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(1/7)+(1/8)+(1/9) is in between?

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(1/7)+(1/8)+(1/9) is in between?  [#permalink]

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New post 04 May 2017, 01:27
3
1
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

76% (01:35) correct 24% (01:39) wrong based on 121 sessions

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\((\frac{1}{7})+(\frac{1}{8})+(\frac{1}{9})\) is in between?

A. \((\frac{1}{6}) and (\frac{1}{5})\)
B. \((\frac{1}{5}) and (\frac{1}{4})\)
C. \((\frac{1}{4}) and (\frac{1}{3})\)
D. \((\frac{1}{3}) and (\frac{1}{2})\)
E. \((\frac{1}{2}) and 1\)

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Re: (1/7)+(1/8)+(1/9) is in between?  [#permalink]

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New post 04 May 2017, 07:10
Option d..
1/7=0.1428
1/8=0.1250
1/9=0.1101

How I remember this.. Well Ive mucked up till 1/20...do let me know if any better approach to this is possible

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Re: (1/7)+(1/8)+(1/9) is in between?  [#permalink]

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New post 04 May 2017, 07:20
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MathRevolution wrote:
\((\frac{1}{7})+(\frac{1}{8})+(\frac{1}{9})\) is in between?

A. \((\frac{1}{6}) and (\frac{1}{5})\)
B. \((\frac{1}{5}) and (\frac{1}{4})\)
C. \((\frac{1}{4}) and (\frac{1}{3})\)
D. \((\frac{1}{3}) and (\frac{1}{2})\)
E. \((\frac{1}{2}) and 1\)


Let's examine EXTREME values.

If all 3 values were equal to 1/9 (the smallest value), then the sum would be 1/9 + 1/9 + 1/9 = 3/9 = 1/3
Of course, the 3 value are not equal to 1/9 (the smallest value), so the original sum must be greater than 1/3

If all 3 values were equal to 1/7 (the biggest value), then the sum would be 1/7 + 1/7 + 1/7 = 3/7 = 3/7
Of course, the 3 value are not equal to 1/7 (the biggest value), so the original sum must be less than 3/7

So, the sum must be BETWEEN 1/3 and 3/7
Hmmm, this is not among the answer choices.

HOWEVER, since 3/7 is less than 1/2, we can also say that the sum must be BETWEEN 1/3 and 1/2

Answer:

Cheers,
Brent
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Re: (1/7)+(1/8)+(1/9) is in between?  [#permalink]

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New post 04 May 2017, 07:22
GMATPrepNow wrote:
MathRevolution wrote:
\((\frac{1}{7})+(\frac{1}{8})+(\frac{1}{9})\) is in between?

A. \((\frac{1}{6}) and (\frac{1}{5})\)
B. \((\frac{1}{5}) and (\frac{1}{4})\)
C. \((\frac{1}{4}) and (\frac{1}{3})\)
D. \((\frac{1}{3}) and (\frac{1}{2})\)
E. \((\frac{1}{2}) and 1\)


Let's examine EXTREME values.

If all 3 values were equal to 1/9 (the smallest value), then the sum would be 1/9 + 1/9 + 1/9 = 3/9 = 1/3
Of course, the 3 value are not equal to 1/9 (the smallest value), so the original sum must be greater than 1/3

If all 3 values were equal to 1/7 (the biggest value), then the sum would be 1/7 + 1/7 + 1/7 = 3/7 = 3/7
Of course, the 3 value are not equal to 1/7 (the biggest value), so the original sum must be less than 3/7

So, the sum must be BETWEEN 1/3 and 3/7
Hmmm, this is not among the answer choices.

HOWEVER, since 3/7 is less than 1/2, we can also say that the sum must be BETWEEN 1/3 and 1/2

Answer:

Cheers,
Brent

Nice approach..

Thanks Brent.

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Re: (1/7)+(1/8)+(1/9) is in between?  [#permalink]

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New post 04 May 2017, 09:01
MathRevolution wrote:
\((\frac{1}{7})+(\frac{1}{8})+(\frac{1}{9})\) is in between?

A. \((\frac{1}{6}) and (\frac{1}{5})\)
B. \((\frac{1}{5}) and (\frac{1}{4})\)
C. \((\frac{1}{4}) and (\frac{1}{3})\)
D. \((\frac{1}{3}) and (\frac{1}{2})\)
E. \((\frac{1}{2}) and 1\)



\(\frac{1}{7} + \frac{1}{8} + \frac{1}{9}\)

\(= \frac{72 + 63 + 56}{504}\)

\(= \frac{191}{504}\)

\(= 0.38\)

Thus, from the given options it is evident that the value ranges from \((\frac{1}{3}) to (\frac{1}{2})\) , hence answer must be (D)
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Re: (1/7)+(1/8)+(1/9) is in between?  [#permalink]

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New post 04 May 2017, 12:53
(17)+(18)+(19)(17)+(18)+(19) is in between?

A. (16)and(15)(16)and(15)
B. (15)and(14)(15)and(14)
C. (14)and(13)(14)and(13)
D. (13)and(12)(13)and(12)
E. (12)and1

+1 d..
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Re: (1/7)+(1/8)+(1/9) is in between?  [#permalink]

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New post 05 May 2017, 20:57
Nice approach by Brent.
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Re: (1/7)+(1/8)+(1/9) is in between?  [#permalink]

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New post 05 May 2017, 21:34
1
MathRevolution wrote:
\((\frac{1}{7})+(\frac{1}{8})+(\frac{1}{9})\) is in between?

A. \((\frac{1}{6}) and (\frac{1}{5})\)
B. \((\frac{1}{5}) and (\frac{1}{4})\)
C. \((\frac{1}{4}) and (\frac{1}{3})\)
D. \((\frac{1}{3}) and (\frac{1}{2})\)
E. \((\frac{1}{2}) and 1\)


\(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}> \frac{3}{9}=\frac{1}{3}\)

\(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}< \frac{3}{7} <\frac{1}{2}\)

The answer is D.
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Re: (1/7)+(1/8)+(1/9) is in between?  [#permalink]

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New post 07 May 2017, 18:02
1
==>The sum of consecutive reciprocal numbers is decided by the first and the last number. In other words, you get
(1/9)+(1/9)+(1/9)< (1/7)+(1/8)+(1/9)< (1/7)+(1/7)+(1/7), and if you reorganize this, from
=1/3=3/9=(1/9)+(1/9)+(1/9)<(1/7)+(1/8)+(1/9)<(1/7)+(1/7)+(1/7)=3/7<3/6=1/2, you get between (1/3) and (1/2).

The answer is D.
Answer: D
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(1/7)+(1/8)+(1/9) is in between?  [#permalink]

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New post 10 Jul 2017, 05:50
Interesting question - can someone share more like these? I felt it was going beyond just series, but more about interpreting the question language as well.Thank you!
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(1/7)+(1/8)+(1/9) is in between?   [#permalink] 10 Jul 2017, 05:50
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