The mean of three numbers is 40. All the three numbers are different natural numbers. If lowest is 19, what coould be the highest possible number of remaining twonumbers.

Answer : 

Given  : The mean of three number  =  40 

And 

two numbers  =  x and y and given third number ( lowest of these three numbers  ) = 19  

So, 

$$\begin{gathered} \frac{x +\hspace{0.17em}y +\hspace{0.17em}19}{3} = 40 \\ \Rightarrow x +\hspace{0.17em}y +\hspace{0.17em}19 = 120 \\ \mathbf{\Rightarrow }\mathit{x}\mathbf{ }\mathbf{+}\mathbf{\hspace{0.17em}}\mathit{y}\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{101} \end{gathered}$$
we get second highest number in order to get highest possible number  =  20 

SO, 

Highest possible number  =  101 - 20 = 81                              ( Ans )