Bunuel wrote:
Which of the following is closest to \(\frac{1}{9} + \frac{1}{99} + \frac{1}{999}\) ?
A. \(\frac{1}{10}\)
B. \(\frac{1}{9}\)
C. \(\frac{1}{8}\)
D. \(\frac{1}{6}\)
E. \(\frac{1}{5}\)
Question: \(\frac{1}{9} + \frac{1}{99} + \frac{1}{999}\) is closest to what?
Since the expression starts with \(\frac{1}{9}\) and adds (something really small) and adds (something even smaller), we can quickly see that the answer will be either B or C. It cant be A because the expression will be larger than \(\frac{1}{9}\), so B is definitely closer than A. And it cant be D or E because those are just too big. So let's check B and C.
\(\frac{1}{9}=\frac{8}{72} and \frac{1}{8}=\frac{9}{72}\)
So, \(\frac{1}{72}\) separates \(\frac{1}{9} and \frac{1}{8}\)
which means that the midpoint between them is \(\frac{1}{144}\)
\(\frac{1}{144}<\frac{1}{99}<\frac{1}{72}\), so \(\frac{1}{9} + \frac{1}{99}\) is closer to \(\frac{1}{8}\) but not greater than \(\frac{1}{8}\)
\(\frac{1}{999}\) is insignificant, so answer is C