Can anyone suggest how would the graph of |y|>|X| look like? I don't have MATLAB with me on my machine. I think that it would be something like :
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I don't have graphing capabilities right now but you could do this.voodoo_child wrote:Changed my mind. I think that it should be :
Draw the lines y=x and y = -x on the graph. Then |y| > |x| is the area between these two lines and the y-axis.
Hence the image you just showed would also have a component (actually a mirror image of what you just shaded) underneath the x-axis in the 3rd and 4th quadrants.
Alternatively:
go to : https://www.desmos.com/calculator
And enter the following two equations:
y > abs(x)
y <-abs(x)
The graph that results is the answer.
Let me know if this helps
The second image posted by you isvoodoo_child wrote:Changed my mind. I think that it should be :
y > |x|
For
|y| > |x|
|y| can be y or -y, so all the values that satisfy y > |x| and the negative value of all these values of y will satisfy the inequality.
For example,
if |5| > |3|, |-5| is also > |3|.
The mirror image of the curve drawn in the second image about the X-axis gives us -y for each y that satisfies y > |x|. So, all that combined region (above and below the X-axis) is the region for |y| > |x|.
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|y| > |x| simply means absolute value of y is greater than that of x.voodoo_child wrote:Can anyone suggest how would the graph of |y|>|X| look like?
Hence, the graph of |y| > |x| will contain those points for which absolute value of y-coordinate is more than that of x-coordinate.
In other words, the graph will contain all the points in the region bounded by y = x and y = -x which contain the y-axis.
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Dear voodoo child,
I've attached an image of the graph of the region |y| > |x|. Basically, that's a mathematical way of saying --- closer to the y-axis than to the x-axis.
Also, you may find this blogpost on related topics interesting.
https://magoosh.com/gmat/2012/gmat-math- ... -line-y-x/
Let me know if you have any further questions.
Mike
I've attached an image of the graph of the region |y| > |x|. Basically, that's a mathematical way of saying --- closer to the y-axis than to the x-axis.
Also, you may find this blogpost on related topics interesting.
https://magoosh.com/gmat/2012/gmat-math- ... -line-y-x/
Let me know if you have any further questions.
Mike
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Mike,
Thanks for your reply. However, I tried plotting the graph using the online tool mentioned by one of the posters above. It's a really good tool. I am not sure why you have only highlighted the second and the third quadrant. HEre's what I plotted. I think that logically this graph makes sense. Can you please explain the difference?
Thanks
Voodoo
Thanks for your reply. However, I tried plotting the graph using the online tool mentioned by one of the posters above. It's a really good tool. I am not sure why you have only highlighted the second and the third quadrant. HEre's what I plotted. I think that logically this graph makes sense. Can you please explain the difference?
Thanks
Voodoo
Dear Voodoo: I don't understand what you mean when you say "I am not sure why you have only highlighted the second and the third quadrant." The region |y| > |x| occupies part of each of the four quadrants. Are you asking why my graph is all of the same color? It really should be all one color, not two, because |y| > |x| is one contiguous region. It's true, some graphing programs, like the one eagleeye recommended, require the input of two different inequalities to graph this single region, but that doesn't change the fact that it's one contiguous thing. I'm not sure what you're asking.voodoo_child wrote:Mike,
Thanks for your reply. However, I tried plotting the graph using the online tool mentioned by one of the posters above. It's a really good tool. I am not sure why you have only highlighted the second and the third quadrant. Here's what I plotted. I think that logically this graph makes sense. Can you please explain the difference?
Thanks
Voodoo
Mike
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Mike@Magoosh wrote:Dear Voodoo: I don't understand what you mean when you say "I am not sure why you have only highlighted the second and the third quadrant." The region |y| > |x| occupies part of each of the four quadrants. Are you asking why my graph is all of the same color? It really should be all one color, not two, because |y| > |x| is one contiguous region. It's true, some graphing programs, like the one eagleeye recommended, require the input of two different inequalities to graph this single region, but that doesn't change the fact that it's one contiguous thing. I'm not sure what you're asking.voodoo_child wrote:Mike,
Thanks for your reply. However, I tried plotting the graph using the online tool mentioned by one of the posters above. It's a really good tool. I am not sure why you have only highlighted the second and the third quadrant. Here's what I plotted. I think that logically this graph makes sense. Can you please explain the difference?
Thanks
Voodoo
Mike
Thanks Mike. I just got confused by the diagram drawn by you because you have shown only the second and the third quadrant. I was wondering why the first and the fourth quadrant are not drawn or shown. Yes, I see that |y| > |x| must occupy all four quadrants. It's a typical "hour glass" type of graph.
Thanks
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You have to start from definition of the absolute value:
absX = X, when X>0
absX = -X, when X<0
apply that to X and Y in each quadrant and you get solution
absX = X, when X>0
absX = -X, when X<0
apply that to X and Y in each quadrant and you get solution
Mike,Mike@Magoosh wrote:Dear voodoo child,
I've attached an image of the graph of the region |y| > |x|. Basically, that's a mathematical way of saying --- closer to the y-axis than to the x-axis.
Also, you may find this blogpost on related topics interesting.
https://magoosh.com/gmat/2012/gmat-math- ... -line-y-x/
Let me know if you have any further questions.
Mike
Here's the question that was troubling me. I am not able to rule out Option A.
If ab#0 and |a| < |b| , which of the following must be negative?
a) (a-b)/b
b) (a-b)/(a+b)
c) a^b - b^a
d) a"-"(b/(a-b)) -- this is a fraction form. e.g. 1"-"2/3 = (3*1 + 2)/3. I don't know how to represent such form on the forum.
e) (b-a)/b
OA - B
After some thought, I was able to understand why C, D and E could be positive. However, I am not sure about a.
