There are three positive numbers. One third of the average of all the three numbers is 8 less than the value of the highest number. The average of the lowest and the second lowest number is 8. What is the highest number?

A. 11

B. 14

C. 10

D. 9

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Solution(By Apex Team)

Let the three positive numbers in increasing order be a, b and c and the average of these numbers be be A. Then, $\begin{array}{l}\left(\Large\frac{a+b+c}{3}\right)=A\ldots.(i)\\ \text{ Given, }\\ c-\frac{A}{3}=8\\ \Rightarrow c-\left(\Large\frac{a+b+c}{9}\right)=8\ldots..(ii)\\ \text{Also given,}\\ \left(\Large\frac{b+a}{2}\right)=8\\ \Rightarrow a+b=16…..\left(iii\right)\end{array}$ Putting the value of (a + b) in equation (ii), we get $\begin{array}{l}\Rightarrow c-\left(\Large\frac{16+c}{9}\right)=8\\ \Rightarrow9c-16-c=72\\ \Rightarrow8c=72+16\\ \Rightarrow8c=88\\ \Rightarrow c=11\\ \therefore\text{ Highest number }=11\end{array}$

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