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

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<math>\textbf{(A) }18\qquad\textbf{(B) }19\qquad\textbf{(C) }20\qquad\textbf{(D) }22\qquad\textbf{(E) }24</math>
 
<math>\textbf{(A) }18\qquad\textbf{(B) }19\qquad\textbf{(C) }20\qquad\textbf{(D) }22\qquad\textbf{(E) }24</math>
  
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== Solution ==
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== Solution 1==
  
 
Look at the first two equations in the problem.
 
Look at the first two equations in the problem.
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Since we know every variable's value, we can substitute it in for <math> x+y+z = 4+6+12 = \boxed{\textbf{(D) }22} </math>.
 
Since we know every variable's value, we can substitute it in for <math> x+y+z = 4+6+12 = \boxed{\textbf{(D) }22} </math>.
 +
 +
== Solution 2 ==
 +
 +
These equations are symmetric, and furthermore, they use multiplication. This makes us think to multiply them all. This gives <math>(xy)(yz)(xz) = (xyz)^2 = (24)(48)(72) = (24 \times 12)^2</math>.
 +
We square root:
 +
<math>xyz = 288</math>.
 +
Aha! We divide each of the given equations into this, yielding <math>x = 4</math>, <math>y = 6</math>, and <math>z = 12</math>. The desired sum is <math>4+6+12 = \boxed{22}</math>, so the answer is <math>\boxed{\text{(D)}}</math>.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2001|num-b=9|num-a=11}}
 
{{AMC10 box|year=2001|num-b=9|num-a=11}}

Revision as of 21:32, 1 June 2011

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

Look at the first two equations in the problem.

$xy=24$ and $xz=48$.

We can say that $2y=z$.

Given $2y=z$, we can substitute $z$ for $2y$ and find

$2y^2=72$ $y^2=36$ $y=6$ $2y=z=12$.

We can replace y into the first equation. $6x=24$ $x=4$.

Since we know every variable's value, we can substitute it in for $x+y+z = 4+6+12 = \boxed{\textbf{(D) }22}$.

Solution 2

These equations are symmetric, and furthermore, they use multiplication. This makes us think to multiply them all. This gives $(xy)(yz)(xz) = (xyz)^2 = (24)(48)(72) = (24 \times 12)^2$. We square root: $xyz = 288$. Aha! We divide each of the given equations into this, yielding $x = 4$, $y = 6$, and $z = 12$. The desired sum is $4+6+12 = \boxed{22}$, so the answer is $\boxed{\text{(D)}}$.

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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