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]]>Once in a while it is fun to take a concept in an official GRE question and run with it. The question below is inspired by a question from the 2^{nd} edition Official Guide. That is not to say it is that similar; I have only extracted a couple of concepts from that problem.

I’ve also made my question far more difficult. The first question was already difficult enough, so it is safe to say that a question of this difficulty would never appear on the GRE (though it could appear on the GMAT). Another of way expressing this: only attempt this question if you are already very strong in Quant. If not, just have fun with it, remembering that questions on the actual GRE will be easier.

1. A semicircle with area of xπ is marked by seven points equally spaced along the half arc of the semicircle, such that two of the seven points form the endpoints of the diameter. What is the probability of forming a triangle with an area less than x from the total number of triangles formed by combining two of the seven points and the center of the diameter?

- 4/5
- 6/7
- 17/21
- 19/21
- 31/35

Again this is a very difficult question, one that requires many steps, and one that has numerous twists. The question is without a doubt expansive, covering in its sweep combinatorics, probability, basic geometry, advanced triangle theory, counting properties and much more. So hang on for the ride!

In order to deal with this whole area of the triangle business, we first have to figure out the area of the circle. That way, we can also derive the radius, a number that be instrumental in helping us in Step #3.

First though we find the area of the circle, which is twice that of the semi-circle . To solve for the radius, we use .

So we have the radius. But how does that help in figuring out the resulting area of any of the inscribed triangles? Well, remember the seven points? If two points are different sides of the diameter and the other five points are evenly spaced along the arc, then the semicircle is broken into equal sixths. Therefore the central angle of each will be 30 degrees (the arc of a semicircle corresponds to 180 degrees).

** **

Using the information above, we can start to play around with triangles, keeping in mind that any two points along the arc will form radii with the center of the diameter (remember that the original question asked us to make triangles using the middle of the diameter).

A good idea is to come up with a triangle that both meets the criteria and is easy to solve area-wise. That way you can see how close you are to an area of x. For instance, you could assume that the triangle is an equilateral, the two points on the arc a distance of 60 degrees away (and thus the central angle will be 60).

A radius of will form the sides of the equilateral. Using the formula for the area of the equilateral triangle, , where s equals the side of the triangle, or in this case the radius, we get . Because equals approximately 1.7, we can see that such an equilateral triangle has an area less than x.

But don’t give up just yet. In trying to increase the area you should keep in mind that skinnier triangles, say with points on the opposite side of the semicircle, will not have an area greater than that of an equilateral. So we want to increase the angle but not too much.

The next most sensible triangle—that is one in which we can use simple formulas to determine the area—is a 45:45:90 triangle, which means that the two points on the semicircle will have to be 90 degrees apart (that way the central angle will be 90 degrees. And with two equal sides (remember all radii are equal), we know have a 45:45:90 triangle with sides of . Using the formula for the area of a triangle, we get .

An important insight: if we bring the points together, we end up with an equilateral, which we know to have a smaller area. Therefore, moving the points on the arc together results in a triangle with less area. Moving the points apart makes the triangle skinnier, something that also results in a triangle with less area.

Therefore, the isosceles right triangle represents the largest possible triangle, area-wise, that you can inscribe given the conditions of the problem (pretty advanced stuff, right?).

Voila! Finally, we have it. The only possible triangle to yield x will be an isosceles right triangle, or a 45-45-90, with a central angle of 90. If we find the number of these triangles and subtract that from 1, we have the probability of choosing a triangle with an area less than 1.

The total number of isosceles right triangle is four, which we can find by choosing points along the circle that are 90 arc degrees apart.

Next we want to find the total number of triangles. Because one point is already fixed (the center of the diameter), we have to determine how many triangles can be formed by using two of the seven points. The ordering of the vertices is not important. Meaning that a triangle with points A, B, and C, is the same as triangle with points B, C, A. Therefore we use 7C2 = 21.

