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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, 05:06, edited 1 time in total.

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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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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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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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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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New post 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.

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New post 04 Nov 2013, 09:00
This Qs appeared in the newly launched GMAT Exam Prep -1.
GMAC said these Qs were never seen before. What a blatant lie!!!

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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

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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: 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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This is how I used to calculate which I think works pretty well:

if you let the average of the 20 other numbers equal a, can you write this equation for sum of the list (S)

n + 20a = S

the question tells us that

n = 4a

plug this back into the first equation and you get that the sum is 24a

4a + 20a = 24a

therefore fraction of n to the total would be

4a/24a or 1/6

answer B

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New post 04 Mar 2016, 16:54
It is always easier for me to plug-in numbers: Let's say that average of 20 numbers is 10 and each of these 20 numbers is 10. So n term would be 40.

Now the sum of all 21 terms is 20*10+40=240 and the ratio of n to sum of all 21 terms is 40/240=1/6

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

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New post 23 Mar 2016, 11:37
I tried to set up an equation and stared at the screen for 1.30 mins. My last try was to use some values and boy it worked :lol:

n= 4(avg of 20#)
let's assume, the avg is 50
n=4(50)
n=200
50(avg of 20#)*20(# in the list) = 1000 + 200(value of n) = 1200

\(\frac{200}{1200}\)

\(\frac{1}{6}\)

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Re: A certain list consists of 21 different numbers [#permalink]

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New post 02 Jun 2016, 19:20
its B

consider sum of 20 numbers = T

so n= 4*(T/20)=T/5.

Therefore total sum of 21 numbers = T/5+T=6T/5

Now n is what part of total sum of 21 numbers = value of n/value of total 21 numbers=(T/5) / (6T/5) = 1/6

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

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New post 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?

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



Suppose sum of 20 nos. (excluding n) is S

Then

n = 4 * (\(\frac{S}{20}\))

n = \(\frac{S}{5}\))
5n = S

If we add n to S, we get sum of all 21 nos.

5n+n = Sum of all 21 Nos.

6n = Sum of all 21 nos.

n = (1/6) S.

B is the answer.

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

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New post 03 Jun 2016, 10:28
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

Final Answer:
[Reveal] Spoiler:
B


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

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New post 03 Jun 2016, 13:48
s=sum of 21 numbers on list
(s-n)/20=n/4
n/s=4/24=1/6

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

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New post 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.

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

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New post 06 Jun 2016, 11:23
Hi Silviax,

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.

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

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New post 06 Jun 2016, 15:47
Yes you are right Rich. I misunderstood what you wrote.

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

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New post 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

Final Answer:
[Reveal] Spoiler:
B


GMAT assassins aren't born, they're made,
Rich


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

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Re: A certain list consists of 21 different numbers. If n is in   [#permalink] 11 Jun 2016, 21:47

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