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

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

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12 Jun 2016, 05:23
He just picked a random number for the average of the numbers which is 10 and then multiplied it my 20 to get the sum, then added the 21st number which is 4 times as big as the average of the other numbers.

vipul1702 wrote:
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

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]

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12 Jun 2016, 09:17
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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) 1/20
(B) 1/6
(C) 1/5
(D) 4/21
(E) 5/21

Let the total of 20 numbers other than 'n' = t

So, n= 4t/20= t/5

total including n= n + t= t/5 +t= 6t/5

ratio of n/total= 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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23 Mar 2017, 16:13
List has 21 numbers, of which n is one.

Setting m as the average of the 20 numbers:

4m/(20m+4m)=4/24=1/6
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Re: A certain list consists of 21 different numbers. If n is in  [#permalink]

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27 Mar 2017, 17:31
1
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

We can let x = the sum of the 21 numbers. Thus, x/21 = the average of the 21 numbers and (x - n)/20 = the average of the 20 numbers when n is removed from the list. Since n is 4 times the average of the other 20 numbers in the list:

n = 4(x - n)/20

n = (x - n)/5

5n = x - n

6n = x

n = (1/6)x

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

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09 Sep 2017, 16:10
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

t=total of 21 numbers
(t-n)/20=n/4
n/t=1/6
B
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Re: A certain list consists of 21 different numbers. If n is in  [#permalink]

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11 Nov 2017, 13:47
VeritasPrepKarishma - Going by your method, assuming if it would have been given in the premise as different integers? Would this method not be suitable to go forward with?

I did try and solve by taking consecutive integers from 1 to 21 but was not able to solve.

P.S Repeatedly kept on trying 6 times but did not succeed.
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Re: A certain list consists of 21 different numbers. If n is in  [#permalink]

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11 Nov 2017, 21:41
2
siddyj94 wrote:
VeritasPrepKarishma - Going by your method, assuming if it would have been given in the premise as different integers? Would this method not be suitable to go forward with?

I did try and solve by taking consecutive integers from 1 to 21 but was not able to solve.

P.S Repeatedly kept on trying 6 times but did not succeed.

It will work for any set of values satisfying all given constraints.

You can assume 20 different integers as 1, 2, 3...20.
Their mean will be the average of middle two numbers 10 and 11 so it will be 10.5.
The 21st number then will be 4*10.5 = 42
Sum of all 21 numbers = 20*21/2 + 42 = 252

Required fraction = 42/252 = 1/6
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Re: A certain list consists of 21 different numbers. If n is in  [#permalink]

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11 Nov 2017, 22:01
Thanks Karishma!
For 6 times i just re-checked my calculations were wrong!

Thanks!

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

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11 Nov 2017, 22:24
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

We need to find n/(sum of the 20 numbers + n)
Sum of the 20 numbers = 20*Ave
We have Ave=n/4
So sum of 20 numbers = 20*(n/4) = 5n
n/(sum of the 20 numbers +n)=n/(5n+n) = 1/6
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Re: A certain list consists of 21 different numbers. If n is in  [#permalink]

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21 Apr 2018, 07:04
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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

A quick solution here is to plug in some values that meet the given criteria.

Aside: I'm going to ignore the part about the numbers being different, since I have a feeling that this is the author's way of eliminating the possibility that all of the values equal zero (which would ruin the question).

So, the first 20 values (excluding n) could all equal 1, in which case their average (mean) would be 1.
Since n is 4 times the average, n would equal 4.
So, the sum of all 21 values is 24.

Question: n is what fraction of the sum of the 21 numbers in the list?
= 1/6

IMPORTANT: Keep in mind that I broke the rule about all of the numbers being different. What's important here is that we COULD replace the 1's with 20 different numbers that still have a mean of 1, in which case the sum of the first 20 numbers would still be 20, which means the answer will remain B.

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

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08 Sep 2018, 10:37
Bunuel 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?

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?
$$\frac{n}{sum_{21}}=\frac{n}{6n}=\frac{1}{6}$$.

For the part highlighted why would this be n/4 versus 4n, given the problem states that n is equal to 4x(Avg of 20 other numbers)?
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Re: A certain list consists of 21 different numbers. If n is in  [#permalink]

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09 Jan 2019, 01:45
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= (4*(sum of 20 no))/ 20
or say sum of 20 no= x
x/5= n

so , x/5/ ( x/5+x)
; x/5 * 5/6x ; 1/6 IMO B
Re: A certain list consists of 21 different numbers. If n is in   [#permalink] 09 Jan 2019, 01:45

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