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

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 $x+y+z = 4+6+12 = \boxed{\textbf{(D) }22}$ .

Solution 2

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

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2001 AMC 10 Problems/Problem 10

Contents

Problem

Solution 1

Solution 2

See Also