A certain list consists of 21 different numbers. If n is in

n = 4x

where 'x' is the average of other 20 numbers

Then sum of other 20 numbers is 20x

So we get 4x/24x = 1/6

B is our answer

Cheers!
J

For this solution, I thought all numbers have to be different?

It usually doesn't matter if the numbers don't need to be integers.
The numbers could be 0.999999999999999, 0.99999998, 1.00000000000000001, 1.000000000002 etc. Effectively, they are all 1 and our solution doesn't change at all.

Given: 21 numbers in a list including n. n = 4 times the average of other 20 numbers.
Assume the average to be x
n = 4x - (i)

Average = Sum/20
Sum of other 20 numbers = 20x - (ii)

n/Sum of other 21 numbers = \(\frac{4x}{(4x + 20x)} = \frac{4x}{24x} = 1/6\)
Option B

Suppose sum of 20 nos. (excluding n) is S

Then

n = 4 * (\(\frac{S}{20}\))

n = \(\frac{S}{5}\))
5n = S

If we add n to S, we get sum of all 21 nos.

5n+n = Sum of all 21 Nos.

6n = Sum of all 21 nos.

n = (1/6) S.

B is the answer.

Hi Silviax,

This question can be solved by TESTing VALUES; you'll have to make sure to take the proper notes and label your work to finish it in an efficient way though.

We're told that a list consists of 21 different numbers (N and 20 others) and that N is equal to 4 times the AVERAGE of the other 20 numbers.

Let's TEST:
Average of 20 numbers = 10
Sum of those 20 numbers = 20(10) = 200
N = 4(10) = 40

We're asked to calculate the fraction N/(sum of all 21 numbers) = 40/(200+40) = 40/240 = 1/6

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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

VeritasPrepKarishma - Going by your method, assuming if it would have been given in the premise as different integers? Would this method not be suitable to go forward with?

I did try and solve by taking consecutive integers from 1 to 21 but was not able to solve.

P.S Repeatedly kept on trying 6 times but did not succeed.

It will work for any set of values satisfying all given constraints.

You can assume 20 different integers as 1, 2, 3...20.
Their mean will be the average of middle two numbers 10 and 11 so it will be 10.5.
The 21st number then will be 4*10.5 = 42
Sum of all 21 numbers = 20*21/2 + 42 = 252

Required fraction = 42/252 = 1/6

We need to find n/(sum of the 20 numbers + n)
Sum of the 20 numbers = 20*Ave
We have Ave=n/4
So sum of 20 numbers = 20*(n/4) = 5n
n/(sum of the 20 numbers +n)=n/(5n+n) = 1/6

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
= 1/6

Answer: B

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 [spoiler]B[/spoiler].

Cheers,
Brent

Bunuel EMPOWERgmatRichC VeritasKarishma chetan2u

Why can't we solve this using AP formulae. I am consistently getting (D) as the answer. Where am I going wrong ?

\(a_{11} (avg) = a_1 + 10d\)
\(n = 4(a_{11}) = 4(a_1 + 10d)\)
\(S (sum) = \frac{21}{2}(2a_1 + 20d)\)

Putting in any dummy values in resulting expression for \(\frac{n}{S}\) will give (D) as the fraction.

Please help.

Hi
As you yourself have mentioned the word AP, so you cannot use the formula here as it is only for AP.
An AP is set where all terms differ by a common number from the previous element.

Here it is not given that it is an AP, that is Arithmetic Progression.

You are not given that the numbers in the list are in arithmetic progression. You are only given that n is 4 times the avg of other 20 numbers.
The numbers could be 1, 5, 6, 6, 6, 6, ... 6, 12
with an average of 6. Then n would be 24.

1, 5, 6, 6, 6, 6, 6, 6, .. 6, 12, 24 is not an AP.

Also, you are given that n = 4 * avg of other 20 numbers.
Even if this were an AP in which n were the last term, a11 is not the avg of first 20 numbers. The avg of first 20 numbers would be avg of 10th and 11th numbers.

hi VeritasKarishma

Thanks for the this quick and effective way to solve.

Just to be sure this won't work if in the question it is given that the numbers are NOT same.
Any guide on how to assume numbers in this type of questions

Thanks

variables :
n,
a=ave of other 20 numbers,
s = sum of the 21 numbers
question :n/s

relationships:
n = 4*a
s= 20a +n

n/s= 4a/(20a+4a) =1/6

Note that the question here mentions that the numbers are distinct but they needn't be integers so we assume them all to be very very close to 1 just to make things easier in our head.
If the question were to talk about distinct integers and then their average and sum etc, one could just assume that the average is still 1 (numbers could just as well be ... -6, -4, -2, 0, 2, 4, 6, 8, ... such that 0 and 2 are the middle two numbers).
Since average of the 20 numbers is 1, their sum would be 20.
n would still be 4 and the sum of 21 numbers would be 24.
4/24 = 1/6