The post Quantitative Comparison and Manipulation appeared first on Magoosh GRE Blog.

]]>Many quantitative questions have variables in both columns. While your first instinct may be to work algebraically, this strategy is not always best. Often the fastest way to a solution is by plugging in different values to see which column is greater.

Developing a sense of when to plug in and when to solve algebraically takes practice. Here are a few helpful guidelines when trying to determine which approach to use.

If you are dealing with a polynomial, simplify. Plugging in may require too much calculation. Instead, work with the familiar algebraic forms shown below:

Now let’s take a look at a question.

Column A | Column B |
---|---|

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

In Quantitative Comparison we can make it so each side is equal. Then we can balance the equation, adding and subtracting, multiplying and dividing, where necessary.

First, note that can be factored into . Now we can set both columns equal to each other:

At this point we have to be careful. While algebra tells us to divide both sides by (x – 2), we need to be aware of the following: if x is between 0 and 2, (x-2) yields a negative. However, (x + 2) yields a positive, meaning (x – 2)(x + 2) gives us a negative. In this case (A) would be bigger.

However, if we divide each side by (x – 2) and solve we get the following:

DIVIDE EACH SIDE BY (x – 2)

SUBTRACT ‘X’ FROM BOTH SIDES

This hardly looks like a solution (in fact it looks like I forgot to go to grade school!). However, what this yields is the important insight: Column B is now 4 greater than Column A. Therefore, the answer is (D). Alternatively you could pick numbers, such as ‘0’ and ‘4’. Each gives us different answers, leading to the same conclusion: Answer (D).

This advice pertains to variables that are not in polynomial form, as seen above. Here come up with easy numbers to plug in to see which values the columns yield.

0 > x > y > z > -1

Column A | Column B |
---|---|

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

.

It would be very nice if we could just stop here and choose answer (B). However, things are not so simple. When plugging in one set of numbers we will inevitably come up with one outcome. Once we’ve plugged in and come up with one answer, whether it is (A), (B), or (C), our job is to disprove that answer.

Can we make Column A larger than Column B? Well what if we plug in a values for x and y that are very close to one another. . Now, for z we can plug in a value close to -1, say -8/9. This gives us . Now you can see (B) is much smaller. Therefore the answer is (D).

You should be adept at both algebra and plugging-in to efficiently—and accurately—answer a quantitative comparison question that contains variables. Typically it is best to simply check for polynomials before plugging in. To test out your skills, try these GRE math questions for practice!

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]]>In my previous post, we examined the pros and cons of two approaches for tackling Quantitative Comparison (QC) questions involving variables. Those approaches are:

- Apply algebraic techniques
- Plug in numbers

In today’s post, we’ll examine a useful strategy that can increase the effectiveness of the plug-in-numbers approach.

To set up today’s strategy, please consider the following QC question:

Column A | Column B |
---|---|

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

You can approach the question algebraically, or by plugging in numbers.

Now, as I mentioned in the last post, the algebraic approach provides the most conclusive results. However, before I begin applying algebraic techniques, I should determine whether it may be faster to use the plug-in-numbers approach. I do this by asking the following question:

“**Is there a nice value for the variable that makes the two columns equal?**”

Notice that I said “nice.” Obviously, we don’t want to spend 2 minutes looking for values of *x* that make the two columns equal. So, I’m going to consider the usual suspects: 1, 0 and -1, and perhaps some other numbers if they look obvious.

Now, when we examine the question above, we see that both expressions have a constant term (the term without any variables) of 3.

Since all of the other terms have *x* in them, we can make those other terms equal zero if we let *x *= 0. So, when *x*=0, we get:

Column A | Column B |
---|---|

When we evaluate this, we get:

Column A | Column B |
---|---|

At this point—within a matter of seconds, and without any significant calculations—we already know that our answer must be either C (the two quantities are always equal), or D (the relationship cannot be determined from the information given).

All we need to do now is try another number. When we plug that new number, we know that if the two quantities are not equal, then the answer must be D. If the two quantities are still equal, then the answer is *leaning towards* C.

Now, which number should we try? Well, we should choose one that makes it easy to evaluate both sides.

How about 1? When *x*=1, we get:

Column A | Column B |
---|---|

When we evaluate this, we get:

Column A | Column B |
---|---|

Since the two columns are not equal when *x*=1, the correct answer is D.

