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by lukaswelker » Thu Apr 17, 2014 5:09 am
Hey Guys

I can't seem to find a way to solve this problem.

Here's the question,

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?

1/20 ; 1/6 ; 1/5 ; 4/21 ; 5/21

Any suggestions?
Many thanks
Lukas
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by Brent@GMATPrepNow » Thu Apr 17, 2014 5:13 am

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?
Answer: 4/24
[spoiler]= 1/6 = B[/spoiler]

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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by Brent@GMATPrepNow » Thu Apr 17, 2014 5:16 am
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average {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

Thanks
Here's an algebraic solution.

Let T = the sum of the 20 different numbers (EXCLUDING n)
So, the average (mean) of those 20 numbers is T/20
Since n is 4 times the average, we can see that n = 4(T/20)
Simplify to get n = 4T/20, or (even better) n = T/5

When we add n to the sum of the first 20 numbers we get: T + n
Since n = T/5, we can see that the sum of ALL 21 numbers = T + T/5
When we simplify T + T/5, we get: the sum of ALL 21 numbers = 6T/5

Question: n is what fraction of the sum of the 21 numbers in the list?
Answer: (T/5)/(6T/5)
Simplify to get [spoiler]1/6 = B[/spoiler]

Cheers,
Brent
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by Brent@GMATPrepNow » Thu Apr 17, 2014 5:19 am
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average {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
Choosing numbers that satisfy the given conditions is still a good way to go.

If we want to adhere to the rule about having DIFFERENT numbers, we could let the first 20 values (excluding n) be the integers from 1 to 20 inclusive.
Since these 20 numbers are all equally spaced, the mean will equal the average of the smallest and biggest values.
So, the mean = (1+20)/2 = 10.5

n is 4 times the average (arithmetic mean) of the other 20 numbers
So, n = (4)(10.5) = 42

Nice rule: the sum of the integers from 1 to k inclusive = k(k+1)/2.
So, the sum of the integers from 1 to 20 inclusive = (20)(20+1))/2 = 210

This means the sum of all 21 values = 210 + 42 = 252

Question: n is what fraction of the sum of the 21 numbers in the list?
Answer: 42/252
[spoiler]= 1/6 = B[/spoiler]

Cheers,
Brent
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by GMATGuruNY » Thu Apr 17, 2014 6:28 am
lukaswelker 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?

1/20 ; 1/6 ; 1/5 ; 4/21 ; 5/21
Slightly different line of reasoning.
The average of 20 different numbers can be ANY VALUE..
Strategy:
For the average of the 20 other numbers, plug in an easy value.

Let the average of the 20 other numbers = 1.
Then the sum of the 20 other numbers = (number)(average) = 20*1 = 20.
Since n is 4 times the average of the 20 other numbers, n = 4*1 = 4.
Thus:
Sum of all 21 numbers = 20+4 = 24.
n/sum = 4/24 = 1/6.

The correct answer is B.
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