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

Latest revision as of 20:41, 15 July 2024

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 $4+6+12 = 22$ , so the answer is $\boxed{\textbf{(D) } 22}$ .

Solution 3(strategic guess and check)

Seeing the equations, we notice that they are all multiples of 12. Trying in factors of 12, we find that $x = 4$ , $y = 6$ , and $z = 12$ work. $4 + 6+ 12 = \boxed{\textbf{(D) } 22}$

~idk12345678

Video Solution by Daily Dose of Math

https://youtu.be/tiDp5E3rwfI?si=n2h6UvQUW-V-bLT2

~Thesmartgreekmathdude

2001 AMC 10 (Problems • Answer Key • Resources)
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.

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