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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. _________________

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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

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.

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.

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

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Here's an approach with algebra Numerator and denominator of this expression must have same signs to be positive

(x + y) z + + - - St1 x<y Not sufficient - no info about z

St2 z<0 , ok, this means x+y<0 Not sufficient (let's say x=2, y=-5 NO or x=-, y=2 YES)

St1+St2 SO we have overall this information: z<0, x+y<0, x<y In order x+y<0 we have 3 Scenarios: a) x and y both negative b) x +,y - and c)y +, x - We have here scenario c) x<y --> x is definitely negative _________________

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Hi guys! While doing this question I was wondering, if the answer we'd get combining the 2 options gave us x>0, would C still be the correct answer? I'm just trying to understand whether data sufficiency in the gmat is confirming the information in the question or also can be used as a negative answer, yet sufficient to give it.

Hope you understand what I mean and can help me out with this.

Hi guys! While doing this question I was wondering, if the answer we'd get combining the 2 options gave us x>0, would C still be the correct answer? I'm just trying to understand whether data sufficiency in the gmat is confirming the information in the question or also can be used as a negative answer, yet sufficient to give it.

Hope you understand what I mean and can help me out with this.

Many thanks and good luck with your preparation!

A definite NO answer to the question is still considered to be sufficient. _________________

Or another way to look at 1+2: St2-->when we know x+y<0 both x and y cant be positive. Possible scenarios: 1) Both are negative: -->Yes 2) x<0 & y>0 ;|x|>|y|-->No 3) x>0 & y<0; |x|<|y|--> Yes Since st1 tells us that x<y scenario3 is not possible

Ans: C _________________

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My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x+yz >0 , is x<0?

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

There are 3 variables (x,y,z) and 1 equation ((x+y)/z>0) in the original condition, and there are 2 more equations given from the 2 conditions, so there is high chance of (C) becoming the answer. Looking at the conditions together, as z<0, x+y<0, x<-y. But as x<y, if both sides are added together, x+x<y-y=0, 2x<0, x<0 this answers the question 'yes' and the conditions become sufficient, making the answer become (C).

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E. _________________

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