naaga wrote:
Is xy > x/y?
(1) xy > 0
(2) y < 0
Is xy > x/y?
Is \(xy > \frac{x}{y}\)? --> is \(\frac{xy^2-x}{y}>0\)? is \(\frac{x(y^2-1)}{y}>0\)? --> is \(\frac{x(y+1)(y-1)}{y}>0\)? This inequality holds true when:
A. \(x>0\) and \(y>1\) or \(-1<y<0\);
B. \(x<0\) and \(0<y<1\) or \(y<-1\).
(1) xy > 0 --> \(x\) and \(y\) have the same sign. Not sufficient.
(2) y < 0. Not sufficient.
(1)+(2) Both \(x\) and \(y\) are negative, so we are in scenario B, though we still need more precise range for \(y\) (if \(y<-1\) then the answer will be YES but if \(-1<y<0\) then the answer will be NO). Not sufficient.
Or: we can reduce our expression by x/y (which is positive since \(x\) and \(y\) have the same sign) and the question becomes: is \((y+1)(y-1)>0\)? and as \(y<0\) then the question reduces whether \(y<-1\). But we don't know that and thus even taken together statements are not sufficient.
Answer: E.
1. How can we multiply both sides by y when we do not know if y is +ve or -ve.
2. I did not understand \(xy > \frac{x}{y}\)? --> is \(\frac{xy^2-x}{y}>0\)? why is there a y in the denominator?