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

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

1/9+1/99+1/999=0.(111)+0.(010)+0.(001)=~0.122=~0.125=1/8

BTW exact fraction would be: 0.(111111)+0.(010101)+0.(001001)=0.(122213).

Answer: C.

For more on repeated decimals check: math-number-theory-88376.html

Hope it helps.
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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...!

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

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

Infact there is another way of getting the answer without even solving it...

logically u r adding \(\frac{1}{99}+\frac{1}{999}\) to \(\frac{1}{9}\), therefore the answer would be more than \(\frac{1}{9}\). Hence \(\frac{1}{10}\) is wrong for sure...

Now \(\frac{1}{8}\) is \(\frac{1}{72}\) more than \(\frac{1}{9}\). This implies the midpoint betwen \(\frac{1}{8}\) & \(\frac{1}{9}\) is \(\frac{1}{144}\)..

Since you are adding \(\frac{1}{99}\) (which is more than \(\frac{1}{144}\)) to \(\frac{1}{9}\), then the sum would be closer to \(\frac{1}{8}\) than \(\frac{1}{9}\) ..... Also note that the \(\frac{1}{99}\) is less than \(\frac{1}{72}\) and hence it would not exceed \(\frac{1}{8}\)

Hence answer would be C.

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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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.

The question can be solved by just looking at the answer choices and minute calculations(if required) inside the brain.

LCM of denominators = 999

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

Dividing 999 by 41 gives 0.8

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

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

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

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

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

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

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

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.

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

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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