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.

@@ Line 5: / Line 5: @@
 <math>\textbf{(A) }18\qquad\textbf{(B) }19\qquad\textbf{(C) }20\qquad\textbf{(D) }22\qquad\textbf{(E) }24</math>
-== Solution ==
+== Solution 1==
-Look at the first two equations in the problem.
+The first two equations in the problem are <math>xy=24 </math> and <math>xz=48</math>. Since <math>xyz \ne 0</math>, we have <math>\frac{xy}{xz}=\frac{24}{48} \implies 2y=z </math>. We can substitute <math> z = 2y </math> into the third equation <math>yz = 72</math> to obtain <math> 2y^2=72 \implies y=6 </math> and <math> 2y=z=12 </math>. We replace <math>y</math> into the first equation to obtain <math> x=4 </math>.
-<math>xy=24 </math> and <math>xz=48</math>.
+Since we know every variable's value, we can substitute them in to find <math> x+y+z = 4+6+12 = \boxed{\textbf{(D) }22} </math>.
-We can say that <math> 2y=z </math>.
+== Solution 2 ==
-Given <math>2y=z</math>, we can substitute <math> z </math> for <math> 2y </math> and find
+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>.
-<math> 2y^2=72 </math>
+==Solution 3(strategic guess and check)==
-<math> y^2=36 </math>
-<math> y=6 </math>
-<math> 2y=z=12 </math>.
-We can replace y into the first equation.
+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>
-<math> 6x=24 </math>
-<math> x=4 </math>.
-Since we know every variable's value, we can substitute it in for $ x+y+z = 4+6+12 = \boxed{\textbf{(D) }22}.
+~idk12345678
+==Video Solution by Daily Dose of Math==
+https://youtu.be/tiDp5E3rwfI?si=n2h6UvQUW-V-bLT2
+~Thesmartgreekmathdude
+== See Also ==
+{{AMC10 box|year=2001|num-b=9|num-a=11}}
+{{MAA Notice}}

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