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If n is an integer and 1/(n+1)<1/31+1/32+1/33<1/n, then what

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If n is an integer and 1/(n+1)<1/31+1/32+1/33<1/n, then what [#permalink] New post 26 Jan 2012, 06:32
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If n is an integer and \frac{1}{n+1}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{1}{n}, then what is the value of n?

A) 9
B) 10
C) 11
D) 12
E) 13
[Reveal] Spoiler: OA
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Re: n=? [#permalink] New post 26 Jan 2012, 08:31
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LM wrote:
If n is an integer and \frac{1}{n+1}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{1}{n}, then what is the value of n?

A) 9
B) 10
C) 11
D) 12
E) 13


Given: \frac{1}{n+1}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{1}{n}.

Now, obviously 3*(\frac{1}{33})<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<3*(\frac{1}{31}), as {3 times the least #}<{given sum}<{3 times the largest #} --> \frac{3}{33}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{3}{31} --> \frac{1}{11}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{3}{31}<\frac{3}{30} --> \frac{1}{10+1}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{3}{31}<\frac{1}{10} --> n=10.

Answer: B.
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Re: If n is an integer and 1/(n+1)<1/31+1/32+1/33<1/n, then what [#permalink] New post 24 May 2013, 03:42
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Re: If n is an integer and 1/(n+1)<1/31+1/32+1/33<1/n, then what [#permalink] New post 25 May 2013, 08:13
LM wrote:
If n is an integer and \frac{1}{n+1}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{1}{n}, then what is the value of n?

A) 9
B) 10
C) 11
D) 12
E) 13



Did it on similar grounds as Bunuel

1/(n+1) < ( 1/31 + 1/32 + 1/33) < 1/n
Substitute ( 1/31 + 1/32 + 1/33) to be 1/a
1/(n+1) < ( 1/a) < 1/n ......... ...................... hence n+1 > a > n--------------------------------- eq 1

1/a > 3/33 ( i.e 1/11) ... Hence a<11
from eq 1 --- n+1 >a>11 ................ n<a<11.. hence n <11

1/a < 3/31 ( or 1/10)..... hence a>10
from eq 1 --- n+1>a>10 .... hence n+1>10 ... n> 9

Ans n=10

Last edited by gmacforjyoab on 26 May 2013, 10:54, edited 1 time in total.
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Re: If n is an integer and 1/(n+1)<1/31+1/32+1/33<1/n, then what [#permalink] New post 25 May 2013, 08:36
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gmacforjyoab wrote:
LM wrote:
If n is an integer and \frac{1}{n+1}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{1}{n}, then what is the value of n?

A) 9
B) 10
C) 11
D) 12
E) 13



Did it on similar grounds as Bunuel

1/(n+1) < ( 1/31 + 1/32 + 1/33) < 1/n ...................... hence n+1 > a > n--------------------------------- eq 1
Substitute ( 1/31 + 1/32 + 1/33) to be 1/a
1/(n+1) < ( 1/a) < 1/n
1/a > 3/33 ( i.e 1/11) ... Hence a<11
from eq 1 --- n+1 >a>11 ................ n<a<11.. hence n <11

1/a < 3/31 ( or 1/10)..... hence a>10
from eq 1 --- n+1>a>10 .... hence n+1>10 ... n> 9

Ans n=10



I did it on the same grounds..

Answers is coming out to be 10.
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Re: If n is an integer and 1/(n+1)<1/31+1/32+1/33<1/n, then what [#permalink] New post 26 May 2013, 04:25
Approximate 1/31 + 1/32 + 1/33 = 0.09+

Now POE. B fits the inequality.

0.09 (1/11) < 0.09+ < 0.1 (1/10)
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Re: If n is an integer and 1/(n+1)<1/31+1/32+1/33<1/n, then what [#permalink] New post 27 May 2013, 01:36
gmacforjyoab wrote:
LM wrote:
If n is an integer and \frac{1}{n+1}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{1}{n}, then what is the value of n?

