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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?

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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.

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?

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

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03 Jun 2016, 01:53

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?

This question can be solved by TESTing VALUES; you'll have to make sure to take the proper notes and label your work to finish it in an efficient way though.

We're told that a list consists of 21 different numbers (N and 20 others) and that N is equal to 4 times the AVERAGE of the other 20 numbers.

Let's TEST: Average of 20 numbers = 10 Sum of those 20 numbers = 20(10) = 200 N = 4(10) = 40

We're asked to calculate the fraction N/(sum of all 21 numbers) = 40/(200+40) = 40/240 = 1/6

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

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05 Jun 2016, 12:09

But I think the problem says that the 21 numbers are all different numbers so they can't all be 10.

I think "test it" is in general a great strategy because it can transform most algebra problems into arithmetic problems which are much easier to solve -at least for me.

In my explanation, I did NOT state that the 20 numbers were all 10s - just that the AVERAGE of those numbers was 10. Whatever the individual numbers were, the SUM of those 20 numbers would be (20)(10) = 200.

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

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11 Jun 2016, 21:47

EMPOWERgmatRichC wrote:

Hi Silviax,

This question can be solved by TESTing VALUES; you'll have to make sure to take the proper notes and label your work to finish it in an efficient way though.

We're told that a list consists of 21 different numbers (N and 20 others) and that N is equal to 4 times the AVERAGE of the other 20 numbers.

Let's TEST: Average of 20 numbers = 10 Sum of those 20 numbers = 20(10) = 200 N = 4(10) = 40

We're asked to calculate the fraction N/(sum of all 21 numbers) = 40/(200+40) = 40/240 = 1/6

Sorry, I am a noob at this and my query could be a stupid one but can you please explain how did you conclude the following: Average of 20 numbers = 10 Sum of those 20 numbers = 20(10) = 200