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A certain list

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by alex.gellatly » Tue Jun 05, 2012 6:01 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
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by Brent@GMATPrepNow » Tue Jun 05, 2012 7:17 am
alex.gellatly wrote: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
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
Last edited by Brent@GMATPrepNow on Wed Oct 30, 2013 6:03 am, edited 1 time in total.
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by Brent@GMATPrepNow » Tue Jun 05, 2012 7:27 am
alex.gellatly wrote: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 the 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


If 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 is 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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