If (x + y)/z > 0, is x < 0 ? (1) x < y (2) z < 0

Hi,

Difficulty: 650

\(\frac{x+y}{z}>0\)

Using (1),
x < y,
Possible values, x= -2, y = -1 & z = -1
or x= 1, y = 2 & z = 1
Both satisfy the given condition, thus Insufficient.

Using (2),
z < 0
or x + y < 0,
Possible values, x = -1, y = -1 or x = 1, y = -2. Insufficient.

Using both the statements,
z < 0 & x < y, we have x + y < 0
if y=2, x < -2 for x+y < 0 to hold true.
or if y=-2, x<-2. Sufficient.

Answer (C)

Regards,

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.

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.

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.

Target question: Is x NEGATIVE?

Given: (x+y)/z > 0
What does tell us?
Not much.
It tells us that (x+y)/z is POSITIVE, which means EITHER (x+y) and z are both positive OR (x+y) and z are both negative

Statement 1: x < y
This statement doesn't FEEL sufficient, so I'll TEST some values.
There are several values of x, y and z that satisfy statement 1 AND the given information. Here are two:
Case a: x = -2, y = -1, and z = -1. In this case x is NEGATIVE
Case b: x = 1, y = 2, and z = 1. In this case x is POSITIVE
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: z < 0
There are several values of x, y and z that satisfy statement 2 AND the given information. Here are two:
Case a: x = -2, y = -1, and z = -1. In this case x is NEGATIVE
Case b: x = 1, y = -2, and z = -1. In this case x is POSITIVE
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 2 tells us that z is NEGATIVE
If z is NEGATIVE and (x+y)/z > 0, then it must be the case that (x+y) is also NEGATIVE
In other words, x + y < 0

Statement 1 tells us that x < y
If we subtract y from both sides of the inequality, we get: x - y < 0

So, we now have the following two inequalities:
x + y < 0
x - y < 0

When we ADD the two inequalities, we get: 2x < 0
Divide both sides by 2 to get: x < 0
In other words, x is NEGATIVE
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

Sir in the 2nd statement i am having doubt. How we can rewrite Z<0 as X+Y <0. Please Explain

\(\frac{x+y}{z}>0\), means that x + y and z have the same sign, either both are positive or both are negative. (2) says that z is positive, so x + y must also be positive.

Hope it's clear.

Hi All,

We're told that (X+Y)/Z is greater than 0. We're asked if X is less than 0. This is a YES/NO question. We can answer it with a mix of Number Properties and/or TESTing VALUES.

To start, for (X+Y)/Z to be greater than 0, one of two options must occur:
(X+Y) > 0 and Z > 0
(X+Y) < 0 and Z < 0

1) X < Y
IF....
X = 1, Y = 2, Z = 1, then the answer to the question is NO.
X = -2, Y = -1, Z = -1, then the answer to the question is YES.
Fact 1 is INSUFFICIENT

2) Z < 0
IF...
X = -2, Y = -1, Z = -1, then the answer to the question is YES.
X = 1, Y = -2, Z = -1, then the answer to the question is NO.
Fact 2 is INSUFFICIENT

Combined, we know:
X < Y
Z < 0

Since Z is NEGATIVE, then we know that (X+Y) must ALSO be negative. Since X < Y, we know that either just X or both X and Y must be NEGATIVE. Either way, the answer to the question is ALWAYS YES.
Combined, SUFFICIENT.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

In statement 2, can't we say z<0, so of course x+y<0, meaning x<-y and so x is < 0? Please someone help me out, what math concept am i getting wrong?

Yes, from z < 0, in (2) it follows that x + y < 0. However, this does not necessarily mean that x < 0. Think of it this way: x + y < 0 means that the sum of two numbers is negative, so at least one of them must be negative. Can we say which one? No, x, y, or both can be negative. If we go the way you are doing, x < -y does not necessarily mean that x is negative. For instance, consider x = 1 and y = -2.

Hope it's clear.

Thank you so much Bunuel, that makes sense. Do you happen to know where I can find similar questions to practice of the same type?

Check DS inequalities questions here: https://gmatclub.com/forum/search.php?s ... tag_id=184 or use filters in DS forum: https://gmatclub.com/forum/data-sufficiency-ds-141/ or use our questions bank: https://gmatclub.com/forum/search.php?view=search_tags

Hope it helps.