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

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

Note that: \(\frac{x+y}{z}>0\) means that \(x+y\) and \(z\) have the same sign - either both positive or both negative.

(1) x < y. No info about z. Not sufficient.
(2) z < 0. No info about y. Not sufficient.

(1)+(2) \(z<0\) means \(x+y<0\). Sum \(x+y<0\) and \(x<y\) (from 1): \(x+y+x<y\) --> \(2x<0\) --> \(x<0\). Sufficient.

Answer: C.

Hope it helps.
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x < y

Since we don’t know anything about z, statement one alone is not sufficient to answer the question.

Statement Two Alone:

z < 0

Since we don’t know anything about x and y, statement two alone is not sufficient to answer the question.

Statements One and Two Together:

Since (x+y)/z is positive and z is negative, (x + y) must be negative also. Since x < y, the only way (x + y) is negative is if x is also negative.

Answer: C
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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

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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