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

Revision as of 15:50, 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 $x+y+z = 4+6+12 = \boxed{\textbf{(D) }22}$ .

Revision as of 15:50, 16 March 2011 (view source) Pidigits125 (talk \| contribs) (Created page with '== Problem == If <math>x</math>, <math>y</math>, and <math>z</math> are positive with <math>xy = 24</math>, <math>xz = 48</math>, and <math>yz = 72</math>, then <math>x + y + z<…')		Revision as of 15:50, 16 March 2011 (view source) Pidigits125 (talk \| contribs) (→‎Solution) Newer edit →
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	<math> x=4 </math>.		<math> x=4 </math>.

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