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(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.

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). _________________

Re: If n is positive, is \sqrt {n} > 100 ? 1. \sqrt {n-1} [#permalink]

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09 Apr 2012, 12:19

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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.

Re: If n is positive, is root(n) > 100 ? [#permalink]

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27 Oct 2015, 01:42

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