Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 AMC 10A Problems/Problem 13"

(added problem and solution)
 
Line 20: Line 20:
  
 
Therefore, the product of all three numbers is <math>xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow A</math>
 
Therefore, the product of all three numbers is <math>xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow A</math>
 +
 +
Alternatively, we can set up the system in matrix form:
 +
 +
<math>x+y+z=20</math>
 +
 +
<math>x=4(y+z)=4y+4z</math>
 +
 +
<math>y=7z</math>
 +
 +
is equivalent to
 +
 +
<math>1x+1y+1z=20</math>
 +
 +
<math>1x-4y-4z=0</math>
 +
 +
<math>0x+1y-7z=0</math>
 +
 +
Or, in matrix form
 +
 +
  \begin{bmatrix}
 +
    1 & 1 & 1 \\
 +
    1 & -4 & -4 \\
 +
    0 & 1 & -7
 +
  \end{bmatrix}
 +
  \begin{bmatrix}
 +
    x \\
 +
    y \\
 +
    z \\
 +
  \end{bmatrix}
 +
=
 +
  \begin{bmatrix}
 +
    20 \\
 +
    0 \\
 +
    0 \\
 +
  \end{bmatrix}
 +
 +
To solve this matrix equation, we can rearrange it thus:
 +
  \begin{bmatrix}
 +
    x \\
 +
    y \\
 +
    z \\
 +
  \end{bmatrix}
 +
=
 +
  \begin{bmatrix}
 +
    1 & 1 & 1 \\
 +
    1 & -4 & -4 \\
 +
    0 & 1 & -7
 +
  \end{bmatrix}
 +
<sup>-1</sup>
 +
  \begin{bmatrix}
 +
    20 \\
 +
    0 \\
 +
    0 \\
 +
  \end{bmatrix}
 +
 +
Solving this matrix equation by using [[inverse matrices]] and [[matrix multiplication]] yields
 +
  \begin{bmatrix}
 +
    x \\
 +
    y \\
 +
    z \\
 +
  \end{bmatrix}
 +
=
 +
\begin{bmatrix}
 +
    0.5 \\
 +
    3.5 \\
 +
    16 \\
 +
  \end{bmatrix}
 +
Which means that x = 0.5, y = 3.5, and z = 16. Therefore, xyz = (0.5)(3.5)(16) = 28
 +
 +
 +
 +
 
== See Also ==
 
== See Also ==
 
*[[2003 AMC 10A Problems]]
 
*[[2003 AMC 10A Problems]]

Revision as of 17:21, 21 November 2007

Problem

The sum of three numbers is $20$. The first is four times the sum of the other two. The second is seven times the third. What is the product of all three?

$\mathrm{(A) \ } 28\qquad \mathrm{(B) \ } 40\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 800$

Solution

Let the numbers be $x$, $y$, and $z$ in that order.

$y=7z$

$x=4(y+z)=4(7z+z)=4(8z)=32z$

$x+y+z=32z+7z+z=40z=20$

$z=\frac{20}{40}=\frac{1}{2}$

$y=7z=7\cdot\frac{1}{2}=\frac{7}{2}$

$x=32z=32\cdot\frac{1}{2}=16$

Therefore, the product of all three numbers is $xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow A$

Alternatively, we can set up the system in matrix form:

$x+y+z=20$

$x=4(y+z)=4y+4z$

$y=7z$

is equivalent to

$1x+1y+1z=20$

$1x-4y-4z=0$

$0x+1y-7z=0$

Or, in matrix form 

 \begin{bmatrix}
   1 & 1 & 1 \\
   1 & -4 & -4 \\
   0 & 1 & -7
 \end{bmatrix}
 \begin{bmatrix}
   x \\
   y \\
   z \\
 \end{bmatrix}

= 

 \begin{bmatrix}
   20 \\
   0 \\
   0 \\
 \end{bmatrix}

To solve this matrix equation, we can rearrange it thus: 

 \begin{bmatrix}
   x \\
   y \\
   z \\
 \end{bmatrix}

= 

 \begin{bmatrix}
   1 & 1 & 1 \\
   1 & -4 & -4 \\
   0 & 1 & -7
 \end{bmatrix}

-1 

 \begin{bmatrix}
   20 \\
   0 \\
   0 \\
 \end{bmatrix}

Solving this matrix equation by using inverse matrices and matrix multiplication yields 

 \begin{bmatrix}
   x \\
   y \\
   z \\
 \end{bmatrix}

= 

\begin{bmatrix}
   0.5 \\
   3.5 \\
   16 \\
 \end{bmatrix}

Which means that x = 0.5, y = 3.5, and z = 16. Therefore, xyz = (0.5)(3.5)(16) = 28



See Also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_13&oldid=19798"