There’s a big advantage in looking for ways to make the two columns equal, rather than just plugging in numbers. Let’s say you plug in a number, and it turns out that Column A is greater than Column B. At this point, we want to find a value for the given variable, such that Column B is greater than Column A. This will allow us to be certain that the answer is D.

Alternatively, if we find a value for the given variable, such that the two columns are equal, we now need only find a value for the given variable such that the two columns are not equal. This is a much easier task than looking for values that make one column greater than the other.

Okay, let’s try another question:

Column A | Column B |
---|---|

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

If we wanted to use the algebraic approach here, we’d have to expand each side and do a lot of work to solve the question.

Alternatively, if we ask the question, “*Is there a nice value for the variable that makes the two columns equal?*”, we can see that, when *x*=2, both columns evaluate to be zero:

Column A | Column B |
---|---|

Since 2-2=0, both products will evaluate to be zero.

At this point, we need only find a value for *x* such that the two columns are not equal.

Well, a nice number to plug in for *x* is 5. When *x*=5, we get:

Column A | Column B |
---|---|

At this point, we can see that the Column A will evaluate to be zero, and Column B will not evaluate to be zero. We don’t really care about which column is greater. All we care about is that the two columns are not equal, which means the answer is D.

So, when tackling QC questions involving variables, be sure to ask, “*Is there a nice value for the variable that makes the two columns equal?*”

Heres’ the whole series of QC tips:

Tip #1: Dealing with Variables

Tip #5: Estimation with a Twist

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]]>The post GRE Math Review Quiz: Plugging In appeared first on Magoosh GRE Blog.

]]>I’ve compiled 7 practice problems (some old, some more recent) from math posts on our blog that are great practice for using the “plugging in” strategy. If you’re already familiar with the approach, test yourself with these problems to review the strategy– it’s a method you’ll definitely be able to use for several different question types in GRE math.

**1. A transcontinental jet travels at a rate of x–100 mph with a headwind and x+100 mph with a tailwind between Wavetown and Urbanio, two cities 3,200 miles apart. If it takes the jet 2 hr 40 minutes longer to complete the trip with a headwind, then what is the jet’s rate flying with a tailwind?**

- 500
- 540
- 600
- 720
- Cannot be determined by the information given.

Answer and explanation: GRE Math Strategies: When to Plug In

**2. Xyla can paint a fence in 6 hrs. Working alone, Yarba can paint the same fence in x hours. Working together they can complete the task in 2.1 hrs.**

Column A | Column B |

x | 3 |

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

Answer and explanation: Does Plugging In work on GRE Quantitative Comparison?

**3. A concession stand sells either hotdogs for $1.75 each or hamburgers for $4 each. If Charlie buys a total of 9 items from the concession stand for a total of 27 dollars, then how many hot dogs did he buy?**

- 3
- 4
- 6
- 7
- 8

Answer and explanation: Part I: The Power of Plugggin In– GRE Math Techniques

**4. If , what is the value of x?**

Answer and explanation: Math Basics: Exponents

**5. Which of the following must be greater than , where -1 < x < 0? Choose ALL that apply. **

Answer and explanation: GRE Math: Difficult Multiple Answer Questions

**6. ^m^ is equal to the digits in positive integer m in reverse order, discounting the zeroes (e.g. ^41^ = 14 but ^3500^ = 53). Which of the following must be true? Select all that apply.**

- ^m^ < ^m+1^
- m = ^(^m^)^
- ^1000m^ = ^m^
- (^m^)(^m^) > ^m^

Answer and Explanation: New GRE Math Strategies

- 200 mph
- 240 mph
- 410 mph
- 480 mph
- 533 mph

Answer and explanation: Math Basics: Distance, Rate, and Time

Let us know if you have any questions, good luck!

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]]>See if you can finish this rate problem in less than two minutes.

*A transcontinental jet travels at a rate of x – 100 mph with a headwind and x + 100 mph with a tailwind between Wavetown and Urbanio, two cities 3,200 miles apart. If it takes the jet 2 hr 40 minutes longer to complete the trip with a headwind, then what is the jet’s rate flying with a tailwind?*

*(A) 500*

*(B) 540*

*(C) 600*

*(D) 720*

*(E) Cannot be determined by the information given.*

**Did you Plug in?**

You may have been tempted to put together an algebraic equation. If you are adept at doing so, and can usually get the answer quickly, then I encourage you to go ahead and make an equation. Most students, however, find this strategy cumbersome and problematic. Even if they set up the right equation, which they can’t be quite sure of unless they get the answer, they may very well make a mistake in solving the equation. If this scenario describes what just happened when you attempted the problem above, know that there is a better strategy: Plugging-In.

