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# If x > 1 and m > 1 , is x m < (m + x + 1) ?

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If x > 1 and m > 1 , is x m < (m + x + 1) ? [#permalink]

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20 Jun 2010, 05:45
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If x > 1 and m > 1, is $$x\sqrt{m} < \sqrt{(m + x + 1)}$$ ?

(1) m > x + 1
(2) m > 1/(x-1)
[Reveal] Spoiler: OA

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Re: Interesting Inequality [#permalink]

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20 Jun 2010, 09:09
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Hussain15 wrote:
If $$x > 1$$ and $$m > 1$$, is $$x\sqrt{m} < \sqrt{(m + x + 1)}$$ ?

(1) $$m > x + 1$$

(2) $$m > \frac{1}{x-1}$$

Is $$x\sqrt{m}<\sqrt{(m+x+1)}$$ --> square both sides (we can safely do this as both sides are positive) --> $$mx^2<m+x+1$$ --> $$mx^2-m<x+1$$ --> $$m(x^2-1)<x+1$$ --> $$m(x-1)(x+1)<x+1$$ --> reduce by $$x+1$$ (we can safely do this as $$x+1>0$$) --> $$m(x-1)<1$$.

So, finally the question becomes: is $$m(x-1)<1$$?

(1) $$m>x+1$$ --> Not sufficient.

(2) $$m>\frac{1}{x-1}$$ --> cross multiply (again we can safely doth is as $$x-1>0$$) --> $$m(x-1)>1$$. Sufficient.

Hope it's clear.
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Re: Interesting Inequality [#permalink]

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20 Jun 2010, 09:48
Bunuel, another feather in your hat. Good explanation. I was eagerly waiting for your answer as I was not able to solve the same Thanks once again.
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Re: Interesting Inequality [#permalink]

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21 Jun 2010, 03:05
Can you explain why statement 1 alone doesnt suffice, it is given both x and m are >1
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Re: Interesting Inequality [#permalink]

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21 Jun 2010, 03:30
ParvezDhamani wrote:
Can you explain why statement 1 alone doesnt suffice, it is given both x and m are >1

Because if $$m$$ is some number, let's say $$m=100$$ and $$x$$ is very small (not violating the condition $$x>1$$), let's say $$x=1.000001$$, then: $$x\sqrt{m}=10.00001<\sqrt{(m + x + 1)}\approx{10.01}$$.

Bit if $$m=9$$ and $$x=6$$, then $$x\sqrt{m}=18>\sqrt{(m + x + 1)}=4$$.
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Re: Interesting Inequality [#permalink]

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21 Jun 2010, 03:47
Thanks a lot for the explanation, but how do we take such hypothetical values in the real test in less than 2 mins, is there a easy way out?
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Re: Interesting Inequality [#permalink]

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21 Jun 2010, 09:22
Perfect Solution Bunuel! +1 .
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Re: If x > 1 and m > 1 , is x m < (m + x + 1) ? [#permalink]

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22 Feb 2014, 12:42
Bumping for review and further discussion.
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Re: If x > 1 and m > 1 , is x m < (m + x + 1) ? [#permalink]

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28 May 2015, 06:10
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If x > 1 and m > 1 , is x m < (m + x + 1) ? [#permalink]

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07 Mar 2016, 21:19
Hussain15 wrote:
If x > 1 and m > 1, is $$x\sqrt{m} < \sqrt{(m + x + 1)}$$ ?

(1) m > x + 1
(2) m > 1/(x-1)

B is sufficient as the question reduces to m(x-1)<1
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Re: If x > 1 and m > 1 , is x m < (m + x + 1) ? [#permalink]

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13 Mar 2016, 06:28
Also the first part of the statement can be proven insufficient by taking examples as x=1.0000000001 and m=100
and x=20 and m = 400
Remember => Do not discard statement 1 as it does not satisfies the inequality directly => rather look for values and prove it both sufficient and insufficient
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Re: If x > 1 and m > 1 , is x m < (m + x + 1) ?   [#permalink] 13 Mar 2016, 06:28
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# If x > 1 and m > 1 , is x m < (m + x + 1) ?

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