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# Is p+q > r+s 1) p>r+s 2) q>r+s

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Is p+q > r+s 1) p>r+s 2) q>r+s [#permalink]  06 Nov 2004, 18:09
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Is p+q > r+s

1) p>r+s
2) q>r+s
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Paul

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Joined: 21 Jun 2004
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Is p+q > r+s

1) p>r+s
2) q>r+s

I think I would go with C.
Although individually, P and Q are greater than r+s, as long as we do not know what Q is in the first case and P in the second, individual statements are not sufficient.Therfore, I will go with C.
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Is the ans D?
as...if we modify this a bit....is p+q>r+s? to is p>r+s-q? or is a>r+s-p?
doesnt each of the choice answers the question?
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Is p+q > r+s

1) p>r+s
2) q>r+s

E is my answer, just because:

from 1&2, we have p+q>2(r+s). If p,q,r,s >0, p+q>r+s

However, we don't know whether q,p,r,s >0 or not. If r+s <0, what happens. Let have a look

p+q>2(r+s)

eg: 2(r+s) = -6 and p+q = -4. Although p+q>2(r+s), but p+q<r+s (p+q = -4 and r+s = -3)
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E for me.

1. p > r+s. No info on q. So, Insuff.
2. q > r+s. No info on p. So, Insuff.

combing we have p+q > 2(r+s) (Note: We can add two different inequalities, PROVIDED, they have the same signs (> or <) in them)

If p+q = 4, r+s = 1 then p+q > 2(r+s) and also p+q > r+s.
If p+q = -4, r+s = -3 then p+q > 2(r+s) but p+q < r+s.

So, cant say. Hence E.
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another vote for E.

very nice question.
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OA is E. Be careful about negative numbers.
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