Question

The average of five consecutive odd integers is 113. What is the second smallest of them

Answer 1

Answer:

let 1st odd number be x

2nd odd number be x+1

3rd odd number be x+2

4th odd number be x+3

5th odd number be x+4

So

(x+x+1+x+2+x+3+x+4)÷5 =113

5x+10=113×5

5x+10=565

5x=565-10

x=555/5

x=111

Answer 2

Answer:

The final answer is 111.

Step-by-step explanation:

Given,

The average of five consecutive numbers is 113. We need to find the second smallest among them.

Let x be the smallest odd integer.

Then the next odd integer will be at x + 2

The next odd integer will be at x + 4

The next odd integer will be at x + 6

The next odd integer will be at x + 8

Average of the numbers = 113

( x + x + 2 + x + 4 + x + 6 + x + 8 ) / 5 = 113

5x + 20 = 565

5x = 565 - 20

5x = 545

x = 545 / 5

x = 109

The second smallest number is x + 2 = 109 + 2 = 111

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