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If |2x| > |3y|, is x > y?

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If |2x| > |3y|, is x > y?  [#permalink]

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New post 07 Oct 2017, 11:57
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If |2x| > |3y|, is x > y?

(1) x > 0
(2) y > 0
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Re: If |2x| > |3y|, is x > y?  [#permalink]

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New post 08 Oct 2017, 14:05
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vitorpteixeira wrote:
If |2x| > |3y|, is x > y?

(1) x > 0
(2) y > 0


Inequalities + absolute values + Data Sufficiency makes this a good problem to use a number line.

Start by understanding the question itself. On a number line, |2x| > |3y| means that 2x is further from the origin than 3y.

'is x > y' means that we want to know whether x is to the right of y, or to the left of y. If we're testing cases, we'll be trying to find a case where x is to the left, and another case where x is to the right. That's the only way we can prove insufficiency.

Statement 1:

x > 0 means that x must be to the right of 0.

So, 2x must be to the right of 0.

Image

From the question stem, we know that 3y must be closer to the origin than 2x is.

However, 3y could be on either side of the origin, as long as it's closer.

It could be on the left, in which case the situation looks like this:

Image

Or it could be on the right, in which case the situation looks like this:

Image

Either way, y is definitely to the left of x. So the answer to the question is 'yes', and this statement is sufficient.

Statement 2:

This time, we know that y is to the right of the origin. 3y will also be to the right of the origin, like this:

Image

2x could be to the left of the origin, like this:

Image

It could also be to the right of the origin, like this. It does have to be closer to the origin than 3y, but let's take it to an extreme. Let's make 2x just a tiny bit closer than 3y.

Image

We've drawn one picture with x to the left of y, and another picture with x to the right. Since the result can go either way, statement 2 is insufficient.

The answer is A.
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Re: If |2x| > |3y|, is x > y?  [#permalink]

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New post 07 Oct 2017, 17:21
1
vitorpteixeira wrote:
If |2x| > |3y|, is x > y?

(1) x > 0
(2) y > 0



Hi...
|2x|>|3y|......|x|>3/2*|y|

So the numeric value( without considering the sign) of x is GREATER than that of y..
If x is NEGATIVE, y>x
If x is POSITIVE, x>y

Let's see the statements..
1) x>0
As can be seen above, x>y
Sufficient
2) y>0
Nothing about x..
Say x is 5 and y is 3, |2*5| > |3*3| and x>y
But if x=-5 and y=3,|2*-5|>|3*3| but y>x
Insufficient

A
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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If |2x| > |3y|, is x > y?  [#permalink]

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New post 07 Oct 2017, 20:11
1
vitorpteixeira wrote:
If |2x| > |3y|, is x > y?

(1) x > 0
(2) y > 0


Square both sides of the question stem to get \(4x^2-9y^2>0\)

or, \((2x-3y)(2x+3y)>0 => x>\frac{3}{2}y\) or \(x<\frac{-3}{2}y\)

So \(x>1.5y\) or \(x<-1.5y\)

Statement 1: this implies \(x\) is positive, so \(x<-1.5y\) is not possible hence only possible solution is \(x>1.5y\). Clearly \(x\) is more than \(y\) by at least \(1.5\) times, hence \(x>y\). Sufficient

Statement 2: in this case both the situations are possible \(x>1.5y\) so \(x>y\) and \(x<-1.5y\) so \(x<y\). Hence Insufficient

Option A

-------------------------------------
Another approach -

\(|x|>\frac{3}{2}|y| => |x|>1.5|y|\)

Statement 1: \(x>0\) so \(x\) is positive

now if \(y>0\) , then \(x>1.5y\) so \(x>y\) and if

\(y<0\), then \(x\) will always be greater than \(y\) because \(x\) is positive. Hence Sufficient

Statement 2: implies \(|x|>1.5y\) (because \(y\) is positive). but nothing mentioned about \(x\). if \(x=-2\) and \(y=1\), then \(x<y\) but \(|x|>1.5y\) will be true and if \(x=2\) & \(y =1\), then \(x>y\) and \(|x|>1.5y\) is also true. Hence No unique solution. Insufficient
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Re: If |2x| > |3y|, is x > y?  [#permalink]

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New post 08 Oct 2017, 19:25
ccooley Great Explanation indeed!!!
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Re: If |2x| > |3y|, is x > y? &nbs [#permalink] 08 Oct 2017, 19:25
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