Just when you thought we were finally done with this monstrosity of the problem, there is one final, and subtle twist. We have to discount one of these twenty-one triangles because it is actually not a triangle. If we choose both end points of the diameter and the center of the diameter, we have a straight line (this line is the diameter of the semicircle).

Therefore, there are twenty triangles, sixteen of which have an area less than x. so the answer is (A) 4/5.

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]]>Each of the math questions below is directly inspired by a question in the on-line Revised GRE test. I’ve provided an easier version of the question (#1) and a more difficult version of the question (#2).

My recommendation is to try the easier version first. Then, if you answer it correctly, click on the link, and take a stab at the actual Revised GRE question.

If you are able to answer that question correctly, then as prize – you get a fiendishly difficult question (#3). Okay, maybe that’s not a prize – but it is great practice for those aiming for the 90% on quant.

The good news is I have explanations. For the Revised GRE question, I have recorded an explanation video you can watch. Finally, it is a good idea to try the easy question before the medium one, and the medium question before the difficult one.

Good luck!

If , where n is a non-negative integer, what is the greatest value of ?

- ½
- 1
- 5
- 32
- 64

**Explanation:** Don’t think big – think small. That is the smaller n becomes the greater ½^n becomes. So what is the smallest value? You may be tempted to say 1, which would give us ½. But remember n = 0, because . Therefore **Answer: B.**

The “hidden zero,” as I like to call it, is a classic GRE math trick. So always keep your eyes open, especially when you see “non-negative integer,” which includes zero.

(5/4)^{-n} < 16^{-1}

What is the least integer value of n?

**Explanation:**

The best place to start here is by getting rid of the unseemly negative signs and translating the equation as follows:

(4/5)^{n} < 1/16

A good little trick to learn using 4/5 taken to some power is that (4/5)^{3} = 64/125, which is slightly — but only slightly — greater than ½. Therefore, we can translate (4/5)^{3} to ½.

(1/2)^{4} = 1/16

That would make (4/5)^{12} a tad larger than 1/16. To make it less than 1/16 we would multiply by the final 4/5, giving us n = 13.

The equation is true for how many unique integer values of n, where n is a prime number?

- 7
- 4
- 2
- 1
- None of the above

This problem can be difficult, indeed downright inscrutable, unless you take your time and process one piece of information at a time. Once you understand what the problem is saying, you should be able to solve the question relatively quickly.

**Explanation: **

The most important piece of info is n is a prime number. So do not start by plugging in zero or one. Neither is a prime. The lowest prime is 2. When we plug in ‘2’ we get:

This is clearly true. Thus we have one instance.

As soon as we plug in other prime numbers a pattern emerges.

is always a negative number if n is odd. Because all of the primes greater than 2 are odd, the number in the middle will always be negative:

Because in each case n is a positive number we can never have the middle of the dual inequality be positive, if n is an odd prime.

Thus the only instance in which the inequality holds true is if we plug in ‘2’, the answer is (D).

If you got that right – congratulate yourself. It’s a toughie.

To find out where exponents sit in the “big picture” of GRE Quant, and what other Quant concepts you should study, check out our post entitled:

What Kind of Math is on the GRE? Breakdown of Quant Concepts by Frequency

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]]>Quite possibly the most intimidating problem on the GRE contains strange symbols: @, #, *, or a black circle often accompany these problems. Many recoil in horror thinking – I’ve never learned that before! (Or perhaps more aptly, what the @#?!)

But don’t despair – the symbols are completely arbitrary and are defined on the spot by the GRE. Here is an example:

#x# = . What is the value of #3# – #2#?

Again, the pound sign surrounding the number has no mathematical meaning outside the problem. For the question, you simply want to follow the rules. Here, wherever we see a number between the pound sign, such as #3#, we want to refer back to #x# = . The 3 essentially is taking the place of the x. So if #x# = , then #3# = .

Now do the same for #2#: .

So #3# – #2# =

Now let’s try another one. This time, though, I am going to put a little spin on it.

n^^ = , where n is a positive integer. For how many values of n is n^^ less than zero?

