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

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Manager
Joined: 03 Oct 2008
Posts: 61
A certain list consists of 21 different numbers. If n is in [#permalink]

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05 Oct 2008, 15:51
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 :
Manager
Joined: 30 Sep 2008
Posts: 111

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05 Oct 2008, 19:51
I go for C

Mean = n/4
=> Sum = 20*n/4 = 5n

Ratio = n / (5n) = 1/5
SVP
Joined: 17 Jun 2008
Posts: 1529

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05 Oct 2008, 23:12
B for me.

If x is the mean of the remaining 20 numbers than n = 4x.

Sum of all the 21 numbers = 20x + n = 24x

Hence, the ratio = 4x/24x = 1/6.
Retired Moderator
Joined: 05 Jul 2006
Posts: 1747

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06 Oct 2008, 02:08
albany09 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 :

sum of 20 = x , sum of 21 = x+n

n = x/5 , sum of 21 = x+x/5 = (5x+x)/5 = 6x/5

(x/5)/ 6x/5 = 1/6
Re: math   [#permalink] 06 Oct 2008, 02:08
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# A certain list consists of 21 different numbers. If n is in

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