2001 AMC 10 Problems/Problem 10

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 $\boxed{\textbf{(D)}}$.

See Also

2001 AMC 10 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS