There are three positive numbers, $${1 \over 3}$$rd of average of all the three numbers is 8 less than the value of the highest number. Average of the lowest and the second lowest number is 8. Which is the highest number?

Let the three positive numbers be $$x,y,z$$     (where $$x < y < z$$)

Average of the three numbers = $$\frac{x + y + z}{3}$$

Acc. to ques, => $$\frac{1}{3} \times (\frac{x + y + z}{3}) = z - 8$$

=> $$x + y + z = 9z - 72$$

=> $$x + y = 8z - 72$$

Dividing both sides by 2, we get :

=> $$\frac{x + y}{2} = 4z - 36$$

Also, average of the lowest and the second lowest number is 8, => $$\frac{x + y}{2} = 8$$

=> $$4z - 36 = 8$$

=> $$4z = 8 + 36 = 44$$

=> $$z = \frac{44}{4} = 11$$