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

(→‎Solution)
(err... matrices really overkill this problem ... wikify)
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<math> \mathrm{(A) \ } 28\qquad \mathrm{(B) \ } 40\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 800 </math>
 
<math> \mathrm{(A) \ } 28\qquad \mathrm{(B) \ } 40\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 800 </math>
  
 +
__TOC__
 
== Solution ==
 
== Solution ==
−
Let the numbers be <math>x</math>, <math>y</math>, and <math>z</math> in that order.  
+
=== Solution 1 ===
 +
Let the numbers be <math>x</math>, <math>y</math>, and <math>z</math> in that order. The given tells us that
  
−
<math>y=7z</math>
+
<cmath>\begin{eqnarray*}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
 +
\end{eqnarray*}</cmath>
  
−
<math>x=4(y+z)=4(7z+z)=4(8z)=32z</math>
+
Therefore, the product of all three numbers is <math>xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow \mathrm{(A)}</math>.
  
−
<math>x+y+z=32z+7z+z=40z=20</math>
+
=== Solution 2 ===
 +
Alternatively, we can set up the system in [[matrix]] form:
  
−
<math>z=\frac{20}{40}=\frac{1}{2}</math>
+
<cmath>\begin{eqnarray*}1x+1y+1z&=&20\\
−
 
+
1x-4y-4z&=&0\\
−
<math>y=7z=7\cdot\frac{1}{2}=\frac{7}{2}</math>
+
0x+1y-7z&=&0\\
−
 
+
\end{eqnarray*}</cmath>
−
<math>x=32z=32\cdot\frac{1}{2}=16</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
 
Or, in matrix form
 
<math>
 
<math>
−
  \begin{bmatrix}
+
\begin{bmatrix}
−
    1 & 1 & 1 \\
+
1 & 1 & 1 \\
−
    1 & -4 & -4 \\
+
1 & -4 & -4 \\
−
    0 & 1 & -7
+
0 & 1 & -7
−
  \end{bmatrix}
+
\end{bmatrix}
−
  \begin{bmatrix}
+
\begin{bmatrix}
−
    x \\
+
x \\
−
    y \\
+
y \\
−
    z \\
+
z \\
−
  \end{bmatrix}
+
\end{bmatrix}
−
=
+
=\begin{bmatrix}
−
  \begin{bmatrix}
+
20 \\
−
    20 \\
+
0 \\
−
    0 \\
+
0 \\
−
    0 \\
+
\end{bmatrix}
−
  \end{bmatrix}
 
 
</math>
 
</math>
 +
 
To solve this matrix equation, we can rearrange it thus:
 
To solve this matrix equation, we can rearrange it thus:
−
<math>
+
 
−
  \begin{bmatrix}
+
<math>\begin{bmatrix}
−
    x \\
+
x \\
−
    y \\
+
y \\
−
    z \\
+
z \\
−
  \end{bmatrix}
+
\end{bmatrix}
−
=
+
= \begin{bmatrix}
−
  \begin{bmatrix}
+
1 & 1 & 1 \\
−
    1 & 1 & 1 \\
+
1 & -4 & -4 \\
−
    1 & -4 & -4 \\
+
0 & 1 & -7
−
    0 & 1 & -7
+
\end{bmatrix}
−
  \end{bmatrix}
 
 
^{-1}
 
^{-1}
−
  \begin{bmatrix}
+
\begin{bmatrix}
−
    20 \\
+
20 \\
−
    0 \\
+
0 \\
−
    0 \\
+
0 \\
−
  \end{bmatrix}
+
\end{bmatrix}
 
</math>
 
</math>
 +
 
Solving this matrix equation by using [[inverse matrices]] and [[matrix multiplication]] yields
 
Solving this matrix equation by using [[inverse matrices]] and [[matrix multiplication]] yields
−
<math>
+
 
−
  \begin{bmatrix}
+
<math>\begin{bmatrix}
−
    x \\
+
x \\
−
    y \\
+
y \\
−
    z \\
+
z \\
−
  \end{bmatrix}
+
\end{bmatrix} =
−
=
+
\begin{bmatrix}
−
\begin{bmatrix}
+
\frac{1}{2} \\
−
    \frac{1}{2} \\
+
\frac{7}{2} \\
−
    \frac{7}{2} \\
+
16 \\
−
    16 \\
+
\end{bmatrix}  
−
  \end{bmatrix}  
 
 
</math>
 
</math>
 +
 
Which means that <math>x = \frac{1}{2}</math>, <math>y = \frac{7}{2}</math>, and <math>z = 16</math>. Therefore, <math>xyz = \frac{1}{2}\cdot\frac{7}{2}\cdot16 = 28</math>
 
Which means that <math>x = \frac{1}{2}</math>, <math>y = \frac{7}{2}</math>, and <math>z = 16</math>. Therefore, <math>xyz = \frac{1}{2}\cdot\frac{7}{2}\cdot16 = 28</math>
  
−
== See Also ==
+
== See also ==
−
*[[2003 AMC 10A Problems]]
+
{{AMC10 box|year=2003|num-b=12|num-a=14|ab=A}}
−
*[[2003 AMC 10A Problems/Problem 12|Previous Problem]]
 
−
*[[2003 AMC 10A Problems/Problem 14|Next Problem]]
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 19:38, 23 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

Solution 1

Let the numbers be $x$, $y$, and $z$ in that order. The given tells us that

\begin{eqnarray*}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 \end{eqnarray*}

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

Solution 2

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

\begin{eqnarray*}1x+1y+1z&=&20\\ 1x-4y-4z&=&0\\ 0x+1y-7z&=&0\\ \end{eqnarray*}

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

$\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \\ \frac{7}{2} \\ 16 \\ \end{bmatrix}$

Which means that $x = \frac{1}{2}$, $y = \frac{7}{2}$, and $z = 16$. Therefore, $xyz = \frac{1}{2}\cdot\frac{7}{2}\cdot16 = 28$

See also

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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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