Bunuel wrote:
Is \(\frac{x}{y} < xy\) ?
(1) \(xy > 0\)
(2) \(y < -1\)
\(\frac{x}{y} < xy\)
\(xy-\frac{x}{y} >0\)
\(\frac{xy^2-x}{y}>0\)
\(\frac{x(y-1)(y+1)}{y}>0\)
This means both the numerator x(y-1)(y+1) and the denominator y have same sign.
We have to check for the following cases
A)
When the denominator or y<0 in \(\frac{x(y-1)(y+1)}{y}>0\), the numerator x(y-1)(y+1)<0, and we get two cases
(i) 0>y>-1, and (y-1)(y+1)<0,
Now x*(y-1)(y+1)<0, that is x*(negative)<0. This means x>0....(I)
(ii) y<-1, and (y-1)(y+1)>0,
Now x*(y-1)(y+1)<0, that is x*(positive)<0. This means x<0.......(II)
B)
When the denominator or y>0 in \(\frac{x(y-1)(y+1)}{y}>0\), the numerator x(y-1)(y+1)>0, and we get two cases
(i) 0<y<1, and (y-1)(y+1)<0,
Now x*(y-1)(y+1)>0, that is x*(negative)>0. This means x<0....(III)
(ii) y>1, and (y-1)(y+1)>0,
Now x*(y-1)(y+1)<0, that is x*(positive)>0. This means x>0.......(IV)
Our statements should clearly give us one of the 4 options as given above.
(1) \(xy > 0\)
Both x and y have same sign.
So case II and IV possible. But we cannot say whether y falls in the ranges y<-1 or y>1.
Insufficient
(2) \(y < -1\)
Nothing about x.
Combined
xy>0 and y<-1.
Exactly as per Case IV above.
Answer is yes.
C
Can you please share your approach. Also, (@cheatan2u) can please make me understand why and how to make the two cases as mentioned in the solution provided