**When and Where to Plug-In**

Once we’ve decided to plug in, where do we start? Do we plug in our own numbers, or the answer choices? First, we want to look at the answer choices. Are they numbers? If so, plug them in. If they are variables, you will need to come up with your own numbers, as long as those numbers conform to the information provided in the question.

Here we have answers, so let’s try plugging them in. The first place to start is the middle. The logic is if the number is too low (or too slow, in this case), you need to pick a larger number. Note: if the middle answer is a weird number like 625, then I would recommend plugging in (B) or (D).

Luckily, we can work with answer choice (C) 600. If the speed with a tailwind is 600 mph, then the speed with the headwind is 400. The distance between the two cities is 3200 miles. Using d = rt, where d stands for distance, r stands for rate, and t stands for time, we find that the time it takes to fly with a tailwind is 5 hr 20 min, and the time with a headwind is 8 hours. The difference in time is 2 hr 40 min. And there is our answer. Just like that.

If that seems too easy, that’s not a bad thing. Plugging-In can look like magic, in that it can make a seemingly intractable problem fall into place, just like that.

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]]>The post Part II: Does Plugging in Work on GRE Quantitative Comparison? appeared first on Magoosh GRE Blog.

]]>**Can Plugging In Work on Quantitative Comparison?**

Plugging in, a great strategy on problem solving, can also be very effective on the current and new GRE’s quantitative comparisons. The ground rules for plugging in on quantitative comparison, however, are a little different. But before I explain how, why don’t you try to crack the following problem.

Xyla can paint a fence in 6 hrs. Working alone, Yarba can paint the same fence in x hours. Working together they can complete the task in 2.1 hrs.

Column A | Column B |

x | 3 |

*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*

**Don’t Dive in Head First with the Algebra**

Solving for x can be very challenging. You’ll be dealing with complex fractions and could easily mess up some of the math. Plugging in, on the other hand, can make things a lot easier.

Wait a second, you’re probably thinking. Wouldn’t it take much longer just plugging in a random value for x? Yes, it definitely would. But with this question, we do not want to plug in any random value. We already have a value staring at us. And that is the 3 in Column B.

Maybe you’ve balked, thinking that this is a GRE quantitative comparison problem and therefore, we’re not looking for a specific answer. You could go on to reason that in problem solving, we can always work with the five answer choices. But in quantitative comparison, we do not want a specific value, we simply want to know which side is greater.

**The Hypothetical **

While it is true that we do not want to solve for a specific value on Quantitative Comparison, let’s still assume that x is 3. Let’s further assume that plugging 3 back into the problem gives us 2.1 hours. Then we could conclude that the x has to equal 3, and, therefore, the answer is (C).

What if we’d plugged in 3 and gotten a total work rate lower than 2.1? Then we would know that a rate of 3 hrs is too fast. We would have to slow down Yarba and we would do so by increasing x, the amount of hours it takes her to finish the job by herself. In this case Column A would be bigger.

The final possibility is if we had plugged in 3 and gotten a number larger than 2.1. Then we would know that x has to be lower than 3, because we need a faster work rate. The answer then would be (B).

**The Approach**

Now that we’ve gone through all the possible scenarios, let’s plug in 3 and see what we get. Click here For a brush up on work rate formula.

1/6 + 1/3 = 1/TWR

Total Work Rate = 6 x 3/9 = 2

So, we’ve plugged in 3 and we’ve gotten a number lower than 2.1. Therefore, 3 hrs is too fast (or to low). We have to slow Yarba down a little. And to slow someone down, we have to increase the amount of time it takes him/her to finish the job. Thus, Column A is bigger.

**Algebra** **– Friend or Foe?**

Had we attempted to solve the problem algebraically, we would have set up an equation that looked like this: 1/6 + 1/x = 10/21. You can try this at home, but don’t be surprised if you get stuck along the way. First off, there are many steps and secondly, when you finally solve for x you get 126/39 – hardly a friendly number. But you can avoid all this if you simply remember that there is an easier way—Plugging In.

Try plugging in with different GRE math question types, as well!

