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2006 Alabama ARML TST Problems/Problem 6

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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}

Solution

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}.

See also

2006 Alabama ARML TST (Problems)
Preceded by:
Problem 5 		Followed by:
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.
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