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Manager  Joined: 14 Jun 2012
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Re: A certain list consists of 21 different numbers. If n is in  [#permalink]

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1
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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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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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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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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A certain list consists of 21 different numbers. If n is in  [#permalink]

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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 s=sum of 21 numbers
n=4(s-n)/20→
n/s=4/24=1/6
B

Originally posted by gracie on 09 Sep 2017, 15:10.
Last edited by gracie on 14 Feb 2020, 10:35, edited 1 time in total.
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Re: A certain list consists of 21 different numbers. If n is in  [#permalink]

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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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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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Thanks Karishma!
For 6 times i just re-checked my calculations were wrong!

Thanks! Sent from my iPhone using GMAT Club Forum
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A certain list consists of 21 different numbers. If n is in  [#permalink]

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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

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

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1. Average of 20 nos = Sum of 20 nos/20.

2. n= 4* Average of 20 nos.

n= 4* (Sum of 20 nos/20) = Sum of 20 nos/5.

5n = Sum of 20 nos.

3. Sum of 21 nos = Sum of 20 nos+n = 5n + n = 6n.

4. n = Sum of 21 nos/6.
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Re: A certain list consists of 21 different numbers. If n is in  [#permalink]

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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

We can let x = the sum of the 21 numbers. Thus, (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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Why can't we solve this using AP formulae. I am consistently getting (D) as the answer. Where am I going wrong ?

$$a_{11} (avg) = a_1 + 10d$$
$$n = 4(a_{11}) = 4(a_1 + 10d)$$
$$S (sum) = \frac{21}{2}(2a_1 + 20d)$$

Putting in any dummy values in resulting expression for $$\frac{n}{S}$$ will give (D) as the fraction.

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

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Why can't we solve this using AP formulae. I am consistently getting (D) as the answer. Where am I going wrong ?

$$a_{11} (avg) = a_1 + 10d$$
$$n = 4(a_{11}) = 4(a_1 + 10d)$$
$$S (sum) = \frac{21}{2}(2a_1 + 20d)$$

Putting in any dummy values in resulting expression for $$\frac{n}{S}$$ will give (D) as the fraction.

Hi
As you yourself have mentioned the word AP, so you cannot use the formula here as it is only for AP.
An AP is set where all terms differ by a common number from the previous element.

Here it is not given that it is an AP, that is Arithmetic Progression.
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Re: A certain list consists of 21 different numbers. If n is in  [#permalink]

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chetan2u wrote:
Hi
As you yourself have mentioned the word AP, so you cannot use the formula here as it is only for AP.
An AP is set where all terms differ by a common number from the previous element.

Here it is not given that it is an AP, that is Arithmetic Progression.

Thanks for the response.

The solutions suggested above where Total = Avg x Number of values, doesn't that signify a set of numbers which is evenly spaced or AP ?
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Re: A certain list consists of 21 different numbers. If n is in  [#permalink]

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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?

let,the average(arithmetic mean) of the other 20 numbers in the list = a
so, sum of the other 20 numbers in the list = 20a
n=4a
now,sum of the other 21 numbers in the list = 20a+4a = 24a
so,n is 4a/24a=1/6 fraction of the sum of the 21 numbers in the list.

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

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chetan2u wrote:

Why can't we solve this using AP formulae. I am consistently getting (D) as the answer. Where am I going wrong ?

$$a_{11} (avg) = a_1 + 10d$$
$$n = 4(a_{11}) = 4(a_1 + 10d)$$
$$S (sum) = \frac{21}{2}(2a_1 + 20d)$$

Putting in any dummy values in resulting expression for $$\frac{n}{S}$$ will give (D) as the fraction.

Hi
As you yourself have mentioned the word AP, so you cannot use the formula here as it is only for AP.
An AP is set where all terms differ by a common number from the previous element.

Here it is not given that it is an AP, that is Arithmetic Progression.

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

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Why can't we solve this using AP formulae. I am consistently getting (D) as the answer. Where am I going wrong ?

$$a_{11} (avg) = a_1 + 10d$$
$$n = 4(a_{11}) = 4(a_1 + 10d)$$
$$S (sum) = \frac{21}{2}(2a_1 + 20d)$$

Putting in any dummy values in resulting expression for $$\frac{n}{S}$$ will give (D) as the fraction.

You are not given that the numbers in the list are in arithmetic progression. You are only given that n is 4 times the avg of other 20 numbers.
The numbers could be 1, 5, 6, 6, 6, 6, ... 6, 12
with an average of 6. Then n would be 24.

1, 5, 6, 6, 6, 6, 6, 6, .. 6, 12, 24 is not an AP.

Also, you are given that n = 4 * avg of other 20 numbers.
Even if this were an AP in which n were the last term, a11 is not the avg of first 20 numbers. The avg of first 20 numbers would be avg of 10th and 11th numbers.
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Veritas Prep GMAT Instructor Re: A certain list consists of 21 different numbers. If n is in   [#permalink] 29 May 2020, 00:12

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