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Number Problem

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redwine01
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Number Problem [#permalink] New post   31 Aug 2011, 07:36
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A
B
C
D
E

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(N/A)

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100% (01:06) correct 0% (00:00) wrong based on 2 sessions
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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 of the other 20 numbers in the list, then n is what fraction of the sum of the 21 number in the list?

A)1/20 B)4/21 C)1/6 d)1/5 e)5/21



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C
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agdimple333
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Re: Number Problem [#permalink] New post   31 Aug 2011, 07:59
redwine01 wrote:
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average of the other 20 numbers in the list, then n is what fraction of the sum of the 21 number in the list?

A)1/20 B)4/21 C)16 d)1/5 e)5/21


the answer that i am getting is not in the options..

n = 4 \(\frac{sum of 20 numbers}{20}\)

n = sum of 20 numbers / 5

5n = sum of 20 numbers
5n+n = sum of 20 numbers + n
6n = sum of all 21 numbers
so n is 1/6 fraction of the sum of 21 numbers.
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TomB
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Re: Number Problem [#permalink] New post   31 Aug 2011, 08:11
Let sum fo 20 numbers be X

Mean is X/20.
N=4(x/20)=X/5
sum of 21 numbers should be =X+X/5=6X/5

X/5/6X/5=1/6.
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Re: Number Problem [#permalink] New post   31 Aug 2011, 12:25
I found the simplest way to solve this problem is to plug in numbers.
For example: assume the sum of the 20 numbers is 100 ---> 100/20 = 5 (average of 20 #'s)
Next, n is equal to 4 times the average of the twenty numbers: n = 4*5 = 20

Next, the sum of n (20) and the twenty numbers (100) is equal to: 20 + 100 = 120.

n is what fraction of the twenty-one numbers???

Since our n equals 20, and the sum of the 21 numbers is equal to 120, 20/120 ----> 1/6 Answer C
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Re: Number Problem [#permalink] New post   31 Aug 2011, 14:35
Let Sum of 20 number = x
so avg of those 20 number = x/20

We are given that

4 * (x/20) = n
so, n = x/5

sum of all 21 numbers = x + n = x + x/5 = 6x/5

faction of n is what faction of sum of 21

(x/5) / (6x/5) = 1/6
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Re: Number Problem [#permalink] New post   31 Aug 2011, 16:37
let N1,N2.....N21 be the 21 number sequence.

and lets say N21 = n

given n = 4(N1+N2+......N20)/20

=> n = (N1+N2+......N20)/5

=> 5n = N1+N2+.....+N20

adding N21 on both sides

=> 5n +N21 = N1+N2+.....N21

=> 6n = N1+N2+.....N21

=> n = (1/6)(N1+N2+...N21)

Answer is C.



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Re: Number Problem [#permalink]
31 Aug 2011, 16:37
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