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"

m (→‎Solution 1)
m (→‎Solution 3)
 
(6 intermediate revisions by 6 users not shown)
Line 4: Line 4:
 
<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 ==
 
=== Solution 1 ===
 
=== Solution 1 ===
Line 20: Line 19:
  
 
=== Solution 2 ===
 
=== Solution 2 ===
−
Alternatively, we can set up the system in [[matrix]] form:
+
Alternatively, we can set up the system in [[equation]] form:
  
 
<cmath>\begin{eqnarray*}1x+1y+1z&=&20\\
 
<cmath>\begin{eqnarray*}1x+1y+1z&=&20\\
Line 27: Line 26:
 
\end{eqnarray*}</cmath>
 
\end{eqnarray*}</cmath>
  
−
Or, in matrix form
+
Or, in [[matrix]] form
 
<math>
 
<math>
 
\begin{bmatrix}
 
\begin{bmatrix}
Line 81: Line 80:
  
 
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>
 
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>
 +
 +
=== 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.
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/3hY9YXqn5zg
 +
 +
~savannahsolver
 +
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=7V2k1WxooJ0  ~David
  
 
== See also ==
 
== See also ==

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_13&oldid=197192"