The average of four consecutive odd natural numbers is eight less than the average of three consecutive even natural numbers. If the sum of these three even numbers is equal to the sum of above four odd numbers, then the average of four original odd numbers is:

Let the four consecutive odd natural numbers are a, a+2, a+4, a+6

Let the three consecutive even natural numbers are b, b+2, b+6

Average of four consecutive odd natural numbers = $$\frac{a+a+2+a+4+a+6}{4}=\frac{4a+12}{4}=a+3$$

Average of three consecutive even natural numbers = $$\frac{b+b+2+b+4}{3}=\frac{3b+6}{3}=b+2$$

The average of four consecutive odd natural numbers is eight less than the average of three consecutive even natural numbers.

$$\Rightarrow$$  $$a+3=b+2-8$$

$$\Rightarrow$$  $$b-a=9$$ .........(1)

The sum of the three even numbers is equal to the sum of four odd numbers.

$$\Rightarrow$$  $$b+b+2+b+4=a+a+2+a+4+a+6$$

$$\Rightarrow$$  $$3b=4a+6$$

$$\Rightarrow$$  $$3b-4a=6$$ ......(2)

Solving 3(1)-(2) we get,  $$a = 21$$

$$\therefore\ $$Average of four consecutive odd natural numbers = $$a+3=21+3=24$$

Hence, the correct answer is Option B