The sum of a series of 5 consecutive odd numbers is 195. The second lowest number of this series is 9 less than the second highest number of another series of 5 consecutive even numbers. What is 40% of the second lowest number of the series of consecutive even numbers?

Let the five consecutive odd numbers in increasing order = $$(x-4) , (x-2) , (x) , (x+2) , (x+4)$$

Sum of these numbers = $$(x-4) + (x-2) + (x) + (x+2) + (x+4) = 195$$

=> $$5x = 195$$

=> $$x = \frac{195}{5} = 39$$

Thus, the odd numbers are = 35 , 37 , 39 , 41 , 43

Let another series of even numbers in increasing order = $$(y-4) , (y-2) , (y) , (y+2) , (y+4)$$

Also, $$37 = (y + 2) - 9$$

=> $$y = 37 + 9 - 2 = 44$$

Thus, second lowest number of the even series = 44 - 2 = 42

$$\therefore$$ 40% of 42 = $$\frac{40}{100} \times 42 = 16.8$$