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Re: Another Inequality Question [#permalink]
31 Oct 2009, 21:36

1

This post received KUDOS

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.

Re: If x and y are positive, is x>y2? (1) y>x2 (2) y=x+1 [#permalink]
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 _________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

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