Also, I took greater than 5 minutes on this question. It will be really helpful if you could provide some shortcut tip to rule other answer choices.
Appreciate your help in advance.
Voodoo
Dear Voodoo_Childvoodoo_child wrote: If ab#0 and |a| < |b| , which of the following must be negative?
a) (a-b)/b
b) (a-b)/(a+b)
c) a^b - b^a
d) a"-"(b/(a-b)) -- this is a fraction form. e.g. 1"-"2/3 = (3*1 + 2)/3. I don't know how to represent such form on the forum.
e) (b-a)/b
I will admit, first of all, I have absolutely no idea what you intend in choice (d) --- normally, you can write any nested fractions with enough brackets and parentheses.
Given ab/= 0, we know neither one is zero.
Given |a| < |b|, we know the absolute value of b is greater than the absolute value of a, but they can be positive and negative in any combination. Basically, we have to consider (a, b) as (3, 5), (-3, 5), (3, -5), and (-3, -5). The first one is the obviously stupid one that everyone will try, so I would not rely too heavily on that. The two mixed sign examples, (-3, 5) and (3, -5), are the two that the unwashed masses are most likely to forget, so the answer choices are somewhat more likely to work for the same sign answers and not work for these mixed sign answers, so I would be likely to test those first. That's a general strategy. With the individual expressions, I would look for ways to adjust the signs to make the expression positive --- as soon as I can make it positive in some possible way, I can eliminate it as an answer.
Now, with (a) as you have it written ----
a) (a-b)/b --->
If b is negative, then the denominator is negative, and subtracting b in the numerator makes it positive; since |a| < |b|, whether we add or subtract a, the different in the numerator will still be positive. Positive over negative --- fraction is negative.
If b is positive, then the denominator is positive, and subtracting b in the numerator makes it negative; since |a| < |b|, whether we add or subtract a, the different in the numerator will still be negative. Negative over positive--- fraction is negative.
(a) as you have it written, is always negative.
Given the fact that I can't even make sense of what you intend for (d), I suspect you may not have all the choices written correctly here. What is the original source for this question? Would there be a way for me to see it in the original?
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Mike,
thanks for your reply. the source is GMATCLub math test. The question is m14 q33.
D) is {[a*(a-b)]+b}/ {a-b}
About A) - there is a typo - it should be a/b - b/a. (https://gmatclub.com/forum/m14-134213.html)
What is the shortcut to do such problems in < 2 minutes? My mind started spinning while doing this problem under test conditions. I missed it because Choice A) seemed good to me.
This was a killer problem.
thanks for your reply. the source is GMATCLub math test. The question is m14 q33.
D) is {[a*(a-b)]+b}/ {a-b}
About A) - there is a typo - it should be a/b - b/a. (https://gmatclub.com/forum/m14-134213.html)
What is the shortcut to do such problems in < 2 minutes? My mind started spinning while doing this problem under test conditions. I missed it because Choice A) seemed good to me.
This was a killer problem.
Dear Voodoo-Childvoodoo_child wrote:what is the shortcut to do such problems in < 2 minutes? My mind started spinning while doing this problem under test conditions. I missed it because Choice A) seemed good to me. This was a killer problem.
First of all, I'll say the way this post began, with the graph of |y|>|x|, is leagues away from the fastest way to answer this question.
If ab≠0 and |a|<|b|, which of the following must be negative?
A. (a/b)− (b/a)
B. (a−b)/(a+b)
C. a^b−b^a
D. a(b/(a−b))
E. (b−a)/b
The tricky thing about moving through this problem quickly is --- it involves number-sense, a certain facility with jumping around among the possibilities -- here, shuffling back and forth quickly between positive and negative values.
For example, let a = +/-3, and b = +/-5. We will swap back and forth between positive and negative possibilities as it serves our interests.
In choice (a), if both are positive, then (3/5 - 5/3), small minus big, is negative. Therefore, I could make either a or b negative, and then it would be (-3/5 + 5/3), minus small plus big, which is positive. Be aware of when changing the sign of one or the other or both changes the sign of the answer. Choice (a) is out.
With (b), I notice that if b is positive in the numerator, then -b is negative in the denominator, and if b is negative, then -b is positive. The bigger positive plus or minus the smaller one (a) is still positive. The bigger negative plus or minus the smaller one (a) is still negative. This seems like it's going to be positive/negative or negative/positive no matter how we swap around the signs.
(c) is easy to make positive ===> make b an positive even number and a a negative number ------ negative to an even will be a big positive number, and even to a negative will be a fraction less than one. For example, let's say a = -2 and b = 4
Then a^b−b^a = (-2)^4 - 4^(-2) = 16 - 1/16, which is positive. Choice (c) is out.
For choice (d):
a(b/(a−b)) = (ab)/(a-b)
If both a & b are positive, or if both are negative, then the numerator is positive and the denominator is negative. If I make a positive and b negative --- remember, the opposite sign possibility which many test takers will overlook --- then
[(3)*(-5)]/[(3) + (-5)} = (-15)/(-2)
That gives negative/negative, which is positive. Choice (d) is out.
For choice (e):
(b−a)/b
Here if both are positive, a = 3 and b = 5, the answer is positive, so we can eliminate this right away. (e) is out.
It took me more than two minutes to type all that out, but I could see each one of those steps very quickly. The more you develop your number sense ---- which is really what this question is testing --- the more fluent you will be in similar quick manipulations of possible choices.
Does this make sense?
Mike
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