Difference between revisions of "2003 AMC 10A Problems/Problem 13"
Stealthhydra (talk | contribs) m (→Solution 2: names) |
|||
Line 80: | 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> | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/3hY9YXqn5zg | ||
+ | |||
+ | ~savannahsolver | ||
== See also == | == See also == |
Revision as of 16:18, 23 June 2021
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,
Video Solution by WhyMath
~savannahsolver
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.