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GMATPrep2: If x != -y, is (x-y)/(x+y) > 1

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GMATPrep2: If x != -y, is (x-y)/(x+y) > 1

by jsl » Sat Oct 04, 2008 6:40 am
If x != -y, is (x-y)/(x+y) > 1

1. x>0
2. y<0

I found this one confusing! Can anyone give an explanation for this with working out?

I tried searching the forum for this but couldn't find it - possibly since there are a tonne of posts with x's and y's in them!
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by namitrajiv » Sat Oct 04, 2008 7:36 am
I feel the ans is A

x-y/x+y>1
i.e x-y-x-y/x+y>0
i.e -2y/x+y>0
or y/x+y <0

nw we nw factorial of of a number is >0 therefore , y hat to be negative
also x therfore will be +ve

nw for y/x+y will be <0 if y<0 and x+y >0
or y>0 and x+y<0

we nw only the first condition will hold i.e y<0 and x+y>0


the first statement says x>0 , that
actually x would in all but two cases 1! and 2! will be less than y in magnitude , in which case x+y <0 but x can be equal to y and then
in either case y/x+y>0

answer will be infinity which is greater than 0



second statement says y <0 , that we already knw ,
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by namitrajiv » Sat Oct 04, 2008 10:47 am
whats the OA
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by jsl » Sat Oct 04, 2008 11:45 am
nope - that's not the OA.... can someone else try?
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by namitrajiv » Sat Oct 04, 2008 12:03 pm
well if not A , it must be E , thats because the case of infinity being greater >0 was ambigious to me ,
anyway let some other try .....even I am confused ......
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by gautamberry » Sat Oct 04, 2008 1:06 pm
IMO the answer is E

Statement 1

we get that X>0 but we do not know the value of Y

If X=3 and Y=-2 then (X-Y)/(X+Y)> 1

But if X=3 and Y=-5 then (X-Y)/(X+Y)<1


Statement 2

Y<0

Same reasoning as for statement 1


When we combine both we don't know the relation between X and Y.........hence we cannot get an answer...........
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by namitrajiv » Sat Oct 04, 2008 1:20 pm
hi gautamberry,
the statement says x! = -y hence the examples that you have taken are incorrect
when x= 3 y = -6 because x! = 6
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by arthi » Sun Oct 05, 2008 4:32 pm
is the answer B?
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by akg_gmat » Mon Oct 06, 2008 10:38 am
The question can be reduced to the form

y/(x+y)<0

IMO: E
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by cubicle_bound_misfit » Mon Oct 06, 2008 10:46 am
question asks
is (x-y)/(x+y) >1

unfortunately we can not simplify this as we do not know sign of (x+y)

stmt 1 and 2 alone is not sufficient for obvious reasons that each omits either x or y

together they are agin insuff

take x =1 y = -2

then (x-y)/(x+y) == 3/(-1) < 1

but if x =10 and y =-2

then (x-y)/(x+y) = 12/8 >1
hence insufficient.

answer is E.

what is OA?
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by cubicle_bound_misfit » Mon Oct 06, 2008 10:49 am
namitrajiv


i think != is not equal to.
blame it on 'C' syntax. :-)
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by eshahid » Mon Oct 06, 2008 4:36 pm
cubicle_bound_misfit wrote:
take x =1 y = -2

then (x-y)/(x+y) == 3/(-1) < 1

but if x =10 and y =-2

then (x-y)/(x+y) = 12/8 >1
hence insufficient.

answer is E.

what is OA?
it can't be anything other than 'E'.
if x>0 and y<0, x-y will always be positive.
now x+y can be positive or negative. When x+y is positive the expression is always greater than 1 and when negative it is obv. less than 1.
Shahid E

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by cramya » Mon Oct 06, 2008 9:29 pm
I think its E too

x not equal to y can imply x and y can be negative or postive(both or either one) . This changes the value we get for x-y / x+y so we cannot definitely say that its > 1
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by msvmuthu » Thu Oct 16, 2008 3:07 am
cubicle_bound_misfit :

I always have this problem : when to simplify the given equation and when not to?

for example, in the above question I simplified the equation to:
y<0

since the first choice is x>0,
and since we have derived y<0;
(x-y) / (x+y) => will be like (x+y) /(x-y)
there fore surely, this will be greater than 1.

So my choice was A.

But you say that: this equation cannot be simplified, since we dont know the sign of (x+y) - denominator.

Can you please throw more light on this?
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by Ian Stewart » Thu Oct 16, 2008 4:01 am
msvmuthu- I replied to your question here (another thread about the same question)

https://www.beatthegmat.com/inequalities-t20825.html

(edited to fix the link)
Last edited by Ian Stewart on Mon May 25, 2009 8:07 pm, edited 1 time in total.
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