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If x > 1 and y > 1 , is x > y ? 1. \sqrt{x} > y [#permalink]
 30 Sep 2008, 10:02
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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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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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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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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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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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Director
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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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A for me - same explanation as bigtreezl.
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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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A to me
1. x > 1 => x > sqrt(x) > y => x > y
2. insufficent
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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
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