Difference between revisions of "2022 AMC 10A Problems/Problem 3"
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<math>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5</math> | <math>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5</math> | ||
| − | == Solution == | + | == Solution 1 == |
Let <math>x</math> be the third number. It follows that the first number is <math>6x,</math> and the second number is <math>x+40.</math> | Let <math>x</math> be the third number. It follows that the first number is <math>6x,</math> and the second number is <math>x+40.</math> | ||
| Line 12: | Line 12: | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
| + | |||
| + | == Solution 2 == | ||
| + | Solve this using a system of equations. Let <math>x,y,</math> and <math>z</math> be the three numbers, respectively. We get three equations: | ||
| + | <cmath>\begin{align*} | ||
| + | x+y+z&=96, \\ | ||
| + | x&=6z, \\ | ||
| + | z&=y-40. | ||
| + | \end{align*}</cmath> | ||
| + | Rewriting the third equation gives us <math>y=z+40,</math> so we can substitute <math>x</math> as <math>6z</math> and <math>y</math> as <math>z+40.</math> | ||
| + | |||
| + | Therefore, we get | ||
| + | <cmath>\begin{align*} | ||
| + | 6z+(z+40)+z&=96 \\ | ||
| + | 8z+40&=96 \\ | ||
| + | 8z&=56 \\ | ||
| + | z&=7. | ||
| + | \end{align*}</cmath> | ||
| + | Substituting 7 in for <math>z</math> gives us <math>x=6z=6(7)=42</math> and <math>y=z+40=7+40=47.</math> | ||
| + | |||
| + | So, the answer is <math>|x-y|=|42-47|=\boxed{\textbf{(E) } 5}.</math> | ||
| + | |||
| + | ~alexdapog A-A | ||
| + | |||
| + | == Solution 3 == | ||
| + | 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> | ||
| + | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
| + | |||
| + | ==Video Solution 1 (Quick and Easy)== | ||
| + | https://youtu.be/v2eJtm4EUkI | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution 2== | ||
| + | https://youtu.be/B-5zSnDFVXs | ||
| + | |||
| + | ~Charles3829 | ||
| + | |||
| + | ==Video Solution 3 (2 minutes) == | ||
| + | https://youtu.be/7yAh4MtJ8a8?si=Xpc2h85yyqEMOnVb&t=322 | ||
| + | |||
| + | ~Math-x | ||
| + | |||
| + | ==Video Solution 4== | ||
| + | https://youtu.be/BmWgwtKJExw | ||
== See Also == | == See Also == | ||
| − | |||
{{AMC10 box|year=2022|ab=A|num-b=2|num-a=4}} | {{AMC10 box|year=2022|ab=A|num-b=2|num-a=4}} | ||
| + | {{AMC12 box|year=2022|ab=A|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 10:35, 24 March 2024
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:
Rewriting the third equation gives us 
so we can substitute as 
and 
as 
Therefore, we get
Substituting 7 in for gives us 
and 
So, the answer is 
~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/1/a/5/1a5a3710cd1e40239ce7df741040ca878a21b393.png)
vladimir.shelomovskii@gmail.com, vvsss
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
Video Solution 2
~Charles3829
Video Solution 3 (2 minutes)
https://youtu.be/7yAh4MtJ8a8?si=Xpc2h85yyqEMOnVb&t=322
~Math-x
Video Solution 4
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. 
