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

Revision as of 23:18, 18 October 2020

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 $xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow \boxed{\mathrm{(A)}\ 28}$ .

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, $xyz = \frac{1}{2}\cdot\frac{7}{2}\cdot16 = 28 \Rightarrow \boxed{\mathrm{(A)}\ 28}$

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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>
-__TOC__
 == Solution ==
 === Solution 1 ===

2003 AMC 10A (Problems • Answer Key • Resources)
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