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

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If (x+y)/z>0, is x<0? [#permalink]  09 Jul 2012, 04:49
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The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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

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

Diagnostic Test
Question: 41
Page: 26
Difficulty: 650

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Re: If (x+y)/z>0, is x<0? [#permalink]  09 Jul 2012, 04:49
1
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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.

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Re: If (x+y)/z>0, is x<0? [#permalink]  09 Jul 2012, 05:38
Bunuel wrote:
If \frac{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.

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Re: If (x+y)/z>0, is x<0? [#permalink]  10 Jul 2012, 08:52
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
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Re: If (x+y)/z>0, is x<0? [#permalink]  10 Jul 2012, 14:46
For me is E the solution.

I wait for an explanation by Bunuel to understand if something went wrong in my reasoning
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Re: If (x+y)/z>0, is x<0? [#permalink]  10 Jul 2012, 16:28
2
KUDOS
Bunuel wrote:

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

x -y < 0
x+y <0
--------
2x <0
So,
x<0

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Re: If (x+y)/z>0, is x<0? [#permalink]  11 Jul 2012, 10:30
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

Please comment and provide an explanation
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Re: If (x+y)/z>0, is x<0? [#permalink]  11 Jul 2012, 11:28
Its C...

statement 1.. if x+y/z>0 that mean numerator or demonitor both have same sign...

x<y..that means x cud be +ve or negative and no any information abt z..so insufficient..

statement 2: .. z<0 .. no any information abt x and y..

state1 and stat 2: .. if x is less than y that means x will be negative and y will be positive or both will be negative.. so both are sufficient ..

ans c..
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Re: If (x+y)/z>0, is x<0? [#permalink]  13 Jul 2012, 03:49
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.

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Re: If (x+y)/z>0, is x<0?   [#permalink] 13 Jul 2012, 03:49
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