The post Part II: Does Plugging in Work on GRE Quantitative Comparison? appeared first on Magoosh GRE Blog.

]]>The post Part I : The Power of Plugging In – GRE Math Techniques appeared first on Magoosh GRE Blog.

]]>With BBQ season nearing, it’s time for hotdogs and burgers. Well, sort of. Try the problem below.

*A concession stand sells either hotdogs for $1.75 each or hamburgers for $4 each. If Charlie buys a total of 9 items from the concession stand for a total of 27 dollars, then how many hotdogs did he buy? *

*(A) **3*

*(B) **4*

*(C) **6*

*(D) **7*

*(E) **8*

On the GRE, you will most likely see a long, tedious word problem. Granted, this problem wasn’t too bad. The question is, how did you go about trying to solve it?

Many begin trying to turn the information from the problem into an equation. Some are very adept at this and get the answer quickly. But, most students I’ve seen struggle as soon as they start thinking about the above problem algebraically. So, I want you to try something that may seem counterintuitive. I want you to turn off the algebra part of your brain. Instead, I want you to work with the answer choices by plugging them back into the problem.

Let’s start with answer choice (C). The reason we start with this answer is because it is in the middle. If it turns out that the total price is less than 27 dollars, then we will need to plug in a lower number. The reason we would choose a lower number is because hotdogs are cheaper; therefore, the fewer hotdogs we have the more burgers we have, and this brings up the price.

Plugging in 6 we get $1.75 x 6 = $10.50. That leaves us 3 hamburgers at $4 each = $12. Adding the two together gives us $22.50. That amount is too low, so we can eliminate answer choices (D) and (E), as well, because they will increase the number of hotdogs and lower the overall price.

Plugging in (B) 4, we get $7 on hotdogs. That leaves us 5 hamburgers or $4 x 5 = $20. Add up $7 + $20 and we get $27, which is the correct answer.

In the next segment on plugging in, I will show you a very different type of problem in which plugging in can be equally effective.

Check out Magoosh’s other GRE math practice problems to try plugging in on different problem types!

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]]>The post GRE Math Strategies Part II of VI: Plug In (Substitution) Method appeared first on Magoosh GRE Blog.

]]>This strategy works best when a symbolic expression (like an equation) is provided and the answer choices are all numerical values.

**Example #1:**

If does not equal 4, then

(A)

(B)

(C)

(D)

(E)

**(C) is the correct answer.**

For the product of two factors to equal zero, one (or both) of the factors must equal zero.

It’s given that x does not equal 4, so let’s examine the first factor. What value for x renders the first factor equal to zero?

Examine the choices.

(A) generates , a positive mixed number. No.

(B) does also: . No.

(C) generates the correct value for the first factor: Yes!

(D) generates , a positive number. No.

(E) generates , a positive mixed number. No.

The Plug In approach also works for hard-to-follow word problems when it’s not easy to represent the unknown quantity and create the equation.

**Example #2:**

Suits at a local clothing store are discounted 20 percent. Despite the discount, the store owner enjoys a net profit (discount price minus cost) equal to 60 percent of the cost of the suits. If the suits cost the clothing store owner $200 each, what was the prediscount retail price of each suit?

(A) $300

(B) $400

(C) $500

(D) $600

(E) $700

**(B) is the correct answer.**

If the suits cost the owner $200 each, then the net profit (discount price minus cost) on each suit is 60 percent times $200, or $120. Thus, the discount retail price of each suit must be $200 (cost) + $120 (net profit) = $320.

Which of the (prediscount) answer choices discounted by 20 percent equals $320?

(A) $300 times 20 percent discount equals $300-$60 = $240 (incorrect).

(B) $400 times 20 percent discount equals $400-$80 = $320 **(correct!)**.

(C) $500 times 20 percent discount equals $500-$100 = $400 (incorrect)

(D) $600 times 20 percent discount equals $600-$120 = $480 (incorrect).

(E) $700 times 20 percent discount equals $700-$140 = $560 (incorrect).

**Notes:**

The Plug In approach demands common sense and confidence in the strategy.

You will have time to use this strategy.

You may choose to abandon your plugging in after securing (B) as the correct answer.

Of course, the Textbook approach is always available:

Let = the prediscount price of a suit, then = discount price.

The net profit (in dollars) is expressed in two ways in the problem.

Set these expressions equal:

Solving for x:

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