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

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

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New post 05 May 2013, 04:40
Expert's post
arpanpatnaik wrote:
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.

Regards,
Arpan


Merging topics. Check here: a-certain-list-consists-of-21-different-numbers-if-n-is-in-103995.html#p810467
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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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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


Waoo thank you! It is a really usufull method!

Thank you!
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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.
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gdediegoi 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


This is a PS question placed in the DS forum. Please relocate accordingly.

________
Done. Thank you.
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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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shashanknitp wrote:
This Qs appeared in the newly launched GMAT Exam Prep -1.
GMAC said these Qs were never seen before. What a blatant lie!!!


I got this question on the Exam prep pack too, I have tagged this question accordingly...
I hope others find it helpful!
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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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Re: A certain list consists of 21 different numbers. If n is in [#permalink]

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New post 17 Dec 2015, 15:26
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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A certain list consists of 21 different numbers. If n is in [#permalink]

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

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New post 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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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
Re: A certain list consists of 21 different numbers   [#permalink] 02 Jun 2016, 19:20

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