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

Latest revision as of 14:45, 19 August 2023

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

Solution 3

Let's denote the 3rd number as $x$ , the 2nd as $7x$ , and the 1st as $4(7x+x)$ according to the information given in the problem. We know that all three numbers add up to $20$ , so $4(8x)+7x+x = 20$ . Solving the equation we get $x = \frac{1}{2}$ . Then substitute this value of x to solve for the other two numbers. Lastly, we obtain $28$ as the product of all three numbers.

Video Solution by WhyMath

https://youtu.be/3hY9YXqn5zg

~savannahsolver

Video Solution

https://www.youtube.com/watch?v=7V2k1WxooJ0 ~David

@@ Line 5: / Line 5: @@
 == 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 \boxed{\mathrm{(A)}\ 28}</math>.
-<math>x+y+z=32z+7z+z=40z=20</math>
+=== Solution 2 ===
+Alternatively, we can set up the system in [[equation]] form:
-<math>z=\frac{20}{40}=\frac{1}{2}</math>
+<cmath>\begin{eqnarray*}1x+1y+1z&=&20\\
+x-4y-4z&=&0\\
+x+1y-7z&=&0\\
+\end{eqnarray*}</cmath>
-<math>y=7z=7\cdot\frac{1}{2}=\frac{7}{2}</math>
+Or, in [[matrix]] form
+<math>
+\begin{bmatrix}
+& 1 & 1 \\
+& -4 & -4 \\
+& 1 & -7
+\end{bmatrix}
+\begin{bmatrix}
+x \\
+y \\
+z \\
+\end{bmatrix}
+=\begin{bmatrix}
+\\
+\\
+\\
+\end{bmatrix}
+</math>
-<math>x=32z=32\cdot\frac{1}{2}=16</math>
+To solve this matrix equation, we can rearrange it thus:
-Therefore, the product of all three numbers is <math>xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow A</math>
+<math>\begin{bmatrix}
+x \\
+y \\
+z \\
+\end{bmatrix}
+= \begin{bmatrix}
+& 1 & 1 \\
+& -4 & -4 \\
+& 1 & -7
+\end{bmatrix}
+^{-1}
+\begin{bmatrix}
+\\
+\\
+\\
+\end{bmatrix}
+</math>
-Alternatively, we can set up the system in matrix form:
+Solving this matrix equation by using [[inverse matrices]] and [[matrix multiplication]] yields
-<math>x+y+z=20</math>
+<math>\begin{bmatrix}
+x \\
-<math>x=4(y+z)=4y+4z</math>
+y \\
+z \\
-<math>y=7z</math>
+\end{bmatrix} =
+\begin{bmatrix}
+\frac{1}{2} \\
+\frac{7}{2} \\
+\\
+\end{bmatrix}
+</math>
-is equivalent to
+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 \Rightarrow \boxed{\mathrm{(A)}\ 28}</math>
-<math>1x+1y+1z=20</math>
+=== Solution 3 ===
+Let's denote the 3rd number as <math>x</math>, the 2nd as <math>7x</math>, and the 1st as <math>4(7x+x)</math> according to the information given in the problem. We know that all three numbers add up to <math>20</math>, so <math>4(8x)+7x+x = 20</math>. Solving the equation we get <math>x = \frac{1}{2}</math>. Then substitute this value of x to solve for the other two numbers. Lastly, we obtain <math>28</math> as the product of all three numbers.
-<math>1x-4y-4z=0</math>
+==Video Solution by WhyMath==
+https://youtu.be/3hY9YXqn5zg
-<math>0x+1y-7z=0</math>
+~savannahsolver
-Or, in matrix form
+==Video Solution==
-<math>
+https://www.youtube.com/watch?v=7V2k1WxooJ0   ~David
-  \begin{bmatrix}
-& 1 & 1 \\
-& -4 & -4 \\
-& 1 & -7
-  \end{bmatrix}
-  \begin{bmatrix}
-    x \\
-    y \\
-    z \\
-  \end{bmatrix}
-=
-  \begin{bmatrix}
-\\
-\\
-\\
-  \end{bmatrix}
-</math>
-To solve this matrix equation, we can rearrange it thus:
-<math>
-  \begin{bmatrix}
-    x \\
-    y \\
-    z \\
-  \end{bmatrix}
-=
-  \begin{bmatrix}
-& 1 & 1 \\
-& -4 & -4 \\
-& 1 & -7
-  \end{bmatrix}
-^(-1)
-  \begin{bmatrix}
-\\
-\\
-\\
-  \end{bmatrix}
-</math>
-Solving this matrix equation by using [[inverse matrices]] and [[matrix multiplication]] yields
-<math>
-  \begin{bmatrix}
-    x \\
-    y \\
-    z \\
-  \end{bmatrix}
-=
- \begin{bmatrix}
-.5 \\
-.5 \\
-\\
-  \end{bmatrix}
-</math>
-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]]
+{{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]]
+{{MAA Notice}}

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