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Source: — Data Sufficiency |
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by patanjali.purpose » Sat Oct 15, 2011 3:38 pm
KItuz wrote:Can you help me solve this problem?

If x!=-y, is [x-y]/[x+y] > 1
[1] x > 0
[2] y < 0
Let's say [x-y]/[x+y] = I
S1: pick numbers

x = 5, y=-1 (I>1) , y=1 (I<1) ==> insufficient

S2: y = -5, x = -1 (I<1), x=1 (I<1)
y=-1, x=5 (I>1)

S1& S2
x=5 & y = -2 ==> yes
x=5 & y=(-1/2) ==> NO

IMO E

Another approach to combine S1 & S2:
x>0; -y>0 ==> add ==> (x-y)>0

But we cannot say anything about (x+y) ==> IMO E
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by KItuz » Sat Oct 15, 2011 3:44 pm
What is wrong with the following?
[x-y]/[x+y] > 1
=> [-2y]/[x+y] > 0
=> either y<0 and x+y>0 OR y>0 and x+y<0
=> either y<0 and x>0 OR y>0 and x<0

Correct ans is E but going by the above why is C incorrect?
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by n@resh » Sat Oct 15, 2011 3:52 pm
KItuz wrote:If x!=-y, is [x-y]/[x+y] > 1
[1] x > 0
[2] y < 0
Note: is it x!= -y? i am afraid, i consider it as x = -y.

From stat1: x > 0;

As x + y = 0, then y must be < 0. so, say x = 0.4, y = -0.4 then [ x - y ] = 0. and [ x + y] = 0
therefore, 0/0 which is an indeterminate form!
say if x = 2 and y = -2, [x - y ] = 4 and [x + y] = 0 => 4/0 = infinity which is also an indeterminate form!

Hence sufficient.

Similarly, stat2 also sufficient.

Answer D!
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by GmatKiss » Sat Oct 15, 2011 4:12 pm
IMO:D
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by KItuz » Sat Oct 15, 2011 4:20 pm
Correct answer is E.
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