Average question that i do not understand

[jfranco23]

A certain list consist 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

[Ian Stewart]

A certain list consist 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 have a list of 21 numbers, one of which is n. Let's say the other 20 numbers (not counting n) add up to some number S. The question tells us that n is four times the average of these 20 numbers:

n = 4*(S/20) = S/5

So n is 1/5 of the sum of the other 20 numbers. That's not quite what the question asks for, however. The question asks "n is what fraction of the sum of the 21 numbers in the list?" The sum of all 21 numbers is equal to

S+n = S + S/5 = 6S/5

We need to work out what fraction n is of this sum; that is, we need to divide n by 6S/5. Since n = S/5, we have:

(S/5)/(6S/5) = 1/6

[logitech]

SUM20/20 = x = av
N = 4x

So = 4x/(4+20)x = 1/6