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If x and y are positive, is x > y^2?

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If x and y are positive, is x > y^2? [#permalink] New post 31 Oct 2009, 21:07
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60% (02:05) correct 40% (01:42) wrong based on 79 sessions
If x and y are positive, is x > y^2?

(1) y > x^2
(2) y = x + 1

M17-32
[Reveal] Spoiler: OA

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Re: Another Inequality Question [#permalink] New post 31 Oct 2009, 21:30
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If x and y are positive, is x > y^2?

(1) y>x2. If x=1 and y=4, then the answer is NO but if x=y=\frac{1}{2}, then the answer is YES. Not sufficient.

(2) y=x+1. The question becomes: is x>(x+1)^2? Which cannot be true for ANY value of x. So, the answer to the question is NO. Sufficient.

Answer: B.
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Re: Another Inequality Question [#permalink] New post 31 Oct 2009, 21:36
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These type of quastions works try to confuse test takers with fractions in power. The best approach is Picking Numbers strategy.

Is x>y^2 given that x and y are positive?

(1) y>x^2

Pick x=\frac{8}{10} and y=\frac{7}{10}, \frac{7}{10}>\frac{64}{100}, and \frac{8}{10}>\frac{49}{100}. But if we try x=\frac{8}{10} and y=1, x will be lower than y^2. So, (1) alone is NOT SUFFICIENT.

(2) y=x+1
given that x>0, therefore y>1. There is y^k>y^1 for all y>1, k>1. So, (2) alone is SUFFICIENT.

The answer is B.
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Re: Another Inequality Question [#permalink] New post 01 Nov 2009, 00:42
Thanks for the explanation guys. I totally missed the "if x and y are positive" language in the question stem :oops:
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Re: Another Inequality Question [#permalink] New post 07 Nov 2009, 21:47
I ignored the fraction possibility..

Nice solution..

slingfox wrote:
Thanks for the explanation guys. I totally missed the "if x and y are positive" language in the question stem :oops:
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Re: If x and y are positive, is X > y^2? [#permalink] New post 17 Jan 2013, 02:43
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slingfox wrote:
How would you guys approach this question:
If x and y are positive, is X > y^2?
1. y > x^2
2. y = x + 1


It is important to note that all you have to test are positive whole numbers and fractions

1.
let x=2, y=5
y > x^2 => 5 > 4 ----- x > y^2 => 2 > 25 NO!

let x=1/2, y=1/2
y > x^2 => 1/2 > 1/4 ----- x > y^2 => 1/2 > 1/4 YES!

The information given in statement 1 is insufficient.

2.
let x = 2, y = x + 1 = 3
x > y^2 ==> 2 > 9 NO!

let x = 1/2, y = 5/2
x > y^2 ==> 1/2 > 25/4 NO!

The information given in statement 2 is sufficient.

Answer: B
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If x and y are positive, is x>y2? (1) y>x2 (2) y=x+1 [#permalink] New post 27 Feb 2014, 06:27
If x and y are positive, is x>y^2?

(1) y>x^2

(2) y=x+1
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Re: If x and y are positive, is x>y2? (1) y>x2 (2) y=x+1 [#permalink] New post 27 Feb 2014, 06:30
i got that statement 2 is sufficient..

But, I tuk around more than 2 minutes to chek whether statement 1 is sufficient or not..

confused when i see x>y^2 or x2>y3 .. these kind of question most of the time i do it correct, bt i tuk long time to prove whether its sufficient or not
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Re: If x and y are positive, is x>y2? (1) y>x2 (2) y=x+1 [#permalink] New post 27 Feb 2014, 07:23
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Re: If x and y are positive, is x>y2? (1) y>x2 (2) y=x+1   [#permalink] 27 Feb 2014, 07:23
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