If only one of the three integers x, y and z is negative, is x

Question

If only one of the three integers x, y and z is negative, is x>yz?
(1)  \frac{x}{y}>z 
(2)  xy>0

New question

PKN · Answer

St1:-
 x/y>z 
This holds true when x and y are non-negative & z is negative.(y can't be zero).
Since magnitude of x,y and z are not known,We have to consider z as negative in order to validate the inequality.
So yz is negative.
Hence x>yz is true.
Sufficient.

St2:-
xy > 0
So both x and y are positive.
Question tells us that only one of x,y, and z is a negative integer.since st2 is not dependent on polarity and magnitude of z,hence,we can consider z as an integer(+/-/0).
So x>yz may or mayn't be true.
So insufficient.

And. A

P.S.-  chetan2u  Sir, Request your kind perusal.

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chetan2u · Answer

PKN , you have correctly found the info but went wrong while solving..

Only one of x, y and z is negative

1)  \frac{x}{y}>z 
If any of x or y is negative, then z is positive.
In that case x/y will be negative and z will be positive, so x/y>z will not hold true
So z is negative and x and y are positive so x>yz as yz <0 and x>0
Suff

(2)  xy>0 
This says either both x and y are positive or both are negative..
But there is only one negative so both x and y are positive and z is negative.
And so x>yz
Sufficient

D

PKN · Answer

New question

PKN , you have correctly found the info but went wrong while solving..

Only one of x, y and z is negative

1)  \frac{x}{y}>z 
If any of x or y is negative, then z is positive.
In that case x/y will be negative and z will be positive, so x/y>z will not hold true
So z is negative and x and y are positive so x>yz as yz <0 and x>0
Suff

(2)  xy>0 
This says either both x and y are positive or both are negative..
But there is only one negative so both x and y are positive and z is negative.
And so x>yz
Sufficient

D

Thank you Sir. I would've considered "if clause" info as a mandatory one since question stem is asked in that condition only.

If only one of the three integers x, y and z is negative, is x

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