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if x>y>0, which of the following must be true? I.

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Manager
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if x>y>0, which of the following must be true? I. [#permalink] New post 29 May 2007, 16:00
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

if x>y>0, which of the following must be true?

I. root(x) - root(y) < root(x-y)
II. x^2 - y^2 > (x-y)^2

1. None
2. I only
3. II only
4. I and II
5. Cannot be determined

If you can, please provide an ALGEBRAIC SOLUTION/DERIVATION/PROOF for the two inequations. I got some answers after testing with several numbers - but it took a while and number plug-in method makes me unsure. Thanks.
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 [#permalink] New post 29 May 2007, 19:18
i don't know of an algebraic/proof way to answer this q- there are some smart people on this board who might have a 'number theory' type of answer/explanation for you... but sometimes i think you just gotta plug in.

i think you can answer this quickly by picking an x and y value that satisfies the inequality (x is bigger than y, y is bigger than zero) and keeping the x and y the same acorss inequalities keeps it consistent

i pikced 100 and 36

10-6<8>16 yup

you can estimate with fractions even:

x=1/4 y=1/9

I: 1/2-1/3 < square root 5/36
1/6 < use 4/36 instead (a smaller # but easy to find root of)

1/6<2/6 yup!
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 [#permalink] New post 29 May 2007, 19:38
Its 4. Both Statements I and II are true.

Since X and Y are positive. We will have to check when X and Y are integers and when they are between 0 and 1 (0<x>1 and 0<Y>1).

When you simplify stat II we get
2XY>y^2, Plug in values below

I used X=4 and Y=9 for integers and X=1/4 and Y=1/9 and found that both statements are valid.
  [#permalink] 29 May 2007, 19:38
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if x>y>0, which of the following must be true? I.

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