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# If (x + y)/z > 0, is x < 0?

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Re: If (x + y)/z > 0, is x < 0? [#permalink]
x + y < 0
Case: x < 0 Must be true
Case: x = 0 thus y < 0 Not possible since x < y
Case: x > 0 thus y < 0 Not possible since x < y
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Re: If (x + y)/z > 0, is x < 0? [#permalink]
Took me a bit of time (shouldn't have).

Question stem is [(x+y)/z]>0. This implies that both num. and den. have the same sign. We need to answer whether x<0

St. 1 -- Not sufficient. x<y. x can be both +ve (2) or -ve here (-2)
St. 2 -- Not sufficient. z<0. This implies that numerator is -ve (since denominator is -ve). But x can be +ve (eg: 3 plus -7) or -ve (eg: -7 plus 3). The answer to the Q stem will be different in each case.

Merging the two statements, we are given that x < y and that x+y is -ve. Therefore, x will always be -ve. Since if y is +ve, x will farther left in the number line from 0 (greater -ve distance). And if y is -ve, x will always be -ve.

Therefore, C is sufficient.
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Re: If (x + y)/z > 0, is x < 0? [#permalink]
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Re: If (x + y)/z > 0, is x < 0? [#permalink]
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