The sum of a series of 5 consecutive odd numbers is 225 The second number of this series is 15 less than the second lowest number of another series of 5 consecutive ever numbers. What is 60% of the highest number of this 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 = $$(x-4) + (x-2) + (x) + (x+2) + (x+4) = 225$$

=> $$5x = 225$$

=> $$x = \frac{225}{5} = 45$$

Thus, the odd numbers are = 41 , 43 , 45 , 47 , 49

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

Also, $$43 = (y - 2) - 15$$

=> $$y = 43 + 15 + 2 = 60$$

Thus, highest number of the even series = 60 + 4 = 64

$$\therefore$$ 60% of 64 = $$\frac{60}{100} \times 64 = 38.4$$