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If x > 1 and y > 1 , is x > y ? 1. \sqrt{x} > y

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VP
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If x > 1 and y > 1 , is x > y ? 1. \sqrt{x} > y [#permalink] New post 30 Sep 2008, 09:02
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If x > 1 and y > 1 , is x > y ?

1. \sqrt{x} > y
2. \sqrt{y} < x
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Re: DS: m01#5 [#permalink] New post 30 Sep 2008, 09:15
A for me.

Since, both x and y are greater than 1, hence if sqrt(x) > y then x will definitely be greater than y.

stmt2 is not sufficient. sqrt(y) < x but y can be greater than, equal to or smaller than x.
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Re: DS: m01#5 [#permalink] New post 30 Sep 2008, 09:21
scthakur wrote:
A for me.

stmt2 is not sufficient. sqrt(y) < x but y can be greater than, equal to or smaller than x.


I don't seem to understand this :( a bit more elaborate explanation will help
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Re: DS: m01#5 [#permalink] New post 30 Sep 2008, 09:27
amitdgr wrote:
scthakur wrote:
A for me.

stmt2 is not sufficient. sqrt(y) < x but y can be greater than, equal to or smaller than x.


I don't seem to understand this :( a bit more elaborate explanation will help


if x is between 0 and 1 then sqrt(x) will be greater than x. However, if x > 1 then sqrt(x) will always be smaller than x. We can check this with any real numbers.

With this logic, if sqrt(x) > y and since x > sqrt(x) as x > 1, hence x > y.
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Re: DS: m01#5 [#permalink] New post 30 Sep 2008, 09:46
How do you rule out the second statement ? I understood the first statement in your first post :) I had asked explanation for the second one alone.

Thanks
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Re: DS: m01#5 [#permalink] New post 30 Sep 2008, 10:11
I get A too

since x and y are greater than 1

1) sqrt x>y
x>y^2
x has to be bigger than y since x is bigger then y*y

2) sqrt y<x
y<x^2
x^2 > y
in this case x is not neccessarily bigger than y

for instance x=2, y=3 and x^2=4, y=3, true in this case
but x=3, y=3 means x^2=9. x^2 is bigger than y, but x is not bigger than y so false in this case
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Re: DS: m01#5 [#permalink] New post 30 Sep 2008, 10:16
A for me - same explanation as bigtreezl.
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Re: DS: m01#5 [#permalink] New post 30 Sep 2008, 10:34
amitdgr wrote:
How do you rule out the second statement ? I understood the first statement in your first post :) I had asked explanation for the second one alone.

Thanks


Ok. take y = 4, x = 5. In this case sqrt(y) < x and also y < x.
Take another example where y = 16, x = 5. In this case, sqrt(y) < x but y > x.
Third example, y = 4, x = 4. Here, sqrt(y) < x, but y = x.

As a rule, if y > 1 then y > sqrt(y)
In the given question x > sqrt(y). Hence, the two cannot be equated. We cannot definitely say that y > x or y < x or y = x.
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Re: DS: m01#5 [#permalink] New post 30 Sep 2008, 23:57
A to me

1. x > 1 => x > sqrt(x) > y => x > y

2. insufficent
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Re: DS: m01#5 [#permalink] New post 01 Oct 2008, 00:00
scthakur wrote:
amitdgr wrote:
How do you rule out the second statement ? I understood the first statement in your first post :) I had asked explanation for the second one alone.

Thanks


Ok. take y = 4, x = 5. In this case sqrt(y) < x and also y < x.
Take another example where y = 16, x = 5. In this case, sqrt(y) < x but y > x.
Third example, y = 4, x = 4. Here, sqrt(y) < x, but y = x.

As a rule, if y > 1 then y > sqrt(y)
In the given question x > sqrt(y). Hence, the two cannot be equated. We cannot definitely say that y > x or y < x or y = x.


yes, examples in this case is the easiest & fastest way to rule out. Try 9 for y and 4 for x
Re: DS: m01#5   [#permalink] 01 Oct 2008, 00:00
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