Problem Solving

imane81 wrote:Your help would be much appreciated on this one...Thank you

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

nataras wrote:How should we know that it's okay to break the rule of "different numbers"?

It's hard to think of every instance in which we can ignore/break/bend rules. However, in this particular case, I know that I COULD meet the given condition by replacing the twenty 1's with twenty different numbers that still have a mean of 1. So, we might say that I'm simply bending the rule in order to make things easier on myself.

Cheers,
Brent

nataras wrote: Is there another way to solve?

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

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

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

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

imane81 wrote:Your help would be much appreciated on this one...Thank you

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

We can let x = the sum of the 21 numbers. Thus, x/21 = the average of the 21 numbers and (x - n)/20 = the average of the 20 numbers when n is removed from the list. Since n is 4 times the average of the other 20 numbers in the list:

n = 4(x - n)/20

n = (x - n)/5

5n = x - n

6n = x

n = (1/6)x

Answer: B