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. \(\frac{1}{5}\)
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To me getting decimal in this case is difficult.
I will multiply 999 with each fraction.
1/9*999 = 111
1/99*999 = 10. (ten point something)
1/999*999 = 1

111+10+1= 122

Now 999* 1/8 = close to 122.

No other option will be close to 122. For that we don't have to divide. By noticing carefully we can find it

I solved it by approximating.

1/9 + 1/99 + 1/999

1/9 + 1/99 = 12/99

Therefore, 12/99 + 1/999 = x

approximate : 12/100 + 1/1000 = 120/1000 + 1/1000 = 121/1000 = 0.121

1/8 = 0.125
1/9 = 0.111

Therefore, 1/8.

This can be easily solved by approximation,

=>1/9+1/99+1/999
=>12/99+1/999
=> ~ 12/100 + ~ 1/1000
=> ~0.121 (Since 12/100 <12/99) +~ 0.001
=> ~0.122
Now 1/8 =0.125, which is the most closest to the value derived
C is the answer.

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