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If x!= -y, is (x-y)/(x+y) > 1? 1. x>0 2. y<0 [Edit: [#permalink]
19 Feb 2008, 23:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
If x!= -y, is (x-y)/(x+y) > 1?
1. x>0 2. y<0 [Edit: Made a mistake posting it previously - its y<0 not y>0)
Answer is E in the Official Guide.
I tried solving it by reducing (x-y)/(x+y) > 1 to 2y<0
In this case, I only have to check if y<0. But this is not the way to solve it. Could someone tell me why. The official guide answer uses substitution of numbers.
Last edited by jjaspirant on 20 Feb 2008, 00:11, edited 1 time in total.
Re: Is (x-y)/(x+y) > 1? [#permalink]
20 Feb 2008, 01:09
jjaspirant wrote:
Yes, my mistake
2. y<0
Still can't I simplify (x-y)/(x+y)>1 down to 2y<0 by cross multiplying the equation? Why is it that we have to move 1 to the other side and simplify?
Incorrect: (x-y)>(x+y)
Correct: (x-y)/(x+y) - 1 > 0
Yes, the reason is that you must sure that x+y is not zero and a negative when you divide or multiple two sides of the inequataion by a variable
(x-y)/(x+y) - 1 = - 2y/(x+y) >0 1. if y<0 and x>-y, ok -2y/(x+y) >0. For example: y<0 so y = -2, x >-y so x> 2 or x = 3. Therefore, -2*(-2)/(3-2) = 4>0
2. if y<0 and x<-y, -2y/(x+y) <0. For example: y=-2, x<-y so x<2 so x = 1. Therefore, -2*(-2)/(1-2) = -4<0
Re: Is (x-y)/(x+y) > 1? [#permalink]
20 Feb 2008, 11:00
Sorry, I'm not totally understanding ppl's responses. Are you saying that the condition x!=-y proves that x-y = -2y?, because I'm not sure that that follows...
Here's how I did it:
(1) if x>0 then y<0. We know x cannot be 1 or 2, since 1! =1 and 2!=2 and this would cause the denominator (x+y) to be zero. In the expression (x-y)/(x+y), the numerator should always be positive (pos - neg), and the denominator will always be negative, i.e. if x=3, y=-(3*2*1)=-6, and 3+(-6)<0. So this establishes that the expression is always negative and therefore not >1. suff.
(2) if y<0 then x>0. Again, x cannot be 1 or 2 without the denominator blowing up. Same situation as before, numerator is always positive and denominator is always negative. suff.
Re: Is (x-y)/(x+y) > 1? [#permalink]
20 Feb 2008, 12:29
prasannar wrote:
It is a clear E
Let us take values
x y -> first case x =2,y=-1 3/1 is > 1 true
second case x=1,y=-3 now 4/-2 =-2
so the answer is a clear E
and for problems where you can take specific values, I would always go for them
regards, Pras
Also X! = -Y, which means the value of X and Y have to such that X! will be equal to Y. As per St1 X> 0, hence X! can't be -ve. that means the value of Y has to be -ve, so as to compensate for the other minus.
Re: Is (x-y)/(x+y) > 1? [#permalink]
20 Feb 2008, 12:36
jjaspirant wrote:
If x!= -y, is (x-y)/(x+y) > 1?
1. x>0 2. y<0 [Edit: Made a mistake posting it previously - its y<0 not y>0)
Answer is E in the Official Guide.
I tried solving it by reducing (x-y)/(x+y) > 1 to 2y<0
In this case, I only have to check if y<0. But this is not the way to solve it. Could someone tell me why. The official guide answer uses substitution of numbers.
The answer has to be A.
As per St1: x > 0 and also x! = -y, since x > 0 ==> x! > 0, Hence y has to be -ve.
Now putting this info into the equation (x - (y) ==> x + y will be > 0. and (x + y) (given y is -ve , from our calculation) will of course be less than x - y, This will lead to (x - y) / (x + y) > 1
Re: Is (x-y)/(x+y) > 1? [#permalink]
20 Feb 2008, 14:50
If x+y > 0, (x-y)/(x+y) > 1 iff x -y > x + y i.e. y < 0. (2) gives us this information, but even with (1), we do not know that x + y > 0. Together not sufficient
Re: Is (x-y)/(x+y) > 1? [#permalink]
21 Feb 2008, 01:26
Given: If both x and y are greater than 0, then we can simplify the solution. But, with the given question stem, let’s evaluate: 1. x > 0 does not provide any clue that y < 0 not sufficient 2. y < 0 does not provide any clue for x – not sufficient Both 1 and 2: x = 1 Y = -5 (x-y)/(x+y) < 1
On the other hand: X = 5 Y = -1 (x-y)/(x+y) > 1
Two different values – not sufficient Answer: E
gmatclubot
Re: Is (x-y)/(x+y) > 1?
[#permalink]
21 Feb 2008, 01:26