A) 9
B) 10
C) 11
D) 12
E) 13



Did it on similar grounds as Bunuel

1/(n+1) < ( 1/31 + 1/32 + 1/33) < 1/n
Substitute ( 1/31 + 1/32 + 1/33) to be 1/a
1/(n+1) < ( 1/a) < 1/n ......... ...................... hence n+1 > a > n--------------------------------- eq 1

1/a > 3/33 ( i.e 1/11) ... Hence a<11
from eq 1 --- n+1 >a>11 ................ n<a<11.. hence n <11

1/a < 3/31 ( or 1/10)..... hence a>10
from eq 1 --- n+1>a>10 .... hence n+1>10 ... n> 9

Ans n=10

hi gmacforjyoab,

I guess i am lacking some mathematics in the highlighted part. Could you please throw some light. It would be great help.

Regards
Atal Pandit
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Re: If n is an integer and 1/(n+1)<1/31+1/32+1/33<1/n, then what [#permalink] New post 27 May 2013, 05:23
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atalpanditgmat wrote:
gmacforjyoab wrote:
LM wrote:
If n is an integer and \frac{1}{n+1}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{1}{n}, then what is the value of n?

A) 9
B) 10
C) 11
D) 12
E) 13



Did it on similar grounds as Bunuel

1/(n+1) < ( 1/31 + 1/32 + 1/33) < 1/n
Substitute ( 1/31 + 1/32 + 1/33) to be 1/a
1/(n+1) < ( 1/a) < 1/n ......... ...................... hence n+1 > a > n--------------------------------- eq 1

1/a > 3/33 ( i.e 1/11) ... Hence a<11
from eq 1 --- n+1 >a>11 ................ n<a<11.. hence n <11

1/a < 3/31 ( or 1/10)..... hence a>10
from eq 1 --- n+1>a>10 .... hence n+1>10 ... n> 9

Ans n=10

hi gmacforjyoab,

I guess i am lacking some mathematics in the highlighted part. Could you please throw some light. It would be great help.

Regards
Atal Pandit


Since (1/n+1) < 1/a < 1/n , we can say that n+1 >a > n
( when u take the reciprocal of two numbers in an Inequality , the inequality flips )
Consider this ---- 1/4<1/3<1/2 , which would mean 4>3>2 ...

Oh and lets say - all the numbers were 1/33 , then the sum would be 3/33 , but all the numbers are not 1/33 , the other two numbers are 1/32 and 1/31 . and these two numbers are greater than 1/33 , hence the sum of 1/31 +1/32 + 1/33 would also be grater than 3/33
hence , 1/a > 3/33 i.e 1/11

HTH
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Re: If n is an integer and 1/(n+1)<1/31+1/32+1/33<1/n, then what [#permalink] New post 08 Aug 2013, 09:27
1/n> 1/31+1/32+1/33> 1/33+ 1/33 + 1/33 = 3/33 = 1/11 ====> n<11

1/(n+1)< 1/31 + 1/32+ 1/33 < 1/31 + 1/31 + 1/31 = 3/31 ====> n>9,3

Then, n=10.

B.
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Re: n=? [#permalink] New post 09 Aug 2013, 04:11
Bunuel wrote:
LM wrote:
If n is an integer and \frac{1}{n+1}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{1}{n}, then what is the value of n?

A) 9
B) 10
C) 11
D) 12
E) 13


Given: \frac{1}{n+1}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{1}{n}.

Now, obviously 3*(\frac{1}{33})<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<3*(\frac{1}{31}), as {3 times the least #}<{given sum}<{3 times the largest #} --> \frac{3}{33}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{3}{31} --> \frac{1}{11}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{3}{31}<\frac{3}{30} --> \frac{1}{10+1}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{3}{31}<\frac{1}{10} --> n=10.

Answer: B.

...............

Amazing solution..... glad to learn this.....
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Re: n=?   [#permalink] 09 Aug 2013, 04:11
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