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If n is an integer and 1/(n+1)<1/31+1/32+1/33<1/n, then what [#permalink]
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26 Jan 2012, 07:32
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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
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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]
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24 May 2013, 04:42



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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]
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25 May 2013, 09:13
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 <111/a < 3/31 ( or 1/10)..... hence a>10 from eq 1  n+1>a>10 .... hence n+1>10 ... n> 9Ans n=10
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Last edited by gmacforjyoab on 26 May 2013, 11:54, 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]
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25 May 2013, 09:36
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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 <111/a < 3/31 ( or 1/10)..... hence a>10 from eq 1  n+1>a>10 .... hence n+1>10 ... n> 9Ans 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]
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26 May 2013, 05:25
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]
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27 May 2013, 02:36
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 <111/a < 3/31 ( or 1/10)..... hence a>10 from eq 1  n+1>a>10 .... hence n+1>10 ... n> 9Ans 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]
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27 May 2013, 06:23
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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 <111/a < 3/31 ( or 1/10)..... hence a>10 from eq 1  n+1>a>10 .... hence n+1>10 ... n> 9Ans 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 Jyothi
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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]
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08 Aug 2013, 10:27
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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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: If n is an integer and 1/(n+1)<1/31+1/32+1/33<1/n, then what [#permalink]
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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]
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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]
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07 Oct 2016, 07:20
solved it the other way...and probably the fastest way... suppose we have 1/33 + 1/33 + 1/33 we have 3/33 or 1/11 since we have 1/31 and 1/32, logically, the result would be slightly more than 1/11. 10 works just fine... we have 1/n+1 => 1/11, and we have 1/10
10 works just fine!



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If n is an integer and 1/(n+1)<1/31+1/32+1/33<1/n, then what [#permalink]
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17 Oct 2016, 02:57
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 <111/a < 3/31 ( or 1/10)..... hence a>10 from eq 1  n+1>a>10 .... hence n+1>10 ... n> 9Ans n=10 Can someone pls explain the highlighted part. TIA




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