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Comment on QC Strategy - Matching Operations
On the next to last example
Plugging in a negative value
Plugging in a negative value for a variable is not the same as multiplying both quantities by a negative number.
with due respect, why not to
That approach can get you
That approach can get you into trouble. If you use the Plugging In Numbers approach, you can't just stop at plugging in one value, since there might be other values of w such that the two quantities are equal or Quantity B is greater (for more on this, watch the part that starts at 2:50 of this video: https://www.greenlighttestprep.com/module/gre-quantitative-comparison/vi...
For example, let's say that we're told that w > 0 and we have the following:
QUANTITY A: w
QUANTITY B: w^2
Using your approach, we might plug in w = 2 (since we're told that w > 0, and 2 > 0). When we do this, we get:
QUANTITY A: 2
QUANTITY B: 4
Here, Quantity B is greater. Does this mean the correct answer is B?
If we plug in w = 1, we get:
QUANTITY A: 1
QUANTITY B: 1
Here, the two quantities are equal.
If we plug in w = 1/2, we get:
QUANTITY A: 1/2
QUANTITY B: 1/4
Here, Quantity A is greater.
So, be careful; plugging in ONE value will not always yield the correct answer.
Also, why not to cancel 'y-3x
We are, indeed, "canceling" y
We are, indeed, "canceling" y - 3x^2 from both quantities. This video is meant to show HOW that works. We do so, by performing the same operation to both quantities.
The term "cancel" can sometimes be misinterpreted. It doesn't mean that, if both quantities share the same algebraic expression, then we can just eliminate those expressions from both quantities.
For example, if we have:
QUANTITY A: 3/x
QUANTITY B: 2/x
We can't just say "Since both expressions have an x term, we just cancel them to get:
QUANTITY A: 3
QUANTITY B: 2
This will yield an incorrect conclusion.
So, we need to be careful to follow the rules/guidelines outlined in the video.
I guess I am confused by the
You're referring to the question that appears at 1:37 in the above video:
Given: w > 0
QUANTITY A: 7w + 4
QUANTITY B: w - 2
For Quantitative Comparison questions involving variables, it's often useful to get all of the variables to just one of the quantities. In most cases, that will make it easier to compare the quantities.
So could I say it is
Are you asking whether we
Are you asking whether we should create the equation 7w + 4 = w - 2?
If so, then the answer is no.
The primary goal of Quantitative Comparison questions is to determine the relationship between the two quantities (Is Quantity A greater? Is Quantity B greater? Do they have equal value?). So, we can never assume that the quantities are of equal value.
Does that help?
In the reinforcement question
I'm happy to help!
I'm happy to help!
Question link: https://gre.myprepclub.com/forum/m-p-and-x-are-positive-integers-3017.html
Here's my solution: https://gre.myprepclub.com/forum/m-p-and-x-are-positive-integers-3017.ht...
I have a question about
given: x < 0
Quant A: x^3
Quant B: x^5
Can't I divide by x^3 on both sides leaving me with
Quant A: 1
Quant B: x^2
leaving me with answer D?
Aren't I dividing by a negative on both sides?
Should i not do this?
I address this concept at 2
I address this concept at 2:45 in the above video.
However, I'll show you why dividing both quantities by a negative can lead to false conclusions.
Let's say we're asked to compare the following:
QUANTITY A: -6
QUANTITY B: 12
Clearly Quantity B is greater. In fact, if we follow any of the allowable operations, Quantity B will always be greater.
However, if we break one of the rules and divide both quantities by -3, we get:
QUANTITY A: 2
QUANTITY B: -4
Now, Quantity A is greater.
I hope that helps.
So the safe method in the example I provided would be to divide by x^2 instead of x or x^3.
Yes, that would be a safer
Yes, that would be a safer approach.
If we divide both quantities by x², we get:
Quant A: x
Quant B: x³
Now let's test some cases....
CASE I: If x = -1, then we get:
Quant A: x = -1
Quant B: x³ = (-1)³ = -1
In this case, the quantities are equal
CASE II: If x = -2, then we get:
Quant A: x = -2
Quant B: x³ = (-2)³ = -8
In this case, Quantitt A is greater
So i solved this question can
x - y=5
X^2 - Y^2 can be written as (X-Y)(X+Y), and X-Y = 5. So we get
Dividing both sides by 5 we get
Let's just deal with one variable so we rewrite the given equation as X = Y + 5 and substitute in A to get
We do not know whats the true value of Y it can be negative or positive so answer is D.
That's a perfectly valid
That's a perfectly valid solution. Nice work!
Hey Brent, at the 3:20
With each step in the process
With each step in the process, I'm manipulating the most recent values.
So, for example, we start with:
QUANTITY A: 2
QUANTITY B: 5
When we add 8 to both quantities we get:
QUANTITY A: 2 + 8 = 10
QUANTITY B: 5 + 8 = 13
Next, we subtract 4 from both quantities to get:
QUANTITY A: 10 - 4 = 6
QUANTITY B: 13 - 4 = 9
Next, we divide both quantities by 3to get:
QUANTITY A: 6 ÷ 3 = 2
QUANTITY B: 9 ÷ 3 = 3
And so on.
Does that help?
Haha, Brent you're the best.
Is there any limitation to applying this strategy because I tried applying it unfortunately it didn't work?
here below is the link to the question
Question link: https:/
Question link: https://gre.myprepclub.com/forum/0-t-7960.html
The strategy won't always work, especially in situations in which we don't know whether the variable(s) is/are negative or positive.
That said, it's a very effective strategy and should be considered for a lot of questions.
The question you noted can be solved using Matching Operations.
Here's my solution: https://gre.myprepclub.com/forum/0-t-7960.html#p14521
I have a question when and
There's no guaranteed rule
There's no guaranteed rule for determining the best strategy.
However, if we're given variables in both quantities, then matching operations is a good strategy to start with.
If both quantities consist of numbers and operations only, approximation is a good place to start.
Some other useful strategies include:
- moving the terms with variables to one side
- arranging the values so that one quantity equals 0
- arranging the values so that one quantity equals 1