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If (x+y)/z>0, is x<0? [#permalink]
09 Jul 2012, 04:49
Question Stats:
51% (01:55) correct
48% (01:11) wrong based on 1 sessions
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Re: If (x+y)/z>0, is x<0? [#permalink]
09 Jul 2012, 04:49
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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}>0Using (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,
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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
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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 Adding the inequalities: x -y < 0 x+y <0 -------- 2x <0 So, x<0C is the answer.
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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
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Re: If (x+y)/z>0, is x<0?
[#permalink]
13 Jul 2012, 03:49
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