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2001 AMC 10 Problems/Problem 10: Difference between revisions

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These equations are symmetric, and furthermore, they use multiplication. This makes us think to multiply them all. This gives <math>(xyz)^2 = (xy)(yz)(xz) = (24)(48)(72) = (24 \times 12)^2 \implies xyz = 288</math>. We divide <math>xyz = 288</math> by each of the given equations, which yields <math>x = 4</math>, <math>y = 6</math>, and <math>z = 12</math>. The desired sum is <math>4+6+12 = 22</math>, so the answer is <math>\boxed{\textbf{(D) } 22}</math>.
These equations are symmetric, and furthermore, they use multiplication. This makes us think to multiply them all. This gives <math>(xyz)^2 = (xy)(yz)(xz) = (24)(48)(72) = (24 \times 12)^2 \implies xyz = 288</math>. We divide <math>xyz = 288</math> by each of the given equations, which yields <math>x = 4</math>, <math>y = 6</math>, and <math>z = 12</math>. The desired sum is <math>4+6+12 = 22</math>, so the answer is <math>\boxed{\textbf{(D) } 22}</math>.
==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 <math>x = 4</math>, <math>y = 6</math>, and <math>z = 12</math> work. <math>4 + 6+ 12 = \boxed{\textbf{(D) } 22} </math>
~idk12345678


== See Also ==
== See Also ==

Revision as of 22:42, 9 April 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

See Also

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

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