Question

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

Answer 1

B.
`(1)/(6)`

Explanation:

Let x be the sum of the 21 numbers,

In which n is one of the numbers,

Since,

`\text{Average}=\frac{\text{Sum of the observations}}{\text{Number of observations}}`

So, the average of 20 numbers excluded n =
`(x-n)/(20)`

According to the question,

`n = 4* (x-n)/(20)`

`n = (x-n)/(5)`

`5n = x - n`

`6n = x`

`\imples n = (1)/(6)x = (1)/(6)\text{ of the sum of the 21 numbers}`

Hence, OPTION 'B' is correct.