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If \(\frac{x+y}{z}>0\), then dividend and divisor must be or positive or negative at the same time. Let's analyze the clues: (1) x < y We don't know whether z is positive or negative. We cannot know whether x is positive or negarive.

(2) z<0 Then x + y <0. However, with this info we cannot claim that x<0.

Combining (1) and (2): (1) x<y ----> x-y<0 Also, based on (2), we know that x+y<0

Adding the inequalities: x -y < 0 x+y <0 -------- 2x <0 So, \(x<0\)

C is the answer. _________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

Re: If (x+y)/z>0, is x<0? [#permalink]
09 Jul 2012, 03:49

1

This post received KUDOS

Expert's post

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SOLUTION

If \(\frac{x+y}{z}>0\), is x<0?

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

(1) x < y. No info about \(z\). Not sufficient. (2) z < 0. This statement implies that \(x+y\) must also be negative: \(x+y<0\). But we cannot say whether \(x<0\). Not sufficient.

(1)+(2) From (1) we have that \(x < y\) and from (2) we have that \(x+y<0\). Sum these two inequalities (remember we can add inequalities with the sign in the same direction): \(x+y+x<y\) --> \(2x<0\) --> \(x<0\). Sufficient.

Re: If (x+y)/z>0, is x<0? [#permalink]
10 Jul 2012, 07:52

1

This post received KUDOS

If (x+y)/z is greater than 0, then it is basically saying that either both z and (x+y) are negative or they are both positive. Question asks if x is negative.

Statement 1 says that x is less than y but does not say anything about z. 3 unknown variables, 2 equations, insufficient data.

Statement 2 says that z is negative. The result is that (x+y) is negative then, but there are a lot of possibilities for x+y to be negative.

If we were to combine the two, then we know that x+y is negative with x being less than y. No matter what combination of numbers we put together, x can never be positive. If y is negative and x < y, x is negative. If y is positive then x must be negative to the point that x + y is still negative.

C.

EDIT: people might not like my explanation since I was taught to simplify or reword questions. Might not be for everyone

Re: If (x+y)/z>0, is x<0? [#permalink]
13 Jul 2012, 02:49

1

This post received KUDOS

Expert's post

SOLUTION

If \(\frac{x+y}{z}>0\), is x<0?

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

(1) x < y. No info about \(z\). Not sufficient. (2) z < 0. This statement implies that \(x+y\) must also be negative: \(x+y<0\). But we cannot say whether \(x<0\). Not sufficient.

(1)+(2) From (1) we have that \(x < y\) and from (2) we have that \(x+y<0\). Sum these two inequalities (remember we can add inequalities with the sign in the same direction): \(x+y+x<y\) --> \(2x<0\) --> \(x<0\). Sufficient.

Re: If (x+y)/z>0, is x<0? [#permalink]
11 Jul 2012, 09:30

if x+y/z > 0, is x<0?

(1) x < y (2) z < 0

Statement 1: if x<y, then necessarily x-y < 0---(1), substitute, x = +ve then always y will have to be a +ve value exceeding x to satisfy the equ. substitute x = -ve then either y will be a -ve value OR y will be a +ve value so that |x| > |y| substitute x = 0, y will always be a +ve value Statement 2: if z < 0, then necessarily x+y < 0--(2) substitute x = + ve then y will be -ve so that |x| < |y| substitute x = -ve then y can be +ve if |x| > |y| OR y wll be -ve substitute x = 0, y will always be a -ve value

Individually nothing can be concluded but evaluating (1) and (2) together we can see that when x < 0 only then can the given equ. be satisfied, Hence C.

I have a question can we go ahead and solve two inequalities like x - y < 0 x + y < 0 ------------ 2x < 0 x < 0

I tried to substitute some values in equations but it didn't fit.. like if we have 2x + 4y < 0 and 2x - 4y < 0, then we cannot conclusively say anything about the value of x or y

Re: If (x+y)/z>0, is x<0? [#permalink]
21 Nov 2013, 10:19

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Re: If (x+y)/z>0, is x<0? [#permalink]
04 Jan 2015, 09:59

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