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 o…

Solution
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 = $\frac{40}{100} \times 42 = 16.8$