- 1^^
- 1^^ – 2^^
- 3^^
- 3^^ + 4^^
- 5^^ – 2^^

First off, note that ! is the factorial sign. It is not a strange symbol, but standard mathematical notation.You should quickly see that after n = 3, n^^ is going to yield a positive result. For example, , so 4^^ . So when n is greater than or equal to 4, n^^ is greater than zero.

Be careful: 1^^ = 0, so it is not less than zero. Therefore, there are two values (2, 3) for which n^^ is less than zero.

When we look at the answers, 2 is not among them. Instead, the strange symbol ^^ has been reintroduced. Therefore you have to figure out which answer choice equals 2, the number of values of n that are less than zero.

The answer is (B), which gives us 1^^ = 0 minus 2^^ = -2, so

Okay, that was a tough one. Let’s make the problem easier, while adding a layer of complexity – the embedded strange symbol.

If x is even, @x = ; if x is odd, @x = . What is the value of @(@(@5)?

- 21
- 40
- 63
- 117
- 140

Notice I’ve used the strange symbol three times. Don’t worry – just follow the operation (the technical name of this process). Taking the problem apart one step at a time, we get @5 = 8. @8 = 21, and @21 = 40. Just like that, B.

Don’t be freaked out by strange symbols on the GRE. The question will always clearly define the symbol for you. Carefully follow the steps to the correct answer.

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]]>The post GRE Quantitative Comparison Tip #3 – Logic over Algebra appeared first on Magoosh GRE Blog.

]]>In previous posts (Tip #1 – Dealing with Variables, Tip #2: Striving for Equality), I have discussed two approaches when tackling Quantitative Comparison (QC) questions involving variables. Those approaches are:

1) Apply algebraic techniques

2) Plug in numbers

In those posts, I noted that the algebraic approach is typically the faster and more reliable approach.

In today’s post, we’ll examine a third strategy that can sometimes be the fastest and easiest approach. We’ll call this the “logical approach.”

To set things up, please consider the following QC question:

A. The quantity in Column A is greater

B. The quantity in Column B is greater

C. The two quantities are equal

D. The relationship cannot be determined from the information given

The algebraic approach to this question looks something like this:

First multiply both sides by 35 (the least common multiple of 5 and 7) to get:

Then add 5x to both sides to get:

Then add 42 to both sides to get:

And, finally, divide both sides by 19 to get:

So, now we’re comparing x and 3, and the given information tells us that x is greater than 3.

This means the correct answer must be A.

Now let’s take the original question and use logic to solve it (in about 5 seconds).

Column A: If x > 3, then 2x > 6, which means that 2x-6 must be positive.

Column B: If x > 3, then 3-x must be negative.

So, the two columns can be rewritten as:

From here, we can see that Column A is always positive and Column B is always negative. As such, Column A will always be greater than Column B. So, the correct answer is A.

Let’s try another one. See if you can solve it in your head.

A. The quantity in Column A is greater

B. The quantity in Column B is greater

C. The two quantities are equal

D. The relationship cannot be determined from the information given

For this question, I’ll leave the algebraic approach to you.

Let’s apply some logic.

First, we’re told that . In order to apply some logic, let’s refer to the denominator as “something.” In other words, 18y divided by “something” equals 3. Well, we know that** **18y divided by **6y** equals 3, so that “something” must equal 6y.

In other words, it must be the case that 7y-x = 6y

Now consider the fact that 7y-x = 6y. If we now refer to “x” as “something,” we can see that 7y minus “something” equals 6y. Since we know that 7y-y=6y, we can see that “something” must equal y.

In other words x = y.

Now that we have concluded that x=y, we’ll return to the original question:

If x=y, we can see that the answer here must be C.

So, although the algebraic approach is typically the superior approach for quantitative comparison questions involving variables, be sure to take a moment to see whether the problem can be solved by applying a little logic.

Heres’ the whole series of QC tips:

Tip #1: Dealing with Variables

Tip #5: Estimation with a Twist

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