If x is not -y, (x-y)/(x+y) 1? 1) x 0 2) y

Question

If x is not -y, (x-y)/(x+y) > 1?
1) x > 0
2) y < 0

Please include your explanations. Thank you!

durgesh79 · Answer

question stem (x-y)/(x+y) > 1 if (x+y) > 0 question can be written as x-y > x+y --> 0 > 2y --> y<0 --> is y negative ----- (1) if (x+y) < 0 question can be written as x-y < x+y --> 2y > 0 --> y>0 --> is y positive ------ (2) First we need to know sign of (x+y) to ask the question. statement 1 : tells nothing about y, Not suff statement 2 : y<0 ; we need to know sign of (x+y) to ask the question. Not Suff. Combine : x>0 and y<0. (x+y) can be positive or negative.. (x+y)>0 answer is YES, if (x+y)<0 answer is NO Option E.

goalsnr · Answer

This DS is testing the concept a/b > 0 if 1. a> 0, b>0 2. a<0, b<0 I get C: Using concept 1. a> 0, b>0 (x-y)> 0 (x+y)> 0 => x> y => x>-y =>if x> y it also implies x> -y x>y Using concept 2. a<0, b<0 (x-y)< 0 (x+y)< 0 => x< y => x<-y =>if x<- y it also implies x 1? we need to know if x>y or x<-y stat1 : 1) x > 0 insuff stat2 : 2) y < 0 insuff From 1 and 2 => x>y SUff C

leonidas · Answer

If x is not -y, (x-y)/(x+y) > 1?
1) x > 0
2) y < 0

Why can't it be (B)?

I agree its not A.

Using 2), Y is negative.
(x-y)/(x+y) > 1
=>(x-y) > (x+y)
Since y is negative, let y=-5,
x can be either +ve or -ve,
Let (x,y)= (3,-5) => 8> -2 
 = (-3,-5) => 2> -8 
 = (8,-5) => 13> 3 
 = (-8,-5) => -3> -13 
 = (0,-5) => 5 > -5

In all the cases, it shows that (x-y)/(x+y) > 1 if y is negative. Am I missing something? What is the OA?

walker · Answer

E

\frac{(x-y)}{x+y}>1  -->  \frac{(x+y-2y)}{x+y}>1  -->  1-\frac{2y}{x+y}>1  -->  \frac{2y}{x+y}<0

For both statements, x+y can be both positive and negative. Therefore, (x+y) will change a sign of the expression and an answer is E.

walker · Answer

It is possible only if x+y is positive. Otherwise, we should change the sign of the inequality.

goalsnr · Answer

Can you please point what's the mistake in my approach?

walker · Answer

a/b>1. You need here different approach.

leonidas · Answer

Thanks Walker. 
Yes, I did not consider (x+y) as a whole.

goalsnr · Answer

Can you please explain the concept tested here?

walker · Answer

we can transform a/b>1 into (a-b)/b>0 and use your reasoning for (a-b) and b.

durgesh79 · Answer

similar question If a > 0, b > 0, and (b + y) does not = 0, is (a + y)/(b + y) < a/b ? 1) a < b 2) y < 0

durgesh79 · Answer

No its E again.... :wink:

try with below numbers.... a=10, b=20 and y=-1 a=10, b=20 and y=-100

alpha_plus_gamma · Answer

ok

we have 2 cases to consider: 1) b + y > 0 => b > -y 2) b + y < 0 => b < -y with this I was inclined to say E because 1) & 2) alone or together don't tell us anything about relationship between b & y But then thought that this may not be correct and started solving for the above two cases,and screwed up somewhere! What will be the right method to solve the problem without plugging in numbers?

durgesh79 · Answer

i also got confused while solving through algebra....

its almost clear that the fight is between C and E.... i actually tried to pick numbers to prove that its E, i fixed a and b so that a/b is same.. and picked extreme -ve values for y, i got different answers and hence its E.

this approach (plugging numbers) is not recommended in reverse (in case of C), becuase if two sets of values are satisfying the question stem doesnt mean that all values will satisfy. But if two sets of values are giving different answers... we can be sure that its E.....

If x is not -y, (x-y)/(x+y) > 1? 1) x > 0 2) y < 0

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