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by zagcollins » Sat Jul 19, 2008 7:03 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
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by pepeprepa » Sat Jul 19, 2008 8:48 am
S: the total sum of the 21 numbers, n included

You have n=4 x ( (S-n) / 20 )
Let's choose n=30
30=4x((S-30)/20)
30=S/5 - 6
36=S/5
S=180

And 30/180=1/6
-->B

You can also make it like that:
n=4 x ( (S-n) / 20 )
n/S=(4 x ( (S-n) / 20 ))/S
n/S=1/5 - n/5S
n/S x (1+1/5)=1/5
n/S x 6/5=1/5
n/S=1/6
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by egybs » Sat Jul 19, 2008 7:55 pm
The question can be solved very quickly by realizing that the sum of a set of numbers is equal to the mean of those numbers * the number of elements.

So the sum of (1,2,3,4,5,6,7,8,9) = 5 (the mean) * 9 (the number of elements).

So we're told that n is equal to 4 times the average of the other elements.

Let's call the average of those other elements x. So n = 4x.

So the sum of those numbers is 20x. The sum of all numbers is 20x + n = 24x.

We now need to find out n/(24x) = (4x)/(24x) which is equal to 1/6.

This is really just a 2 step problem when you know that piece of info.
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