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If (x+y)/z>0, is x<0? [#permalink]
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09 Jul 2012, 04:49
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Re: If (x+y)/z>0, is x<0? [#permalink]
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09 Jul 2012, 04:49
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Re: If (x+y)/z>0, is x<0? [#permalink]
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09 Jul 2012, 05:38
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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. Answer (C) Regards,



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Re: If (x+y)/z>0, is x<0? [#permalink]
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10 Jul 2012, 08:52
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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



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Re: If (x+y)/z>0, is x<0? [#permalink]
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10 Jul 2012, 14:46



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Re: If (x+y)/z>0, is x<0? [#permalink]
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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 > xy<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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Re: If (x+y)/z>0, is x<0? [#permalink]
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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 xy < 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]
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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]
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23 Jun 2015, 13:14
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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Re: If (x+y)/z>0, is x<0? [#permalink]
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10 Oct 2015, 04:20
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!



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Re: If (x+y)/z>0, is x<0? [#permalink]
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11 Oct 2015, 05:28



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Re: If (x+y)/z>0, is x<0? [#permalink]
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06 Dec 2015, 13:28
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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Re: If (x+y)/z>0, is x<0? [#permalink]
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09 Dec 2015, 07:14
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<yy=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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If (x+y)/z>0, is x<0? [#permalink]
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Updated on: 17 Apr 2018, 08:05
Bunuel wrote: If \(\frac{x+y}{z}>0\), is x<0?
(1) x < y (2) z < 0
Target question: Is x NEGATIVE? Given: (x+y)/z > 0What 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 NEGATIVECase b: x = 1, y = 2, and z = 1. In this case x is POSITIVESince we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT Statement 2: z < 0There 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 NEGATIVECase b: x = 1, y = 2, and z = 1. In this case x is POSITIVESince 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 < 0Statement 1 tells us that x < y If we subtract y from both sides of the inequality, we get: x  y < 0So, we now have the following two inequalities: x + y < 0x  y < 0When we ADD the two inequalities, we get: 2x < 0 Divide both sides by 2 to get: x < 0 In other words, x is NEGATIVESince we can answer the target question with certainty, the combined statements are SUFFICIENT Answer: C Cheers, Brent
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Re: If (x+y)/z>0, is x<0? [#permalink]
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04 Oct 2017, 22:44
Bunuel wrote: If \(\frac{x+y}{z}>0\), is x<0? (1) x < y (2) z < 0 Diagnostic Test Question: 41 Page: 26 Difficulty: 650 both statements are insufficient alone as the sign of x depends on those of y and z. Combine both and they are sufficient. i.e in (2), z<0, z is negative, then x+y must be negative for (x+y)/z to be positive and this is true from (1) x<y (x must be negative for x+y to be negative). Hence C



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Re: If (x+y)/z>0, is x<0? [#permalink]
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29 Oct 2017, 05:34
Sir in the 2nd statement i am having doubt. How we can rewrite Z<0 as X+Y <0. Please Explain



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Re: If (x+y)/z>0, is x<0? [#permalink]
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29 Oct 2017, 05:39



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Re: If (x+y)/z>0, is x<0? [#permalink]
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09 Dec 2017, 17:20
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
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