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# M17-29

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Math Expert
Joined: 02 Sep 2009
Posts: 47983

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16 Sep 2014, 01:02
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Difficulty:

75% (hard)

Question Stats:

43% (00:52) correct 57% (00:49) wrong based on 241 sessions

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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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Math Expert
Joined: 02 Sep 2009
Posts: 47983

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16 Sep 2014, 01:02
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:

$$\frac{1}{9}=0.111111...$$

$$\frac{1}{99}=0.010101...$$

$$\frac{1}{999}=0.001001...$$

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

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Joined: 11 Sep 2013
Posts: 156
Concentration: Finance, Finance

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27 Nov 2014, 16:09
2
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
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Joined: 29 Apr 2014
Posts: 122
Location: Viet Nam
Concentration: Finance, Technology
GMAT 1: 640 Q50 V26
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17 Nov 2015, 16:46
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)
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Joined: 28 Dec 2016
Posts: 88
Location: United States (IL)
Concentration: Marketing, General Management
Schools: Johnson '20 (M)
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21 Jan 2017, 22:34
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.
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Joined: 18 May 2017
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19 May 2017, 05:15
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 .
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Joined: 25 Jul 2017
Posts: 96

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17 Jun 2018, 20:39
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. $$\frac{1}{5}$$

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
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Joined: 12 May 2018
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05 Jul 2018, 01:57
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
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Joined: 30 Mar 2017
Posts: 136
GMAT 1: 200 Q1 V1

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05 Aug 2018, 05:22
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
Re: M17-29 &nbs [#permalink] 05 Aug 2018, 05:22
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# M17-29

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