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A certain list consists of 21 different numbers

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by rsarashi » Sat Sep 02, 2017 10:34 pm
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

OAB
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by Alv » Sun Sep 03, 2017 12:32 am
Solution:
Let S be the sum of the 21 different numbers.
Then (S - n) / 20 is the average of the other 20 numbers in the list.
Hence,

n = 4 ((S - n) / 20) = (S - n) / 5.

Solving for n:

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

Therefore, n is 1 / 6 of the sum (S) of the 21 numbers in the list. ANS: B) 1/6
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by GMATGuruNY » Sun Sep 03, 2017 2:45 am
rsarashi wrote: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
Let the average of the other 20 numbers = 1, implying that the sum of the other 20 numbers = (count)(average) = 20*1 = 20.
Since n is equal to 4 times the average, n = 4*1 = 4.
n/(sum of all 21 numbers) = 4/(20 + 4) = 4/24 = 1/6.

The correct answer is B.
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by Brent@GMATPrepNow » Sun Sep 03, 2017 5:47 am
rsarashi wrote: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

OAB
Notice that it doesn't really matter what the first 20 numbers are.
For example, we can arbitrarily say that the average of the first 20 numbers is 1 or 10 or 113.5 etc and the go from there, since it will always be possible to find 20 different numbers that have a given average.

So, let's just say that the average of the first 20 numbers is 2 (since Mitch already used 1 as his average)
This means the SUM of the first 20 numbers is 40

GIVEN: n is 4 times the average(arithmetic mean) of the other 20 numbers in the list
So, n = (4)(2) = 8

n is what fraction of the sum of the 21 numbers in the list?
Sum = (sum of first 20 numbers) + n
= 40 + 8 = 48
So, the fraction = 8/48 = 1/6

Answer: B
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by [email protected] » Sun Sep 03, 2017 11:52 am
Hi rsarashi,

Brent's first explanation took a clever approach with the question. In the event that you don't "see" things the way that Brent explained them, you can still follow the instructions and get the correct answer without too much trouble.

This is a perfect question for TESTing VALUES. We have a list of 21 different numbers, so let's choose...
1 through 20 and the number "N"

Since the first 20 values are consecutive, the sum = 10(21) = 210 and the average will be the "middle" of the group. Here, the average = 10.5

We're told that N = 4 times the average of the other 20 values, so N = 4(10.5) = 42

We're asked to determine the value of N/(sum of the list)

N = 42
Sum of the list = 210 + 42 = 252

42/252 might take a little work to reduce, but we can use the answer choices to our advantage....42(6) = 252, so 42/252 = 1/6

Final Answer: B

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Re: A certain list consists of 21 different numbers

by Scott@TargetTestPrep » Sat Feb 08, 2020 12:49 pm
rsarashi wrote: ↑
Sat Sep 02, 2017 10:34 pm
 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

OAB
We can let x = the sum of the 21 numbers. Thus, (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

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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