Official Solution:

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. \(\frac{1}{5}\)


Probably the easiest way would be if we just add the fractions. Notice that these fractions are repeated decimals:

\(\frac{1}{9}=0.111111...\)

\(\frac{1}{99}=0.010101...\)

\(\frac{1}{999}=0.001001...\)

\(\frac{1}{9}+\frac{1}{99}+\frac{1}{999}=0.(111111)+0.(010101)+0.(001001)=0.(122213) \approx 0.125=\frac{1}{8}\)


Answer: C
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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

My solution for this question (although I first did it wrongly):

1/9 + 1/99 = 12/99, which is smaller than 12/96 (1/8) (consider 1/999 equals to 0)

- 12/99 is 1/99 greater than 1/9
- 12/99 + 1/99 = 13/99, which is greater than 13/104 (1/8), so the amount 1/8 greater than 12/99 is less than 1/99

=> the sum is closest to 1/8 (C)

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 is how I rationalized my answer:

1/9 ( 11.11) , 1/10( 1) , 1/999( .1) = 12.21

A= 1 or 1/10
B= 11.11 or 1/9
C=12.5 or 1/8
D=16.67 or 1/6
E=20 or 1/5

12.21 is closest to 12.5 which is C . I'm really bad at adding decimals for some odd reason so I'm probably wrong .

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.

My process is by elimination.
1/10 not possible since 1/9 is already larger than 1/10 so obviously sum will be greater than this
1/9 same reason because of 2 more number
1/6 far greater than 1/9 so even adding 2 more no it wont catch 1/6
1/5 same reason like 1/6

correct me if i am wrong

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

1/9+1/99+1/999>1/10+1/100+1/1000

By solving RHS,
1/10+1/100+1/1000=1/9

1/9+1/99+1/999>1/9


1/9+1/99+1/999 should be 1/8(because that is the next closest value)