Question

the average of 10 consecutive odd numbers is 120. what is the average of the first 5 smallest numbers among them

Answer 1

average of 10 no.s is 120
so total is 120*10=1200
let first no be x
so the nos are x,x+1,x+2,x+3,,,,,,,,,x+9.
Sum of these nos.is 10x+45
So,10x+45=1200
so,x=115.5
So,the five nos are
115.5,116.5,117.5,118.5,119.5
so their average is (115.5,116.5,117.5,118.5,119.5)/5=587.5/5=117.5

Answer 2

Let us assume, 2n – 9, 2n – 7, 2n – 5, 2n – 3, 2n – 1, 2n + 1, 2n + 3, 2n + 5, 2n + 7, 2n + 9 are the 10 consecutive odd numbers for any positive integer n.

 

Given:

(2n – 9 + 2n – 7 + 2n – 5 + 2n – 3 + 2n – 1 + 2n + 1 + 2n + 3 + 2n + 5 + 2n + 7 + 2n + 9) / 10 = 120

20n / 10 = 120

2n = 120

n = 60

Hence, 2n – 9 = 120 – 9 = 111. This is first odd number

Therefore, the 10 consecutive odd numbers are 111, 113, 115, 117, 119, 121, 123, 125, 127, 129

 

Therefore, the average of the smallest five odd numbers in the above series is 115.

 

Average =  (111 + 113 + 115 + 117 + 119) / 5 = 575 / 5 = 115.