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xy will be positive if x and y have the same sign. Statement 1 gives no information about y, and from Statement 2, it is possible that x and y are both positive or both negative, or that x is negative and y is positive.

Using both statements, we know x is positive, so 1/x is positive. From Statement 2, if y > 1/x, then y is greater than something positive, so y is also positive, and xy is positive, so the answer is C.
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xy will be positive if x and y have the same sign, when both are negative or both are positive.

(1) x > 0. No info about \(y\). Not sufficient.

(2) 1/x < y. If \(x=-1\) and \(y=1\) then the answer is NO but if \(x=2\) and \(y=1\) then the answer is YES. Not sufficient.

(1)+(2) Since from (1) \(x>0\), then \(\frac{1}{x}>0\), hence from (2) we have that \(0<\frac{1}{x}<y\) --> \(y>0\). Thus both \(x\) and \(y\) are positive. Sufficient.

Sorry but I don't understand why the answer can not be B.

B) 1/x < y This can be transformed in xy > 1. So if xy>1 it should be also true xy > 0.

Please could you tell me where I am wrong?

Thank you

Never multiply an inequality by variable (or expression with variable) unless you know the sign of variable (or expression with variable). Because if \(x>0\) you should write \(xy>1\) BUT if \(x<0\), you should write \(xy<1\) (flip the sign when multiplying by negative expression).

Re: If x is not equal to zero is xy > 0 ? [#permalink]

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19 Jan 2016, 00:51

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Re: If x is not equal to zero is xy > 0 ? [#permalink]

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20 Jan 2016, 09:37

hi bunuel,

Cant we write the option B as y>1/x= (x-y)/x>0 and if you put both the statements together and try few values as mentioned below it is satisfying for neg/pos value of Y.

X=10 Y=8 and X=10 and Y=-8 in both cases above equation holds.

Cant we write the option B as y>1/x= (x-y)/x>0 and if you put both the statements together and try few values as mentioned below it is satisfying for neg/pos value of Y.

X=10 Y=8 and X=10 and Y=-8in both cases above equation holds.

Kindly explain!! Thanks

y is NOT greater than 1/x as per highlighted set of values so they can't be taken as they violate the information os Second statement.

Cant we write the option B as y>1/x= (x-y)/x>0and if you put both the statements together and try few values as mentioned below it is satisfying for neg/pos value of Y.

X=10 Y=8 and X=10 and Y=-8 in both cases above equation holds.

Kindly explain!! Thanks

You can not perform the calculations you have mentioned above.

y>1/x \(\neq\) (x-y)/x>0

You can although, rewrite statement 2 as y-1/x > 0 --->(xy-1)/x>0 and as x>0 ---> for this fraction to be >0 --> xy-1>0 ---> xy>1 . Again, 2 cases possible, x>0 and y>0 and x<0 and y<0. Thus this statement is NOT sufficient.
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