if x does not equal -y

[guest612]

If x does not equal -y, is (x-y)/(x+y) > 1?

1. x>0
2. y>0

Answer: E.

I got B. I multiplied both sides by the denominator and then subtracted x from both sides leaving me with:
-y>y? So my rephrase was, is y negative?

Can you please tell me how this is wrong? The official answer is E.

[Sudhan]

If x does not equal -y, is (x-y)/(x+y) > 1?

1. x>0
2. y>0

Rephrase:
x-y > x+y (x not equal to -y)

ADBCE Grid:-p
1) X>0; Does not say about y. Hence Insuff
2) Y>; Does not say about x. Hence Insuff
1 and 2: x>0, y>0 x= 1, y=1; 0> 2. No
x= 4, y=1; 3 > 5 No
x=1 y=-4; 1-(-4) > 1-4
= 5 >-3 yes.

Insuff.

Hence E.

Thanks

[RonPurewal]

Rephrase:
x-y > x+y (x not equal to -y)

...

2) Y>; Does not say about x. Hence Insuff

nope.

if that were a proper rephrase, then you could subtract x from both sides, giving the same rephrase found in the original post. in this case, (2) would be sufficient, because y>0 necessarily means -y is not > y.

but ... that's not a proper rephrase. see below.

[RonPurewal]

I got B. I multiplied both sides by the denominator and then subtracted x from both sides leaving me with:
-y>y? So my rephrase was, is y negative?

yeah, you can't do that. this is a common rookie mistake, but it's a BIG mistake. to wit:
do not ever multiply or divide an inequality by a variable or expression, UNLESS you KNOW THE SIGN of that variable or expression.
the reason is something you undoubtedly know, but just haven't connected to this situation: if you multiply or divide by a positive quantity, then the sign doesn't flip; if you multiply or divide by a negative quantity, then the sign does flip.

the issue here, then, is that you don't know the sign of (x + y) in this problem.

you can verify that (1) and (2) are each insufficient alone with the following plug-ins (x, y):
(1, 1) --> no
(2, -1) --> yes
(-2, 1) --> yes
the first and second statements show that (1) is insufficient; the first and third show that (2) is insufficient.

but here's my big problem: (1) and (2) together ARE sufficient, so the answer should be c.
justification:
if x + y is positive, then you can definitively multiply by it, giving 'is x - y > x + y ?' then, as in the original post, you can subtract x, giving 'is -y > y ?', which is equivalent to 'is y negative?'

another way of arriving at the same result:
if x and y are both positive, then x + y is a positive denominator. if a fraction with a positive denominator is to be greater than 1, there's only one way to accomplish that: the numerator must also be positive, and it must be BIGGER than the denominator. but, since y is positive, that's impossible, because we have left numerator = x - y < x + y = right denominator.

so answer = c, unless something in the problem has been copied incorrectly.

[VV]

I agree with Ron the answer would be C; Unless the following was asked

The answer would be E, if the qustion was Is (x-y)/x+y >0 and xis not equal to -y
I agree with the first two statements why Stat 1 and stat 2 alone would be insufficient.

But together 1 and 2 are still insufficient because:
If x and y >0 and x>y
Then (x-y)/(x+y)
If x=4 and y=1
3/5 >0

If x<y,
x=1 and y=4
Then (x-y)/(x+y) =-3/5 <0 So insufficient

[StaceyKoprince]

I find this one in my set and statement 2 is supposed to say "y<0" NOT "y>0" as typed above! Try it again - this should clear things up. :)

[Aragorn]

A humble request to post the official answer as a reply to the question, or at the far bottom of the question. It is of little use to try a problem knowing the correct answer right away.
Thanks

[rfernandez]

That's a good point, Aragorn. I'll raise this issue to the other instructors on the forum. Thanks.

Rey

[Guest]

I got B. I multiplied both sides by the denominator and then subtracted x from both sides leaving me with:
-y>y? So my rephrase was, is y negative?

yeah, you can't do that. this is a common rookie mistake, but it's a BIG mistake. to wit:
do not ever multiply or divide an inequality by a variable or expression, UNLESS you KNOW THE SIGN of that variable or expression.
the reason is something you undoubtedly know, but just haven't connected to this situation: if you multiply or divide by a positive quantity, then the sign doesn't flip; if you multiply or divide by a negative quantity, then the sign does flip.

the issue here, then, is that you don't know the sign of (x + y) in this problem.

you can verify that (1) and (2) are each insufficient alone with the following plug-ins (x, y):
(1, 1) --> no
(2, -1) --> yes
(-2, 1) --> yes
the first and second statements show that (1) is insufficient; the first and third show that (2) is insufficient.

