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\(\frac{1}{9} + \frac{1}{99} + \frac{1}{999} = \frac{10-9}{9} + \frac{100-99}{99} + \frac{1000-999}{999} = \frac{10}{9} + \frac{100}{99} + \frac{1000}{999} - 3 = 3.3 - 3 = 0.3\)

among the options given, \(\frac{1}{5}\) is the highest which is 0.2.

Therefore answer IMO is E

Cheers!
JT
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\(\frac{1}{9} + \frac{1}{99} + \frac{1}{999} = \frac{10-9}{9} + \frac{100-99}{99} + \frac{1000-999}{999} = \frac{10}{9} + \frac{100}{99} + \frac{1000}{999} - 3 = 3.3 - 3 = 0.3\)

among the options given, \(\frac{1}{5}\) is the highest which is 0.2.

Therefore answer IMO is E

Cheers!
JT

Oops.... committed a blunder... Please ignore...!
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jeeteshsingh
\(\frac{1}{9} + \frac{1}{99} + \frac{1}{999} = \frac{10-9}{9} + \frac{100-99}{99} + \frac{1000-999}{999} = \frac{10}{9} + \frac{100}{99} + \frac{1000}{999} - 3 = 3.3 - 3 = 0.3\)

among the options given, \(\frac{1}{5}\) is the highest which is 0.2.

Therefore answer IMO is E

Cheers!
JT

Oops.... committed a blunder... Please ignore...!

\(\frac{1}{9} + \frac{1}{99} + \frac{1}{999} = \frac{10-9}{9} + \frac{100-99}{99} + \frac{1000-999}{999} = \frac{10}{9} + \frac{100}{99} + \frac{1000}{999} - 3 = 3.122 - 3 = 0.122\)

among the options given, \(\frac{1}{8}\) is the highest which is 0.12.


Therefore answer IMO is C... Wonder if there is an easy way to solve :!:
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hi iwud do it in this way ..since it is an approx ans i wud take 990 instead of 990...eq becomes:
1/9+1/99+1/990=1/990(110+11+1)=122/990=121/990=11/90=1/8.2=1/8
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GMATPASSION
Which of the following is closest to \(\frac{1}{9} + \frac{1}{99} + \frac{1}{999}\) ?

\(\frac{1}{10}\)
\(\frac{1}{9}\)
\(\frac{1}{8}\)
\(\frac{1}{6}\)
\(\frac{1}{5}\)

You can quickly eliminate A and B as these less than or equal to 1/9.

1/9 = 0.1111
1/99 = 0.0101
1/999 = 0.0010
Total = 0.1222

You can eliminate 1/6 and 1/5 as they are obviously high.
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The question can be solved by just looking at the answer choices and minute calculations(if required) inside the brain.
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LCM of denominators = 999

So, \(\frac{(111+11+1)}{999} = \frac{123}{999} = \frac{41}{999}\)

Dividing 999 by 41 gives 0.8

Answer = \(C = \frac{1}{8}\)
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hrish88
Which of the following is closest to \(\frac{1}{9}+\frac{1}{99}+\frac{1}{999}\)

A. 1/10
B. 1/9
C. 1/8
D. 1/6
E. 1/5

\(\frac{1}{9}+\frac{1}{99}+\frac{1}{999}\)

= \(\frac{111+11+1}{999}\)

= \(\frac{123}{999}\)

= \(\frac{41}{333}\)

~ \(\frac{40}{320}\)

= \(\frac{1}{8}\)

Thus, answer will be (C) \(\frac{1}{8}\)
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Which of the following is closest to 1/9+1/99+1/999

A. 1/10
B. 1/9
C. 1/8
D. 1/6
E. 1/5

1/9 ≈ 0.111
1/99 ≈ 1/100 = 0.01
1/999 ≈ 1/1000 = 0.001

--> sum ≈ 0.111 + 0.01 + 0.001 = 0.122 ≈ 0.125 = 1/8 --> C
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The sum clearly cannot be less than 1/9.
Eliminate A.
The difference between B and C = 1/8 - 1/9 = 1/72.
Since 1/99 is more than 1/2 of this difference -- (1/2)(1/72) = 1/144 -- the sum is closer to 1/8 than to 1/9.
Eliminate B.
Since 1/999 has an relatively insignificant effect on the sum, the sum cannot be as great at 1/6 or 1/5.
Eliminate D and E.
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1/9 = 0.11111
1/99 = 0.0101010101
1/999= 0.001001001001

Adding these,we will get 0.122~ 0.125 = 1/8

Correct Answer- C
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hrish88
Which of the following is closest to 1/9+1/99+1/999

A. 1/10
B. 1/9
C. 1/8
D. 1/6
E. 1/5

1/9 + 1/99 + 1/999
—> 111/999 + 1/999 + 1/99
—> 112/999 + 1/99
—> 112/1000 + 1/100 (approx)
—> 0.112 + 0.010
—> 0.122

Close to 1/8 (0.125)

IMO Option C
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