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?

Answer: Option A

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