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Q: Is x+y > 0 if (1) x / (x+y) > 0 (2) y / (x+y) >

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Q: Is x+y > 0 if (1) x / (x+y) > 0 (2) y / (x+y) > [#permalink] New post 09 Mar 2007, 19:39
Q: Is x+y > 0 if

(1) x / (x+y) > 0
(2) y / (x+y) > 0

My answer is (C), both together are SUFF.

From eq. (1) we get to know that x > 0 but no info abt y and from eq. (2) y > 0 but no info abt x.

Taking them together, we know x and y both > 0, so x+y > 0.

Am I correct?
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 [#permalink] New post 10 Mar 2007, 01:52
Dear Summer,

Answer should be (e) and not (c). Following is the explanation

for 1) assuming that x/(x+y) >0, which can be possible in the following conditions: x and y both positive that means x+y>0, x and y both negative that means x+y<0>Y, then X + Y >0 or X is negative and Y is positive but [X]> [Y]. so its clear that 1 is not sufficient

Similarly 2) is also not sufficient. from this you can discard option number (d) as well.

Now you have either (c) or (e).


Lets take both into consideration. on combining both you have either both x and y are positive or both x and y are negative. which means either x+ y >0 or x+y<0, so we can eliminate option c as well.

Answer is (e)

rgds

Amardeep :)
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 [#permalink] New post 10 Mar 2007, 01:55
(E) for me :)

x+y > 0 ?

From 1
x / (x+y) > 0

Implies that:
Sign (x) = Sign(x+y)

Thus, x+y could be either positive or negative.

INSUFF.

From 2
y / (x+y) > 0

Implies that:
Sign (y) = Sign(x+y)

Thus, x+y could be either positive or negative.

INSUFF.

(1) combined with (2)
o x / (x+y) > 0 (A)
o y / (x+y) > 0 (B)

(A) + (B) <=> (x+y) / (x+y) > 0 <=> 1 > 0 >>> Always truth... Brings us nothing more :)

To be sure, we can plugs some values :)
o If x=y=-1, then -1 / (-2) = 1/2 > 0 and x+y < 0
o If x=y=1, then 1 / (2) = 1/2 > 0 and x+y > 0

INSUFF.
  [#permalink] 10 Mar 2007, 01:55
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