but here's my big problem: (1) and (2) together ARE sufficient, so the answer should be c.
justification:
if x + y is positive, then you can definitively multiply by it, giving 'is x - y > x + y ?' then, as in the original post, you can subtract x, giving 'is -y > y ?', which is equivalent to 'is y negative?'

another way of arriving at the same result:
if x and y are both positive, then x + y is a positive denominator. if a fraction with a positive denominator is to be greater than 1, there's only one way to accomplish that: the numerator must also be positive, and it must be BIGGER than the denominator. but, since y is positive, that's impossible, because we have left numerator = x - y < x + y = right denominator.

so answer = c, unless something in the problem has been copied incorrectly.

one small doubt, the question doesn't say x>y, therefore both can be positive (according to 1 & 2), but if y>x, wouldn't my answer be E?

[Guest]

I guess everyone agrees that we can rule out answers A and B.

Combining statements 1 and 2, where x and y are both positive we can evaluate the question with x=y=1 and you will arrive at 0 > 1 which is FALSE. Therefore with statements 1 and 2 TRUE, there is still a possibility that the answer to the question can be TRUE or FALSE which means statements 1 and 2 are insufficient together.

My answer is E.

[RonPurewal]

one small doubt, the question doesn't say x>y, therefore both can be positive (according to 1 & 2), but if y>x, wouldn't my answer be E?

nope. if y > x, the number is negative, and so it's still less than 1.
the way the problem is currently written on this thread, it's always less than 1 if you take both statements together.

it turns out there actually has been a transcription error - statement 2 is supposed to say y < 0, not y > 0. in that case, the answer is indeed (e).

[RonPurewal]

I guess everyone agrees that we can rule out answers A and B.

Combining statements 1 and 2, where x and y are both positive we can evaluate the question with x=y=1 and you will arrive at 0 > 1 which is FALSE. Therefore with statements 1 and 2 TRUE, there is still a possibility that the answer to the question can be TRUE or FALSE which means statements 1 and 2 are insufficient together.

My answer is E.

nope.

to give answer choice (e), the QUESTION PROMPT must still be able to generate both YES and NO answers (so that the overall answer is "MAYBE"), even if both statements are taken to be true.

as you have demonstrated, it's possible for the answer to the question to be "NO" if both statements are true.
however, with the problem as written in this thread (with "greater than" in both statements), it's actually impossible for the answer to be "YES" if both statements are true. try it - you won't be able to do it (read my longer post above if you aren't convinced).

you can't count the truth of statements (1) and (2) as a Yes answer to the question!!!!

[Guest]

If X not equal to -Y; is X-Y/X+Y>1
a. X>0
b. Y<0
Ans : E

this is the right question? i encountered it in GMATPrep

[RonPurewal]

If X not equal to -Y; is X-Y/X+Y>1
a. X>0
b. Y<0
Ans : E

this is the right question? i encountered it in GMATPrep

yes.
that's the right original question.

i use this problem in class for "overtime" (i.e., enrichment problems after normal class time ends) to teach an important point: IF YOU'RE GOING TO TEST NUMBERS FOR A DATA SUFF PROBLEM, ALWAYS TEST NUMBERS COMMON TO BOTH CHOICES FIRST (if you can come up with such numbers without a lot of trouble).

so, if you're going to test, say, statement (1) on this problem, you should START by picking numbers that are COMMON to statements (1) and (2).
such as
x = 3, y = -1
and
x = 1, y = -3
(notice the difference: in the first pair, x has the greater magnitude; in the second pair; y has the greater magnitude)
for the first pair, (x - y)/(x + y) = 4/2 = 2. that's a "yes" to the question.
for the second pair, (x - y)/(x + y) = 4/-2 = -2. that's a "no".
insufficient.
BUT, since these values are common to the two statements, you're DONE! insufficient overall! e! done, in like twenty seconds!

this is why you should plug in common values first:
* if they turn out to give different answers, then you're DONE and the answer is (e)
* even if they don't, you're already covered if you have to plug in for (c) vs. (e); the common values you've just plugged in will form the basis for that decision.

[vinnu.m558]

Hi,

Even after going through all comments on this particular question I still feel that B is the right choice.

The actual ques after rephrasing is :
IS x-y>x+y ??

a. x>0
b. y>0

Let's try putting some values in x-y>x+y?:
(x,y) = (1,1)
Is 0>2? No

(x,y) = (-2,1)
Is -3>-1? No

(x,y) = (0,1)
Is -1>1? No

When we are aware that the value of Y is +ve, obviously x-y i.e, subtracting of a +ve number from x will always be less than x+y i.e, adding of a +ve number to x, irrespective of the value of x.

Please correct me if I' am on a wrong track.