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13 Apr 2021, 03:45
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Difficulty:
35%
(medium)
Question Stats:
72% (02:04) correct
28%
(01:51)
wrong
based on 93
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A set consists of 20 numbers. If n is a number in the list and is four times the average (arithmetic mean) of the numbers in the list other than itself, then n is what fraction of the sum of the 20 numbers in the list?
(A) 1/20
(B) 4/23
(C) 1/5
(D) 1/10
(E) 5/11
(A) 1/20
(B) 4/23
(C) 1/5
(D) 1/10
(E) 5/11
13 Apr 2021, 04:07
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Total 20 numbers
Assume Sum = 100
So, Average of 20 numbers = 5
Given: n is a number in the list and is four times the average (arithmetic mean) of the numbers in the list other than itself.
This is the tricky part.
n = 4*(100-n)/19
19n = 400 - 4n
n = 400/23
So, n is 4/23 times the sum of the 20 numbers (100) in the list.
Assume Sum = 100
So, Average of 20 numbers = 5
Given: n is a number in the list and is four times the average (arithmetic mean) of the numbers in the list other than itself.
This is the tricky part.
n = 4*(100-n)/19
19n = 400 - 4n
n = 400/23
So, n is 4/23 times the sum of the 20 numbers (100) in the list.
Bunuel
Let the sum of the 20 numbers be = Sum
\(Average = \frac{Sum}{20}\)
Given, n = 4 * Average other than itself.
\(n = \frac{4 * (Sum-n)}{19}\)
\(19n = 4sum - 4n\)
\(23n = 4sum\)
\(n = 4 sum/23\)
Asked, n is what fraction of the sum.
So,\( \frac{n}{sum} = \frac{4sum}{23}*\frac{1}{sum} = \frac{4}{23}\)
