Difference between revisions of "2001 AMC 10 Problems/Problem 10"

Revision as of 15:20, 16 March 2011

If $x$ , $y$ , and $z$ are positive with $xy = 24$ , $xz = 48$ , and $yz = 72$ , then $x + y + z$ is

$\textbf{(A) }18\qquad\textbf{(B) }19\qquad\textbf{(C) }20\qquad\textbf{(D) }22\qquad\textbf{(E) }24$

Look at the first two equations in the problem.

$xy=24$ and $xz=48$ .

We can say that $2y=z$ .

Given $2y=z$ , we can substitute $z$ for $2y$ and find

$2y^2=72$ $y^2=36$ $y=6$ $2y=z=12$ .

We can replace y into the first equation. $6x=24$ $x=4$ .

Since we know every variable's value, we can substitute it in for $x+y+z = 4+6+12 = \boxed{\textbf{(D) }22}$ .

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

@@ Line 25: / Line 25: @@
 Since we know every variable's value, we can substitute it in for <math> x+y+z = 4+6+12 = \boxed{\textbf{(D) }22} </math>.
+== See Also ==
+{{AMC10 box|year=2001|num-b=9|num-a=11}}