1. 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?
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By: anil on 05 May 2019 02.53 pm
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, => $$frac{1}{3} imes (frac{x + y + z}{3}) = z - 8$$ => $$x + y + z = 9z - 72$$ => $$x + y = 8z - 72$$ 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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