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

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31 Oct 2010, 04:06

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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?

Helpful Geometry formula sheet: http://gmatclub.com/forum/best-geometry-93676.html I hope these will help to understand the basic concepts & strategies. Please Click ON KUDOS Button.

Last edited by monirjewel on 31 Oct 2010, 05:06, edited 1 time in total.

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 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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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?

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]

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04 May 2013, 13:22

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.

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.

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

Re: A certain list consists of 21 different numbers. If n is in [#permalink]

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05 Nov 2013, 09:13

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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.

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.

Re: A certain list consists of 21 different numbers. If n is in [#permalink]

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30 Dec 2013, 14:17

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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?

Re: A certain list consists of 21 different numbers. If n is in [#permalink]

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23 Nov 2015, 22:16

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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

Given: 21 numbers in a list including n. n = 4 times the average of other 20 numbers. Assume the average to be x n = 4x - (i)

Average = Sum/20 Sum of other 20 numbers = 20x - (ii)

n/Sum of other 21 numbers = \(\frac{4x}{(4x + 20x)} = \frac{4x}{24x} = 1/6\) Option B
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A certain list consists of 21 different numbers [#permalink]

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02 Jun 2016, 19:08

(couldn't find this problem by using the search bar. Please add to existing post if it already exists.)

A certain list consists of 21 different numbers. If n is on the list and if n is 4 times the average (arithmetic mean) of the other 20 numbers of the list then n is what fraction of the sum of the 21 numbers on the list.

A. \(\frac{1}{20}\) B. \(\frac{1}{6}\) C.\(\frac{1}{5}\) D. \(\frac{4}{21}\) E. \(\frac{5}{21}\)

I don't have the slightest idea how to even get started with this problem.... Help anyone? Please?

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