Difference between revisions of "2022 AMC 10A Problems/Problem 3"
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~alexdapog A-A | ~alexdapog A-A | ||
== Solution 3 == | == Solution 3 == | ||
| − | In accordance with | + | In accordance with Solution 2, |
<cmath>y = z+40, x = 6z \implies |x-y| = |6z - z - 40| = 5|z - 8| \implies \boxed{\textbf{(E) } 5}</cmath> | <cmath>y = z+40, x = 6z \implies |x-y| = |6z - z - 40| = 5|z - 8| \implies \boxed{\textbf{(E) } 5}</cmath> | ||
Revision as of 13:30, 8 October 2023
Contents
Problem
The sum of three numbers is 
The first number is 
times the third number, and the third number is 
less than the second number. What is the absolute value of the difference between the first and second numbers?
Solution 1
Let 
be the third number. It follows that the first number is 
and the second number is 
We have ![\[6x+(x+40)+x=8x+40=96,\]](http://latex.artofproblemsolving.com/a/8/2/a82848a0e0a81667b270afbd8560aba712f745c6.png)
from which 
Therefore, the first number is 
and the second number is 
Their absolute value of the difference is 
~MRENTHUSIASM
Solution 2
Solve this using a system of equations. Let , 
, and 
be the three numbers, respectively. We get three equations:
![\[x+y+z=96\]](http://latex.artofproblemsolving.com/7/b/1/7b1a37d92f3fed67554c9131cb214b74cca94762.png)
![\[x=6z\]](http://latex.artofproblemsolving.com/b/0/1/b01547616152adc55ed1fa525c67c11f50e5c7c4.png)
![\[z=y-40\]](http://latex.artofproblemsolving.com/0/c/3/0c38de59f105c823063b23e7f7bff60f0d250b91.png)
Rewriting the third equation gives us 
, so we can substitute as 
and as 
.
Therefore, we get
![\[6z+(z+40)+z=96\]](http://latex.artofproblemsolving.com/e/5/8/e58ab559d1a251776e8b7f56a5ce02bd77595515.png)
![\[8z+40=96\]](http://latex.artofproblemsolving.com/6/e/6/6e63292dc055589dee5f08bb6834f3583a56b460.png)
![\[8z=56\]](http://latex.artofproblemsolving.com/b/7/2/b72684856a44f3af1e4477b01dda45f3ade44ea1.png)
![\[z=7\]](http://latex.artofproblemsolving.com/f/8/7/f87bae25a67f81988ba1c9c7c92db6951fdef056.png)
Substituting 7 in for gives us
![\[x=6z=6(7)=42\]](http://latex.artofproblemsolving.com/7/a/7/7a7d6e8f5d00e35a078a72e7c223f7713ec52fe8.png)
and
![\[y=z+40=7+40=47\]](http://latex.artofproblemsolving.com/e/4/1/e410843a538ae1e8b348bedaf018725f0b2078e8.png)
So 
~alexdapog A-A
Solution 3
In accordance with Solution 2,
![\[y = z+40, x = 6z \implies |x-y| = |6z - z - 40| = 5|z - 8| \implies \boxed{\textbf{(E) } 5}\]](http://latex.artofproblemsolving.com/2/2/9/2294d5bfcf243342e380dabc498264417649663f.png)
vladimir.shelomovskii@gmail.com, vvsss
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
Video Solution 2
~Charles3829
See Also
| 2022 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 | ||
| 2022 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
