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Is x+y > 0? a) x/(x+y) > 0 b) y/(x+y) > 0

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Manager
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New post 11 Mar 2004, 11:37
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Is x+y > 0?

a) x/(x+y) > 0
b) y/(x+y) > 0
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New post 11 Mar 2004, 11:58
I think both are required.

Assuming both x and y to be negative,
a) and b) independently are not suffiecient.

using both .

I just added a) and b) which is 1 always greater than 0.

cheers.
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New post 11 Mar 2004, 12:07
I would guess E!

The more I prepare, the more questions I'm getting incorrect.
I'm not sure, if my answer is correct!
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New post 11 Mar 2004, 12:12
C.

Is x + y > 0?


Statement a) x/(x+y) > 0
multiplying both sides by x+y, x>0. This is insufficient because we know nothing about y.

Statement b) y/(x+y) > 0
multiplying both sides by x+y, y>0. This is insufficient because we know nothing about x.

a) and b) together, x is positive, y is positive so x + y > 0.
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New post 11 Mar 2004, 12:28
I will go with E.

1) both x and y could be -ve ot +ve
2) both x and y could be -ve ot +ve
Even if they r combined the same thing holds good.
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New post 11 Mar 2004, 13:36
it should be E.

Virtual is assuming
x/(x+y) + y/(x+y) will give x+y/ x+y.

Still, x+y/x+y is not equal to x+y.
if we take x= -2 and y=-6. x+y/x+y will give a +ive
while x+y will give a -ive
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New post 11 Mar 2004, 13:39
I agree with Anand here. My original answer is incorrect.
God !!! what was i thinking !!!!


cheers.
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New post 11 Mar 2004, 14:13
What is wrong with the way I solved it? Is it wrong to multiply both sides of x/(x+y) > 0 by (x+y)? This has always confused me when solving these types of questions. Especially since it's easy to forget to test all possible combinations for x and y. Is there another way to solve this problem without picking numbers?

Please anyone who knows the answers to my questions should please reply.

Thanks.
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New post 11 Mar 2004, 14:22
ndidi204,

I think .......

multiplying both sides by x+y, x>0. and
multiplying both sides by x+y, y>0.

will hold possible with the assumption that (x+y) > 0.

If (x+y) < 0, in that case it will be

multiplying both sides by x+y, x<0. and
multiplying both sides by x+y, y<0.

So u r assuming that (x+y) is greater than zero.

HTH


HTH
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New post 11 Mar 2004, 14:24
Hmm. You're right. I never considered that (x+y) could be negative.

You know what they say about assumptions! It's the chief of all F-ups!

Thanks Virtual.
  [#permalink] 11 Mar 2004, 14:24
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