE split. If we try to plug each of these three answers into the sentence, we can see that E is the only one that is grammatically correct and therefore we get E as an answer.

The “splits” technique is especially useful in helping you narrow down the choices and find the right answer more easily.

Learn the most common GMAT idioms

In order to do well on the Sentence Correction GMAT questions, you need to have a good command of idioms. If you have already started preparing you might have come across a GMAT idiom list in the prep materials. So, you might be wondering why it is important to learn them and how they will be tested on the GMAT.

First, let us begin by explaining what an idiom is. Chances are, if you are not a “grammar freak” you might not be sure what the exact meaning of an idiom is. An idiom is a common expression or a grammatical structure in a given language, in this case – English. Oftentimes, the term idiom is used to describe a saying such as “let the cat out of the bag” or “a piece of cake”. Even though these are important to know if you want to sound more fluent and natural in English, they are not tested on the GMAT. In the context of the GMAT, an idiom is a formation of two or more words that are often used together, such as “invest in” or “indicate that”.

So, now that we have gotten the definition out of the way, you might be wondering why it is important to learn some of the most common GMAT idioms, and how they will be tested. In the GMAT Sentence Correction problems, oftentimes you will come across an incorrectly used idiom. The mistake can take several different forms. 

  • Preposition

Take, for instance, the expression invest on. Here, the preposition used is on when in fact it should be in. Even though it could be apparent in this case, on the GMAT the mistake can often be subtle and a little more difficult to spot. 

  • Word choice 

This is also a common mistake, especially when it comes to words that are close in meaning. Examples of such words are among/between, fewer/less, whether/if, like/as, and so on.

  • Correlatives

Correlatives are words that are used together to serve a single function in a sentence. Some examples include both/and, either/or and neither/nor. A mistake in correlative pairs is also common, especially when it comes to longer and more complex sentences, as these mistakes could be more difficult to spot in those cases.

Conclusion

Here’s a summary of all of the techniques that we discussed here:

gmat sentence correction

These techniques are not mutually exclusive and they can be used in combination with one another. Applying them and putting them into practice can save you a whole lot of work and help you do better on the GMAT Sentence Correction problems. And if you feel like you could use some more guidance, please make sure to check out our highly personalized one-on-one GMAT tutoring. Our tutoring sessions are delivered by 770+ scoring tutors and are available both online and in-person, no matter where in the world you are.

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GMAT Combinations with Restrictions Article
Posted on
04
Mar 2021

Combinations with Restrictions

By: Rich Zwelling, Apex GMAT Instructor
Date: 4th March, 2021

In our previous post, we discussed how GMAT combinatorics problems can involve subtracting out restrictions. However, we discussed only PERMUTATIONS and not COMBINATIONS.

Today, we’ll take a look at how the same technique can be applied to COMBINATION problems. This may be a bit more complicated, as you’ll have to use the formula for combinations, but the approach will be the same.

Let’s start with a basic example. Suppose I were to give you the following problem:

The board of a large oil company is tasked with selecting a committee of three people to head a certain project for the following year. It has a list of ten applicants to choose from. How many potential committees are possible?

This is a straightforward combination problem. (And we know it’s a COMBINATION situation, because we do not care about the order in which the three people appear. Even if we shift the order, the same three people will still comprise the same committee.)

We would simply use the combination math discussed in our Intro to Combination Math post:

                         10!
 10C3 =       ————-
                     3! (10-3!)

 

   10!
———
3! (7!)

 

10*9*8
———
3!

 

10*9*8
———
3*2*1

= 120 Combinations 

However, what if we shifted the problem slightly to look like the following? (As always, give the problem a shot before reading on…):

The board of a large oil company is tasked with selecting a committee of three people to head a certain project for the following year. It has a list of ten applicants to choose from, three of whom are women and the remainder of whom are men. How many potential committees are possible if the committee must contain at least one woman?

A) 60
B) 75
C) 85
D) 90
E) 95

In this case, there’s a very important SIGNAL. The language “at least one” is a huge giveaway. This means there could be 1 woman, 2 women, or 3 women which means we would have to examine three separate cases. That’s a lot of busy work. 

But as we discussed in the previous post, why not instead look at what we don’t want and subtract it from the total? In this case, that would be the case of 0 women. Then, we could subtract that from the total number of combinations without restrictions. This would leave behind the cases we do want (i.e. all the cases involving at least one woman). 

We already discussed what happens without restrictions: There are 10 people to choose from, and we’re selecting a subgroup of 3 people, leading to 10C3  or 120 combinations possible. 

But how do we consider the combinations we don’t want? Well, we want to eliminate every combination that involves 0 women. In other words, we want to eliminate every possible committee of three people that involves all men. So how do we find that?

Well, there are seven men to choose from, and since we are choosing a subgroup of 3, we can simply use 7C3 to find the number of committees involving all men:

                       7!
7C3 =       ————-
                 3! (7-3!)

 

  7!
———
3! (4!)

 

7*6*5
———
    3!

= 7*5 = 35 Combinations involving all men

So, out of the 120 committees available, 35 of them involve all men. That means 120-35 = 85 involve at least one woman. The correct answer is C. 

Next time, we’ll return to probability and talk about how the principle of subtracting out elements that we don’t want can aid us on certain questions. Then we’ll dovetail the two and talk about how probability and combinatorics can show up simultaneously on certain questions.

Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements
Permutations With Restrictions
Combinations with Restrictions
Independent vs Dependent Probability
GMAT Probability Math – The Undesired Approach
GMAT Probability Meets Combinatorics: One Problem, Two Approaches

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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: 25th 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
Permutations With Restrictions
Combinations with Restrictions
Independent vs Dependent Probability
GMAT Probability Math – The Undesired Approach
GMAT Probability Meets Combinatorics: One Problem, Two Approaches

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Triangle Inequality Rule on the GMAT
Posted on
09
Feb 2021

Triangle Inequality Rule

By: Rich Zwelling, Apex GMAT Instructor
Date: 9th February, 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

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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: 14th January, 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:

Equilateral 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:

Equilateral 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:
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

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5-12-13 and 7-24-25 triangles on the gmat
Posted on
12
Jan 2021

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

By: Rich Zwelling, Apex GMAT Instructor
Date: 12th January, 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:

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
45-45-90 triangles on the gmat
Posted on
06
Jan 2021

45-45-90 Right Triangle – GMAT Geometry Guide

By: Rich Zwelling, Apex GMAT Instructor
Date: 6th January, 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:


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