I'm sure there is a smarter way to do this, but it's clear that \(\sqrt{119}\) is something slightly less than \(11\)

However, the answer choices are not spread far enough apart to be done at this point.

So then I thought about solving \((10+x)^2 = 119\) where I know \(x<1\)
Trying \(x=0.9\) gives us \(10^2+2*10*0.9+0.9^2\)
Bingo, this is \(100+18+0.81\) which is very close to \(119\) but a little less, so our x was a little too low

Now to solve \(100*(11-10.9) = 100* 0.1 = 10\)

Well shoot we know it's a little less but is it \(9.9\) or \(9.2\)?

To have gotten \(9.9\), we would have needed \(x = .901\). That would not have been enough to get all the way to \(119\) in \(10^2+2*10*0.901+0.901^2\)

Therefore, best guess is that B is the right answer

\(?\,\,\,:\,\,\,\,100\left( {11 - \sqrt {119} } \right)\,\,\,{\rm{approx}}{\rm{.}}\)


\(\left( {11 - \sqrt {119} } \right)\left( {11 + \sqrt {119} } \right) = {11^2} - 119 = 2\,\,\,\,\, \Rightarrow \,\,\,\,\,11 - \sqrt {119} = {2 \over {11 + \sqrt {119} }}\)

\(100 < 119 < 121\,\,\,\, \Rightarrow \,\,\,\,10 < \sqrt {119} < 11\,\,\,\,\mathop \Rightarrow \limits^{ + 11} \,\,\,\,21 < 11 + \sqrt {119} < 22\,\,\,\, \Rightarrow \,\,\,\,{1 \over {22}} < {1 \over {11 + \sqrt {119} }} < {1 \over {21}}\)


\({2 \over {22}} < {2 \over {11 + \sqrt {119} }} < {2 \over {21}}\,\,\,\,\, \Rightarrow \,\,\,\,100 \cdot {1 \over {11}} < \underbrace {100\left( {11 - \sqrt {119} } \right)}_{{\rm{focus}}\,{\rm{!}}} < 100 \cdot {2 \over {21}}\)

\(\left. \matrix{
{{100} \over {11}} = {{99 + 1} \over {11}} = 9{1 \over {11}}\,\, \cong \,\,9.1 \hfill \cr
{{200} \over {21}} = {{210 - 10} \over {21}} = 10 - {{10} \over {21}} = 9{{11} \over {21}}\,\, \cong \,\,9.5 \hfill \cr} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\left( {\rm{B}} \right)\)


The correct answer is (B).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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