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# If n is positive, is root(n) > 100 ?

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If n is positive, is root(n) > 100 ? [#permalink]  11 Jan 2011, 16:57
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If n is positive, is $$\sqrt {n} > 100$$?

(1) $$\sqrt {n-1} > 99$$

(2) $$\sqrt {n+1} > 101$$

Can someone write out the algebra on this one, I just want to double check work. Thanks.
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Re: Quant Rev DS #51 [#permalink]  11 Jan 2011, 17:08
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If n is positive, is $$\sqrt {n} > 100$$?

Is $$\sqrt {n} > 100$$? --> is $$n>100^2$$?

(1) $$\sqrt {n-1} > 99$$ --> $$n-1>99^2$$ --> $$n>99^2+1$$: $$99^2+1$$ is less than $$100^2$$ (as $$100^2=(99+1)^2=99^2+2*99+1$$), so $$n$$ may or may not be more than this value. Not sufficient.

(2) $$\sqrt {n+1} > 101$$ --> $$n+1>101^2$$ --> $$n>101^2-1=(101-1)(101+1)=100*102$$, so $$n>100*102>100^2$$. Sufficient.

 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/No posting of PS/DS questions is allowed in the main Math forum.

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Re: Quant Rev DS #51 [#permalink]  11 Jan 2011, 18:30
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tonebeeze wrote:
If n is positive, is $$\sqrt {n} > 100$$?

1. $$\sqrt {n-1} > 99$$

2. $$\sqrt {n+1} > 101$$

Can someone write out the algebra on this one, I just want to double check work. Thanks.

Even though you asked for algebra, let me point out that you don't really need algebra to solve this.

When you are considering big numbers, square roots of consecutive numbers differ by very little. e.g. $$\sqrt {10000} = 100$$ and $$\sqrt {9999} = 99.995$$... Square roots of even small positive numbers differ by less than 1 e.g. $$\sqrt {1} = 1$$ and $$\sqrt {2} = 1.414$$ - Difference of just 0.414 $$\sqrt {3} = 1.732$$ - Difference of just 0.318. The difference just keeps getting smaller and smaller.

So if $$\sqrt {n-1} > 99$$, $$\sqrt {n}$$ will be very close to $$\sqrt {n-1}$$ and will also be greater than 99. But will it be greater than 100, we cannot say. So not sufficient.

If $$\sqrt {n+1} > 101$$, then $$\sqrt {n}$$ may not be greater than 101, but it will definitely be greater than 100 since between two consecutive integers, the square root difference will not reach 1 (as shown above).
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 15 Apr 2011 Posts: 70 Followers: 0 Kudos [?]: 13 [0], given: 45 Re: If n is positive, is \sqrt {n} > 100 ? 1. \sqrt {n-1} [#permalink] 09 Apr 2012, 11:07 I didn't understand why statement 1 is not sufficient. Can someone please explain? thanks. _________________ Math Expert Joined: 02 Sep 2009 Posts: 30437 Followers: 5102 Kudos [?]: 57613 [1] , given: 8819 Re: If n is positive, is \sqrt {n} > 100 ? 1. \sqrt {n-1} [#permalink] 09 Apr 2012, 11:19 1 This post received KUDOS Expert's post mymbadreamz wrote: I didn't understand why statement 1 is not sufficient. Can someone please explain? thanks. Question asks whether $$n>100^2$$. From (1) we have that $$n>99^2+1$$. Now, since $$100^2>99^2+1$$, then it's possible that $$n>100^2>99^2+1$$, which would mean that the answer is YES, as well as that $$100^2>n>99^2+1$$, which would mean that the answer is NO. Two different answers, hence not sufficient. Hope it's clear. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6062 Location: Pune, India Followers: 1605 Kudos [?]: 8971 [1] , given: 195 Re: If n is positive, is \sqrt {n} > 100 ? 1. \sqrt {n-1} [#permalink] 09 Apr 2012, 21:05 1 This post received KUDOS Expert's post mymbadreamz wrote: I didn't understand why statement 1 is not sufficient. Can someone please explain? thanks. Stmnt 1 just tells you that $$\sqrt{n-1} > 99$$ Think of 2 diff cases: $$\sqrt{n-1} = 99.2$$ $$\sqrt{n} = 99.205$$ or $$\sqrt{n-1} = 112$$ $$\sqrt{n} = 112.015$$ Can you say whether $$\sqrt{n}$$ is greater than 100? _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: If n is positive, is root(n) > 100 ? [#permalink]  27 Oct 2015, 00:42
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