(x-y)/(x+y)>1------> (x-y)/(x+y)-1>0------->-2y/(x+y)>0
we need to know relationship b/n x and y and whether y >0??
s1) only about x not suff
s2) only about y we need to know relationship b/n x and y not suff
Inequality prob
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thephoenix
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raviki8208
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re-writing problem statement:selango wrote:If x is not equal to y,is (x-y)/(x+y)>1 ?
A)x>0
B)y<0
OA E
(x-y)/(x+y)>1
(x+y-2y)/(x+y)>1
1 - (2y/(x+y)) >1
2y/(x+y) < 0
using 1 alone, x>0
x = 3, y = -2 statement is true
x = 1, y = 2 statement is false. so insufficient
using 2 alone, y<0
x = 3, y = -2 statement is true
x = 1, y = -2 statement is false, so insufficient
using both x>0 and y<0
x = 3, y = -2 statement is true
x = 1, y = -2 statement is false, so insufficient
so,
E
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liferocks
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question is
is (x-y)/(x+y)>1
or [-y/(x+y)]>0
this is possible when -y and (x+y) will have same sign
both conditions mentions about the sign of either x or y but not both ,hence either alone is not sufficient
combining
y<0..so -y>0
x>=..but no information about if |x|>|Y| which is necessary to deduce whether (x+y)>0 ..hence not sufficient
Ans option E
is (x-y)/(x+y)>1
or [-y/(x+y)]>0
this is possible when -y and (x+y) will have same sign
both conditions mentions about the sign of either x or y but not both ,hence either alone is not sufficient
combining
y<0..so -y>0
x>=..but no information about if |x|>|Y| which is necessary to deduce whether (x+y)>0 ..hence not sufficient
Ans option E
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Geva@EconomistGMAT
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The problem with inequalities is that when you multiply or divide by a negtive number, you need to flip the sign. If (x+y) is negative, then you would need to flip the sign and getTOPGMAT wrote:Guys,
Need your help on this one...
Can you tell me what is the flaw in the logic that I am using ?
(x-y)/(x+y) >1
= (x-y) > (x+y)
= 0 > 2y
or y<0
Thanks
(x-y)/(x+y) >1 --> (x-y) < (x+y)
but if (x+y) is positive, then the sign remains as is, and your logic holds.
the answer is E because even if you know that x>o and y<0, you still don't know whether the sum x+y is positive or negative - that will depend on the relative absolute value of x and y:
If x=2, y=-3, then x+y=2-3=-1 is negative, and you need to flip the sign when multiplying by (x+y)
But
If x=10 and y=-3, then x+y=10-3=7, and the sign remains the same when multiplying by (x+y)
