Difference between revisions of "2003 AMC 10A Problems/Problem 13"
(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> | ||
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
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\end{eqnarray*}</cmath> | \end{eqnarray*}</cmath> | ||
− | Therefore, the product of all three numbers is <math>xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow \mathrm{(A)}</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>. |
=== Solution 2 === | === Solution 2 === | ||
− | Alternatively, we can set up the system in [[ | + | Alternatively, we can set up the system in [[equation]] form: |
<cmath>\begin{eqnarray*}1x+1y+1z&=&20\\ | <cmath>\begin{eqnarray*}1x+1y+1z&=&20\\ | ||
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\end{eqnarray*}</cmath> | \end{eqnarray*}</cmath> | ||
− | Or, in matrix form | + | Or, in [[matrix]] form |
<math> | <math> | ||
\begin{bmatrix} | \begin{bmatrix} | ||
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</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 \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 == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 15:45, 19 August 2023
Contents
Problem
The sum of three numbers is . 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?
Solution
Solution 1
Let the numbers be ,
, and
in that order. The given tells us that
Therefore, the product of all three numbers is .
Solution 2
Alternatively, we can set up the system in equation form:
Or, in matrix form
To solve this matrix equation, we can rearrange it thus:
Solving this matrix equation by using inverse matrices and matrix multiplication yields
Which means that ,
, and
. Therefore,
Solution 3
Let's denote the 3rd number as , the 2nd as
, and the 1st as
according to the information given in the problem. We know that all three numbers add up to
, so
. Solving the equation we get
. Then substitute this value of x to solve for the other two numbers. Lastly, we obtain
as the product of all three numbers.
Video Solution by WhyMath
~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.