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A certain list consists of 21 different numbers. If n is in

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A certain list consists of 21 different numbers. If n is in [#permalink] New post 31 Oct 2010, 03:06
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A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average(arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?

(A) 1/20
(B) 1/6
(C) 1/5
(D) 4/21
(E) 5/21
[Reveal] Spoiler: OA

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Last edited by monirjewel on 31 Oct 2010, 04:06, edited 1 time in total.
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Re: A certain list consists [#permalink] New post 31 Oct 2010, 07:08
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monirjewel wrote:
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average(arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?

(A) 1/20
(B) 1/6
(C) 1/5
(D) 4/21
(E) 5/21


Given: average_{20}=\frac{sum_{20}}{20}=\frac{n}{4} --> sum_{20}=5n --> sum_{21}=sum_{20}+n=5n+n=6n;

Question: n is what fraction of the sum of the 21 numbers in the list?
\frac{n}{sum_{21}}=\frac{n}{6n}=\frac{1}{6}.

Answer: B.
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Re: A certain list consists [#permalink] New post 09 Dec 2010, 04:03
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monirjewel wrote:
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average(arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?

(A) 1/20
(B) 1/6
(C) 1/5
(D) 4/21
(E) 5/21


When you have your answer in percentages/fractions and there are no values in the data given to you, assume values.
Here you do not have the value of any number. So assume 20 numbers are all 1. Their average is 1 and sum is 20.
Then n is 4 and sum of 21 numbers is 24.
4/24 = 1/6
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Re: A certain list consists [#permalink] New post 09 Apr 2013, 08:02
VeritasPrepKarishma wrote:
monirjewel wrote:
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average(arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?

(A) 1/20
(B) 1/6
(C) 1/5
(D) 4/21
(E) 5/21


When you have your answer in percentages/fractions and there are no values in the data given to you, assume values.
Here you do not have the value of any number. So assume 20 numbers are all 1. Their average is 1 and sum is 20.
Then n is 4 and sum of 21 numbers is 24.
4/24 = 1/6


For this solution, I thought all numbers have to be different?
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Re: A certain list consists [#permalink] New post 09 Apr 2013, 22:18
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gmatfighter12 wrote:
VeritasPrepKarishma wrote:
monirjewel wrote:
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average(arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?

(A) 1/20
(B) 1/6
(C) 1/5
(D) 4/21
(E) 5/21


When you have your answer in percentages/fractions and there are no values in the data given to you, assume values.
Here you do not have the value of any number. So assume 20 numbers are all 1. Their average is 1 and sum is 20.
Then n is 4 and sum of 21 numbers is 24.
4/24 = 1/6


For this solution, I thought all numbers have to be different?


It usually doesn't matter if the numbers don't need to be integers.
The numbers could be 0.999999999999999, 0.99999998, 1.00000000000000001, 1.000000000002 etc. Effectively, they are all 1 and our solution doesn't change at all.
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A certain list consists of 21 different numbers. If n is in the [#permalink] New post 04 May 2013, 12:22
udaymathapati wrote:
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average (arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?
A. 201
B. 61
C. 51
D. 214
E. 215


Umm 1/6 does seem the answer...udaymathapati could u please confirm the answer and change the choices? well 201 or 61 are not really the proper fractions the question expects.

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Re: A certain list consists of 21 different numbers. If n is in  [#permalink] New post 05 May 2013, 03:40
Expert's post
arpanpatnaik wrote:
udaymathapati wrote:
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average (arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?
A. 201
B. 61
C. 51
D. 214
E. 215


Umm 1/6 does seem the answer...udaymathapati could u please confirm the answer and change the choices? well 201 or 61 are not really the proper fractions the question expects.

Regards,
Arpan


Merging topics. Check here: a-certain-list-consists-of-21-different-numbers-if-n-is-in-103995.html#p810467
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Re: A certain list consists of 21 different numbers. If n is in [#permalink] New post 04 Nov 2013, 08:00
This Qs appeared in the newly launched GMAT Exam Prep -1.
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Re: A certain list consists [#permalink] New post 04 Nov 2013, 14:34
VeritasPrepKarishma wrote:
monirjewel wrote:
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average(arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?

(A) 1/20
(B) 1/6
(C) 1/5
(D) 4/21
(E) 5/21


When you have your answer in percentages/fractions and there are no values in the data given to you, assume values.
Here you do not have the value of any number. So assume 20 numbers are all 1. Their average is 1 and sum is 20.
Then n is 4 and sum of 21 numbers is 24.
4/24 = 1/6


Waoo thank you! It is a really usufull method!

Thank you!
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Re: A certain list consists of 21 different numbers. If n is in [#permalink] New post 05 Nov 2013, 08:13
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monirjewel wrote:
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average(arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?

(A) 1/20
(B) 1/6
(C) 1/5
(D) 4/21
(E) 5/21


This is a PS question placed in the DS forum. Please relocate accordingly.
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Re: A certain list consists of 21 different numbers. If n is in [#permalink] New post 05 Nov 2013, 08:22
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gdediegoi wrote:
monirjewel wrote:
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average(arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?

(A) 1/20
(B) 1/6
(C) 1/5
(D) 4/21
(E) 5/21


This is a PS question placed in the DS forum. Please relocate accordingly.

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Done. Thank you.
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COLLECTION OF QUESTIONS:
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Re: A certain list consists of 21 different numbers. If n is in [#permalink] New post 30 Dec 2013, 13:17
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monirjewel wrote:
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average(arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?

(A) 1/20
(B) 1/6
(C) 1/5
(D) 4/21
(E) 5/21


n = 4x

where 'x' is the average of other 20 numbers

Then sum of other 20 numbers is 20x

So we get 4x/24x = 1/6

B is our answer

Cheers!
J :)
Re: A certain list consists of 21 different numbers. If n is in   [#permalink] 30 Dec 2013, 13:17
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