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

Revision as of 10:10, 4 July 2013

Problem

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

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

Solution 1

The first two equations in the problem are $xy=24$ and $xz=48$ . Since $xyz \ne 0$ , we have $\frac{xy}{xz}=\frac{24}{48} \implies 2y=z$ . We can substitute $z = 2y$ into the third equation $yz = 72$ to obtain $2y^2=72 \implies y=6$ and $2y=z=12$ . We replace $y$ into the first equation to obtain $x=4$ .

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

Solution 2

These equations are symmetric, and furthermore, they use multiplication. This makes us think to multiply them all. This gives $(xyz)^2 = (xy)(yz)(xz) = (24)(48)(72) = (24 \times 12)^2 \implies xyz = 288$ . We divide $xyz = 288$ by each of the given equations, which yields $x = 4$ , $y = 6$ , and $z = 12$ . The desired sum is $4+6+12 = \boxed{22}$ , so the answer is $\boxed{\textbf{(D)}}$ .

2001 AMC 10 (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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

Revision as of 10:10, 4 July 2013

Contents

Problem

Solution 1

Solution 2

See Also