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

Latest revision as of 17:14, 18 October 2025

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

Solution 4

From the first equation, we can get $y=\frac{24}{x}$ . (Note that $x$ cannot be $0$ , because then it would be $xy=0$ ). Substituting this into the third equation gives $\frac{24z}{x}=72$ , so $24z=72x$ meaning that $z=3x$ . Substituting our equation for $z$ into the 2nd equation gives $3x^2=48$ , so $x=4$ (note that $x$ is positive), which means that $z=12$ , and $y=\frac{24}{x}=6$ . This means that $x+y+z=4+6+12=\boxed{\textbf{(D) } 22}$

~Baihly2024

Video Solution by Daily Dose of Math

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

~Thesmartgreekmathdude

@@ Line 22: / Line 22: @@
 == Solution 4 ==
-From the first equation, we can get <math>y=\frac{24}{x}</math>. (note that <math>x</math> cannot be <math>0</math>, because then it would be <math>xy=0</math>) Substituting this into the third equation gives <math>\frac{24z}{x}=72</math>, so <math>24z=72x</math> meaning that <math>z=3x</math>. Substituting our equation for <math>z</math> into the 2nd equation gives <math>3x^2=48</math>, so <math>x=4</math> (note that <math>x</math> is positive), which means that <math>z=12</math>, and <math>y=\frac{24}{x}=6</math>. This means that <math>x+y+z=4+6+12=22</math>.
+From the first equation, we can get <math>y=\frac{24}{x}</math>. (Note that <math>x</math> cannot be <math>0</math>, because then it would be <math>xy=0</math>). Substituting this into the third equation gives <math>\frac{24z}{x}=72</math>, so <math>24z=72x</math> meaning that <math>z=3x</math>. Substituting our equation for <math>z</math> into the 2nd equation gives <math>3x^2=48</math>, so <math>x=4</math> (note that <math>x</math> is positive), which means that <math>z=12</math>, and <math>y=\frac{24}{x}=6</math>. This means that <math>x+y+z=4+6+12=\boxed{\textbf{(D) } 22} </math>
 ~Baihly2024
@@ Line 36: / Line 36: @@
 {{AMC10 box|year=2001|num-b=9|num-a=11}}
 {{MAA Notice}}
+[[Category: Introductory Algebra Problems]]

2001 AMC 10 (Problems • Answer Key • Resources)
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