Question 134

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?

Solution

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, => $$\frac{x + y}{2} = 8$$

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

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

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


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