2006 Alabama ARML TST Problems/Problem 6

Problem

Let $\lfloor a \rfloor$ be the greatest integer less than or equal to $a$ and let $\{a\}=a-\lfloor a \rfloor$ . Find $10(x+y+z)$ given that

$\begin{align} x+\lfloor y \rfloor +\{z\}=14.2,\\ \lfloor x \rfloor+\{y\} +z=15.3,\\ \{x\}+y +\lfloor z \rfloor=16.1. \end{align}$

Let's add all three equations:

$x+\lfloor y \rfloor +\{z\}+\lfloor x \rfloor+\{y\} +z+\{x\}+y +\lfloor z \rfloor=x+x+y+y+z+z=2(x+y+z)=45.6$

And thus $10(x+y+z)=5\cdot 45.6=\boxed{228}$ .