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vitorpteixeira
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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ccooley Great Explanation indeed!!!
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What if y=0 and x is negative say we we take the value of x as -1 then x<y and if we take the value of x as 1 then x>y so should 'nt the answer be C
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ccooley
in the last diagram the 2x should be to the right of 3y. Very well explained.
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