contestada

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

Respuesta :

Answer:

B. [tex]\frac{1}{6}[/tex]

Step-by-step explanation:

Let x be the sum of the 21 numbers,

In which n is one of the numbers,

Since,

[tex]\text{Average}=\frac{\text{Sum of the observations}}{\text{Number of observations}}[/tex]

So, the average of 20 numbers excluded n = [tex]\frac{x-n}{20}[/tex]

According to the question,

[tex]n = 4\times \frac{x-n}{20}[/tex]

[tex]n = \frac{x-n}{5}[/tex]

[tex]5n = x - n[/tex]

[tex]6n = x[/tex]

[tex]\imples n = \frac{1}{6}x = \frac{1}{6}\text{ of the sum of the 21 numbers}[/tex]

Hence, OPTION 'B' is correct.