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
A certain list consists of 21 different numbers
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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
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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Let the average of the other 20 numbers = 1, implying that the sum of the other 20 numbers = (count)(average) = 20*1 = 20.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
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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Notice that it doesn't really matter what the first 20 numbers are.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
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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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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Rich
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
GMAT assassins aren't born, they're made,
Rich
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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:rsarashi wrote: ↑Sat Sep 02, 2017 10:34 pmA 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
n = 4(x - n)/20
n = (x - n)/5
5n = x - n
6n = x
n = (1/6)x
Answer: B
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