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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 03 Oct 2004, 07:27
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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
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 [#permalink] New post 03 Oct 2004, 07:41
C. 1/5

n=(4)*(K/20), where (K/20) = AM of the 20 numbers.
=> n= 1/5.
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 [#permalink] New post 03 Oct 2004, 08:16
B. 1/6.

1.X+n = 21Y

2. also, n=4X/20 = (1/5)X

substituting 2 in 1 => X+X/5 = 21Y =>n=(21/6)Y

we need to find n/21Y =?

=>(21/6)Y/21Y = 1/6
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 [#permalink] New post 03 Oct 2004, 09:36
Either yu r posting questions or yu r the first one to answer them venkune :).
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Re: PS: A.M [#permalink] New post 08 Oct 2004, 10:26
linker 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


Let X be the sum of the other 20, then it means that n= 4X/20 = X/5
Total of 21 numbers then become X + X/5 = 6x/5

Fraction of n = X/5 divided by 6x/5 = 1/6
Re: PS: A.M   [#permalink] 08 Oct 2004, 10:26
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