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2001 AMC 10 Problem 10

Problem 10 of 25EasierAlgebra

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

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Solution

Dividing xz=48xz=48 by xy=24xy=24 gives z=2y.z=2y. Then yz=2y2=72,yz=2y^2=72, so y=6,y=6, z=12,z=12, and x=24y=4.x=\frac{24}{y}=4. Hence x+y+z=22.x+y+z=22. Thus, the correct answer is D.
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