If x not equal to -y, is (x-y)/(x+y) > 1?
(1) x>0
(2) y<0
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daretodream
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thephoenix
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simplifying (x-y)/(x+y)-1>0daretodream wrote:If x not equal to -y, is (x-y)/(x+y) > 1?
(1) x>0
(2) y<0
-2y/(x+y)>0
if we know relation b/n x and y and whether its +ve or -ve we can find it
s1) we know only x>0 for x>y and for x<y we get diff ans
s2) same
combine suff
x>0>y
suff to tell whether the exp is> or not than 0
thephoenix wrote:simplifying (x-y)/(x+y)-1>0daretodream wrote:If x not equal to -y, is (x-y)/(x+y) > 1?
(1) x>0
(2) y<0
-2y/(x+y)>0
if we know relation b/n x and y and whether its +ve or -ve we can find it
s1) we know only x>0 for x>y and for x<y we get diff ans
s2) same
combine suff
x>0>y
suff to tell whether the exp is> or not than 0
Fellas, this is how I see this
simplifying (x-y)/(x+y)-1>0
(x-y)/(x+y)>1
x-y>x+y
Or 2y<0
Or y<0
So question boils down to is Y<0
A is Insufficient, B is sufficient.
Am I doing anything wrong here?
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harsh.champ
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Well you have transposedtata wrote:thephoenix wrote:simplifying (x-y)/(x+y)-1>0daretodream wrote:If x not equal to -y, is (x-y)/(x+y) > 1?
(1) x>0
(2) y<0
-2y/(x+y)>0
if we know relation b/n x and y and whether its +ve or -ve we can find it
s1) we know only x>0 for x>y and for x<y we get diff ans
s2) same
combine suff
x>0>y
suff to tell whether the exp is> or not than 0
Fellas, this is how I see this
simplifying (x-y)/(x+y)-1>0
(x-y)/(x+y)>1
x-y>x+y
Or 2y<0
Or y<0
So question boils down to is Y<0
A is Insufficient, B is sufficient.
Am I doing anything wrong here?
TO(x-y)/(x+y)>1
This can only be done if we know that the sum(x+y) is A +ve no.x-y>x+y
Otherwise,the inequality sign will change.
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shashank.ism
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x=/= -ydaretodream wrote:If x not equal to -y, is (x-y)/(x+y) > 1?
(1) x>0
(2) y<0
St.1) x>0 --> but we don't know about y so insuff.
St.2) y<0 --> but we don't know about x so insuff.
combined) : x>0,y<0 --> (x-y)> x and x+y <x
so (x-y)/(x+y) > 1 Hence Suff. Ans C
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girish3131
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